ࡱ> y{y   !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~Root Entry F87yWordDocument CompObj^re separated by distances of a few light years to a few hundred thousand light years. It is always convenient to use a unit of measure that isapproximately the same size as the object or distance being measured. Hence, we use centimeters or inches when measuring a book or countertop and kilometers or miles when measuring distances between towns. Most of us would never consider measuring the distance to even the nearest town in inches because the number would be very large and cumbersome to work with. Also, we probably won't know the distance to the nearest town to within one centimeter or one inch. Consequently, when we measure the distances to stars, the light-year is convenient because stars are separated by a few light-years and we probably won't be able to determine the distance to the stars more accurately than a few tenths of a light-year. light year The distance that light, moving at a constant speed of 300,000 km/s, travels in one year. One light year is about 10 trillion kilometers. 7. Why is it difficult to see planets orbiting other stars? 7. Seeing planets around other star systems is very difficult for three reasons. First, they will appear as point sources. Planets are at the same distance from Earth, smaller in radius than the stars they orbit. Their angular diameter (cf. Chapter 3) would be even smaller. Second, the planets will also be much fainter than the stars they orbit. The light received from a planet is startlight reflected off the planet's surface. The planet's surface probably isn't perfectly reflecting and is smaller in size than the star; therefore, the light from the planet will be much fainter than that from the star. Finally, we are trying to look for this very faint light from the planet right next to the much brighter light from the star, making the planet even more difficult to notice. A planet around another star would be very hard to observe because it would be so dim. A planet shines only by star light reflected off of the planet. A star emits its own light, so it is much brighter than one of its planets. The light from the planet would be swamped by the light from the star. Those planets and stars are so far away. 8. Why can't we measure the diameters of stars from the size of the star images on photographs? What does the diameter of a star image really tell us about the star? 8. Most stars are so far away that we see them as points of light even with the largest telescopes. People looking through a telescope for the first time are often surprised that the stars don't look any bigger through the scope than with an unaided eye. The size of star images on a photograph depends on the camera exposure and, for a given photograph, the relative sizes indicate relative apparent brightnesses of the stars. Almost all stars are so far away that their diameters as viewed from the earth are below the detectable limit, even when the world's largest telescopes are used. The size of a star's image on a photograph tells us something about the brightness of the star, not about its size. The diameter of a star is what the size of the star is. 9. How long does it take light to cross the diameter of our solar system? of our galaxy? 9. Light from the Sun reaches Earth in 8 minutes. A little more accurate value is 8 minutes and 20 seconds (=500 seconds). Thus light takes 8 min. 20 sec. to travel the distance of 1 AU. The size of Pluto's orbit (one end to the other) is about 80 AU. After doing the arithmetic, we find that light will take about 40,000 seconds, or about 11 hours, to cross the diameter of our solar system (although the size increases thousand-fold if we include all the comets and debris that are out there). The size of the field of view of Figure 1-10 is given as 1700 ly; therefore, that of Figure 1-11, which is dominated by a spiral galaxy like our own, is 170,000 ly (a factor of 100 larger). We can thus infer that the diameter of our galaxy is about 100,000 ly, which by definition means that light takes 100,000 years to cross our galaxy.ܥe# gd,l,l  o(T4|oTimes New Roman Symbol ArialTimes New RomanCourier NewTimes New Roman Example The radius of the moon is 1738 km. What is this in miles? Table A-3 indicates that 1 mile equals 1,609 km, so 1738 km X 1 mile = 1080 miles. 1,609 km 1 angstrom A 10-8 cm light-year (ly) Review Questions 1. What is the largest dimension you have personal knowl- edge of? Have you run a mile? hiked 10 miles? run a marathon? If by personal knowledge it is meant that we traveled the distance under our own power, then most everyone has covered a distance of a few miles at some time. A few may well have covered in excess of 100 miles on an extended backpacking or bicycling trip. I have hiked more than 10 miles on the Oregon coast. Coos Bay to Charleston in southern Oregon. 2. Why are astronomical units more convenient than miles or kilometers for measuring some astronomical distances? astronomical unit (A.U.) The average distance of Earth from the Sun. Precise radar measurements yield a value for the A.U. of 149,603,500 km. The orbits of the planets are not perfect circles apparent for Mercury Its orbit carries it as close to the sun as 0.307 AU and as far away as 0.467 AU. 3. In what ways is our planet changing? 3. Our planet changes daily due to the effects of wind and water erosion, fire, and seismic and volcanic activity. These all affect the structure of Earth's surface. Also the atmosphere changes over long climatic cycles. Additionally, the type of life forms found on Earth have changed and continue to change. Even the relative population of the human species on Earth has greatly changed even within the last 100 years. Our planet environment changes with seasonal weather changes. Some years are warmer temperatures than in other years or colder temperatures in some years. Their is beach sand erosion on the beaches. Colder temperatures Warmer temperatures 4. What is the difference between our solar system, our gal- axy, and the universe? 4. Our solar system is the collection of gas, dust, and larger bodies gravitationally bound to the Sun. This includes the Sun, all planets and their satellites, asteroids, comets, and the gas and dust found in interplanetary space. Our galaxy, on the other hand, includes our solar system and other planetary systems, billions of other stars, and clouds of gas and dust known as nebulae in interstellar space. The galaxy is held together by the mutual gravitational attraction of all of these objects. It is billions of times the size of a typical planetary orbit. The universe is the collection of all matter, including billions of galaxies like our own. Anything that has mass or energy is part of the universe, whether we can observe it or not. Our Solar system is our sun and all of the objects gravitationally bound to it. This includes planets, moons, asteroids, comets, gas, and dust. Our galaxy (The Milky Way) is a huge collection of gas, dust, and billions of stars all gravitationally bound to each other. The universe is all of the physical objects that exist, including all matter and energy. The Solar System consists of 9 planets Our galaxy contains 9 planets, and the stars beyond planet Pluto. universe The totality of all space, time, matter, and energy. The universe contains all the solar systems, galaxies, planets, and stars, including asteroids, comets, and meteorites. Dark Matter is in the universe. 5. Why do all stars, except for the sun, look like points of light as seen from Earth? Its the brightness of the stars light. 5. All of the stars appear as points of light because they are so far away. (See #8) 6. Why are light-years-rather than miles or kilometers- more convenient units for measuring some astronomi- cal distances? 6. Note that objects in the solar system are separated by distances of a few AUs, and stars in our galaxy a It takes light about 11 hours to travel the diameter our solar system and about 100,000 years to travel the diamter of our galaxy. 10. What are the largest known structures in the universe? Filaments (i.e., large strings of superclusters of galaxies). 11. How many planets inhabited by intelligent life do you think the universe contains? Explain your answer? 11. Estimates will vary depending on who you talk to. Within our galaxy alone there are approximately 100 billion stars. If 1% of these stars have planets, and 1% of these systems contain one planet that has water in its liquid and gaseous form, and 1% of these have life, and 1% of them have intelligent life, then, within our own galaxy, the number of planets with intelligent life would be. Planets with intelligent life in our galaxy = (1 x 1011) x 0.01 x 0.01 x 0.01 x 0.01 = 1000 This is a simplified view of the problem and none of the values have been selected on scientific grounds. They are random estimates that give an indication of how the problem is often approached. The actual value may be much smaller, and may well be 1, in which case we would be the only intelligent species in the entire galaxy. 1 since Earth is the only known planet to be inhabited by the humans of intelligent life due to oxygen, and water on Earth. Problems 1. If 1 mile equals 1,609 km and the moon is 2160 miles in diameter, what is its diameter in kilometers? in meters? 3475440 3 5 2160 x 1,609 18 44 0000 18660 +2160 3475440 2. If sunlight takes 8 minutes to reach Earth, how long does moonlight take? The distance d from the moon to the earth is the velocity v of light times the time t it takes to travel from the moon to the earth. That is, d = v t. Consequently the travel time for light in this problem is given by t = d / v = (3.84 x 108 m) / (3 x 108 m/s) = 1.3 s 8 4 3. How many suns would it take, laid edge to edge, to reach the nearest star? There is no formula that you can look up in order to do this problem. You just have to reason it out. We need to have the distance d to the nearest star equal to twice the radius R of the sun times number n of suns. That is, d = 2R n. Consequently, the number of suns is given by n = d / (2R) = (4.28 ly)(9.46 x 1015 m/ly) / (2 x 6.96 x 108 m) = 2.8 x 107 The answer is 28 million suns. millions of suns. 4. How many kilometers are there in a light-minute? (Hint: The speed of light 3 x 10 5 km/s) 31.5 10.5 x 3 31.5 5. How many galaxies like our own, laid edge to edge, would it take to reach the nearest large galaxy (which is 2 x 10 6 ly away)? Just as in problem 3, there is no formula to look up in order to do this problem. Just think about it. The number n of galaxies like ours that you would have to lay end to end to reach the nearest large galaxy is the distance d to the large galaxy divided by the diameter D of our own galaxy. That is, n = d/D. Chapter 1 tells us that the diameter of our galaxy is approximately 75,000 ly in diameter. I think it might be closer to 100,000 ly, but we will use the book's number in this problem. We are told in Problem 5 that the distance to the nearest large galaxy is 2 x 106 ly. Therefore, n = d/D = (2,000,000 ly)/(75,000 ly) = 80/3 = 27 approximately The answer is approximately 27 galxies, laid end to end. 10.6 x 2 21.2 Exploring The Sky 1. Locate and center one example of each of three differ- ent types of objects: a. A planet, such as Saturn. Find its rising and setting time. Such objects have distances measured in astro- nomical units (AU). How to proceed: Decide on the object you want to lo- cate Then find and center the object by pressing the Find button on the Object Toolbar. The second method is to press the F key. The third is to click Edit, then Find. Once you have the Object Information window, press the center button. b. A star. All stars in The Sky belong to our Milky Way Galaxy. Give the star's name, its magnitude, and its distance in light-years. How to proceed: Click on any star, which brings up an Object Information window containg a variety of information about the star. c. A galaxy; give its name and/or its designation. How to proceed: To show galaxies, click on the Galax- ies button in the Object Toolbar, then click on any galaxy. Distances to galaxies are on the order of mil- lions and billions of light years. 2. Look at the solar system from beyond Pluto by clicking on View and then on 3D Solar System Mode. Tip the solar sydtem edge on and then face on. Zoom in to see the inner planets. Under Tools, set the Time Skip Incre- ment to 1 day and then go forward in time to watch the planets move. 3. Identify some of the brightest constellations located along the Milky Way. (Hint: See View, Reference Lines.) Chapter 1 8 Part 1 The Sky Review Questions 1. Why have astronomers added modern constellations to the sky? Answers to Review Questions Most of the constellations that were not handed down from ancient civilizations were added during the 15th to 17th centuries. Some of the added constellations were very small constellations composed of faint stars located in the Northern Hemisphere. These constellations filled in gaps between larger and brighter constellations. Also added were constellations in the Southern Hemisphere that had not been observed by western civilization. When sailors and explorers began to sail south of the tropics, new star patterns were observed and named to help in remembering them for navigation. Examples of asterisms include the Big Dipper (part of Ursa Major), the Great Square (part of Pegasus), the Water Jug (part of Aquarius), the Summer Triangle (composed of three bright stars in the constellations of Lyra, Cygnus, and Aquilla), the Tea Pot (part of Sagittarius), and the Northern Cross (part of Cygnus). The stars in a constellation or an asterism are generally close to each other in the sky and have a shape that suggests a particular object, person, or animal to the people of a given culture. the human constructed "patterns" help cultures encode valuable information about the year: for example, the proper time to plant and reap. Constellations are visual mnemonics to help remember the near random pattern of the sky. Our modern (1928 International Astronomical Union certified) constellations have descended from the ancient Greek culture Why have astronomers added modern constellations to the sky? The constellations indicate where stars are or how they are viewed from a distance. 2. What is the difference between an asterism and a con- stellation? Give some examples. An asterism is a group of stars that is not formally recognized as a constellation by the International Astronomical Union. Many asterisms are part of larger constellations. There are 88 constellations officially recognized by the IAU. Examples of asterisms include the Big Dipper (part of Ursa Major), the Great Square (part of Pegasus), the Water Jug (part of Aquarius), the Summer Triangle (composed of three bright stars in the constellations of Lyra, Cygnus, and Aquilla), the Tea Pot (part of Sagittarius), and the Northern Cross (part of Cygnus). The stars in a constellation or an asterism are generally close to each other in the sky and have a shape that suggests a particular object, person, or animal to the people of a given culture. A group of stars with a local name, that are not officially constellations, are called asterisms. Big Dipper is a constellation 3. What characteristic do stars in a constellation or aster- ism share? A constellation is A human grouping of stars in the night sky into a recognizable pattern. 4. Do people from other cultures on Earth see the same stars, constellations, and asterisms that you see? People from other cultures on Earth see the same stars, but they probably group them in their minds in different ways than you do. So, they probably see different constellations and asterisms. People on planets circling other stars (if there are such people) would see the stars in different places from their vantage point, if they were tens of light years from us. If they were thousands of light years from us, they would see different stars than we see. In either of these last two cases, the people would see different constellations than we would see. People from different cultures all see the same stars, but the asterisms and constellations are different. Technically, we should now all see the same constellations, because these have official definitions and borders. However, this designation might not be well accepted by people of various cultures. The asterisms are certainly dependent on the culture. The images we see in the sky depend on how we view different objects and the value we place on them. Even within a culture we can have different asterisms. & People& on planets circling other stars will not see the same constellations or asterisms, and may not see the same stars. The stars will not form the same patterns for them because they will be looking at them from a different location (perspective). In fact one star (the Sun) that we don't see as a part of any constellation may well be a very bright star to them that plays a key role in an asterism or constellation. If the planets orbit very distant stars, then the & people; may not even be able to see some of the stars that we observe as very bright. The Greek letter designations generally indicate the brightnesses because the stars in a given constellation were given Greek letter designations running in alphabetical order from brightest to faintest within that constellation. However, this system does not allow us to compare the relative brightnesses of stars in different constellations with certainty since the brightest star in each constellation is generally designated a, but not all constellations contain a really bright star. Do people from other cultures on Earth see the same stars, constellations, and asterisms that you see? 5. How does the Greek-letter designation of a star give us a clue to its brightness? Generally speaking, the brighter a star is in a constellation, the closer to the start of the Greek alphabet is the star's label. By the way the first four letters of the Greek alphabet are α, β, γ, and δ. If your browser just printed a "&" followed by the name of the Greek letter followed by ";", then you might want to update to a new browser which will actually display the Greek letter rather than its name. The brighter a star is in a constellation, the closer to the start of the Greek alphabet is the star's label. By the way the first four letters of the Greek alphabet are alpha;, beta;, gamma;, and delta How does the Greek-letter designation of a star give us a clue to its brightness? 6. Give two features of the magnitude scale that might seem confusing. What is the origin of these two features? 7. What does the word apparent mean in apparent visual magnitude? apparent magnitude The apparent brightness of a star, expressed using the magnitude scale. The word "apparent" means "as it appears to observers on earth." A star's apparent magnitude is a measure of the star's brightness as it appears to observers on earth. The phrase "apparent visual magnitude" emphasizes that we are referring to the brightness of the object's visible light (as opposed to other kinds of radiation such as radio waves or ultraviolet radiation) as it appears to observers on earth. There are three aspects of the brightness scale that most students find confusing. These are that the magnitude system is a logarithm scale, its apparent reverse order (bright stars have small magnitudes), and that the term magnitude is related to the brightness of a star, not its size. The logarithm scale comes from the response of the human eye. The eye detects brightness changes in a logarithmic fashion. This logarithmic response provides us with an ability to see over a large range of brightness. The reverse order of the system comes about from placing the stars in order of brightness. The brightest stars were thrown into the first group, magnitude 1, the next brightest stars were thrown into the second group, magnitude 2 and so on. Consequently, bright stars have small numerical magnitude values, while faint stars have very large numerical magnitude values. The word apparent in apparent visual magnitude means simply that it is the magnitude of the star as it appears to us when viewing the star from here on Earth. Apparent visual magnitude does not take into account any corrections for the stars distance or size or temperature or the amount of dust between us and the star. It is simply the brightness as it appears to us in the night sky. 8. In what ways is the celestial sphere a scientific model? The celestial sphere is an excellent scientific model. It is an accurate representation of what we observe when we view the night sky. Note that as we look out at the night sky, all the stars appear to be at equal distance away as if they were dots painted on a giant ceiling. Consequently, the celestial sphere does represent what we see, and permits us to discuss what would happen if Earth was at the center and Earth rotated on its axis, and/or revolved around the Sun. It provides us with a way to step back and picture in our minds what is going on as Earth rotates on its axis and revolves around the Sun. The celestial poles and celestial equator exist because Earth rotates on an axis. If Earth did not rotate we could define the ecliptic and the poles of the ecliptic, but there would not be a separate set of celestial poles and celestial equator. Under such a circumstance, we would have difficulty defining a geographic equator and geographic poles. The geographic equator would most likely be defined as the intersection between the ecliptic and Earths surface and the geographic poles would be 90° from this equator. celestial sphere Imaginary sphere surrounding the Earth, to which all objects in the sky were once considered to be attached. 9. Why do astronomers use the word on to describe angles on the sky rather than angles in the sky? The use of the word on instead of the word in when referring to angular distance between celestial objects comes about because all of the objects appear to be on the celestial sphere and at an indeterminable distance. While we know that objects are at different distances in the sky, their distance from Earth is irrelevant in determining the angular distance between the two objects as viewed from Earth. Measure the angle between the north point on the horizon and the north star Polaris. That angle is your latitude. Why do astronomers use the word on to describe angles on the sky rather than angles in the sky? 10. If Earth did not rotate, could we define the celestial poles and celestial equator? To see the north and south celestial poles at the same time, one needs to be at Earth&s equator. Due to the refraction of light by the atmosphere, an observer on the equator would observe the north celestial pole approximately 0.5° above the northern horizon and the south celestial pole about 0.5° above the southern horizon. An observer at latitudes between 0.5° S and 0.5° N could see both celestial poles above their horizon. 11. Where would you go on Earth if you wanted to be able to see both the north celestial pole and the south celes- tial pole at the same time? A celestial pole will be on your zenith if you are at a latitude of 90° N, the north geographic pole, or at 90 degrees, the south geographic pole. 12. Where would you go on Earth to place a celestial pole at your zenith? Your latitude can be determined by observing the angle between your northern horizon and the North Celestial Pole. Since Polaris, the North Star, is within 1° of the North Celestial Pole, Polaris can be used as a fairly accurate marker of the North Celestial Pole. Determining latitudes in the southern hemisphere is more difficult because there is no bright star within a few degrees of the South Celestial Pole. 13. Explain how to make a simple astronomical observa- tion that would determine your latitude. Measure the angle between the north point on the horizon and the north star Polaris. That angle is your latitude. 14. Why does the number of circumpolar constellations de- pend on the latitude of the observer? Circumpolar constellations are those constellations close enough to a celestial pole so that they never pass below an observers horizon, but instead pass directly between the observer's celestial pole and northern or southern horizon at their lowest points in the sky. At different latitudes the celestial pole will be at different distance above an observer's horizon. If the observer is at a latitude of 60° N, then all constellations within 60° of the North celestial pole will be circumpolar. However, if an observer is at a latitude of only 30° N, then only those constellations within 30° of the North Celestial Pole will be circumpolar. 15. How could we detect Earth's precession by examing star charts from ancient Egypt? One of the easiest ways to detect the existence of precession by examining ancient Egyptian star charts would be to look at which stars they show as circumpolar and which are circumpolar in Egypt now. Since the location of the north celestial pole moves relative to the stars because of precession, the stars that appear within the circumpolar zone also changes. If on the ancient charts, Thuban is listed as “nearest the pole”, then at the latitude of approximately 30° N, Polaris would have been circumpolar, but would have been very near the horizon at its lowest point. Additionally, all of the Big Dipper asterism would have been circumpolar, while today only one of the seven bright stars forming the Big Dipper is circumpolar at Egypt&s latitude. Discussion Questions 1. All cultures on Earth named constellations. Why do you suppose this was such a common practice? Nearly every culture on Earth has seen patterns in the stars. But, not surprisingly, very few have seen the same patterns. Take, for example, the Big Dipper, perhaps the most recognizable star pattern in the sky. The Big Dipper is not actually a constellation itself, but is part of a larger pattern known to the Greeks as Ursa Major, the Great Bear. Stars have existed for billions of years, aiding navigation, inspiring storytellers and even being relied upon to predict the future. But these "little diamonds in the sky" also form constellations. Constellations, in essence are patterns of stars in the sky interpreted differently from culture to culture, an assistant professor of astronomy at the University. There are a total of 88 officially recognized and recorded constellations in the entire sky. Forty-eight of them were derived from the ancient Greeks and the other 40 were more or less made up by modern astronomers.. Constellations fill night skies with myths and legends Some constellation identities can be traced to specific cultures and civilizations. To be able to see where all the stars are located in each group of each of the constellations. 2. If you were lost at sea, you could find your approximate latitude by measuring the altitude of Polaris. But Polaris isn't exactly at the celestial pole. What else would you need to know to measure your latitude more accurately? You need your approximate longitude. Problems 1. If light from one star is 40 times more intense than light from another star, what is their difference in magnitudes? This problem uses Table 2-1. If one star has light 40 times more intense than another star, then this is the same as saying that the intensity ratio between the two stars is 40. An intensity ratio of 40 implies a magnitude difference of 4 (see the table). From this data we cannot tell which of the two stars is brighter, only that there is a magnitude difference of 4 between the stars. P1. In this problem IA = 40 IB, and we want to find mB - mA. The relevant formula is mB - mA = (5/2) log (IA/IB) = (5/2) log (40) = (5/2) (1.6) = 4.0 The two stars differ by four magnitudes. 2. If two stars differ by 8.6 magnitudes, what is their in- tensity ratio? P2. In this problem mB - mA = 8.6, and we want to find IA/IB. The relevant formula is IA/IB = 10(2/5)(mB - mA) = 10(2/5)(8.6) = 103.44 = 2750 The brighter star is approximately 2750 times brighter than the dimmer star. 3. Star A has a magnitude of 2.5; Star B, 5.5; and Star C, 9.5. Which is brightest? Which are visible to the unaided eye? Which pair of stars has an intensity ratio of 16? The brightest star has the smallest magnitude value; Star A is brightest. The naked eye limit is about magnitude 6, so Star A and Star B are visible to the unaided eye. The difference in magnitude between Star B and Star A is 5.5 - 2.5 = 3. That implies an intensity ratio of 16, as determined by Table 2-1. Chapter 2 The Sky page 19 4. By what factor is sunlight more intense than moonlight? see figure 2-6 5. If you are at a latitude of 35 degrees north of Earth's equator, what is the angular distance from the northern horizon up to the north celestial pole? from the south- ern horizon down to the south celestial pole? For observers in the northern hemisphere, the NCP is always your latitude above the northern horizon. So, at 35 degrees north latitude, the NCP is 35 degrees above the northern horizon. Same reasoning applies to the SCP. Critical Inquiries for the Web 1. We've all heard of the "North Star," but is there a "South Star" for southern hemisphere observers? As Earth pre- cesses on its axis, Polaris will cease to be our pole star and others will eventually hold that title. Search for star maps available online and for information on preces- sion, and determine what bright southern hemisphere stars might be future pole stars for people living south of the equator. Yes. Inhabitants of the Southern Hemisphere can't see Polaris, even if they just live one degree south of the Equator, and they don't have a pole star they can call their own. There's nothing even close to the South Pole. A couple of faint stars are about 10 degrees from the South Pole, but the nearest star of equivalent brightness to Polaris is nearly 15 degrees from the pole. Finding Polaris is a snap. You can simply use the "pointer stars" in the bowl of the Big Dipper to point the way. However, trying to locate the spot in the night sky over the South Pole is a chore. For someone in South Africa, for instance, it's first necessary to find south. The Southern Cross can be used for this purpose. The northern most and southern most stars in this 4 star constellation points toward the South Pole. To zero in on the actual location of the spot above the South Pole, a line can be drawn from the first order magnitude star, Canopus, to the line extending south from the Southern Cross. Canopus can be found to the lower right or southeast of the Southern Cross and is about the same distance to the pole as is the cross. These two lines intersect at an approximately 90 degree angle (right angle) and fairly close to the South Pole. Both the Northern and Southern Hemisphere have nearly the same amount of bright stars and both have bright constellations in the shape of a cross. www.gsfc.nasa.gov/scienceques2001/20020322.htm what bright southern hemisphere stars precession 2. Would the stars of a familiar constellation like Pegasus or Orion look similar to our view from Earth if we lived on planets orbiting other stars? Find a table of distance data for the bright stars in a familiar constellation and construct a diagram that visualizes the positions of these stars in space similar to Figure 2-3. 3. Most cultures around the world developed their own constellations. Search the Web to find constellation mythology from other cultures. Native Americans have a rich mythology of the sky, as do Polynesian, Asian, and African peoples. Mythology of the Constellations ancient Greek and Roman mythology associated with the major constellations. www.emufarm.org/~cmbell/myth/myth.html constellation mythology http://einstein.stcloudstate.edu/Dome/clicks/constlist.html Origin of the Constellation Camelopardus is a modern constellation. According to Ian Ridpath, it was invented by the Dutch theologian and astronomer Petrus Plancius. Both Ridpath and Julius D. W. Staal state that the German astronomer Jakob Bartsch included the constellation on a star map in 1614 (Ridpath) or 1624 (Staal). Exploring The Sky 1. Describe the apparent rotation of the sky in the north, south, east, and west sky from your location. (Hint Use Site Information under the Data menu to set location and date. Use Look under the Orientation menu and use the time skip arrows in the toolbar to go forward in time.) North America Seattle, Washington north, Move Up arrow goes up to the North. south, move down arrow goes down to the south mars goes to north and then south. east, move left arrow comes from west to east and west sky move left sky arrow east to west movements from your location 2. What constellations are visible in your evening sky? (Hint Use Reference Lines and Labels under the View menu to turn on the constellations and their labels. Use Site Information under the Data menu to set your loca- tion, date, and time.) The Big Dipper 3. What would the sky look like if you could not see stars fainter than second magnitude? (Hint Use Filters under the View menu to change the magnitude limits.) The sky would be more black at night hardly any visible light from stars. 4. Locate the constellation Orion as seen from your loca- tion, from Earth's north pole, and from Earth's south pole. Orion Abbreviation: Ori Genitive Form: Orionis Description: Orion, the Hunter Pronunciation: oh RY' uhn Genitive Pronunciation: OH' rih OH' nis Sky Database: Constellation Labels RA: 05h 35m 37.2s Dec: +0435'02" RA: 05h 35m 24.0s Dec: +0434'48" (Epoch 2000) Azm: 35312'23" Alt: +3734'54" Rise: 17:57 Transit: 23:39 Set: 05:25 From Cursor position: Angular separation: 8600'46" Position angle: +3115' Part 1 The Sky page 20 Chapter 2 Review Questions 1. What is the difference between the daily and annual motions of the sun? What is the difference between the daily and annual motions of the sun? 2. If Earth did not rotate, could we still define the eclip- tic? Why or why not? If the Earth did not rotate on its axis but did revolve around the sun, then we would still define the ecliptic as the apparent path of the sun around the sky, with respect to the background stars. In one year, the sun would appear to go once around the sky, with respect to the background stars. 3. What would our seasons be like if Earth were tipped 35 degrees instead 23.5 0? What would they be like if Earth's axis were perpendicular to its orbit? 4. Why are the seasons reversed in the southern hemi- sphere? In the Southern Hemisphere, the climatic seasons differ by about six months. This hemisphere has summer when the Northern Hemisphere has winter. The changing seasons are caused by the changing position of the Earth in relation to the sun. When the North Pole slants toward the sun, the Northern Hemisphere receives the most sunlight and it is summer there. When the pole slants away from the sun, the Northern Hemisphere receives the least sunlight and it is winter. Spring begins when the pole starts to slant toward the sun, and autumn begins when the pole starts to slant away again. EQUINOX is either of the two days of the year when the sun is directly above Earth's equator. At these times, the days and nights are of nearly equal length everywhere on Earth. The term equinox comes from a Latin word meaning equal night. The equinoxes occur on March 19, 20, or 21 and on September 22 or 23. In the Northern Hemisphere, the March equinox marks the start of spring and is often called the vernal equinox. The position of the vernal equinox is called the first point of Aries. The word vernal means of spring. The September equinox marks the beginning of autumn and is called the autumnal equinox. The seasons are reversed in the Southern Hemisphere. The time interval from the March equinox to the September equinox is longer than that between the September equinox and the next March equinox. This time difference results from Earth's elliptical (oval-shaped) orbit around the sun. Earth moves faster in its orbit when it is closer to the sun. The distance between Earth and the sun is shortest in January. Therefore, Earth completes the semicircle from the September equinox to the March equinox faster than it does the opposite semicircle. Why are the seasons reversed in the southern hemisphere? 5. Do the phases of the moon look the same from every place on Earth, or is the moon full at different times in different locations? 6. What phase would Earth be in if you were on the moon when the moon was full? at first quarter? at waning crescent? 7. How does the moon slow Earth's rotation, and how does Earth change the moon's orbit? 8. Why have most people seen a total lunar eclipse, while few have seen a total solar eclipse? 9. Why isn't there an eclipse at every new moon and at every full moon? To get an eclipse the sun, moon, and Earth must lie along a straight line. Since the moon's orbit is inclined 5 degrees with respect to the eliptic, at a new moon or full moon the moon may be as much as five degrees above or below the ecliptic where the sun is. So, the moon could miss the needed straight-line allignment by as much as 5 degrees. 10. Why is the moon red during a total lunar eclipse? 11. Why should the eccentricity of Earth's orbit make win- ter in the northern hemisphere different from winter in the southern hemisphere? The earth's orbit is eccentric; the orbit is an ellipse not a perfect circle. During winter in the northern hemisphere the Earth is slightly closer to the sun than on average. During summer in the northern hemisphere (which is winter in the southern hemisphere), the Earth is slightly farther from the sun than on average. Consequently the northern hemisphere winters are expected to be a little milder than the southern hemisphere winters. 12. How could small changes in the inclination of Earth's axis affect world climate? Discussion Questions 1. Do planets orbiting other stars have ecliptics? Could they have seasons? Yes some planets have ecliptics and seasons, and depends on the mass, and size of planets, and what the planets are made of. 2. Why would it be difficult to see prominences if you were on the moon during a total lunar eclipse? Yes it would be difficult. The moon is smaller in size. rosy pink flames called Prominences around the sun Problems 1. If Earth is about 4.6 billion (4.6 x 10 9) years old, how many precessional cycles have occurred? 10 9 x4.6 654 +446 1100 P 1. This is really a chapter 2 problem, since chapter 2 talked about precession. One precessional cycle of the earth lasts tP = 26,000 years. That's the time it takes the Earth to precess around once like a top. Essentially all astronomers think the Earth is approximately tE = 5 x 109 years old. Although there is no formula given in the textbook for this problem, we can easily figure out what the formula should be. A little thought shows that the number of cycles n must be given by n = tE / tP = 5 x 109 / 26,000 = 190,000. 2. Identify the phases of the moon if on March 21 the moon were located at (a) the vernal equinox, (b) the au- tumnal equinox, (c) the summer solstice, (d) the winter solstice. 3. Identify the phases of the moon if at sunset the moon were (a) near the eastern horizon, (b) high in the south, (c) in the southeast, (d) in the southwest. 4. About how many days must elapse between first-quarter moon and third-quarter moon? 5. Draw a diagram showing Earth, the moon, and shadows during (a) a total solar eclipse, (b) a total lunar eclipse, (c) a partial lunar eclipse, (d) an annular eclipse. P 5. See answers in our texbook on page A-1, near the end of the book. 6. Phobos, one of the moons of Mars, is 20 km in diameter and orbits 5982 km above the surface of the planet. What is the angular diameter of Phobos as seen from Mars? (Hint See By the Numbers 3-1). answer is P 6. Phobos has a linear size of s = 20 km. Its distance from Mars is d = 5982 km. The angular size of Phobos is given by the angle formula. theta; = s / d = (20 km) / (5982 km) = 0.00334 radians To convert this answer from radians to degrees remember that there are (360 degrees) per (2 π radians) = (360 degrees) / (2 π radians). That is the conversion factor. Note there are approximately 60 degrees per radian. So, θ = 0.00334 radians = (0.00334 radians) (360 degrees) / (2 π radians) = 0.192 degrees This answer is about one-fifth of a degree, which is smaller than the size of our moon as seen from earth. If we consider each degree to be divided into 60 equal parts, each part is called a minute of arc. The angle θ in this problem is then 11.5 minutes of arc. If we consider each minute of arc to be divided into 60 equal parts, each part is called a second of arc. The 11.5 minutes of arc then becomes 690 seconds of arc. 7. A total eclipse of the sun was visible from Canada on July 10, 1972. When did this eclipse occur next? From what part of Earth was it total? 8. When will the eclipse described in Problem 7 next be total as seen from Canada? Critical Inquiries for the Web 1. There are many Web sites on astrology, but nearly all accept the old superstition as real. Can you find Web sites that analyze astrology logically? Begin at the Web site for the Committee for the Scientific Investigation of Claims of the Paranormal. Committee for the Scientific Investigation of Claims of the Paranormal (CSICOP) contains publications, mailing lists, and links to other paranormal sites. www.csicop.org/ http://www.csicop.org/ CSICOP encourages the critical investigation of paranormal and fringe-science claims from a responsible, scientific point of view and disseminates factual information about the results of such inquiries to the scientific community and the public. 2. Find photographs of recent lunar and solar eclipses. Where did the observers have to go to see the eclipse? Solar Eclipses: 2001 - 2008 The last total solar eclipse occurred on 2002 December 04. It was visible from southern Africa and Australia. Please see our special web site for maps, diagrams, tables, etc.: http://sunearth.gsfc.nasa.gov/eclipse/eclipse.html Solar Eclipses: 2001 - 2008 Date Eclipse Type Eclipse Magnitude Central Duration Geographic Region of Eclipse Visibility 2001 Jun 21 Total 1.050 04m57s e S. America, Africa [Total: s Atlantic, s Africa, Madagascar] 2001 Dec 14 Annular 0.968 03m53s N. & C. America, nw S. America [Annular: c Pacific, Costa Rica] Lunar Eclipses: 2001 - 2008 Date Eclipse Type Umbral Magnitude Total Duration Geographic Region of Eclipse Visibility 2001 Jan 09 Total 1.195 62m e Americas, Europe, Africa, Asia 2001 Jul 05 Partial 0.499 - e Africa, Asia, Aus., Pacific 3. What is the latest news concerning the Milankovitch hypothesis? Is there new evidence to test the theory? Milankovitch hypothesis http://www.geo.arizona.edu/palynology/geos462/21climastro.html THE ASTRONOMICAL THEORY OF CLIMATIC CHANGE (The Milankovitch Hypothesis) Observations on the Earth's orbit: REVIVAL OF Milankovitch Hypothesis: 1955 long records: C. Emiliani 18O/16O noted general correspondence to Milankovitch's curves 1956 accurate chronology: Erickson et al.(1956) Barbados raised beaches dated at 82, 105, and 125 K 1969 terrestrial conformation: Kukla (1970) published loess chronology dated by Uranium-Thorium the dates of for terminations match 45o N insolation. Exploring The Sky 1. How long is daylight on the day of the summer solstice and on the day of the winter solstice for your location? Use Site Information under the Data menu to set your lo- cation and the date. Find the time of sunrise and sunset. Seattle, Washington summer solstice 15 hours winter solstice 10 hours 2. How many days must elapse between new moons? Use the Moon Phase Calendar under the Tools menu to lo- cate new moons. 30 days in December 2003. 3. Where would you have to go to see a total solar eclipse in the next year or two? Use the Eclipse Finder under the Tools menu to select a total solar eclipse. Check Show Path of Totality to see a map of Earth. 3 28 06 Total Solar Eclipse Africa, Europe, Russia. 4. View the total solar eclipse you located in Activity 3. In the Eclipse Finder, click View to see the sun and moon from your present location. Then use Site Information under the Data menu to change your location to a spot in the path of totality. Set the Time Step to 5 minutes and click the arrows to move the moon across the sun. Europe 5. Locate and observe an annular and a total lunar eclipse. annular and a total lunar eclipse Part 1 The Sky page 40 Chapter 3 Review Questions 1. Why did Greek astronomers conclude that the heavens were made up of perfect crystalline spheres moving at constant speeds? This idea can ultimately be traced back to Plato and his student Aristotle. The two big ideas of the time were that the circle was the perfect geometrical shape and the only perfect motion was uniform motion. These two ideas, when coupled with the belief that the heavens are the place of perfection, naturally lead to the idea of giant crystalline spheres in the heavens moving at constant speed as the best explanation for the observed motions in the sky. The philosopher Plato taught that heavens were perfect and unchangeable. To Plato, the most perfect geometrical figure was the sphere and the natural motion of a sphere is rotation about an axis. Additionally, perfect motion was uniform, that is, at constant speed. Therefore, all heavenly bodies being perfect must be composed of perfect geometrical figures (spheres) moving in perfect motion (uniform circular motion). 2. Why did classical astronomers conclude that Earth had to be motionless? It was believed that the Earth was motionless because there was no observed parallax in the stars or planets. Actually, the parallax is too small to observe with the naked eye. Classical astronomers thought the earth was motionless for several reasons. Two reasons mentioned in the text were a) no one felt the earth moving the way you feel motion while riding a chariot or a horse, and 2) no one saw any parallax in the position of any stars. Some classical astronomers also argued that the earth must be at rest because if it were moving fast, we would constantly feel the air rush by us as we do when we travel fast. The classical astronomers concluded that Earth had to be motionless because no parallax was detected and no motion was felt. If Earth moved, they reasoned, the vibrations generally associated with motion should be felt. Additionally, if Earth orbited the Sun, then nearby stars should show parallax. 3. How did the Ptolemaic model explain retrograde motion? Retrograde motion of the planets was explained by using epicycles moving along a deferent. If the planet rides along the epicycle, the planet's apparent retrograde motion closely matches that observed in the sky. The Ptolemaic model explained retrograde motion by saying the planet undergoing retrograde motion was on the near side of its epicycle, traveling east to west as we view it from earth. The Ptolemaic model explained retrograde motion using a complicated system of wheels within wheels. Each planet moved in a small circle called an epicycle, and the center of the epicycle moved along a larger circle around Earth called a deferent. Earth was placed slightly off center of the deferent. As the planet is carried by both the deferent and the epicycle, it will appear to go through retrograde motion when the planet is on the inside half of its motion of the epicycle. (See the Concept Art). Both the Ptolemaic and Copernican system assumed uniform circular motion and needed to employ equants and epicycles. Ptolemys system required them to account for the variation in the orbital speed of a given planet and retrograde motion. Copernicus system required them to account only for the variation in the orbital speed of the planets. Both also assumed a celestial sphere at a great distance. a) Greek philosophy taught that & new stars had to be closer to Earth than the Moon because the heavens were perfect and unchangeable. This new star of 1572 should therefore be close to Earth and show daily parallax as the Moon and planets do. However, Tycho observed no parallax and reasoned that it was farther away than the Moon and probably as far away as the distant stars. 4. In what ways were the models of Ptolemy and Coperni- cus similar? 5. Why did the Copernican hypothesis win gradual accep- tance? 6. Why is it difficult for scientists to replace an old para- digm with a new paradigm? Paradigms are difficult to replace because paradigms structure our view of nature. Our ideas and concepts of how nature works are based on well accepted scientific ideas. Seeing the limitations of these accepted ideas is very difficult for anyone who works within them. Generally, the ideas and concepts that form a paradigm are so apparently clear and universal, that their limitations are difficult to identify. A scientific paradigm is powerful because it shapes our perceptions. It determines what we judge to be important questions and what we judge to be significant evidence. Overthrowing an outdated paradigm is not easy, because we must learn to see nature in an entirely new way. 7. Why did Tycho Brahe expect the new star of 1572 to show parallax? Why was the lack of parallax evidence against the Ptolemaic model? a) Tycho's planetary system incorporated three elements of the Ptolemaic system: 1) solid celestial sphere, 2) motionless Earth at the center of the celestial sphere, and 3) the Sun orbits Earth. b) Tycho's system incorporated one of Copernicus ideas in that all planets, except Earth, orbited the Sun (but Tycho had the Sun orbit Earth). Tycho expected the "new star" of 1572 to show parallax because he expected it to be much closer than all of the real stars on the distant sphere of fixed stars. Tycho, and just about everybody else, believed that set of real stars on the distant sphere of fixed stars doesn't change; stars there don't disappear and new stars there don't suddenly appear. 8. How was Tycho's model of the universe similar to the Ptolemaic model? How did it resemble the Copernican model? b) The Ptolemaic system taught that the heavens were perfect and unchanging. Since a new star appeared to be located on the celestial sphere, the basic assumptions of the Ptolemaic system were not valid. laws of planetary motion Three laws, based on precise observations of the motions of the planets by Tycho Brahe, which summarize the motions of the planets about the Sun. Copernican revolution The realization toward the end of the sixteenth century that Earth is not at the center of the universe. 9. Explain how Kepler's laws contradict uniform circular motion? In uniform circular motion the objects move along circular paths at constant speed. Kepler's first law states that the planets follow elliptical paths, not circular ones. Secondly, Kepler's second law states that the planets move faster near perihelion than they do near aphelion. Therefore the planets are not always moving at the same speed at all points in their orbits. Kepler's first law says planets move in ellipses not circles. Kepler's second law says that a planet moves at different speeds at different points in its orbit. These two laws thus contradict the notion that planets move in perfect circles at uniform (i.e. constant) speed. 10. What is the difference between a hypothesis, a theory, and a law? A hypothesis is a single assertion or conjecture that can and must be tested. The hypothesis can be either true or false. A theory is a system of rules and principles that can be applied to a wide variety of circumstances. A scientific law is a theory that almost everyone accepts as true. Such a theory is usually very specific and has been tested many times. One can think of a hypothesis as a single statement of how one believes nature works, but with little evidence for the validity of that belief. A theory is a developed set of ideas and mathematical descriptions of how nature works; it has been tested only in a very limited sense. A law is a theory that has been tested in nearly all conceivable manners so that nearly everyone is willing to accept its description of nature and the predictions it makes. These are words that mean whatever I want them to mean. But Seeds says that a hypothesis is a single assertion or conjecture that must be tested. A theory is a system of rules and principles that can be applied to a wide variety of circumstances. A law is a theory that has been refined, tested, and confirmed so often that scientists have great confidence it. These words are probably more important in a course on the philosophy of science than they are to scientists. evolutionary theory A theory which explains observations in a series of gradual steps, explainable in terms of well-established physical principles. 11. How did the Alfonsine Tables, the Prutenic Tables, and the Rudolphine Tables differ? The Alphonsine Tables were based on the Ptolemaic model and used epicycles and deferents to predict the locations of the planets. The Prutenic Tables were based on the Copernican model and used uniform circular motion around the Sun to predict the locations of the planets. The model used epicycles to account for the variation in the speed of the planets at different locations in their orbits. They were not significantly more accurate than the Alphonsine Tables. The Rudolphine Tables were based on the heliocentric model incorporating Kepler's three laws of planetary motion. The Rudolphine Tables were much more accurate than the others. 12. Review Galileo's telescope discoveries, and explain why they supported the Copernican model and contra- dicted the Ptolemaic model. The observation of moons orbiting Jupiter contradicted the Ptolemaic idea of the Earth as the center of everything. Obviously, something could orbit about something other than the Earth. The observations of the craters of the moon, spots on the sun, and stars in the Milky Way brought into question the idea of the perfection of the heavens. However, the observation of the phases of Venus was something that did disprove the geocentric model. This one observation proved that Venus did not orbit about the earth, but instead orbited about the sun. This supported the Copernican model and contradicted the Ptolemaic model. Galileo made three significant telescopic discoveries that argued against the Ptolemaic model. First he observed that the surface of the Moon was not perfect, but had in fact mountains and valleys, much like those on Earth. Earth was a very changeable body, while heavenly bodies including the Moon, were seen as perfect in the Ptolemaic model. Second, Galileo found four small bodies that orbited Jupiter. Here was clear evidence that not all objects orbited Earth, for at least these four were centered on Jupiter. Lastly, Galileo observed that Venus displayed phases consistent with it orbiting the Sun and inconsistent with it riding on an epicycle located between the Sun and Earth. 13. Galileo was condemned by the Inquisition, but Kepler, also a Copernican, was not. Why not? Kepler was spared primarily because he was not a Catholic. He was a Lutheran, and an ordained Lutheran at that. Kepler also worked in northern Germany and Denmark, far from Italy and Rome. Galileo on the other hand was Catholic, lived near the seat of Catholic power, and wrote articles in Italian for the common people of his day. He was a much bigger threat to the doctrines of the Catholic Church than was Kepler. Newton concluded that gravitation had to be universal because the exact same mathematical description of gravity that was applicable to falling objects on Each could be used to describe the orbit of the Moon and an the known planets. 14. Why did Newton conclude that gravitation had to be universal? Newton showed that Earths mass controlled both the falling of objects near Earths surface and the orbit of the Moon. He then showed that by using the Sun’s mass, this description of gravity applied to the motions of all the planets in their orbits about the Sun. Therefore, gravity and Newton’s description of it were applicable to all objects with mass.
The lunar orbit is contrlled by the gravitational force between Earth and the Moon. Everyday experience tells us, along with Newton's law of gravity, that Earth will cause objects to fall toward it. 15. Explain why we might describe the orbital motion of the moon with the statement, "The moon is falling." More significantly, the Moon would move in a straight line and leave Earth if gravity did not cause it to orbit. Gravity causes the Moon to move (fall) toward Earth at the same rate that Earth's surface, being spherical, curves away from it. Therefore, the Moon is continuously falling toward Earth, but its horizontal (tangential) motion is keeping it in an orbit around Earth. The moon is attracted to the earth by gravity. If the moon were placed anywhere in its orbit at rest, it would fall directly to the earth. In reality, the moon is moving fast in a nearly circular orbit around the earth. If the earth's gravity were suddenly turned off, the moon would no longer circle around the earth, but rather the moon would move in a straight line, getting farther and farther from the earth. It is the earth's gravity which keeps the moon from moving out away from the earth. The earth's gravity keeps bending the moon's path so it stays in a circular orbit around the earth. The earth's gravity keeps making the moon fall away from a straight line path and curving the moon's path around the earth. Discussion Questions 1. Historian of science Thomas Kuhn has said that De Revo- lutionibus was a revolution-making book, but not a rev- olutionary book? How was it classical? 2. Why might Tycho Brahe have hesitated to hire Kepler? Why do you suppose he appointed Kepler his scientific heir? 3. How does the modern controversy over creationism and evolution reflect two ways of knowing about the physi- cal world? This is actually a fairly deep question that deserves a long answer, but at least one of the points I think Seeds is trying to get at by asking the question concerns Cause and Effect. Evolution seeks the cause for the observation (i.e., the effect). In fact, as Seeds says, "All of science is focused on understanding the causes of the effects we see." Although scientific theories, including evolution, are almost never complete and from time to time require trashing and re-molding, they are all ultimately trying to explain the observed effects by tracing backwards to causes. On the other hand, creationism, in its strictest sense, starts with an assumption; i.e., that there is a creator. And while many creationist theories may attempt to explain observed effects by tracing backwards to their causes; ultimately any creation theory must end at the creator. So, does a scientists (creationist or evolutionist) start with observed effects and try to build models to explain causes; or does a scientist start with the cause and try to build a model to explain the effect? In reality, it is never so neat. Science is brutal and messy and goes both ways. Problems 1. If you lived on Mars, which planets would describe ret- rograde loops? Which would never be visible as cres- cent phases? The planets farther from the sun than Mars would describe retrograde loops. These would be Jupiter and Saturn. The other outer planets would be too dim to see with the unaided eye, although they would also describe retrograde loops. These same planets would never be visible as a crescent phase. 2. Galileo's telescope showed him that Venus has a large angular diameter (61 seconds of arc) when it is a cres- cent and a small angular diameter (10 seconds of arc) when it is nearly full. Use the small-angle formula to find the ratio of its maximum distance to its minimum distance. Is this ratio compatible with the Ptolemaic universe shown on page 45? P 2. The small angle formula is θ = s / d. The units of the right hand side are meters divided by meters, which in this context is a radian. Recall earlier in class we said that 1 radian = 360 degrees / (2 π). However if you want to find θ not in radians and not degrees but in seconds of arc where one second of arc is 1/3600 of a degree, then you need to know that 1 radian = 206,265 seconds of arc. With this conversion factor the θ equation becomes (see page 31 of our text): theta; = (s / d) (206,265 seconds of arc / radian). In this problem we are asked about distance, so solve this equation for d. d = (s / θ) (206,265 seconds of arc / radian) When Venus is at its maximum distance d max, then its angular size is at its minimum value θ min, and d max = (s / θ min) (206,265 seconds of arc / radian). When Venus is at its minimum distance d min, then its angular size is at its maximum value θ max, and d min = (s / θ max) (206,265 seconds of arc / radian). The problem asks for the ratio of d max to d min, so divide the second to last equation by the last equation. This gives d max / d min = θ max / θ min = 61 / 10 = 6.1 Notice that the size of Venus s and the constant 206,265 both algebraically cancel out. This result is not compatible with the Ptolemaic model. Figure 4-7 in our text shows that in the Ptolemaic model the maximum distance of Venus is only approximately 1.2 times the minimum distance of Venus, not 6.1 times. 3. Galileo's telescopes were not of high quality by modern standards. He was able to see the rings of Saturn, but he never reported seeing features on Mars. Use the small- angle formula to find the angular diameter of Mars when it is closest to Earth. How does that compare with the maximum diameter of Saturn's rings? 4. If a planet had an average distance from the sun of 10 AU what would its orbital period be? Using Kepler's Third Law, P squared equals a cubed, where P is expressed in units of years and a is expressed in units of AU. In this case, a = 10AU. Solving for P gives 32 years. 5. If a space probe were sent into an orbit around the sun that brought it as close as 0.5 AU to the sun and as far away as 5.5 AU, what would its orbital period be? In this problem the semi-major axis of the probe is the average of its maximum and minimum distances. This gives the value a = (0.5 AU + 5.5 AU) / 2 = 3 AU. Now solve Kepler's third law (P / 1 yr)2 = (a / 1 AU)3 for P, and plug in the probe's semi-major axis. P = (a / 1 AU)3/2 yr = (3 AU / 1 AU)3/2 yr = 5.2 yr 6. Pluto orbits the sun with a period of 247.7 years. What is its average distance from the sun? Using Kepler's Third Law again. In this case, P = 247.7 years. Solving for a gives 39.4 AU. Solve Kepler's third law for the semi-major axis of Pluto's orbit, which is its average distance from the sun. (P / 1 yr)2 = (a / 1 AU)3 Therefore, a = (P / 1 yr)2/3 AU = (247.7 yr / 1 yr)3/2 AU = 39.4 yr 7. Calculate the circular velocity of Venus and Saturn around the sun. Hint: The mass of the sun is 2 x 10 30kg. 8. The circular velocity of Earth around the sun is about 30 km/s. Are the arrows for Venus and Saturn correct in figure 4-4b? Hint: see problem 7. 9. What is the orbital velocity of an Earth satellite 42,000 km from Earth? How long does it take to circle its orbit once? The orbital velocity of an object in circular orbit is given by a formula based on Newton's study of motion: Vc = sqrt (G M / r), where sqrt means square root. The G is a universal constant, M is the mass of the object at the center of the orbit, and r is the radius of the circular orbit (see page 62 of our text). In our problem, Vc = sqrt (G M / r) = sqrt ( [6.67 x 10-11 m3 / kg-s2] [5.98 x 1024 kg] / [42,000,000 m] ) = 3080 m/s. To find out how long it takes to circle the orbit once, we need to use the distance equals velocity times elapsed time formula: d = v t. Since we are asked for time, solve this equation for t to get t = d / v. The distance d is just the circumference of the circular orbit which is given by 2 π r. Putting this all together, we get t = d / v = (2 π r) / Vc = (2 π 42,000,000 m) / (3080 m/s) = 85,700 s This answer is close to 24 hours. An object which orbits in a time equal to the time it takes the earth to rotate would stay stationary in the sky as seen from earth. This is what a telecommunications satellite does. Critical Inquiries for the Web 1. The trial of Galileo is an important event in the history of science. We now know, and the Church now recog- page 64 chapter 4 nizes, that Galileo's view was correct, but what were the arguments on both sides of the issue as it was un- folding? Research the Internet for documents chroni- cling the trial. Galileo's observations and publications, and the position of the Church. Use this information to outline a case for and against Galileo in the context of the times in which the trial occurred. http://www.facingthechallenge.org/galileo1.htm The trial of Galileo: A modern myth? Whenever people talk about the (supposed) conflict between science and religion, it is not long before two trials come up in the conversation: One is the Scopes trial in the USA in 1925. The other is the trial of Galileo in 1633. Both of these have been misrepresented by fiction, misunderstanding, and occasionally downright lies. The Italian astronomer Galileo Galilei lived from 1564 to 1642. The classic picture is of Galileo as the humble but heroic scientist who looked through his telescope and discovered that the Earth went round the Sun, rather than the Sun going round the Earth. This brought him into conflict with the anti-scientific and ignorant authorities of the Church (because the Bible supposedly teaches that the Earth is at the center and the Sun goes round the Earth). Because of this, the Church persecuted Galileo, brought him to trial, condemned, imprisoned, and tortured him. this is a modern 'science versus faith' myth: 2. It's hard to imagine that an observatory could exist be- fore the invention of the telescope, but Tycho Brahe's observatory at Hveen was a great astronomical center of its day. Search the Web sites on Tycho and his instru- ments and describe what an observing session at Hveen might have involved. http://csep10.phys.utk.edu/astr161/lect/history/brahe.html The instruments of Brahe allowed him to determine more precisely than had been possible the detailed motions of the planets. http://www.hao.ucar.edu/public/education/sp/images/tycho.3.html Here are other noteworthy instruments built by Tycho and his team of instrument builders, listed in chronological order: 1576: Brass Azimuthal Quadrant Brass azimuthal quadrant, 65 centimeters in radius. Built in 1576 or 1577, this was one of the first instrument built at Hveen, and was used for observations of the 1577 comet. Estimated accuracy of 48.8 seconds of arc. 1580: Computational Globe Tycho's great globe, about 1.6 meter in radius. Over 10 years in the making, this instrument came in service in late 1580. Most of the work involved making the hollow wooden globe as perfectly spherical as possible, after which it was covered in brass plates. The globe had two primary scientific uses; it came to be used to record the position of stars observed by Tycho. By 1595 he had 1000 accurately observed stars inscribed on the globe. However, it was originally intended as a computational device. By means of auxiliary circles, the local azimuth/altitude coordinates, as measured with Tycho's instruments, were converted into the conventional celestial coordinates used to record stellar and planetary positions. 1581: Armillary Sphere 1582: Triangular Sextant 1585: Great Equatorial Armillary 1586: Revolving Quadrant 1588: Revolving Steel Quadrant Large instruments such as these, with improved sighting devices and measuring scales, as well as Tycho's advanced procedures to correct for atmospheric refraction, allowed him to compute stellar and planetary positions consistently accurate to within seconds of arc. Tycho's determination of the tropical year was too small by about one second, and his determination of the Earth's orbital tilt (which Tycho, committed to the Earth's fixity as he was, referred to as the angle between the ecliptic and the celestial equator) by half a minute of arc. His observations of planetary motion, particularly that of Mars, provided the crucial data for later astronomers like Kepler to construct our present model of the solar system. Brahe made careful observations of a comet in 1577. By measuring the parallax for the comet, he was able to show that the comet was further away than the Moon. observing session at Hveen Tycho Brahe's observatory at Hveen Exploring The Sky 1. Go to Stonehenge in southern England and watch the sun rise on the morning of the summer solstice. Where does it rise on the morning of the winter solstice? Northwest 2. Observe Mars going through its retrograde motion. Hint Use Reference Lines under the View menu to turn on the ecliptic. Be sure you are in Free Rotation under the Orientation menu. Locate Mars and use the time skip arrows to watch it move.) Mars Mars Set: 12:21 AM on 12/21/03 Rise: 12:22 PM on 12/21/03 Transit: 6:20 PM on 12/21/03 RA: 00h 10m 33.9s Dec: +0054'56" Azm: 27346'32" Alt: +0212'06" (with refraction: +0229'22") Phase: 87.033%, Apparent magnitude: 0.02 Heliocentric ecliptical coordinates: l: 4500'10.9" b: -0008'52.3" r: 1.461125 Geometric geocentric ecliptical coordinates: l: +0247'43" b: -0012'44" r: 1.017665 Mean geometric ecliptical coordinates: l: +0247'29" b: -0012'45" r: 1.017717 True equatorial coordinates: RA: 00h 10m 34s Dec: +0054'50" Physical Data Dec Earth: -26.08, Dec Sun: -26.08 Position angle: 331.68 Longitude of central meridian: 88.86 Defect of illumination: 1.19, Position angle: 66.57 Apparent angular diameter: 9.20 From SAO 109373: Angular separation: 0750'34" Position angle: +26351' 3. Compare the size of the retrograde loops made by Mars, Jupiter, and Saturn. Mars is smaller retrograde loops. 4. Can you recognize the effects of Kepler's second law in the orbital motion of any of the planets? Hint Use 3D Solar System Mode under the View Menu. Yes they all seem to have effects The terrestrial planets appeared small orbital motions while Jovian planets were bigger orbital motions. more outward type. 5. Can you recognize the effects of Kepler's third law in the orbital motion of the planets? Yes it was. Chapter 4 page 65 Review Questions 1. Why would you not plot sound waves in the electro- magnetic spectrum? Because sound waves are acoustic and not electromagnetic in nature. Sound waves require a medium to propagate. EM waves propagate through a vacuum by means of the oscillating electric and magnetic fields. Thats why in space no one can hear you scream. Sound waves are a mechanical disturbance that travels through the air from source to ear. Sound requires a medium, so on the moon, where there is no air, there can be no sound. 2. If you had unlimited funds to build a large telescope, which type would you choose, a refractor or a reflector? Why? A reflector, because you get more aperture for your buck. Also large refractors suffer from sag due to the fact that they are very thick and are supported only by the edges. This sag practically limits the diameter of a refractor. 3. Why do nocturnal animals usually have large pupils in their eyes? How is that related to astronomical telescopes? The larger the lens is the more light is collected. 4. Why do optical astronomers sometimes put their tele- scopes at the tops of mountains, while radio astrono- mers sometimes put their telescopes in deep valleys? Optical astronomers put their telescopes at the tops of tall mountains for two reasons a. The atmosphere absorbs light from space. At the tops of tall mountains the telescope is above much of the atmosphere. b. Turbulence in the atmosphere degrades the images of astronomical objects. At the tops of tall mountains the air is often less turbulent than lower down. Radio astronomers put their telescopes in deep valleys to shield them from radio waves produced on earth by sources such as radio transmitters and electrical equipment that produces radio noise. Nearly all major observatories are located far from major cities and usually on high mountains. Astrono- mers avoid cities because light pollution, the bright- ening of the night sky by light scattered from artificial outdoor lighting can make it impossible to see faint objects. Optical Telescopes Astronomers build optical telescopes to gather light and focus it into sharp images. This requires sophisti- cated optical and mechanical designs, and it leads astronomers to build gigantic telescopes on the tops of high mountains. Radio Telescopes in deep valleys detect weak radio sources. Fortunately, radio astronomers can combine two or more radio telescopes to improve the resolving power. Such a linkup of radio telescopes is called a radio interferometer and has the resolving power of a radio telescope whose diameter equals the separation of the radio telescopes. For example the Very Large Array radio interferometer uses multiple radio dishes spread across the New Mexico-Arizona desert to simulate a single radio telescope with a diameter of 40 km. 25 miles. It can produce radio maps with a resolution better than 1 second of arc. Radio telescopes poor resolution, low intensity, and interference. 5. Optical and radio astronomers both try to build large telescopes but for different reasons. How do these goals differ? Radio telescopes It also depends on the wavelength of the radiation. A very long wavelengths, like those of radio waves, im- ages become fuzzy because of the large diffraction fringes. As with an optical telescope the only way to improve the resolving power is to build a bigger tele- scope. Consequently radio telescopes must be quite large. Because there are two ways to focus light, there are two kinds of astronomical telescopes. Refracting telescopes use a large lens to gather and focus the light. Reflecting telescopes use a concave mir- ror. The advantages of the reflecting tele- scope have made it the preferred design for modern observatories. 6. What are the advantages of making a telescope mirror thin? What problems does this cause? Thinner mirros also cool faster at nightfall and suffer less distortion from uneven expansion and contraction. These mirrors are cheaper, and they are quicker to make because they don't take as long to grind into shape. They are also lighter which affords several advantages. Thin mirrors can be supported with a lighter support system and moved with smaller motors. Thin mirrors cool down to the evening temperature more quickly than thicker mirrors. Thin mirrors can be adjusted active or adaptive optics. A disadvantage to thin mirrors is that they are more delicate. 7. Small telescopes are often advertised as "200 power" or "magnifies 200 times." As someone knowledgeable about astronomical telescopes, how would you improve such advertisements? Add the diameter of the primary to the ad. Also add something about what kind of telescope - refractor or reflector. 8. An astronomer recently said, "Some people think I should give up photographic plates." Why might she change to something else? They can detect both bright and faint objects in a single exposure, are much more sensitive than a photographic plate, and can be read directly into computer memory. 9. What purpose do the colors in a false-color image or false- color radio map serve? Astronomers also manipulate images to produce false-color images in which the colors represent different levels of intensity and are not related to the true colors of the object. 10. How is chromatic aberration related to a prism spectro- graph? Analyze light in detail we need to spread the light out according to wavelength into a spectrum, a task performed by a spectrograph. When placing a prism in a thin beam of light that is streaming into a darkened room a beautiful spectrum appears on the far wall. 11. Why would radio astronomers build identical radio telescopes in many different places around the world? Radio astronomers can combine two or more radio telescopes to improve the resolving power. A linkup of radio telescopes is called a radio interferometer. 12. Why do radio telescopes have poor resolving power? Because the radio telescopes are small in size diameter. The only way to improve the resolving power is to build a bigger telescope. Resolving power depends on the ratio of wavelength of the waves detected divided by the diameter of the telescope. The smaller the ratio, the better the resolving power. Radio telescopes may have large diameters compared to optical telescopes, but the wavelengths they observe at are gigantic compared to the wavelength of visible light. Consequently the wavelength-over-diameter ratio for radio a radio telescope is much larger than that ratio for an optical telescope. Therefor the resolution of a radio telescope is poor. 13. Why must telescopes observing in the far infrared be cooled to low temperatures? If a telescope observes at far-infrared wavelengths, then it must be cooled. Infrared radiation is emitted by heated objects, and if the telescope is warm it will emit many times more infrared radiation than that coming from a distant object. 14. What might we detect with an X-ray telescope that we could not detect with an infrared telescope? The obvious answer is that we would detect X-rays. X-rays are emitted by very hot objects such as accretion disks around black holes or neutron stars. Such objects emit most intensely in the X-ray region and not as intensely in the IR region of the EM spectrum. X-ray images can give us information about the heavens that we can get in no other way. Exploding stars and massive black holes. 15. If the Hubble Space Telescope observes at visual wave- lengths, why must it observe from space? Avoid turbulence in the Earth's atmosphere. The Hubble Space Telescope wouldn't have to observe from space. It would work from the ground, however from the ground it would suffer from the problems mentioned in Review Question 4 above. Discussion Questions 1. Why does the wavelength response of the human eye match so well the visual window of Earth's atmosphere? I suppose one could argue that biological systems seem to optimize themselves to take maximum advantage of the niche they have. In this case the niche is being able to take the best advantage of the available daytime light, just as 'growing long necks' was an adaptation to eating leaves off tree tops. Also, I think it is true that photons in the optical range of wavelengths are at energies that are compatible with the widest range of biochemical reactions involving carbon compounds, and the flux of these photons from the Sun is maximal in the visual spectrum at the surface of the earth. I do not believe that the atmospheric window has exactly the same detailed transmission as the responsivity of our retinal photoreceptors, but I think the bandwidth from blue to red is very similar. The small size of the human eye. 2. Basic research in chemistry, physics, biology, and similar sciences is supported in part b industry. How is as- tronomy different? Who funds the major observatories? The Federal Governments. Federally funded ground-based observatories provide extremely capable and expensive telescopes such as the Very Large Array, the Gemini Infrared Observatory, etc. 1. The thickness of the plastic in plastic bags is about 0.001 mm. How many wavelengths of red light is this? s another problem where the text doesn't give you a formula to mindlessly plug numbers into. You have to figure out a fr The ula for the answer. If the thickness ofthe plastic is t = 0.001 mm = 1 x 10-6 m and the wavelength of red light is λ = 700 x 10-9 m, then the number n of wavelengths that can fit is n = t / lambda; = (1 x 10-6 m) / (700 x 10-9 m) = 1.4 2. Measure the actual wavelength of the wave in Figure 5-1. In what portion of the electromagnetic spectrum would it belong? The wavelength of the wave in Figure 5-1 is 3cm. From Figure 5-2, the EM spectrum, 3cm is located somewhere between the microwave and UHF regions. 3. Compare the light-gathering powers of a 5-m telescope and a 0.5-m telescope. 4. How does the light-gathering power of the largest tele- scope in the world compare with that of the human eye? Hint: Assume that the pupil of your eye can open to about 0.8 cm. When the problem asks us to compare two light gathering powers, they want us to find the ratio of the light gathering power of the telescope LGPT divided by the light gathering power of the eye LGPE, as in the examples on page 76 of our textbook. The required ratio depends on the diameter DT of the telescope and the diameter DE of the eye. The formula from page 76 is LGPT / LGPE = (DT / DE)2 = (11 m / 0.008 m)2 = 1,900,000 5. What is the resolving power of a 25-cm telescope? What do two stars 1.5 seconds of arc apart look like through this telescope? 6. Most of Galileo's telescopes were only about 2 cm in di- ameter. Should he have been able to resolve the two stars mentioned in Problem 5? 7. How does the resolving power of the 5-m telescope compare with that of the Hubble Space Telescope? Why does the Hubble Space Telescope outperforms the 5-m telescope? The resolving power of a telescope optics is given by α = 11.6 cm sec of arc / D, where is the diameter of the telescope. For the Hubble space telescope, D is 240 cm, and so the resolving power α is 0.048 sec of arc. For the 5 m telescope, D is 500 cm, and so its resolving power α is 0.023 sec of arc, which is about twice as good as the resolving power of Hubble (remember a smaller α is better). However, the 5 m telescope on the ground cannot actually observe with a 0.023 sec of arc resolution; the earth's atmosphere limits the resolution to about 0.5 sec of arc. Consequently, Hubble will have a resolution which is about ten times better then the resolution of the 5 m telescope, even though Hubble is smaller. 8. If we build a telescope with a focal length of 1.3 m, what focal length should the eyepiece have to give a magnification of 100 times? Page 76 tells us that the magnification M = FO / FE, where FO is the focal length of the objective and FE is the focal length of the eyepiece. In this problem we want to find the focal length of the eyepiece, so solve our equation for FE. FE = FO / M = (1.3 m) / (100) = 0.013 m = 1.3 cm 9. Astronauts observing from a space station need a tele- scope with a light-gathering power 15,000 times that of the human eye, capable of resolving detail as small as 0.1 second of arc, and having a magnifying power of 250. Design a telescope to meet their needs. Could you test your design by observing stars from Earth? 10. A spy satellite orbiting 400 km above Earth is suppos- edly capable of counting individual people in a crowd. What minimum-diameter telescope must the satellite carry? Hint Use the small-angle formula. The small angle formula from earlier in the textbook and from class is θ = (s / d) (206265 sec of arc / radian). Let's first calculate the angular size θ of a person as seen from a distance d of 400 km. The linear size of a person as seen from above is the person's shoulder width which is about 0.5 m. Therefore θ = (s / d) (206265 sec of arc / radian) = (0.5 m / 400,000 m) (206265 sec of arc / radian) = 0.26 seconds of arc. Now let's use our resolving power formula from page 76, α = 11.6 cm sec of arc / D, to find D the diameter of the telescope needed. To find D, solve this last formula for D. D = 11.6 cm sec of arc / α = (11.6 cm sec of arc) / (0.26 sec of arc) = 45 cm. This solution assumes that the atmosphere will allow the telescope to achieve it full 0.26 seconds of arc resolving power. Critical Inquiries for the Web 1. How do professional astronomers go about making ob- servations at major astronomical facilities? Visit sev- eral observatory Web sites to determine the process an astronomer would go through to secure observing time and make observations at the facility. Depends on what the weather is on certain days or nights for viewing the objects in the night sky. Astronomers work through a schedule before using a telescope to know when to use one on a certain day or night. http://www.noao.edu/kpno/ Kitt Peak National Observatory Instruments recently is use SPECTROSCOPY Mayall 4-m Telescope 2.1-m Telescope 2. NASA is in the process of completing a fleet of four space-based "Great Observatories." (The Hubble Space Telescope is one: what are the others?) Examine the cur- rent state of these missions by visiting their home pages on the Internet. What advantages would these facilities have over ground-based observatories? fleet of four space-based "Great Observatories gamma ray radio telescopes http://legislative.nasa.gov/hearings/goldin2-16.html We started off the year with the launch of Deep Space One, a mission to test 12 revolutionary technologies including spacecraft autonomy and ion propulsion. The Submillimeter Wave Astronomy Satellite (SWAS), a small explorer mission, was launched to study the chemical composition of interstellar gas clouds. We launched Stardust on February 7, 1999 to rendezvous with comet Wild -2 in 2004, and bring back to Earth a sample of comet dust in 2006. In Earth Science, we launched Landsat-7, the continuation of the successful Landsat program and Terra, our flagship mission to study the Earth as a system. FUSE, the Far Ultraviolet Spectroscopic Explorer, was launched on June 24, 1999, to observe the universe in the ultraviolet and try to answer questions such as what conditions existed in the universe a few minutes after the Big Bang. Exploring The Sky 1. Astronomical telescopes using equatorial mountings must be aligned precisely with the north celestial pole. Locate Polaris and determine how far it is from the north celestial pole. Hint: Use Reference Lines under the View menu, and check Grid under Equitorial. Be sure the spacing is set to auto/fine. Then locate the Little Dipper and zoom in on Polaris.) Not very far. 04 hours 00m gride line. Polaris The North Star SAO 308 GSC 4628:237, HIP 11767, PPM 431, HD 8890, B+88 8 Flamsteed-Bayer: 1-Alpha Ursae Minoris Spectral: F7:Ib-IIv SB **** Data from Hipparcos Catalog **** Proper motion (mas/yr): RA = 44.22, Dec = -11.74 Magnitudes Bt: 2.756, Vt: 2.067 Parallax: 7.560 mas, 132.2751 pc Distance: 431.42 light-years, 27283753.74 astronomical units Magnitude: 1.97 RA: 02h 36m 35.884s Dec: +8917'14.749" RA: 02h 31m 49.084s Dec: +8915'50.794" (Epoch 2000) Azm: 0054'49" Alt: -4759'23" Always below horizon. Transit: 18:15 Position error: 0.60 mas From Cursor position: Angular separation: 0006'46" Position angle: +30805' SAO 623 GSC 4625:120, HIP 19454, PPM 696, HD 22701, B+86 51 Spectral: F5IV **** Data from Hipparcos Catalog **** Proper motion (mas/yr): RA = 142.70, Dec = -87.62 Magnitudes Bt: 6.294, Vt: 5.878 Parallax: 24.440 mas, 40.9165 pc Distance: 133.45 light-years, 8439655.16 astronomical units Magnitude: 5.84 RA: 04h 11m 37.769s Dec: +8638'33.113" RA: 04h 10m 01.638s Dec: +8637'34.099" (Epoch 2000) Azm: 35850'16" Alt: -5053'37" Always below horizon. Transit: 19:38 Position error: 0.65 mas From Polaris: Angular separation: 0243'18" Position angle: +14900' SAO 460 GSC 4628:68, HIP 17195, PPM 624, HD 14369, B+88 9 Spectral: F0 **** Data from Hipparcos Catalog **** Proper motion (mas/yr): RA = -52.23, Dec = -31.20 Magnitudes Bt: 8.564, Vt: 8.146 Parallax: 11.350 mas, 88.1057 pc Distance: 287.36 light-years, 18173142.71 astronomical units Magnitude: 8.10 RA: 03h 46m 31.501s Dec: +8907'21.839" RA: 03h 40m 54.020s Dec: +8906'17.577" (Epoch 2000) Azm: 35914'25" Alt: -4654'02" Always below horizon. Transit: 19:37 Position error: 0.80 mas From SAO 209: Angular separation: 0031'17" Position angle: +6439' appeared very near to the north pole by the first circle on top using The Sky program software. how far is Polaris from the north celestial pole Polaris is so far away that its starlight is coming in parallel to the north celestial pole. Since the star Polaris is near the celestial north pole it does not move very far from one spot as the Earth turns so the camera can be fixed. page 89 Chapter 5 Astronomical Tools Review Questions 1. Why might we say that atoms are mostly empty space? 1. Sound waves are not electromagnetic waves and do not travel at the speed of light. Consequently, the wavelength and frequency are each vastly different from those of electromagnetic waves and related through a different speed. The energy associated with a sound wave does not depend only on its wavelength, as it does for electromagnetic waves. 1. Atoms are composed of neutrons, protons and electrons. The proton and neutron each have a diameter of about 1.6 x 10 -6 nm, and the electron cloud has a diameter of about 0.4 nm. Therefore, most of the region inside the electron cloud is empty. 2. What is the difference between an isotope and an ion? 2. An ion is an atom that has unequal numbers of electrons and protons. An ion can be positive or negative. Ionization means removing one or more electrons, leaving a positive ion behind.) Isotopes are determined by the relative number of neutrons they have compared to the number of protons. Oxygen always has 8 protons, but can have 7 or 8 neutrons. This changes the configuration of its nucleus. Oxygen 15 and oxygen 16 are both isotopes of oxygen. A isotope is Atoms that have the same number of protons but a different number of neutrons are isotopes. Carbon has two stable isotopes. During a process called ionization and the atom that has lost one or more electrons is an ion. 3. Why is the binding energy of an electron related to the size of its orbit? 3. The binding energy is determined by the Coulomb force between the positive nucleus and the negative electrons. The Coulomb force is an inverse square law force; the closer two oppositely charged particles are to each other, the stronger the Coulomb force, and hence the stronger the binding energy. Therefore, electrons that are closer to the nucleus are more tightly bound than electrons farther from the nucleus. An electron bound to the nucleus with a lot of binding energy experiences a strong pull by the nucleus. This strong pull keeps the electron in an orbit near the nucleus. Therefore, if an electron has a large binding energy it is in a small orbit. An electron may orbit the nucleus at various dis- tances. If the orbit is small, the electron is close to the nucleus, and a large amount of energy is needed to pull it away. Therefore its binding energy is large. An electron orbiting farther from the nucleus is held more loosely, and less energy will pull it away. It therefore has less binding energy. The size of an electron's orbit is related to the energy that binds it to the atom. The electrons are bound to the atom by the attrac- tion between their negative charge and the positive charge on the nucleus. 4. Explain why ionized calcium can form absorption lines but ionized hydrogen cannot? 4. Neutral (unionized) calcium contains 20 electrons while neutral hydrogen contains one electron. If you ionize calcium and remove one of the electrons, one of the remaining electrons will fill the outer most shell and be capable of producing absorptions. On the other hand, hydrogen, once ionized, has no electrons left. Therefore, there are no electrons to make transitions and energy can not be absorbed by ionized hydrogen. Ionized calcium appears around the 400 nm on the spectra wavelength The two ultraviolet lines of ionized calcium are strong in cooler stars. Hydrogen appears around at 660 nm on the spectra wavelength. 5. Describe two ways an atom can become excited? 5. Atoms can become excited from the absorption of a photon or from collisions. If a photon has an energy that is equal to the energy between any two of the energy levels in an atom and the electron of that atom is in the lower of these two levels, the atom can absorb the energy of that photon and the electron will jump to the upper energy level. In a collision some of the energy associated with the speed of the two particles (i.e. the kinetic energy) can be absorbed by an electron in a lower energy level and cause the electron to move to an excited level. One way an atom can become excited (i.e., energized) is by colliding with another atom. The collision can add energy to one of the atom's electrons. A second way an atom can become excited is by absorbing a passing photon. The photon disappears and an electron of the atom gains energy. The absorption of the photon can occur only if the photon's energy is equal to the energy difference between the electron's initial energy and the energy of a higher energy level to which the electron can move. An atom can become excited by collision. If two atoms collide, one or both may have electrons knocked into a higher energy level. This happens very com- monly in hot gas, where the atoms move rapidly and collide often. Another way an atom can become excited is that moves an electron to a higher energy level is to absorb a photon. Only a photon with exactly the right amount of energy can move the electron from one level to an- other. A hydrogen atom can absorb only those photons that move the atom's electron to one of the higher-energy orbits. Here three different photons are sown along with the change they would produce if they were absorbed. in Figure 6-4. 6. Why do different atoms have different lines in their spectra? 6. Different atoms have different lines in their spectra because different elements have different energy levels for the electron. The electron energy levels are determined by the total number of protons, neutrons, and electrons that the individual atom contains. And these energy levels control the energies of the transitions (jumping between electron energy levels) that can occur. The lines tell us things like the atoms temperature. Not only can a star's spectrum tell us the surface temperature of the star, but it can also tell us how the star is moving through space. The spectra run from the hot O-star at the top to the cool M2 star at the bottom. 7. Why does the amount of black body radiation emitted depend on the temperature of the object? 7. The temperature of a heated object determines how fast the particles (atoms and molecules) vibrate or move within that material. The particles in a hot object are moving very rapidly. The greater the temperature, the greater the number of collisions that occur and the greater the amount of energy released in each collision. Therefore, hot objects should produce more energy than cooler objects. The hotter an object is the more black body radiation it emits. Hot objects emit more radiation because their agitated particles collide more often and more violently with electrons. 8. Why do hot stars look bluer than cool stars? 8. Hot stars look bluer than cool stars because the atoms and electrons in hot stars are moving much faster. As atoms collide with each other, photons with high energies will be produced. High energy photons have short wavelengths and appear blue to the human eye. In the cooler stars, the atoms and electrons have smaller speeds and the collisions produce lower energy photons. These photons are of longer wavelengths. The blue stars produce many more collisions each second and the proportion of high energy (blue) photons produced is much greater than in the cooler stars. Hot stars have hotter temperatures. Cool stars have cooler temperatures. Hot stars look bluer than cool stars because the wavelength at which their radiation peaks depends on their temperature. The radiation from hot stars peaks near the blue end of the spectrum. The radiation from cool stars peaks near the red end of the spectrum. This is a consequence of the fact that stars behave like black bodies. 9. What kind of spectrum does a neon sign produce? 9. A neon sign produces an emission spectrum. Electrons are stimulated by the electric current in the tube. These energetic electrons collide with and excite the neon atoms, which then spontaneously move back to lower energy states by emitting photons characteristic of neon. 10. Why are Balmer lines strong in the spectra of medium- temperature stars and weak in the spectra of hot and cool stars? 10. The visible Balmer (absorption) lines of the hydrogen spectrum occur as a result of transitions from the first excited level (n = 2) to higher energy levels. In cool stars, nearly all of the hydrogen is in the ground state so that only transitions from n = 1 to higher levels are observed (and not the Balmer lines). In hot stars much of the hydrogen is ionized by the violent (because stars are hot) collisions between the atoms. Since the atoms now contain no electrons at all, hydrogen cannot produce an absorption spectrum. In the medium temperature stars, the temperature is high enough that collisions keep a large number of the hydrogen atoms excited to the first excited level (n = 2). In these stars there is a sufficient number of hydrogen atoms in the first excited level to readily absorb the visible photons associated with the Balmer spectrum. The strength of the Balmer lines de- pends on the temperature of the star's surface layers. The Balmer Thermostat We can use the Balmer lines as a thermostat to find the temperatures of stellar surfaces. Red stars are cool. Blue stars are hot. We will use the Kelvin tem- perature scale to refer to stellar temperatures. 40,000 to 2000 K surface temperature of a star. The spectra tell us about the surface layers from which the light originates. An absorption spectrum results when radiation passes through a cool gas. 11. Why are titanium oxide features visible in the spectra of only the coolest stars? 11. Titanium Oxide (TiO) is a molecule and is easily broken apart by energetic collisions. At temperatures above 4000 K, collisions between TiO molecules and other atoms are sufficient to dissociate TiO into atomic titanium and atomic oxygen. Therefore, at temperatures in excess of 4000 K, TiO no longer exists and cannot produce special lines. Titanium oxide bands are strong in the spectra of the coolest stars. 12. Explain the similarities among Table 6-1, Figure 6-7c, Figure 6-8, and Figure 6-9 12. Figure 6-9 is a graphical representation of stellar spectra and Figure 6-8 is a photographic image of stellar spectra. Table 6-1 describes which absorption lines should be strong (thick and dark) for various spectral types. Figure 6-7c plots the relative strengths of absorption lines due to different atoms for various spectral types. All of these show the variation in the spectra of stars as a function of spectral type (or equivalently, temperature). Table 6-1 Spectral Classes Spectral Class Approximate temperature K Hydrogen Balmer Lines Other Spectral Features O 40,000 Weak Ionize helium B 20,000 Medium Neutral helium A 10,000 Strong Ionized calcium weak F 7500 Medium Ionized calcium weak G 5500 Weak Ionized calcium medium K 4500 Very weak Ionized calcium strong M 3000 Very weak Titanium oxide strong 6-7c The strengths of the spectral lines produced by each atom, ion, and molecule depend on the temperature of the star. Astronomers can compare the strength of lines in a stellar spectrum with a diagram such as this to find the temperature of the star. Ionized helium black line Helium green line Hydrogen red line ionized iron purple line Ionized calcium blue line Titanium oxide purple line lines in u shapes in figure 6-7c 10,000 6000 4000 Temperature K In Figure 6-9 Their appeared to be more hotter stars 05 B0 A1 F0 G1 cooler stars K0 M0 M5 Blue Yellow Red colors at top of figure If a star is moving toward Earth the lines in its spectrum will be shifted slightly toward shorter wave- lengths. blue of the spectrum blue shift If a star is moving away from Earth the lines are shifted slightly toward the red end of the spectrum a red shift. . Figure 6-9 Modern digital spectra are often represented by graphs of intensity versus wavelength. Dark absorption lines are dips in intensity. The hottest stars are at the top and the coolest at the bottom. Hydrogen Balmer lines are strongest at about A0, white lines of ionized calcium Call are strong in K stars. Titanium oxide TIO bands are strongest in the coolest stars. Compare these spectra with Figure 6-7c and 6-8. 13. Why does the doppler effect detect only radial velocity? 13. The Doppler effect is produced because the relative motion of the source and observer affects the time of arrival between the crests of adjacent waves in any wave phenomenon. If the observer is moving away from the source, the time between the arrival of one crest and the arrival of the next crest of the wave will be lengthened. As the first crest is received the observer continues to move farther from the source and so the next crest must travel a little farther than the previous one. This causes the observer to measure the wavelength to be a little longer than it would be if they were not moving. If the observer and source are moving parallel to each other, the time between the arrival of the two waves will not be altered since the observer does not move away from or towards the second crest following the arrival of the first. When an object emitting waves is moving only perpendicular to our line of sight, its waves are neither stretched nor compressed, so no Doppler effect is observed. To observe a Doppler effect the wave source must have part of its velocity directed toward us or away from us; then its waves will be compressed or stretched as in Figure 6-18. The Doppler shift is only sensitive to the part of the velocity directed away from us or to- ward us. This part of the velocity is called the radial velocity. We cannot use the Doppler effect to de- tect any part of the velocity that is perpendicular to our line of sight. 14. How can the Doppler effect explain shifts in both light and sound? 14. The Doppler effect is produced when the source and observer of a wave are moving relative to each other, regardless of the type of waves produced. If an observer is receding from a source of waves the time it takes adjacent crests to reach the do will be greater than if the observer were not moving. Whether those waves are sound eaves, light waves, or water waves is not important; their observed frequency and wavelength will be altered. We can detect the Doppler shift in sound. Sound is not electromagnetic radiation of course it is a mechanical wave transmitted through air. Be- cause it is a wave however, it is subject to the Dop- pler effect. Sounds with long wavelengths have low pitches, and sounds with short wavelengths have higher pitches. Radio astronomers for instance observe the Doppler effect when they measure the wavelength of radio waves coming from objects mov- ing towa rd or away from Earth. The Doppler Effect Astronomers can measure the wavelengths of lines in a star's spectrum and find the velocity of the star. The Doppler effect is the apparent change in the wave- length of radiation caused by the motion of the source. The Doppler effect tells us how rapidly the dis- tance between us and the source of light is increasing or decreasing. Only the relative velocity is important. 15. Explain why the presence of spectral lines of a given ele- ment in the solar spectrum tells us that element is pres- ent in the sun, but the absence of the lines would not mean the element was absent from the sun? 15. The absorption lines of an element are unique to that element, so detecting the set of absorption lines due to a particular element in the solar spectrum proves that the element is in the Sun. Not finding the absorption lines of an element in the solar spectrum does not imply that the element is absent in the solar atmosphere. The visible absorption lines will be produced by transitions from one specific energy level to levels above it. The absorption lines may not be present if the temperature of the Sun is either so hot that the element is ionized or so cool that the atoms are all in an energy level below that necessary to produce an absorption spectrum. Chapter 6 Atoms and Starlight page 107 Discussion Questions 1. In what ways is our model of an atom a scientific model? In what ways is it incorrect? Scientific model - ia a mental picture, or idealization, based on physical concepts and aesthetic notions, that accounts for what scientists see regarding particular phenomena and that allows predictions of the course of future events for the phenomena. Examples Bohr model of the atom - solar system-like model of atoms heliocentric model of Solar System - sun-centered arrangement for planets and other bodies stellar-structure models - idealized mathematical models of the structure of the interior of stars use to predict their evolution cosmological models by Friedmann - mathematical models of the universe as to its geometry and evolution 2. Can you think of classification systems we use to sim- plify what would otherwise be complex measurements? Consider foods, movies, cars, grades, and clothes. Spectral classification systems Distances Temperatures Measurements Wavelength Asteroids are classified using a lettering system which, in my opinion, is one of the more non-intuitive classification systems in astronomy. There are usually 14 classifications (A,B,C,D,E,F,G,M,P,Q,R,S,T,and V) http://curious.astro.cornell.edu/question.php?number=13 Problems 1. Human body temperature is about 310 K (98.6 F). At what wavelength do humans radiate the most energy? What kind of radiation do we emit? Wiens Law is, Lambda(max) = 3x10^6 / Temp, where Lambda(max) is in nm and Temp is in K. Plugging in 310K for Temp and converting to meters gives an answer of 10^-5 meters. From Figure 5-2, this is in the Infrared region of the EM spectrum. 2. If a star has a surface temperature of 20,000 K, at what wavelength will it radiate the most energy? 3. Infrared observations of a star show that it is most in- tense at a wavelength of 2000 nm. What is the tempera- ture of the star's surface? Using Wiens Law, plugging in 2000nm for Lambda(max), and solving for Temp gives an answer of 1500K. The law which relates a black body's temperature T to the wavelength λmax at which its radiation peaks is Wien's Law: λmax = 3,000,000 nm K / T Solving this for T, we obtain the surface temperature of the star: T = 3,000,000 nm K / λmax = (3,000,000 nm K) / (2000 nm) = 1500 K This is a cool surface. 4. If we double the temperature of a black body, by what factor will the total energy radiated per second per square meter increase? 5. If one star has a temperature of 6000 K and another star has a temperature of 7000 K, how much more energy per second will the hotter star radiate from each square meter of its surface? The problem requires us to use the Stefan-Boltzmann Law which gives the energy radiated by a star per second per square meter of its surface. This quantity is not properly called the energy of the star, but it is the intensity I of the star at its surface (not the intensity of the star measured at the earth). So I will write the Stefan-Boltzmann Law with an I on the left side, not an E as our textbook does: I = σ T4 What the question is really asking for is the intensity IH of the hotter star divided by the intensity IC of the cooler star. This is IH / IC = (σ TH4 / σ TC4) = (TH/TC)4 = [(7000 K)/(6000 K)]4 = 1.85 Thus at their surfaces, the hotter star is 1.85 times more intense than the cooler star. 6. Transition A produces light with a wavelength of 500 nm. Transition B involves twice as much energy as A. What wavelength light does it produce? 7. Determine the temperatures of the following stars based on their spectra. Use Figures 6-8 and 6-9. a. medium-strength Balmer lines, strong helium lines B b. medium-strength Balmer lines, weak ionized-calcium lines. F c. strong TiO bands. M d. very weak Balmer lines, strong ionized calcium lines. K 8. To which spectral classes do the stars in Problem 7 belong? a. 20,000K b. 7,500K c. 3,000K d. 4,500K 9. In a laboratory, the Balmer beta line has a wavelength of 486.1 nm. If the line appear in a star's spectrum at 486.3 nm, what is the star's radial velocity? Is it ap- proaching or receding? The Doppler Effect formula is: delta_Lambda/Lambda(nought) = Vr/c, where Vr is the radial velocity of the star and c is the speed of light. delta_Lambda is Lambda minus Lambda(nought). Remember that Lambda is the wavelength observed in the stars spectrum and Lambda(nought) is the wavelength measured in the laboratory. Since Lambda is greater than Lambda(nought) then the spectrum is red-shifted, thus the star is receding. Plugging the numbers into the formula gives an answer of about 123, 431.4 meters/sec. So the star is moving away with a radial velocity of 123,431.4 m/s. The radial velocity of an object is given by the Doppler shift formula: Vr = (Δ λ / λ0) c = ( 0.2 nm / 486.1 nm) (3 x 108 m/s) = 123,000 m/s 10. The highest-velocity stars an astronomer might observe have velocities of about 400 km/s. What change in wave- length would this cause in the Balmer gamma line? Hint: Wavelengths are given on page 99. Critical Inquiries for the Web 1. The name for the element helium has astronomical roots. Search the Internet for information on the discovery of helium. How and when was it discovered, and how did it get its name? Why do you suppose it took so long for helium to be recognized? Janssen obtained the first evidence of helium during the solar eclipse of 1868 when he detected a new line in the solar spectrum. Except for hydrogen, helium is the most abundant element found through out the universe. Helium is extracted from natural gas. In fact, all natural gas contains at least trace quantities of helium. It has been detected spectroscopically in great abundance, especially in the hotter stars, and it is an important component in both the proton-proton reaction and the carbon cycle, which account for the energy of the sun and stars. http://pearl1.lanl.gov/periodic/elements/2.html 2. How was the model of the atom presented in the text you read developed? Search the Web for Information on historical models of the atom, and compile a time line of important developments leading to our current under- standing. What evidence exists that supports our model? historical models of the atom http://www.nyu.edu/pages/mathmol/textbook/atom.html STRUCTURE OF THE ATOM Matter has mass and takes up space. Atoms are basic building blocks of matter, and cannot be chemically subdivided by ordinary means. The word atom is derived from the Greek word atom which means indivisible. The Greeks concluded that matter could be broken down into particles to small to be seen. These particles were called atoms Atoms are composed of three type of particles: protons, neutrons, and electron. Protons and neutrons are responsible for most of the atomic mass e.g in a 150 person 149 lbs, 15 oz are protons and neutrons while only 1 oz. is electrons. The mass of an electron is very small (9.108 X 10-28 grams). Both the protons and neutrons reside in the nucleus. Protons have a postive (+) charge, neutrons have no charge --they are neutral. Electrons reside in orbitals around the nucleus. They have a negative charge (-). It is the number of protons that determines the atomic number, e.g., H = 1. The number of protons in an element is constant (e.g., H=1, Ur=92) but neutron number may vary, so mass number (protons + neutrons) may vary. atoms are made up of electrons, protons, and neutrons Structure of the Atom (grades 6-8) Helium is an example of a noble (inert) gas. It is not present in organisms because it is not chemically reactive. Historical Models of the atom. BOHR MODEL. ... www.nyu.edu/pages/mathmol/textbook/atom.html Exploring The Sky 1. Locate the following stars, click on them, and deter- mine their spectral types: Antares in Scorpius, Betel- geuse in Orion, Aldebaran in Taurus, Sirius in Canis Major, Rigel in Orion. Antares in Scorpius, Betelgeuse in Orion, M Spectral Type Aldebaran in Taurus, Sirius in Canis Major, is A Spectral Type Rigel in Orion. is B Spectral Type. 2. How are spectral types correlated with the colors of stars? Hint: Locate Orion and choose Spectral Colors under the View menu.) Temperatures and luminosities of stars are observed to be highly correlated Stars of Orion's Belt temperature range >30,000 helium <97nm ultraviolet color. Spectral Type 0. The Hertzsprung-Russell diagram (1911-1913) is a plot of temperature (or color -- basically the same thing as color, for black bodies) vs. luminosity. Every star can be plotted somewhere on the\ H-R diagram. It is one of the most important graphs in astronomy, and shows the complete life cycles of stars of all different masses, ages, temperatures, colors, and luminosities of stars. 90% of all stars have luminosities (or absolute magnitudes) and temperatures (or colors) that place them in a narrow diagonal band in the HR diagram, called the Main Sequence. Sequence dwarf, giant, and supergiant stars spectral types. observed to be highly correlated (like weight, masses, ages, temperatures, colors, and luminosities some stars that are dim and blue, and others that are brilliant and red. page 100 Chapter 6 Table 6-2 The Most Abundant Elements in the Sun Element Percentage by Number of Atoms Percentage by Mass Hydrogen 91.0 70.9 Helium 8.9 27.4 Carbon 0.03 0.3 Nitrogen 0.008 0.1 Oxygen 0.07 0.8 Neon 0.01 0.2 Magnesium 0.003 0.06 Silicon 0.003 0.07 Sulfur 0.002 0.04 Iron 0.003 0.3 a violent collision can produce a short wavelength high-energy photon. Black body radiation from three bodies at different temperatures demonstrates that a hot body radiates more total energy and that the wavelength of maximum intensity is shorter for hotter objects. The hotter object here will look blue to our eyes whereas the cooler object will look red. Black body radiation is made up of photons with a distribution of wavelengths and very short and very long wavelengths are rare. wavelength of maximum intensity 2 max. The coldest gas drifting in space has a tempera- ture only a few degrees above absolute zero, but it too emits black body radiation. An atom can absorb a photon only if the photon has the correct amount of energy. The excited atom is unstable and within a fraction of a second returns to a lower energy level, reradiat- ing the photon in a random direction. zero degrees Kelvin is absolute zero -459.7 F. the temperature at which an object contains no heat energy that can be extracted. Water freezes at 273 K and boils at 373 K. The Kelvin Temperature scale is useful in astronomy because it is based on absolute zero. page 100 Chapter 6 Atoms and Starlight Part 2 The Stars Review Questions 1. Why can't we see deeper than the photosphere? We can't see deeper than the photosphere because photons emitted from regions of the sun below the photosphere don't escape the sun. Those photons get absorbed on their way out of the sun by the dense gasses they are traveling through. Photons emitted in the photosphere can escape (if they are heading away from the center of the sun) because there isn't much gas they have to travel through to get away from the sun there isn't much gas they have to travel through to get away from the sun. The photosphere is the thin layer of gas from which we receive most of the sun's lght. The photosphere is not a solid surface, and is the visible surface of the sun. Below the phtosphere the gas is denser and hotter and therefore radiates plenty of light. 2. What evidence do we have that granulation is caused by convection? Convection is the rising of a hot fluid. Granules are thought to be convection cells for several reasons. The light from the center of a granule is Doppler shifted showing the gas at the center is rising. The light from the center of the granule shows that the gas their is at a hotter temperature than at the edges of the granule; this is consistent with the idea that hot gas rises at the center and falls back down at the edges of the granule. Figure 7-2 A visibe-light photo of the sun's surface shows granulation. This model explains granulation as the tops rising convection currents just below the photosphere. Heat flows upward as rising currents of hot gas and sinking currents of cool gas. Astronomers recognize granulation as the surface effect of convection just below the photosphere. Convection occurs when hot fluid rises and cool liquid sinks, . 3. How are granules and supergranules related? How do they differ? Supergranules are caused by larger convection currents circulating deeper below the photosphere. 4. How can a filtergram reveal structure in the chromo- sphere? Yes An H, filtergram shows that the chromosphere is not uniform. 5. What evidence do we have that the corona has a very high temperature? Some of the coronal light consists of a continous spectrum, which appears to form when sunlight from the photosphere is scattered off free electrons in the ionized coronal gas has a high temperature over 1 million k and the electrons travel very fast. Superimposed on the corona's continous spectrum are emission lines of highly ionized gases. Solar Wind in the sun 6. What heats the chromosphere and corona to high tem- perature? The sun's turbulent magnetic field is thought to be responsible for heating chromosphere and corona to high temperatures. Magnetic energy is converted to thermal energy. coronal gases low density gas is in constant motion where Heat flows from hot regions to cool regions Turbulene below the surface maybe whipping these fields about and churning the chromosphere and corona. That could heat the gas. Gas follows the magnetic fields pointing outward and flows away from the sun in a breeze of ionized atoms called the solar wind. 7. How are astronomers able to explore the layers of the sun below the photosphere? Looking at the sun through a telescope can burn your eyes, but you can view the sun safely with a small telescope by pro- jecting the image on a white screen. small telescope is safe to observe the sun. By choosing the proper wavelength solar astronomers can observe structures in the sun's chromosphere. does not work below the photosphere. Ultraviolet and X-ray observations. Satellites A white-light image of the solar corona was obtained by the SOHO satellite. The Global Oscillation Network Group GONG uses a network of telescopes spread around the world to observe the sun continuosly as Earth turns. 8. What evidence do we have that sunspots are magnetic? At far- ultraviolet wavelengths the TRACE satellite can detect the hot gas trapped in the mag- netic fields arching above the sunspot group. Sunspots are great magnetic storms. The magnetic fields in sunspots can be measured through the Zeeman effect the splitting of single spectral lines into multiple components through the influence of a magnetic field. The dynamo effect occurs when a rapidly rotating conductor is stirred by convection to produce a magnetic field. Dark spots on the photosphere are called sunspots. 9. How does the Babcock model explain the sunspot cycle? The babcock model explains the reversal of the sun's magnetic field from cycle to cycle. the Babcock model explains the magnetic cycle as a progressive tangling of the solar magnetic field. Because the electrons in an ion- ized gas are free to move the gas is a very good con- ductor of electricity and any magnetic field in the gas is frozen into the gas. If the gas moves the magnetic field must move with it. 10. What does the spectrum of a prominence tell us? What does it shape tell us? A prominence is composed of hot ionized gas trapped in a magnetic arch rising up through the photosphere and chromosphere into the lower corona. Arched shape. prominence s a pink color. 11. How can solar flares affect Earth? Flares can add a lot of high speed, high energy particles to the solar wind. When these charged particles reach the earth they can cause auroras, surges in high voltage power lines, and radiation hazards to airline passengers and astronauts traveling above the earth's atmosphere. Flares are a major problem for astronauts in space and can be lethal in terms of radiation doses. For humans on Earth it is not a problem unless you rely on satellites, which can sometimes get damaged. 12. Why does nuclear fusion require high temperatures? High temperatures are required for hydrogen gas. The star ignites hydrogen burning. 13. Why does nuclear fusion in the sun occur only near the center? The sun is a star and it is powered by nuclear re- actions that occur near its center. the gas is totally ionized 14. How can astronomers detect neutrinos from the sun? To catch neutrinos in the act of oscillating scien- tists used the Super Kamiokande neutrino detector. The Davis experiment can only detect electron neutrinos. The Davis Solar neutrino experiment counts neutrinos from the sun. Neutrinos can be counted using the using 100,000 gallons of cleaning fluid held in a tank nearly a mile underground. A miniscule fraction of the solar neutrinos that pass through the cleaning fluid convert chlorine atoms into argon atoms that can be counted because they are radioactive. 15. How can neutrino oscillation explain the solar neutrino problem? 15. The solar neutrino problem is a discrepancy between the number of neutrinos that models of the solar interior predict the Sun should produce and the number that are actually detected in neutrino collection apparati. The number that is predicted in one day is about one, while the number measured in only about one every three days. One recent theory suggests that the neutrino may oscillate between three different states &flavors), only one of which the neutrino detectors can detect. This theory leaves unchanged the highly successful theory of stellar structure, only what we would expect to detect in our measurement experiments. Discussion Questions 1. What energy sources on Earth cannot be thought of as stored sunlight? Hydro Electric Power Nuclear Power Solar Panels. Rechargeable batteries for night use while recharged by sunlight during the day. 2. What would the spectrum of an auroral display look like? Why? Purple, green colors in the auroral display. From May 10-12, 1999, the solar wind that blows constantly from the Sun virtually disappeared -- the most drastic and longest-lasting decrease ever observed. Dropping to a fraction of its normal density and to half its normal speed, the solar wind died down enough to allow physicists to observe particles flowing directly from the Sun's corona to Earth. This severe change in the solar wind also changed the shape of Earth's magnetic field and produced an unusual auroral display at the North Pole. 3. What observations would you make if you were ordered to set up a system that could warn astronauts in orbit of dangerous solar flares? Such a system exists. a satellite carrying a 10-cm diameter telescope could identify dangerous solar flares on the sun's surface, warning satellite operators before the flares reach Earth. Problems 1. The radius of the sun is 0.7 million km. What percent- age of the radius is taken up by the chromosphere? 2. The smallest detail visible with ground-based solar telescopes is about 1 second of arc. How large a region does this represent on the sun? Hint: Use the small- angle formula. The equation to use is θ = (s / d) (206265 sec of arc / radians), where θ is the angular resolution of the ground based telescope, s is the size of the region on the sun, and d is the distance to the sun. We need to solve this equation for s. s = (θ d) / (206265 sec of arc / radians) = (1")(1.5 x 1011 m) / (206265") = 730,000 m = 730 km 3. What is the angular diameter of a star like the sun lo- cated 5 ly from Earth? Is the Hubble Space Telescope able to resolve detail on the surface of such a star? 4. If a sunspot has a temperature of 4200 K and the solar surface has a temperature of 5800 K, how many times brighter is the surface compared with the sunspot? Hint: Use the Stefan-Boltzmann Law, By the Numbers 6-1 . The Stefan-Bolzmann law for intensity at the surface of an object radiating black-body radiation is I = σ T4. To find the ratio of the sun's normal intensity IS to the sunspot's intensity ISS, we need to use the above formula twice, once for the sun and once for the sunspot. Dividing the two formulas, we get IS / ISS = (σ T4 S) / (σ T4 SS) = (TS/TSS)4 = (5800 K/ 4200 K)4 = 3.64 By the Numbers 6-1 Black Body Radiation Wien's law 3,000,000 is divided by the temperature in degress kelvin. 5.67 X 10 -8 J/m2s degree 4. A max= 3,000,000       !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~ T 5. A solar flare can release 10 25 J. How many megatons of TNT would be equivalent? Hint: A 1-megaton bomb produces about 4 X 10 15 J. 6. The United States consumes about 2.5 X 10 19 J of en- ergy in all forms in a year. How many years could we run the United States on the energy released by the solar flare in Problem 7? 7. Neglecting energy absorbed or reflected by our atmo- sphere, the solar energy hitting 1 square meter of Earth's surface is 1360 J/s the solar constant. How long does it take a baseball diamond 90 ft on a side to receive 1 megaton of solar energy? Hint: See Problem 5. 8. How much energy is produced when the sun converts 1 kg of mass into energy? We need to use Einstein's famous formula: E = m c2 = (1 kg) (3 x 108 m/s2)2 = 9 x 1016 J As an additional problem, it is interesting to calculate the number of years this amount of energy would keep a million 100-watt lightbulbs burning. To find how long of a time t a million 100-watt light bulbs could burn using this much energy, we need to divide the energy E by the luminosity L (wattage) of a million 100-watt light bulbs. The luminosity is the rate at which energy is transformed. So to get the time t, let's divide the amount E by the rate L.) t = E / L = (9 x 1016 J) / (108 J/s) = 9 x 108 s In the above line, I have used the fact that a watt is a joule per second. To convert this time from second to years, use the fact that there are approximately 3.2 x 107 second in a year. Therefore t = (9 x 108 s) / (3.2 x 107 s/yr) = 28.5 yr 9. How much energy is produced when the sun converts 1 kg of hydrogen into helium? Hint: How does this problem differ from Problem 8? 10. A 1-megaton nuclear weapon produces about 4 X 10 15 J of energy. How much mass must vanish when a 5-megaton weapon explodes? Critical Inquiries for the Web 1. Do disturbances in one layer of the solar atmosphere produce effects in other layers? We have seen that fil- tergrams are useful in identifying the layers of the solar atmosphere and the structures within them. Visit a Web site that provides daily solar images, choose today's date or one near it, and examine the sun in several wave- lengths to explore the relation between disturbances in various layers. Yes. they do produce effects. He I 10830 spectroheliograms from the U.S. National Solar Observatory at Kitt Peak (Arizona) [ 2003/05/01 17:20:31 UT ] http://umbra.nascom.nasa.gov/images/latest_nsoHe.gif Orange, red, and black colors were in picture of the sun. SOHO Extreme ultraviolet Imaging Telescope (EIT) full-field Fe XII 195 images from NASA Goddard Space Flight Center [ 2003/05/01 23:24:10 ] http://umbra.nascom.nasa.gov/images/latest_eit_195.gif dark green, green. 2. Page 123 shows the auroral displays caused by interac- tion of particles from the sun with our magnetosphere and ultimately our atmosphere. Explore the Web to find out how auroral activity is affected as solar activity rises and falls through the solar cycle. What changes in auroral visibility occur during this cycle? In what other ways can the increased activity associated with a solar maximum affect Earth? http://edmall.gsfc.nasa.gov/99invest.Site/science-briefs/ace/ed-solar.html Solar Activity: The output of the Sun in all forms, light, solar wind, and energetic particles, is not constant. It varies with both time (seconds to centuries!) and position on the Sun. These changes are called solar activity and are probably reflections of changes below the Sun's surface. (Scientists can study the output and how it varies to study the workings of the Sun). http://sprg.ssl.berkeley.edu/fast/aurora.html Earth's aurora occurs at polar latitudes in an oval-shaped region. Its properties are under investigation by space physicists; the Fast Auroral SnapshoT Explorer has provided a wealth of data that is currently being used to revise theories about its nature. The level of global auroral activity is roughly related to the magnetic activity on the sun, which increases and decreases periodically every 11 years. The year 2000 included the peak of the solar cycle's 11-year period. Earth's magnetic field is affected by solar events (solar wind shocks), which can cause auroral displays. When an aurora becomes more active, often the auroral oval expands and widens with aurora moving both towards the pole and towards the equator. The visible aurora is created by energetic electrons colliding with the molecules and atoms low in the upper atmosphere (100 - 200 km), causing the excitation of these gases. Electronic communications are affected. 3. What can you find on the Web about Earth-based efforts to generate energy through nuclear fusion? How do nu- clear fusion power experiments attempt to trigger and control nuclear fusion? So-called cold fusion has been largely abandoned as a false trail. How did it resemble nuclear fusion? http://www.iclei.org/EFACTS/FUSION.HTM the only practical application of fusion technology to date has been the hydrogen. Exploring The Sky 1. Locate the six photos of the sun provided in The Sky and attempt to draw in the sun's equator in each photo. Hint: In the sun's information box, choose More Infor- mation and then Multimedia. What features are visible in these images that help us recognize the orientation of the sun's equator? dark red spots latitude and longitude lines. Sun Rise: 4:07 AM on 12/21/03 Transit: 12:07 PM on 12/21/03 Set: 8:07 PM on 12/21/03 RA: 17h 58m 18.9s Dec: -2326'19" Azm: 30719'31" Alt: +5704'07" (with refraction: +5704'44") Geometric geocentric ecliptical coordinates: l: +26937'26" b: -0000'00" r: 0.983762 Mean geometric ecliptical coordinates: l: +26937'26" b: -0000'00" r: 0.983762 True equatorial coordinates: RA: 17h 58m 19s Dec: -2326'23" Physical Data LO: 297.44 BO: -1.73 P: 7.21 Apparent angular diameter: 0032'31" Vernal equinox: 3/20/03 5:01 PM Summer solstice: 6/21/03 12:06 PM Autumnal equinox: 9/23/03 3:47 AM Winter solstice: 12/21/03 11:05 PM page 129 13. The temperature and density are high enough only at the center of the Sun for fusion reactions to proceed and produce enough energy needed to meet the Suns energy needs. Chapter 7 The Sun - Our Star Almost no light emerges from below the photosphere, so we can't see details in the solar interior. However, solar astronomers can use the vibrations in the sun to explore its depths in a process called helioseismology. Random motions in the sun constantly produce vibrations. periods range to 3 to 20 minutes sound waves. Astronomers can detect these vibrations by ob- serving Doppler shifts in the solar surface. As a vibrational wave travels down into the sun the increasing density and temperature curve its path and it returns to the surface, where it makes the photo- sphere heave up and down. Review Questions 1. Why are Earth-based parallax measurements limited to the nearest stars? Parallax measurements are limited because we measure the apparent motion of a star due to the motion of Earth around the Sun. Earth's orbit is so small compared to the distances to stars that even the nearest stars show barely measurable apparent motions. Therefore, we are limited to only the nearest stars. If Earth's orbit were larger, we could measure the parallax of stars at greater distances. Because the stars are so distant, their parallaxes are very small angles 2. Why was the Hipparcos satellite able to make more ac- curate parallax measurements than are ground-based telescopes? The primary benefit of a telescope orbiting Earth, such as Hipparcos, is the improvement in the resolution of the images. The resolution would be much better because the atmosphere would not smear the images out as the light passes through it. Therefore the positions of the stellar images could be determined more precisely. the Satellite Hipparcos to measure stellar parallaxes from orbit above the blurring effects of Earth's atmo- sphere. The blurring caused by Earth's atmosphere makes star images about 1 second of arc in diameter, and that makes it difficult to measure parallax. 3. What do the words absolute and visual mean in the de- finition of absolute visual magnitude? Absolute refers to the value of the magnitude being independent of distance. The absolute magnitude is the magnitude that the stars would have if they were all at a common distance of 10 pc from the observer. In this manner the absolute magnitude provides a way to compare the intrinsic brightness (~ luminosity) of two stars. Visual refers to the value in the visible portion of the spectrum. The visual magnitude is related to the amount of light received only between the wavelengths of 400 nm and 700 nm. Therefore the visual magnitude neglects radiation at all other wavelengths, e.g., gamma-ray, x-ray, ultraviolet, infrared, and radio. The absolute visual magnitude is then related to the intrinsic brightness of the star in the visible portion of the electromagnetic spectrum. Absolute means that the magnitude refers to the brightness the star would appear to have if it were 10 parsecs from the earth. Visual means that the magnitude only takes into account the visible light emitted by the star. The absolute magnitude tells us how bright the star really is. The visual magnitude based only on the wavelengths of light we can see. 4. What does luminosity measure that is different from what absolute visual magnitude measures? The luminosity is the total amount of energy emitted by a star in one second throughout the entire electromagnetic spectrum. This contrasts with the visual absolute magnitude, which is described in #3 above. The luminosity of the star tells us how much energy per second each star is emitting. The absolute magnitudes can tell us the luminosities of the stars. 5. Why does the luminosity of a star depend on both its radius and its temperature? The temperature of a star tells us how much energy is emitted from each square meter of the stars surface (Stefan-Boltzmann law). The higher the temperature, the greater the amount of energy emitted from each square meter of the stars surface. To determine the total amount of energy emitted by the star (i.e., its luminosity) we must then multiply the surface area of the star. This surface area of the star depends on the square of the stars radius. The luminosity of the star is therefore dependent upon the stars surface temperature and radius. One of the best ways is to compare the temperatures and luminosities of the giant stars to those of main sequence stars. Stars with similar temperatures, but of different luminosities, must have different radii. The temperature tells us that both stars are producing an equal amount of energy from each square meter of their surfaces. Therefore, for one star to produce more total energy, it must have more square meters of surface, which implies that it has a larger radius. The luminosity is the total amount of energy emitted by the star per second. By the Stefan-Boltzman Law, the temperature of the surface tells you how much energy per second is emitted by each square meter of the star's surface. To get the luminosity you need to know this, but you also need to know the total surface area of the star. The radius of the star lets you calculate the total surface area of the star. The radius of a star The temperature tells us how cold a star is or how hot a star is. A cold star tells us how faint or dim the star is. A hot star tells us how bright the star is. 6. How can we be sure that giant stars really are larger than main-sequence stars? A star's luminosity is proportional to its surface area. In the H-R diagram the stars at the top are larger in diameter and the stars at the bottom are smaller. Heavy Light The temperature and size determine a star's luminosity. The hot main-sequence stars are more luminous than the cool main-sequence stars. 7. Why do we conclude that white dwarfs must be very small? White dwarfs have temperatures that are equal to those of spectral type B and A stars, 10,000 K to 30,000 K, yet have luminosities that are 1000 to 10,000 times smaller than the spectral type B and A main-sequence stars. Therefore, the white dwarfs must have much smaller surface areas and thus be much smaller than the spectral type B and A main sequence stars. White dwarfs must be very small because they emit a lot of light through each square meter of their surface (because they are hot stars), yet their total luminosity is small. The only way to reconcile these two facts is to conclude that white dwarfs are very small. the lowest -mass stars are the coolest, faintest main-sequence stars. All white dwarfs have about the same mass, somewhere in the narrow range of 0.5 to about 1 solar mass. The bright star Sirius A has a faint companion Sirius B (arrow) a white dwarf. The arrow shows Sirius B is on the upper right side of Sirius A Star. 8. What observations would we make to classify a star ac- cording to its luminosity? Why does that method work? In order to classify a star based on its luminosity, we need to obtain a spectrum of the star. The widths of some of the absorption lines in the spectrum can be used to determine its luminosity class. Stars whose surface layers are very dense will have broad absorption lines, while stars with low density surface layers will have very narrow absorption lines. The difference in the width of the absorption lines is produced because collisions between atoms within a gas can alter slightly the energy levels of the atoms. This means that some of the atoms in the gas will make transitions that are slightly larger in energy than normal and other will make transitions that are slightly smaller than normal. The absorption line we observe is a combination of all of these transitions, and the line gets smeared out (i.e., broadened). The amount of broadening will be greatest when the rate at which collisions occur is the greatest. The greatest rate of collisions will occur in those stars with the greatest density in their surface layers. The stars with the greatest density are those with the smallest radii. Therefore, we will find that the smallest stars should have the broadest absorption lines. We can look at a star's spectrum and tell roughly how big it is. These are called luminosity classes, because the size of a star is the dominating factor in determining luminosity. Supergiants for example are very luminous because they are very large. Luminosity Classes Ia Bright supergiant Ib Supergiant II Bright giant III Giant IV Subgiant V Main-sequence star 9. Why does the orbital period of a binary star depend on its mass? Since the orbital period of a binary star system is controlled by the gravitational force, which in turn is dependent upon the masses of the two stars, the orbital period is dependent upon the masses of the two stars. 4HE + energy? 7. In the model shown in Figure 9-14, how much of the sun's mass is hotter than 12,000,000 K? 12.65 Figure 9-14 A stellar model is a table of numbers that represents conditions inside a star. Surface Convective zone RR T 10 6 K Density g/cm 3 M/M L/L 1.00 0.006 0.00 1.00 1.00 0.90 0.60 0.009 0.999 1.00 0.80 1.27 0.035 0.996 1.00 0.70 1.80 0.12 0.990 1.00 0.60 2.42 0.40 0.97 1.00 0.50 3.42 1.3 0.92 1.00 0.40 4.74 4.1 0.82 1.00 0.30 6.65 13. 0.63 0.99 0.20 9.35 36 0.34 0.91 0.10 12.65 85. 0.073 0.40 0.00 14.62 134. 0.000 0.00 Center Radiative Zone dM = 4ttr2P dr dl = 4ttr2 pe de dP =- GM dr r2 P dT = -3 Rp L dr 16ttac T3 r2 8. If a brown dwarf has a surface temperature of 1500 K, at what wavelength will it emit the most radiation? Hint: See By the Numbers 6-1. By the Numbers 6-1 Black Body Radiation Wien's law 3,000,000 is divided by the temperature in degress kelvin. 5.67 X 10 -8 J/m2s degree 4. A max= 3,000,000 T 9. What is the life expectancy of a 16-solar-mass star? The approximate time on the main sequence is equal to the 1/(M^2.5). You read this as one over mass to the 2.5 power. Plugging in 16 solar masses for M gives an approximate time on the main sequence of 1/1024 solar lifetimes, or 0.001 solar lifetime. Since the solar lifetime is 10^10 years, the approximate life expectancy of a 16 solar mass star is about 10 million years. 10. If the 06 V star in the Orion Nebula is magnitude 5.4, how far away is the nebula? Hint: Use spectroscopic parallax. From the HR diagram, the approximate absolute magnitude of the O6V star is about -5.6. That means the distance modulus is 5.4 - (-5.6) = 11.0. From the distance modulus chart, this translates into a distance of 1600 pc. 11. The hottest star in the Orion Nebula has a surface tem- perature of 40,000 K. At what wavelength does it radi- ate the most energy? Hint: See By the Numbers 6-1. We need to use Wien's law again. λmax = 3,000,000 nm K / T = (3,000,000 nm K) / (30,000 K) = 100 nm This is in the ultraviolet portion of the spectrum. This is a Weins Law problem where Lambda_max (in nm) = 3x10^6/(Temperature in K) = 3x10^6/4x10^4 (by plugging in 40,000K) = 75 nm (this is in the X-ray region) We need to use Wien's law again. lambda;max = 3,000,000 nm K / T = (3,000,000 nm K) / (30,000 K) = 100 nm This is in the ultraviolet portion of the spectrum. By the Numbers 6-1 Black Body Radiation Wien's law 3,000,000 is divided by the temperature in degress kelvin. 5.67 X 10 -8 J/m2s degree 4. A max= 3,000,000 T Roughly 1000 young stars with hot chromoshpheres appear in X-ray image of the Orion Nebula. Of all the stars in the Orion Nebula only one is hot enough to ionize the gas. Only photons with wavelengths shorter than 91.2 nm can ionize hydrogen. The second-hottest stars in the nebula are B1 stars, and they emit little of this ionizing radiation. The hottest star however is an 06 star 30 times the mass of the sun. At a temperature of 40,000 K it emits plenty of photons with wavelengths short enough to ionize hydrogen. Remove that one star, and the nebula would turn off its emission. The more massive a protostar is the faster it contracts. A 1-M star requires 30 million years to reach the main sequence. Stars begin contracting from the interstellar gas, which is very cold. very low temperatures 32 K. page 178 Chapter 9 Critical Inquiries for the Web 1. Use the Web to supply additional details concerning the evolution of protostars and T Tauri stars. the evolution of protostars and T Tauri stars http://zebu.uoregon.edu/~imamura/208/feb15/tauri.html T Tauri Stars The evolutionary picture of low mass protostars (T Tauri stars, M less than 2 M(sun)) is thought to be somewhat under control, i.e., very slowly rotating cloud + trigger ===> collapse ===> star + disk formation ====> slow contraction and accretion + intense stellar winds (bipolar flows) ===> ignition of hydrogen burning and appearance of star on the Main Sequence. http://casswww.ucsd.edu/physics/ph7/StevI.html Lecture Summary #9 Stellar Evolution I - Solar ... mass clouds which continue to condense forming protostars. images of young stars called T Tauri stars in the 2. Astronomers have been searching for brown dwarfs for years. How is the search going? About how many have been found so far? How do astronomers distinguish brown dwarfs from low-mass red dwarfs? http://zebu.uoregon.edu/~imamura/208/feb15/tauri.html Lower Mass Limit As the forming stars slowly contract trying to reach densities and temperatures in their cores which are high enough to ignite fusion, an interesting thing happens to low mass clouds. Before their T's get high enough, their densities actually become large enough to make the electrons in the cores degnerate. Once the electrons become degenerate, it becomes very hard to compress them. Effectively what this means is that the cloud stops contracting (stops compressing). The core of the star thus stops getting hotter and nuclear fusion is never ignited ===> the cloud never formally becomes a star. We refer to such objects as brown dwarfs planets (e.g., Jupiter) An interesting comment is is that there is something strange about the star formation process. We might expect there to be a smooth distribution of objects (in mass) between stars and planet-sized things. That is, we might expect to see objects with mass spanning the range between Jupiter and the stars on the Main Sequence. Further, we know that the number of stars for a given mass is larger the smaller the mass of the star, e.g., M stars are much more common than O stars. If this trend continued, we would expect to see many more planets and brown dwarfs than M stars. Surprisingly, there appears to be a gap between the lowest mass stars and brown dwarfs and planets in terms of mass. There is not a smooth distribution in terms of their mass. The process of star apparently is different for star-sized things than it is for planet-sized things. 3. What is BM Orionis? W here is it located and what does it do? Why might it be interesting to observe with even a small telescope? http://www-2.cs.cmu.edu/~zhuxj/astro/html/bmorionis.html BM Orionis This is the "B" star, or the Northern-most star of the famous Orion Trapezium in the famous Orion Nebula M42. You will need a small telescope to see it. I actually noticed its brightness change before I know it is a variable star. On 1998/12/27 4:20 UT, BM was so dim that it disappeared from the Trapezium. According to Burnham's Celestial Handbook, BM Orionis is an eclipsing binary with a period of 6.471 days, and the magnitude ranges from 8.0 -- 8.7. Yes, you can feel the 0.7 magnitude difference! 1. BM Orionis BM Orionis. According to Burnham's Celestial Handbook, BM Orionis is an eclipsing binary with a period of 6.471 days, and the magnitude ranges from 8.0 -- 8.7. www.cs.cmu.edu/~zhuxj/astro/html/bmorionis.html Exploring The Sky 1. The following nebulae are all star-formation regions: M42, M20, M8, M17. What kind of nebulae are they? Hint: To center on an object use Find under the Edit menu. Choose Messier Objects, and pick from the list. M42 is a Great Nebula in Orion Orion Nebula M42 NGC 1976 Other description: Nebula. Constellation: Ori Dreyer description: A magnificent (or otherwise interesting) object! Theta1 Ori and the great nebula; = M42. M20 Trifid Nebula M20 NGC 6514 Other description: Nebula with dust. Constellation: Sgr Dreyer description: A magnificent (or otherwise interesting) object! Very bright, very large, trifid (three-lobed), double star involved; = M20. Magnitude: 6.3 M8 Hour Glass Nebula Lagoon Nebula M8 NGC 6523 Other description: Nebula with dust and cluster. Constellation: Sgr Dreyer description: A magnificent (or otherwise interesting) object! Very bright, extremely large, extremely irregular figure, with large cluster; = M8. Magnitude: 5.8 M17 Omega Nebula Swan Nebula M17 NGC 6618 Other description: Nebula. Constellation: Sgr Dreyer description: A magnificent (or otherwise interesting) object!, bright, extremely large, extremely irregular figure, 2 hooked; = M17. Magnitude: 6.0 2. Locate M8 in the sky, zoom in, and identify other nebu- lae in the region. Study the photo of NGC 6559. Hour Glass Nebula Lagoon Nebula M8 NGC 6523 From GSC 6846:618: GSC 6846:618 HIP 88496, PPM 733664, HD 315032, P-24 6156 Spectral: B2Vne SAO 186061 GSC 6841:1403, HIP 88116, PPM 267632, HD 163955, P-23 6707 Flamsteed-Bayer: 4 Sagittarii Spectral: B9V From Omega Nebula: page 179 Chapter 9 The Formation and Structure of Stars Review Questions 1. Why does helium fusion require a higher temperature than hydrogen fusion? 1. Helium fusion requires a greater temperature than hydrogen fusion because the Coulomb barrier (the electrical repulsion) is greater between fusing nuclei. Helium Fusion Hydrogen fusion in main-sequence stars leaves behind helium ash, which cannot begin fusing because the temperature is too low. Helium nuclei have a positive charge twice that of a proton, and so to overcome the repulsion between nuclei, they must collide at a ve- locity higher than that at which hydrogen fuses. The temperature for hydrogen fusion is too cool to fuse he- lium. As the star becomes a giant star fusing hydrogen in a shell, the inner core of helium contracts and grows hotter.It may even become degenerate, when it fi- nally reaches a temperature of 100,000,000 K it begins to fuse helium nuclei to make carbon. 2. How can the contraction of an inert helium core trigger the ignition of a hydrogen-fusion shell? 2. Just outside the inert helium core, hydrogen that has not undergone fusion is present. As the helium core contracts along with the hydrogen just outside of it, the temperature increases in both regions as gravitational potential energy is converted into thermal energy. When the temperature of the hydrogen shell reaches 10 million K it will initiate hydrogen fusion in this shell around the inert helium core. As the inert helium core contracts, gravitational potential energy is converted into thermal energy. Thus the core and the hydrogen right around the core heat up. As the hydrogen reaches 10 million K, the hydrogen in a spherical shell around the core starts to fuse into helium. Stars more massive than 3 solar masses ignite helium before their contracting cores become degenerate. Whether a star experiences a helium flash or not, the ignition of helium in the core changes the structure of the star. The star now makes energy in its helium- fusion core and in its hydrogen-fusion shell. The en- ergy flowing outward from the core can halt the con- traction of the core, and the distended envelope of the star contracts and grows hotter. 3. Why does the expansion of a star's envelope make it cooler and more luminous? 3. The outer layers of a star will cool as they expand because much of the energy produced goes into expanding the outer layers and not into the thermal motions of the gas. Therefore, the star cools as it expands. Recall that the luminosity of an object depends on both its radius and surface temperature. As the star expands, its surface area increases rapidly, while its temperature decreases rather slowly, resulting in a net luminosity increase. As gases expand, they typically cool off. As the star's outer gases cool, each square meter of the star's surface decreases in intensity (since surface intensity depends on surface temperature as given by the Stefen-Boltzman law). This effect would make the star less luminous if it were not for the fact that the star is much larger than it had been. Although each square meter of the star is less intense, the star actually becomes brighter since the enlarged star has so many more square meters of surface area radiating light. The envelope could be keeping it cooler temperature. the distended envelope of the star contracts and grows hotter. all of the energy is of the explosion is absorbed by the distended envelope. 4. Why is degenerate matter so difficult to compress? 4. Degenerate matter is difficult to compress because nearly all of the energy levels (states) available to the electrons are filled. An energy state of an electron contains information on both the electron’s position and its speed. The moving electrons confined in a stars core can only be in certain energy states, and no two electrons can be in the same state. If the density is very high, the electrons in the gas are forced very close together, and the only way they can occupy the same proximity is if they had different speeds, so they will take all possible energies, starting at the lowest levels. But, in order to further compress the electron gas, the electrons & speeds must be altered. If they are slowed down they would move to lower energy states, but in a degenerate gas, these are all full. They could speed up only if they can absorb enough energy to leap to the top of the energy ladder. Therefore, because it is extremely difficult to change the energies of the particles in a degenerate gas, and because we must change the energies of the particles to further compress the gas, it is extremely difficult to compress the gas. The degenerate gas resists compression. To compress the gas, we must push against the moving electrons, and changing their motion means changing their energy. That requires tremendous effort, because we must boost them to the top of the energy ladder. Degenerate matter, though still a gas, takes on the consistency of hardened steel. speed of the electrons. 5. How does the presence of degenerate matter in a star trigger the helium flash? 5. Helium flash occurs when helium fusion is initiated while the electrons in the core are degenerate. If the electrons in the core are degenerate, the pressure in the core does not depend on the temperature of the core. When the core temperature reaches 100 million K, helium will start to fuse. However, since the core is degenerate, the core can not expand and regulate the rate at which the nuclear reactions occur. The fusion of helium causes the temperature of the core to increase rapidly, the core can not expand and cool off so the rate of nuclear reactions and the amount of energy that they produce increases very rapidly, resulting in a runaway explosion called the helium flash. In this degenerate matter, the pressure does not depend on temperature, and that means the pressure-temperature thermostat does not regulate energy production. When the temperature becomes hot enough, helium fusion begins to make energy and the temperature rises pressure does not increase because the gas is degen- erate. The higher temperature increases the helium fu- sion even further and the result is a runaway explosion called the helium flash in which for a few minutes, the core of the star can generate more energy per sec- ond than does an entire galaxy. 6. How can star clusters confirm our theories of stellar evolution? 6. A star cluster presumably contains stars of a wide variety of masses that all formed at about the same time. Since our stellar models tell us that massive stars should evolve faster than low mass stars, we expect to see differences in the star clusters based on the mass of the stars. One clear indication that our models are at least on the right track is that we find no star clusters with O and B main sequence stars and many giants and supergiants. Observatins show us that when giants and supergiants are present in the clusters, there are few if any massive stars (O and B stars) on the main sequence. We can take models of many different masses of stars and determine what a cluster should look like after a given time. We can then compare the H-R diagram from our model with that of the observations of actual clusters to see how well they agree. For the most part, the agreement is quite good. Since all stars in a cluster form at about the same time but with different masses, we should be able to find stars at almost every stage of evolutionary development consistent with the age of the cluster. Stellar evolutionary theory makes predictions about the track stars take on the HR diagram. Plotting all cluster stars on an HR diagram should help confirm whether our theories of stellar evolution are correct. High mass stars should turn-off the main sequence before low mass stars. The stars in a star cluster all form at about the same time. massive stars evolve faster than low-mass stars. When we look at the H-R diagram of a star cluster we can see stellar evolution made visible. the most massive stars have reached the main sequence consumed the hydrogen in their cores, and moved off to become supergiants. The medium -to low- mass stars have not yet reached the main sequence. The most massive stars are the most luminous. 7. Why don't red dwarfs become giant stars? 7. Red dwarfs, because of their small mass, cannot squeeze their cores sufficiently under their own weight to reach a temperature of 100 million K and ignite helium fusion which is necessary to cause the outer layers to expand to become a giant.

Red dwarf stars are not massive enough to ignite helium. It is part of the helium fusion process that causes a main sequence star to become a red giant. Because a red dwarf is mixed by convection, it can- not develop an inert helium core surrounded by un- processed hydrogen. Then it can never ignite a hydro- gen shell and cannot become a giant star. red dwarfs less massive than about 0.4 solar mass They have very small masses, and they have very little weight to support. Their pressure-temperature thermostats are set low, and they consume their hydrogen fuel very slowly. low-mass red dwarfs, medium-mass sunlike stars, and massive upper-main-sequence stars. These stars lead dramatically different lives and die in different ways. 8. What causes an aging giant star to produce a planetary nebula? 8. A planetary nebula is produced as the carbon-oxygen core of a star contracts and its outer envelope expands. The intense radiation from the hotter inner layers ionizes the innermost gas of the envelope and drives it outward from the star. As this gas overtakes mass lost earlier in the stars life as a giant, material piles up to form a dense shell of ionized gas. The gas is continually ionized by the radiation from the hot exposed core. When a medium-mass star like the sun becomes a dis- tended giant, it can expel its outer atmosphere to form one of the most interesting objects in astronomy, a planetary nebula, so called because it looks like the greenish-blue disks of planets such as Uranus and Neptune. In fact, a planetary nebula has nothing to do with a planet. It is compossed of ionized gases ex- pelled by a dying star. 9. Why can't a white dwarf contract as it cools? What is its fate? 9. A white dwarf can not contract as it cools because the electrons in the white dwarf are degenerate. Degenerate gasses are very difficult to compress because all of the low energy levels for the electrons are full. This makes it is very difficult to change the motion of the electrons and cause the white dwarf to collapse. The ultimate fate of a white dwarf is to become a cold lump of degenerate electrons and carbon and oxygen nuclei. Such an object is referred to as a black dwarf. A white dwarf can't contract because the force of gravity in a white dwarf isn't strong enough to overcome the electron degeneracy force which supports the star. The fate of a white dwarf is to become a black dwarf as it cools and its luminosity approaches zero. a white dwarf cannot generate energy by nuclear fu- sion. It has exhausted its hydrogen and helium and cannot get hot enough to ignite carbon. contains no gas. white dwarfs may become cold and dark. The death of a star leads invariably to one of three final states. Most stars, including stars like the sun, be- come white dwarfs, stars about the size of Earth with no usable fuels. 10. Why can't a white dwarf have a mass greater than 1.4 solar masses? 10. Equations predict that the radius of a white dwarf is inversely proportional to its mass. That is, the more massive a white dwarf is the smaller its radius. At a mass of about 1.4 solar masses the radius of a white dwarf would go to zero. At masses greater than 1.4 solar masses, the degenerate electrons no longer have a sufficient pressure to support the weight of the star and the object will collapse further. If we added enough mass to raise its total mass to about 1.4 solar masses, its radius would shrink to zero. 11. How can a star of as much as 8 solar masses form a white dwarf when it dies? 11. Stars with masses on the main sequence of up to 8 solar masses can become white dwarfs if they lose most of their mass. We see the effects of mass loss in red giants and supergiants, indicating that these stars are shedding mass. Additionally, the formation of the planetary nebula releases a significant amount of mass. So if a star’s core is less than 1.4 solar masses when it begins to release its outer layers, the result will be a white dwarf, regardless of the star’s initial mass while on the main sequence. The only way it can form a white dwarf is if the star looses mass so that it ends its life with less than 1.4 solar masses. A couple of means by which this mass loss can be accomplished is by strong stellar winds throughout the lifetime of the star (this generally occurs in high mass stars), and by extensive mass loss in the planetary nebula stage. An eight solar mass star can form a white dwarf if the star can lose enough mass to reduce its mass to a value less than 1.4 solar masses by the time its planetary nebula phase is over. If it cannot, then it is too massive for electron degeneracy to support it, and it will turn into a neutron star or a black hole. A star more massive than 1.4 solar masses could not become a white dwarf unless it shed mass in some way. Theoretical models show that an 8-solar-mass star should be able to reduce its mass to 1.4 solar masses before it collapses, and slightly more massive stars may also be able to get under the limit. A wide range of medium-mass stars even- tually die as white dwarfs. 12. How can we understand the Algol paradox? 12. Algol is a binary system of a five-solar-mass main-sequence star and a one-solar-mass giant. This presents a paradox because high-mass stars evolve faster than low-mass stars, so the five-solar-mass star should leave the main sequence stage before the one-solar-mass star. The resolution of this paradox lies in mass transfer. Suppose that the system is originally composed of a 5-solar-mass star and a one-solar-mass star. After about 200 million years the five-solar-mass star will run out of hydrogen in the core and will expand to become a red giant. If, as it expands, the star fills its Roche lobe, material will be drawn off of the five-solar-mass star and deposited onto the one-solar-mass star still on the main sequence. This can continue until four solar masses of material is transferred from the giant to the main-sequence star, resulting in a five-solar-mass main sequence star and a one-solar-mass giant. Evolution with Mass Transfer Mass transfer between stars can affect the evolution of the stars in surprising ways. In fact, this explains a problem that puzzled astronomers for many years. In some binary systems, the less massive star has become a giant, while the more massive star is still on the main sequence. If higher-mass stars evolve faster than lower-mass stars, how do the lower-mass stars in such binaries manage to leave the main sequence first? This is called the Algol paradox after the binary system Algol. Mass transfer explains how this could happen. Imagine a binary system that contains a 5-solar-mass star and a 1-solar-mass companion. The two stars formed at the same time, so we expect the higher- mass-star to evolve faster and leave the main sequence first. When it expands into a giant, however, it can fill its Roche lobe and transfer matter to the low-mas companion. The higher-mass star could lose mass and evolve into a lower-mass star, and the companion could gain mass and become a high-mass star still on the main sequence. We might find a sstem such as Algol containing a 5-solar-mass-main-sequence star and a 1-solar-mass giant. 13. How can the inward collapse of the core of a massive star produce an outward explosion? 13. As the core contracts inward at a great rate, material from the outer regions also begins to fall to the core. The collapse is so quick that a shock wave develops in the collapsing gas that propagates outward like a sonic boom from a jet, causing the outer layers to expand. If the shack wave is strong enough the outward expansion is violent and completely overcomes gravity, ripping the star apart in a great supernova explosion. To understand how the inward collapse of the core can produce an outward explosion. The collapse of the innermost part of the degenerate core allows the rest of the core to fall inward, and this creates a tremendous traffic jam as all of the nuclei fell toward the center. Something happens suddenly. The inner core falls inward, a shock wave develops and begins to move outward. The shock wave moves outward through the star aided by two additional sources of energy. First when the iron nuclei in the core are disrupted they produce a flood of neutrinos tremendous turbulence and intensely hot gas rushes outward from the interior. Again this rising hot gas carries energy out into the envelope and helps drive the shock wave outward. Within a few hours, the shock wave bursts outward through the surface of the star and blasts it apart. 14. Neutrinos pass through ordinary matter unimpeded, so they are extremely hard to detect, but certain nuclear reactions can be triggered by a neutrino of the right energy. The first such experiment, the Davis experiment, used 100,000 gallons of cleaning fluid perchloroethylene to count the number of neutrinos from the Sun. A miniscule fraction of the solar neutrinos passng through the tank convert chlorine atoms into argon atoms that can be counted because they are radioacitve. Later experiments are refinements of this first one. 14. What is the difference between type I and type II super- novae? 14. Type I supernovae are more luminous than Type II supernovae and decline in brightness in a more regular manner. The Type I supernovae also show almost no hydrogen lines in their spectra, while Type II supernovae show very strong hydrogen lines in their spectra. Type II supernovae are believed to be produced when the iron core of a massive star collapses. Type I supernovae, on the other hand, are believed to be produced when mass from a binary companion accumulates on a white dwarf and pushes it over the Chandrasekhar limit. The white dwarf collapses and produces a supernova, incinerating the whole star. type I supernovae have spectra lackng hydrogen lines, and the evidence suggests that these supernovae are pro- duced by the collapse of a whit dwarf. type II supernovae have spectra rich in hydrogen and appear to be produced by the collapse and explosion of a massive star. 15. What is the difference between a supernova explosion and a nova explosion 15. A nova occurs when hydrogen fuses on the surface of a white dwarf. This generally happens as mass is transferred from a giant to a white dwarf in a binary system; neither star is greatly affected by this process. A supernovae can occur if the mass transfer causes the white dwarf to exceed the Chandrasekhar limit resulting in a sudden collapse of the entire star. A supernova can also occur when the iron core of a massive star rapidly collapses. In both types of supernova shock waves rip apart and drastically alter the stars, whereas in a nova no collapse occurs, just the fusion of hydrogen on the surface of the white dwarf. supernova explosion fades to ob- scurity in a year or two, an expanding shell of gas marks the site of the explosion. The gas originally ex- pelled at 10,000 to 20,000 km/s, may carry away a fifth of the mass of the star. The collision of that ex- panding gas with the surrounding interstellar medium can sweep up even more gas and excite it to produce a supernova remnant, the nebulous remains of a su- pernova explosion. white dwarf supernova produced by the collapse and total destruction of a white dwarf. Discussion Questions 1. How do we know the helium flash occurs if it cannot be observed? Can we accept something as real if we can never observe it? The helium flash comes about because of what we know of the fusion reactions in the suns core and what we know about the behavior of degenerate matter. While it is theoretical, it does seem to match well with our current understanding of stellar evolution. Further study of the He flash may allow for experiments that lend support to the theory. A helium flash occurs because the core of the star is in what is known as a "degenerate" state. This means that the core has contracted so much that the pressure of the electron shells of the atoms making up the core prevent the core from contracting further. Under normal gas conditions (i.e. NOT a degenerate state), an increase in the temperature of the core would cause an increase in core pressure resulting in the core expanding and the temperature then dropping. This state is known as hydrostatic equilibrium. With a degenerate core, the temperature increases but the pressure doesn't. This extra energy ignites the helium creating run-away nuclear reactions. This is what is referred to as a helium flash. 2. False-color radio images and time exposure photographs of astronomical images show us aspects of nature we can never see with our unaided eyes. Can you think of common images in newspapers or on television that re- veal phenomena we cannot see? we have X-rays taken of our broken bones. X-rays images. The entire range of energies of light, including both light we can see and light we cannot see, is called the electromagnetic spectrum. It includes, from highest energy to lowest: gamma-rays, X-rays, ultraviolet, optical, infrared, microwaves, and radio waves. Because we can only see visible light, we are put at a disadvantage when we study the Universe because, because it is actively emitting light at all these different energies. Problems 1. About how long would a 0.4-M star spend on the main sequence? Hint: See By the Numbers 9-1. The approximate time on the main sequence is equal to the 1/(M^2.5). You read this as one over mass to the 2.5 power. Plugging in 0.4 solar masses for M gives an approximate time on the main sequence of about 10 solar lifetimes. Since the solar lifetime is 10^10 years, the approximate life expectancy of a 0.4 solar mass star is about 100 billion years. By the Numbers 9-1 T= fuel rate of consumption T 1 X (10 X 10 9 yr) 32 =310 X 10 6 years. 2. The Ring Nebula in Lyrae is a planetary nebula with an angular diameter of 72 seconds of arc and a distance of 5000 ly. What is its linear diameter? Hint: See By the Numbers 3-1. We are given the angular diameter & theta; of the Ring Nebula and its distance d. We are asked to find its linear diameter (or "size") s. We should use the small angle formula: & theta; = (s / d) (206,265 seconds of arc / radian) Let's solve this for s. s = d θ / (206,265 second of arc / radian) = (5000 ly) (72") / (206,265") = 1.74 ly 3. If the Ring Nebula is expanding at a velocity of 15 km/s, typical of planetary nebulae, how old is it? This is a problem in which we have the planetary nebula expanding at a constant speed v = 15 km/s, and it has reached a distance of one half of 1.74 ly (the radius of the nebula). We need to find the time t that this motion has taken. Use the distance formula, d = v t, solved for t. t = d / v = (0.87 ly) / (15 km/s) = (0.87 ly)(9.5 x 1015 m/ly) / (15,000 m/s) = 5.5 x 1011 seconds = 17,000 years To convert this to years, I divided 5.5 x 1011 seconds by the numbers of seconds in a year which is 3.16 x 107 s/yr. 4. Suppose a planetary nebula is 1 pc in radius. If the Doppler shifts in its spectrum show it is expanding at 30 km/s, how old is it? Hints: 1 pc equals 3 X 10 13 km, and 1 year equals 3.15 X 10 7 seconds. 5. If a star the size of the sun expands to form a giant 20 times larger in radius, by what factor will its average density decrease? Hint: The volume of a sphere is (4ttr3.) 3 6. If a star the size of the sun collapses to form a white dwarf the size of Earth, by what factor will its density page 202 The definition of density is mass M divided by volume V. The volume V of a sphere is (4/3) π R3. Putting this together we get: density = M / V = M / ( 4/3 π R3) Let's write this formula once for the white dwarf WD and once for the star S before it became a white dwarf. densityWD = MWD / ( 4/3 π R3WD) densityS = MS / ( 4/3 π R3S) Now divide the second to last equation by the last equation. (densityWD / densityS) = (MWD / MS) (R3S / R3WD) = (1) (100)3 =106 I used the fact that the sun's radius is about 100 times the earth's radius. Therefore, the white dwarf is approximately one million times denser than the original star. increase? Hints: The volume of a sphere is (4)ttr3. See 3 Appendix A for the radii of the sun and Earth. 7. The Crab Nebula is now 1.35 pc in radius and is ex- panding at 1400 km/s. About when did the supernova occur? Hint: 1 pc equals 3 X 10 13 km. 8. If the Cygnus Loop is 40 pc in diameter and is 20,000 years old, with what average velocity has it been ex- panding? Hints: 1 pc equals 3 X 10 13 km, and 1 year equals 3.15 X 10 7 seconds. 9. Observations show that the gas ejected from SN1987A is moving at about 10,000 km/s. How long will it take to travel one astronomical unit? one parsec? Hints: 1 AU equals 1.5 X 10 8 km, and 1 pc equals 3 X 10 13 km. 10. If the stars at the turnoff point in a star cluster have masses of about 4 M, how old is the cluster? All we need to do for this problem is to find the lifetime of stars which have a mass of four solar masses. We can estimate this from table 9-2 on page 172. That table tells us that the lifetime of a 3.5 solar mass star is 440 million years. Therefore, the lifetime of a four solar mass star is a little shorter, maybe around 300 or 350 million years. Critical Inquiries for the W eb 1. As seen in on pages 190 and 191 there is an incredible diversity of appearance for planetary nebulae. Browse the Web for images and information on these dying stars, and discuss why there is a range of shapes of planetary nebulae that we see. The gravity from the material in the outer part of the star takes its inevitable toll on the structure of the star condensation and heat planetary nebulae http://www.noao.edu/jacoby/pn_gallery.html A planetary nebula forms when a star can no longer support itself by fusion reactions in its center. The gravity from the material in the outer part of the star takes its inevitable toll on the structure of the star, and forces the inner parts to condense and heat up. The high temperature central regions drive the outer half of the star away in a brisk stellar wind, lasting a few thousand years. When the process is complete, the remaining core remnant is uncovered and heats the now distant gases and causes them to glow. Despite the name, these objects are totally unrelated to "planets". It is commonly thought that they may represent the final episode of the Sun's existence as a star. 2. Naked-eye supernovae in our galaxy are rare, but astron- omers have noted supernovae in other galaxies for years. Look for summaries of observations of recent supernovae. Are similar numbers of type I and type II being seen? Compare the number of supernovae seen during the last few years with that of two decades ago. Why are we find- ing so many more supernovae in recent years than in the past? Type I Type IIs have shown somewhat less promise as distance indicators. They are considerably fainter http://nedwww.ipac.caltech.edu/level5/Willick/Willick6.html The use of supernovae as distance indicators has grown dramatically in the last few years. Supernovae have been applied to the Hubble Constant problem, to measurement of the cosmological parameters 0 and , and even, in a preliminary way, to constraining bulk peculiar motions. There is every reason to believe that in the next decade supernovae will become still more important as distance indicators. It is certain that many more will be discovered, especially at high redshift. Supernovae come in two main varieties. Type Ia supernovae (SNe Ia) are thought to result from the nuclear detonation of a white dwarf star that has been overloaded by mass transferred from an evolved (Population II) companion. (Recall that a white dwarf cannot have a masss above the Chandrasekhar limit, 1.4 M When mass transfer causes the white dwarf to surpass this limit, it explodes.) Type II supernovae result from the imploding cores of high-mass, young (Population I) stars that have exhausted their nuclear fuel. (1) Of the two, it is the Type Ias that have received the most attention lately. Type IIs have shown somewhat less promise as distance indicators. They are considerably fainter (~ 2 mag), and thus are detected less often in magnitude limited surveys (although their intrinsic frequency of occurrence is in fact greater than that of Type 1as). The discussion to follow will be restricted to Type Ias. Exploring The Sky 1. Locate the planetary nebulae M 57, M 97, and M 27. How does their shape distinguish them from the star formation nebulae such as M 42 and M 8? Hint: To find an object use Find under Edit. Choose Messier Objects, and pick from the list. M 57, M 97, and M 27 Ring Nebula in Lyra M57 Magnitude: 9.0 RA: 18h 53m 42.6s Dec: +3302'15" RA: 18h 53m 36.0s Dec: +3302'00" (Epoch 2000) Azm: 34757'35" Alt: +0820'10" Rise: 10:00 Transit: 13:03 Set: 16:06 Size: 2.5' Owl Nebula M97 Magnitude: 11.2 RA: 11h 15m 01.8s Dec: +5459'31" RA: 11h 14m 48.0s Dec: +5501'00" (Epoch 2000) Azm: 30238'43" Alt: -5759'53" Always below horizon. Transit: 05:26 Size: 3.2' Dumbbell Nebula M27 NGC 6853 Other description: Planetary nebula irregular. Constellation: Vul Dreyer description: A magnificent (or otherwise interesting) object!, very bright, very large, binuclear, irregular extended (Dumbbell); = M27. Magnitude: 8.1 RA: 19h 59m 44.2s Dec: +2243'36" RA: 19h 59m 36.0s Dec: +2243'00" (Epoch 2000) Azm: 0212'22" Alt: +1937'35" Rise: 09:55 Transit: 14:09 Set: 18:23 Size:15.2' From Owl Nebula: Angular separation: 9149'52" Position angle: +13601' M 42 and M 8 Great Nebula in Orion Orion Nebula M42 NGC 1976 Other description: Nebula. Constellation: Ori Dreyer description: A magnificent (or otherwise interesting) object! Theta1 Ori and the great nebula; = M42. Magnitude: 4.0 RA: 05h 35m 36.3s Dec: -0526'47" RA: 05h 35m 24.0s Dec: -0527'00" (Epoch 2000) Azm: 14038'39" Alt: -2910'49" Rise: 17:17 Transit: 23:43 Set: 06:13 Size:66.0' Hour Glass Nebula Lagoon Nebula M8 NGC 6523 Other description: Nebula with dust and cluster. Constellation: Sgr Dreyer description: A magnificent (or otherwise interesting) object! Very bright, extremely large, extremely irregular figure, with large cluster; = M8. Magnitude: 5.8 RA: 18h 04m 00.2s Dec: -2423'04" RA: 18h 03m 48.0s Dec: -2423'00" (Epoch 2000) Azm: 30817'26" Alt: +5835'55" Rise: 04:11 Transit: 12:13 Set: 20:15 Size:90.0' 2. The Crab Nebulae is M 1. Locate it, zoom in, measure its angular size in seconds of arc, and compute its diame- ter, assuming it is about 6000 ly from Earth. Crab Nebula M1 NGC 1952 Other description: Nebula. Constellation: Tau Dreyer description: Very bright, very large, extended 135 +/-, very gradually little brighter middle, resolvable, but mottled; = M1. Magnitude: 8.4 RA: 05h 34m 44.9s Dec: +2201'15" RA: 05h 34m 30.0s Dec: +2201'00" (Epoch 2000) Azm: 12116'34" Alt: -5239'13" Rise: 19:24 Transit: 23:42 Set: 04:04 Size: 6.0' From Hour Glass Nebula: Angular separation: 17251'53" 3. Locate the supernova remnant called the Cygnus Loop just south of e Cygni. How big is this object in angular diameter compared to the full moon? Hint: Under the View menu choose Labels and Setup. Check Bayer Designation go to Cygnus and zoom in on e Cygni until the Cygnus Loop appears. The canonical distance for the Cygnus Loop, used by nearly all astronomers for the last 40 years or so, is 770 pc. 1 pc = 1 parcsec = 3.26 light-years. So 770 pc = 2510 ly. The Cygnus Loop supernova remnant is the expanding leftovers of a star about ten times the mass of the sun that exploded some 10,000 years ago in the direction of the constellation Cygnus the Swan (a.k.a. the Northern Cross). How big is Cygnus Loop object in angular diameter compared to the full moon? Cygnus Abbreviation: Cyg Genitive Form: Cygni Description: The Swan Pronunciation: SIG' nuhs Genitive Pronunciation: SIG' ny Sky Database: Constellation Labels RA: 20h 37m 18.3s Dec: +4202'40" RA: 20h 37m 12.0s Dec: +4201'48" (Epoch 2000) Azm: 0838'20" Alt: -0016'05" Rise: 13:49 Transit: 14:46 Set: 15:44 From SAO 186038: Angular separation: 7630'14" Position angle: +2906' page 203 Chapter 10 The Deaths of Stars A Visual Binary Star System The bright star Sirius A has a faint companion Sirius B arrow a white dwarf. 1960 Over the years astronomers can watch the two move and map their orbits. 1970 A line between the stars always passes through the center of mass of the system. 1980 The star closest to the center of mass is the most massive. The elliptical orbits are tipped at an angle to our line of sight. Beginning of Figure 8-17 An Eclipsing Binary Star System m t A small hot star orbits a large cool star and we see their total light. As the hot star crosses in front of the cool star we see a decrease in brightness. As the hot star uncovers the cool star, the brightness returns to normal. When the hot star is eclipsed behind the cool star the brightness drops. The depth of the eclipses depends on the surface temperatures of the stars. Figure 8-17. From Earth, an eclipsing binary looks like a single point of light, but changes in brightness reveal that two stars are eclipsing each other. Doppler shifts in the spectrum combined with the light curve shown here as magnitude versus time, can reveal the size and mass of the individual stars. page 146 Chapter 8 Use luminosity and temperature to calculate the radii of stars. Eclipsing binary systems give us a way to measure the sizes of stars directl. For example if it took 300 seconds for the small star to disappear while traveling 500 km/s relative to the large star then it must be 150,000 km in diameter. 500 x 300 000 000 =150,000 Copernicus system finally won acceptance because it described the apparent motions of the planets in a more simplistic or elegant manner. Additionally, several of the philosophical underpinnings of the Ptolemaic system (perfection of heavens, all celestial bodies revolve around Earth) were shown to be incorrect by Galileos telescopic discoveries. http://nedwww.ipac.caltech.edu/level5/Glossary/Glossary_O.html Death can come between 36 and 72 hours after exposure. It is estimated that one milligram of ingested ricin can kill an adult, the bulletin said. The toxin could be used in ventilation systems, drinking water or food supplies, it would be most effective as a poison through individual injection or as a food contaminant. Use of ricin in an attempted mass attack in water supplies, would require the processing of an extremely large amount of castor beans. Chapter 11 Answers to Chapter 11 Review Questions Seeds Horizons 2002 Chapter 11 Neutron Stars and Black Holes Answers to Review Questions Review Questions 1. How are neutron stars and white dwarfs similar? How do they differ? Neutron stars and white dwarfs are similar in that both are at the end points of stellar evolution, have high surface temperatures, no longer produce energy via thermonuclear fusion, are composed of degenerate matter, have very small radii and are very dense. Neutron stars and white dwarfs are different in that neutron stars are the cores of massive stars that went through a supernova, while white dwarfs are the cores of intermediate- mass stars that produced planetary nebulae. (White dwarfs can also be formed by low mass stars whose outer layers sort of evaporate, but the universe isn't old enough for any of this to have formed in this way.). Additionally, neutron stars are hotter and more compact than white dwarfs. Consequently they have stronger magnetic fields, rotate faster, and radiate larger amounts of short wavelength energy (e.g. x-rays) and can produce synchrotron radiation. The inside of a neutron star would have to be much denser than the inside of a white dwarf. A neutron star is a star of a little over 1 solar mass com- pressed to a radius of about 10 km. Its density is so high that the matter is stable only as a fluid of neu- trons. 2. Why is there an upper limit to the mass of neutron stars? The upper limit on the mass of a neutron star is set by the maximum pressure that a degenerate neutron gas can provide. At extremely high pressure the neutrons will become degenerate and the pressure that supports the weight of the star comes from the degenerate neutrons. If the mass is greater than 2 or 3 solar masses, then the degenerate pressure of the neutrons is not sufficient to support the outer layers and the neutrons will be driven into each other. There is no known method to keep the material from collapsing to a point when this occurs. The theory of neutron degeneracy says that neutron degeneracy pressure will not be able to support a star against gravitational collapse if the mass of the star is more than about two or three solar masses. A neutron star should have a powerful magnetc field. Whatever magnetic field a star has is frozen into the star. 3. Why do we expect neutron stars to spin rapidly? Conservation of angular momentum tells us that as rotating objects are drawn closer to the rotation axis, the object will spin faster. In producing a neutron star, the core of a star is spinning and begins to collapse. The collapsing core must conserve angular momentum, and since it will decrease its radius by a large amount, it will spin much faster than a normal star. A neutron star is a star of a little over 1 solar mass com- pressed to a radius of about 10 km. Its density is so high that the matter is stable only as a fluid of neu- trons. Theory predicts that such an object would spin a number of times a second, be nearly as hot at its sur- face as the inside of the sun, and have a magnetic field a trillion times stronger than Earths. The neutron was discovered in the laboratory in 1932, and its proper- ties suggested something fantas- tic. Neutrons spin in much the way that electrons do, which means that neutrons must obey the Pauli ex- clusion principle. In that case, if neutrons are packed together tightly enough, they can become degen- erate just as electrons do. White dwarfs are supported by degener- ate electrons 4. If neutron stars are hot, why aren't they very luminous? Neutron stars are very hot but not very luminous because they have very small surface areas. Recall that the luminosity of an object depends on both its temperature and its size. The radius of a neutron star is typically 100,000 times smaller than that of a main sequence star. If a neutron star were 10 times hotter than a main sequence star (e.g. 100,000 K) and 100,000 times smaller, then the luminosity of the neutron star would be 1 million times less than that of a main sequence star. It might be because of the small size a neutron star is. 5. Why do we expect neutron stars to have a powerful mag- netic field? Theory predicts that the magnetic field of the star will be frozen into the star so that as it collapses it becomes much stronger because it is packed into a smaller and smaller region. Because some stars have magnetic fields over 1000 times stronger than the sun's, we might expect a neutron star to have a magnetic field as much as a trillion times stronger than the sun's. Basic theory also predicts that a neutron star should have a powerful magnetic field. Whatever mag- netic field a star has is frozen into the star. The gas of the star is ionized, and that means the magnetic field cannot move easily through the gas. Neutron stars are very hot and the small surface areas of neutron stars mean that they will be faint objects. 6. Why did astronomers conclude that pulsars could not be pulsating stars? Astronomers were able to show that pulsars could not be pulsating normal stars because the duration of the pulse (that is, the period of pulsation) was so short that the stars had to be smaller than 300 km in diameter. Not even white dwarfs are this tiny. Additionally, normal stars, including white dwarfs, do not posses a gravitational field that is strong enough to hold them together under the violent (rapid) pulsations that would be required to produce the observed pulses. Normal stars could also not rotate fast enough and remain intact to produce the observed pulses. Finally, neutron stars would vibrate much too fast and with too small an amplitude to produce the observed pulses. The only option that could produces the pulses observed and not destroy the star was a rotating neutron star. Pulsars are pulsing with different periods. Pulsars are natural, and are objects in the sky. Pulsing radio source Periods ranged from 0.033 to 3.75 seconds. 7. What does the short length of pulsar pulses tell us? The very short period of pulsar pulses tell us that the objects must be very small. If an object changes its brightness rapidly, the object can not be bigger than the distance traveled by light in the period of time that the variations occur. That is, if a pulse lasts only 0.001 seconds, then the pulsar has to have a diameter of 0.003 light seconds (approximately 300 km) or smaller. Otherwise, any distinct variations would be smeared out traveling greater distances. Thus the variations must occur over relatively small distances. A pulsar's quick pulses tell us that the pulsar must be small. If an object changes brightness quickly, the object can't be bigger than the time it takes light to travel across the object. If a pulsar pulse lasts only 0.001 s, then the size of the pulsar must be no larger than approximately this time interval times the speed of light, i.e., (0.001 s)(3 x 108 m/s) = 300 km. For other reasons, we think pulsars are actually about ten times smaller than this. 8. How does the lighthouse model explain pulsars? The lighthouse theory suggests that pulsars are the result of a rapidly rotating neutron star with a strong magnetic field that is not aligned with the star’s rotation axis. The rotating magnetic field produces an electric field near the neutron star that pulls charged particles off the surface near the magnetic poles. These accelerated charged particles, spiraling through magnetic fields, produce synchrotron radiation that forms two beams emanating from each magnetic pole. As the neutron star rotates about an axis that is different from that of the magnetic field, these two beams sweep through space like a lighthouse beacon. An observer located in a direction near the path of one of these beams will see the neutron star turn on and off rapidly as the beams sweep past them. The Lighthouse Model of a Pulsar A pulsar does not pulse but rather emits beams of radiation that sweep around the sky as the neutron star rotates. The mechanism that produces the beams involves extremely high energies and is not fully understood. We tend to see only those pulsars whose beams sweep over Earth. A white light revolves around inside of a lighthouse. Neutron Star Rotation with Beams As in the case of Earth, the magnetic axis of a neutron star could be inclined to its rotational axis. The rotation of the neutron star will sweep its beams around like beams from a lighthouse. While a beam points roughly toward Earth, we detect a pulse. While neither beam is pointed toward us, we detect no pulse. Beams may not be as exactly symmetric as in this model. 9. What evidence do we have that pulsars are neutron stars? The evidence that supports pulsars as neutron stars is their pulsation period, synchrotron radiation, strong magnetic fields, and the nebulae that surround several of them. The short pulsation period can only be achieved by a very small radius star that is extremely dense. The synchrotron radiation indicates the presence of both a very strong magnetic field and a very hot surface. The magnetic fields and number of high energy charged particles required to produce the observed synchrotron radiation indicates that the star must have an extremely large magnetic field, larger than in white dwarfs and normal stars. Finally, that some of the neutron stars lie at the center of visible supernova remnants and power those remnants indicates that they are the core of the star that produced the supernova remnant. Theoretical studies link supernovae with neutron stars or black holes, not normal stars or white dwarfs. The pulse period of a pulsar equals the rotation period of the neutron star. If a neu- tron star 10 km in radius spins 642 times a second as does PSR 1937 +21 then the period is 0.0016 seconds and the equator of the neutron star must be trav- eling about 40,000 km/s. 10. Why would astronomers at first assume that the first millisecond pulsar was young? The millisecond pulsar was first assumed to be young because it had a very short period. This short period suggested that it was rotating very rapidly, more rapidly than any pulsar known at that time. All other pulsars were known to slow down with time, as they gradually convert energy of rotation into energy of radiation. The oldest pulsars were expected to have the longest pulse periods and the youngest were expected to have the shortest. 11. How can a neutron star in a binary system generate X rays? X-rays can be produced when matter falls on the surface of a neutron star. The strong gravitational field of a neutron star will accelerate the material to very high speeds and the energy released when the material impacts with the surface and with each other is sufficient to produce x-rays. X-rays can also be produced by neutron stars as fresh hydrogen piles up on the surface and eventually becomes hot enough to ignite thermonuclear fusion of this material. Since the nuclear reaction occurs on the surface of the neutron star, it releases a tremendous amount of energy directly. Finally, x-rays can be created by a neutron star as material piles up in an accretion disk around a neutron star and interacts with the strong magnetic field. X-ray beams can be produced out of the magnetic poles of the neutron star in this manner. Notice that, in all three cases, matter had to be added to the neutron star from another source. In binary systems material leaving the neutron stars companion can be drawn into an accretion disk around the neutron star and produce x-rays by one of the mechanisms described above. 12. If the sun has a Schwarzschild radius, why isn't it a black hole? The Schwarzschild radius of an object is a theoretical construct. The Schwarzschild radius is the distance from an object of a given mass at which the escape velocity is equal to the speed of light. A black hole is an object that has all of its mass inside the Schwarzschild radius. So we can calculate the size of the Schwarzschild radius for the Sun, a one solar mass object, but since not all of this mass is inside the Schwarzschild radius, the Sun is not a black hole. All objects have a Schwarzchild radius. Only if an object has shrunk to a radius smaller than its Schwarzchild radius will the object be a black hole. Fortunately Earth will not collapse spontaneously into a black hole because its mass is less than the critical mass of about 3 solar masses. Schwarzschild Black Holes A black hole forms when an object collapses to a small size per- haps to a singularity, and the escape velocity becomes so great light cannot escape. The boundary of the black hole is called the event horizon because any event that occurs inside is invisible to outside observes. The radius of the black hole R is the Schwarzschild radius. 13. How can a black hole emit X rays? A black hole itself can not emit x-rays, the material around it can. If a black hole is near another star or gas cloud, then it can draw some of that matter into itself. As this matter approaches the black hole it will form an accretion disk around the black hole. The matter in the accretion disk will become very hot as it is squeezed together by the gravitational field of the black hole. The temperature of this accretion disk can reach over 1 million K. Wiens law tells us that such an object would radiate most strongly near 3-nm, X-ray wavelengths. How to find a black hole: Look for a strong source of X rays. It may be a black hole into which matter is flowing. 14. What evidence do we have that black holes really exist? The best evidence that black holes exist is from the x-ray binaries that contain a compact object. Several of the x-ray binaries contain a normal star and a compact star whose mass is greater than 3 solar masses. There are other objects that can produce x-rays, most of these should also be very luminous at ultraviolet and visible wavelength. Current theories suggest that the mass of a neutron star cannot exceed 3 solar masses, so the compact object is postulated to be a black hole. Until we understand the physics of degenerate neutron material, we cannot be sure of the upper limit on the mass of a neutron star. Thus, these compact objects are not definitively proven to be black holes. 15. How can mass transfer into a compact object produce jets of high-speed gas? X-ray bursts? gamma-ray bursts? As material falls toward a compact object the gravitational field of the compact object will accelerate that material. The material must also conserve angular momentum, so that as it falls toward the compact object an accretion disk will form. As the material orbits in the accretion disk and falls toward the compact object, friction will heat the material to extremely high temperatures and the material will radiate at X-ray wavelengths. Thermal and magnetic processes can confine this radiation energy into narrow cones along the magnetic poles of the compact object, resulting in bipolar jets. X-ray bursts occur as material accretes around a neutron star or a black hole. Soft (low-energy) gamma-ray bursts seem to come from the accreted material processed by a magnetar, a neutron star with unusually strong magentic field. Hard (high-energy) gamma-ray bursts seem to be related to much more violent events, and not mass transfer in a binary system. X-ray nova outbursts as matter flows rapidly into the accretion disk. A year or so after an outburst the flow has stopped, the accretion disk has dimmed and astronomers can detect the spectrum of the main-sequence star. gamma-ray bursts Astronomers realized that the bursts might be coming from neutron stars or black holes. These objects are now known as gamma-ray bursters. Table 11-1 The Schwarzschild Radius Mass Rs Star 10 30 km Star 3 9 km Star 2 6 km Star 1 3 km Earth 0.000003 0.9 cm Table 11-2 Six Black Hole Candidates Object Location Companion Star Orbital Period Mass of Compact Object Cygnus X-1 Cygnus O supergiant 5.6 days >3.8M LMC X-3 Dorado B3 main sequence 1.7 days ~10M V616 Mon Monocerotis K main sequence 7.75 hours 10 + 5 M V404 Cygni Cygnus k main sequence 6.47 days 12 + 2 M J1655-40 Scorpius F-G main sequence 2.61 days 6.9 + 1 M QZ Vul Vulpecula K main sequence 8 hours 10 + 4 M How can a black hole emit X rays? Once a bit of matter falling into a black hole crosses the event horizon, no light or other electromagnetic radia- tion it emits can escape from the black hole. The matter becomes lost to our view. But if it emitted radiation be- fore it cross the event horizon, that radiation could escape, and we could detect it. Wein's law By the numbers 6-1 tells us that matter at such a high temperature should emit X-ras We can search for black holes by looking for X-ray binaries. By the Numbers 6-1 Black Body Radiation Wien's law 3,000,000 is divided by the temperature in degress kelvin. 5.67 X 10 -8 J/m2s degree 4. A max= 3,000,000 T page 223 Chapter 11 Neutron Stars and Black Holes Discussion Questions 1. In your opinion, has the existence of neutron stars been sufficiently tested to be called a theory, or should it be called a hypothesis? What about the existence of black holes? Neutron stars have been called a hypothesis. Black holes exist and are called a hypothesis, 2. Why wouldn't an accretion disk orbiting a giant star get as hot as an accretion disk orbiting a compact object? Problems 1. If a neutron star has a radius of 10 km and rotates 642 times a second, what is the speed of the surface at the neutron star's equator in terms of the speed of light? Hint: The circumferences of a circle is 2tt r. Since we are asked for the speed of a point on the surface of a neuton star, and since we are told something about distance traveled and time, let's use the distance formula d = v t. Solving this for speed v, we get v = d / t. The distance a point on the surface travels in one revolution is the circumference 2 π R, where R is the radius of the star. The time the star takes to make one revolution is the reciprocal of N, the number of revolutions per second the star makes. Let's put this all together. v = d / t = (2 π R) / (1/N) = 2 π R N = 2 π (10,000 m) (642/s) = 4 x 107 m/s Since the speed of light is 3 x 108 m/s, our answer is 13% of the speed of light. 2. A neutron star and a white dwarf have been found or- biting each other with a period of 11 minutes. If their masses are typical, what is the average distance be- tween them? Hint: See By the Numbers 8-4. By The Numbers 8-4 the total mass of two stars orbiting each other is related to the average distance a between them and their orbital period p. M + M = a3 p2 a in AU P in years.and mass in solar masses. Example A: If we observe a binary system with a period of 32 years and an average separation of 16 AU what is the total mass? Solution: The total mass equals 16 3 / 32 2, or 4 solar masses. 3. If Earth's moon were replaced by a typical neutron star, what would the angular diameter of the neutron star be as seen from Earth? Hint: See By the Numbers 3-1. 4. What is the Schwarzschild radius of Jupiter (mass = 2 X 10 27 kg)? of a human adult (mass = 75 kg)? Hint: See Appendix A for the values of G and c. For Jupiter, RSch = 2 G M / c2 = 2 (6.7 x 10-11 m3/[s2 kg]) (2 x 1027kg) / (3 x 108m/s)2 = 3 m For the adult human, RSch = 2 G M / c2 = 2 (6.7 x 10-11 m3/[s2 kg]) (75 kg) / (3 x 108m/s)2 = 1.1 x 10-25 m 5. If the inner accretion disk around a black hole has a temperature of 10 5 K, at what wavelength will it radiate the most energy? Hint: See By the Numbers 6-1. By the Numbers 6-1 Black Body Radiation Wien's law 3,000,000 is divided by the temperature in degress kelvin. 5.67 X 10 -8 J/m2s degree 4. A max= 3,000,000 T 6. What is the orbital period of a bit of matter in an accre- tion disk 2 X 10 5 km from a 10-M black hole? Hint: See By the Numbers 8-4. By The Numbers 8-4 the total mass of two stars orbiting each other is related to the average distance a between them and their orbital period p. M + M = a3 p2 a in AU P in years.and mass in solar masses. Example A: If we observe a binary system with a period of 32 years and an average separation of 16 AU what is the total mass? Solution: The total mass equals 16 3 / 32 2, or 4 solar masses. 7. If SS 433 consists of a 20-M star and a neutron star or- biting each other every 13.1 days, then what is the av- erage distance between them? Hint: See By the Num- bers 8-4. By The Numbers 8-4 the total mass of two stars orbiting each other is related to the average distance a between them and their orbital period p. M + M = a3 p2 a in AU P in years.and mass in solar masses. Example A: If we observe a binary system with a period of 32 years and an average separation of 16 AU what is the total mass? Solution: The total mass equals 16 3 / 32 2, or 4 solar masses. 8. What is the orbital velocity at a distance of 7400 meters from the center of a 5-solar-mass black hole? What kind of particles could orbit at this distance? Hint: See By the Numbers 4-1. We need to use the equation for the orbital velocity for an ob      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~ject in a circular orbit. VC = square root of (G M / r) = square root of ( (6.7 x 10-11 m3/[s2 kg]) (5 x 2 x 1030 kg) / (7400 m) ) = 3 x 108 m/s = the speed of light. Only massless particles like light and maybe neutrinos can move at that speed. 9. Compare the orbit in Problem 8 with an orbit having the same velocity around a 2-solar-mass neutron star. Why is this orbit impossible? Hint: See By the Numbers 4-1. Critical Inquiries for the Web 1. Imagine that you are on a mission to explore one of the pulsar planets noted in the chapter. What would you find there? Look for information about pulsars and the known pulsar planets on the Web and describe what you might encounter on such a mission. 2. What would you experience if you were to pilot a space- craft near a black hole? Visit black-hole-related Internet sites to determine what the gravitational effects and general environment would be. Also use the Internet to find the limits of human tolerance to strong gravitational forces. Hint: Look for information about astronaut train- ing and find out how many g's a human can withstand. Use these sources to give a brief account about what your voyage would be like. Exploring The Sky 1. The Crab Nebula pulsar is located in the Crab Nebula, also known as M 1. Locate it, zoom in, and compare its shape and size in Figure 11-3. appears to be very small like a size of a period in Figure 11-3. page 224 Chapter 11 Neutron Stars and Black Holes Seeds Horizons 2002
Chapter 12 The Milky Way Galaxy Answers to Review Questions Review Questions 1. Why isn't it possible to tell from the appearance of the Milky Way that the center of our galaxy is in Sagittaurius? The appearance of the Milky Way in the direction of Sagittarius is not noticeably different than it is in most other directions. It is a little brighter and wider, but not significantly. Additionally, clouds of gas and dust obscure our view of the disk of the Milky Way. Consequently, there is not a nice even distribution of visible light that is significantly brighter in one direction and significantly fainter in the opposite direction which would better indicate the location of the center. Our Milky Way Galaxy is over 75,000 ly in diameter and contains over 100 billion stars. 2. Why is there a period-luminosity relation? Massive stars cross the instability strip of the HR diagram at a greater luminosity than lower-mass stars. High-mass Cepheids have greater luminosities and radii than low-mass Cepheids. Because high-mass stars are both greater in mass and radius, they will pulsate more slowly than the lower-mass Cepheids. Therefore, high-luminosity Cepheids pulsate more slowly and have longer periods than the lower-luminosity Cepheids. A period-luminosity relation exists for the Cepheid and RR Lyra variable stars. This relationship between a star's luminosity and its pulsation period allows astronomers to determine the luminosity (absolute magnitude) of the star by simply measuring the pulsation period of the star. The apparent magnitude of the star can be determined by simply measuring the brightness of the star. Then using the equation in By the Numbers 8-2, which is based on the inverse square relation of light intensity and the magnitude scale, the distance to the star can be determined. Late in their lives stars pulse, and consequently their light output varies in a regular way. More massive stars are brighter on the average and take more time to pulse because they are more massive. Less massive stars are dimmer on the average and can pulse more quickly because they are less massive. So there is a period-luminosity relationship for these kinds of variable stars--the more massive the star is the greater is its luminosity and the greater is the period for its variability cycle. 3. How can astronomers use variable stars to find distance? Shapley knew that he could find the distance to the globular clusters if he could find the absolute magnitude of the variable stars in the clusters. find their average distance and their average absolute magnitude. 4. Why is it difficult to specify the thickness or diameter of the disk of our galaxy? The disk of the galaxy does not have a well defined boundary. Additionally the dimensions of the disk depend on the type of object you use to determine the boundary. For example, O stars and gaseous nebulae are confined to within 100 pc of the disk, while sun-like stars extend out to a thickness of about 1000 pc. Finally, the disk contains a large amount of gas and dust that obscures our view at visible and ultraviolet wavelengths. Observations before Shapley relied on counting the number of stars in a given direction and some also used their brightness. The galaxy contains giant molecular clouds that absorb light from distant objects, which then cannot be observed. Therefore, these early surveys of the galaxy only recorded stars out to the first giant molecular clouds. This made the galaxy appear much smaller than it actually is. The interstellar medium dims the more distant stars and makes them look farther away than they really are. Space is filled with gas and dust that dims our view of distant stars. 5. Why didn't astronomers before Shapley realize how large the galaxy is? Space is filled with gas and dust that dims our iews of distant stars. When we look toward the band of the Milky Way, we can see only the neighborhood near the sun. Most of the star systen is invisible. Leavitt could not find the absolute magnitude of any of the variable stars. 6. How do we know how old our galaxy is? We can determine the age of our galaxy by studying the turn-off points of the globular clusters. The turn-off points of a cluster diagram tell us the mass of the stars just ready to leave the main sequence. Since we can calculate the time that a star of a given mass will spend on the main sequence, we can determine how long the stars at the turn-off point have been around. Since all of the stars of a cluster presumably formed at about the same time, the cluster, and all of the stars within it, must have the same age. Therefore, the age of the cluster is equal to the lifetime of a star that is at the turn-off point of the cluster diagram. We know how old our galaxy is from the turn-off points of the globular clusters. Where the stars in the globular clusters turn-off the main sequence on an H-R diagram indicates the age of the clusters. 7. Why do we conclude that metal-poor stars are older than metal-rich stars? Metal-poor stars are found in globular clusters which have ages in excess of 10 billion years. The metal-rich stars are found in open clusters and associations with ages less than 10 billion years, and the most metal-rich stars, extreme population I stars, are found in star forming regions. Metal-rich stars are not found in old clusters. Also, on theoretical grounds, metal-poor stars are first ones to form because there had been no stars to manufacture metals (elements heavier than helium) to form metal-rich stars. This leads to the conclusion that metal-poor stars are older than the metal-rich stars. 8. How can astronomers find the mass of the galaxy? Astronomers can determine the mass of our galaxy by measuring the orbital velocities of stars at different distances from the center of the galaxy. Even determining the orbital velocity of the Sun gives a quick estimate of the mass of the galaxy located within a radius of the center equal to the orbital distance of the Sun. Determining the orbital velocity of objects farther from the center of the galaxy than the Sun allows us to determine the mass enclosed by that objects orbit. Astronomers can find the mass of our galaxy by using the generalized form of Kepler's third law. Plugging into this equation the time (P in the formula) it takes a star (like the sun) to orbit the center of the galaxy and the radius (a in the formula) of the star's orbit, one can solve for the mass of that part of the galaxy within the orbit of the star. We can use the orbital motion of the sun to find the mass of the galaxy. By observing the radial veloc- ity of other galaxies in various directions around the sky, astronomers can tell that the sun moves about 220 km/s in the direction of Cygnus. it's orbit is a circle with a radius of 8.5 kpc. divide the circumference of the orbit by the velocity and find the sun completes a single orbit in about 240 million years. 9. What evidence do we have that our galaxy has an ex- tended corona of dark matter? Our galaxy is believed to have a galactic corona because the rotation curve of the galaxy increases beyond the orbit of the Sun. The increase in the orbital velocity of stars farther from the center of the galaxy than the Sun suggests that the galaxy contains a great deal of mass in its outer reaches. This suggests a large amount of unseen material (dark matter) stretching beyond the extent of the halo. Some of this dark matter is located in the disk of our galaxy. the shape of the rotation curve and theoretical models of the rotation of the disk suggest that much of the dark matter is located in an extended dark halo also called the galactic corona. 10. How do the orbits of stars around the Milky Way Galaxy help us understand its origin? The orbits of stars around the Milky Way provide evidence of how the galaxy may have formed when we look carefully at how stars with different metallicities and ages orbit. The stars of the halo are metal-poor, old stars and have randomly inclined highly elliptical orbits. The stars of the disk on the other hand are young, metal-rich stars and have circular orbits that are confined to the disk. This suggests that the early galaxy was composed of metal-poor stars and the shape of this early galaxy was primarily spherical. Most of the gas and dust has now settled into the disk and stars formed from this material are younger and metal-rich. That their orbits are confined to the disk indicates that collisions between atoms, molecules, and gas clouds in the early galaxy caused random motions to cancel out with only the dominant rotational motion we now observe remaining. Astronomers try to reconstruct our galaxy's past from the fossil it left behind as it formed and evolved. That fossil is the spherical component of the galaxy. The stars we see in the halo formed when the galaxy was young. The chemical composition and the distribution of these stars can give us clues to how our galaxy formed. 11. What evidence contradicts the traditional theory for the origin of our galaxy? Four contradictions are presented. First, the traditional theory predicts that the globular clusters and single stars of the halo should all be of about the same age and metallicity as the oldest stars in the galaxy. If any real difference in age of globular clusters exists, the oldest clusters should be at the outer edges of the halo. A few very metal-poor stars have been found in the halo. They are significantly more metal poor than the average halo star. Secondly, some of the oldest stars are located not in the halo, but in the central bulge, again counter to the predictions of the traditional theory. Thirdly, studies have shown that the globular clusters of the halo range from about 13 billion years old to 17 billion, which is a much larger spread in age than expected from the traditional theory. Finally, the younger globular clusters are found at the outer reaches of the halo, instead of nearer the disk where they would be expected in the traditional theory. The spiral tracers are young because the spiral arms are young. The spiral arms are places in the disk of the galaxy that appear significantly brighter than their surroundings. The spiral arms contain the most luminous stars, the largest fraction which are O and B main sequence stars. These stars do not live very long and hence are always identified as young objects. Since these objects formed recently, we also expect to see gaseous nebulae associated with star formation near them. These include giant molecular clouds and HII regions, both of which are young objects. Globular cluster ages average about 11 billion years. Data tells us the halo of our galaxy is at least 11 billion years old. The ages of star clusters 12. Why do spiral tracers have to be short-lived? The primary observational evidence against the spiral density wave theory is the existence of spurs and branches and the existence of multi-armed spiral galaxies. Computer simulations reveal that spiral density waves lead to two well defined spiral arms that radiate outward smoothly from the nuclear bulge. The Milky Way and many other spiral galaxies have spurs and branches sticking off of the spirals. Additionally, many of the spirals contain more than two well-defined arms. Therefore, though the overall idea of the density wave theory may be on the right track, it is clear that the theory is incomplete until it can explain these features. spiral tracers have to be short-lived because other tracers include young open clusters clouds of hydrogen ionized by hot star Notice that all spiral tracers are young objects. O stars for example live for only a few million years. If their orbital velocity is about 250 km/s, they cannot have moved more than about 500 pc since they formed. This is less than the width of a spiral arm. Because they don't live long enough to move away from the spiral arms, they must have formed there. Objects used to map spiral arms are called spiral tracers. 13. What evidence do we have that the density wave the- ory is not fully adequate to explain spiral arms in our galaxy? The most popular theory since the 1950s is called the density wave theory. It proposes that spiral arms are waves of compression, rather like sound waves, that moves around the galaxy, triggering star formation. Because these waves move slowly, orbiting gas clouds overtake the spiral arms from behind and create a moving traffic jam within the arms. 14. What evidence do we have that the center of our galaxy is a powerful source of energy? The core of the galaxy appears to be a very powerful source of energy for several reasons. First, the gas very close to the core is highly ionized, indicating that extremely high temperatures exist in this region. This suggests that the core produces large amounts of X-ray, ultraviolet and possibly gamma-ray radiation. Secondly, radio observations have shown the presence of large powerful jets emanating from the core. Such jets require high-energy phenomena to produce them. Finally, the motions of the hot gas and cool clouds and stars in the vicinity of the core suggest that it is extremely massive and very compact. We cannot see the center of our galaxy in visible light because that light is absorbed by interstellar dust long before it gets to us. However we can observe the radio waves and some infrared waves the center emits. From that information we see that the atoms near the center of the galaxy are highly ionized, suggested a powerful energy source is there. Furthermore we observe jets from the center, also suggesting a powerful energy source. Finally the motions of gas near the center suggests a very massive compact object is there; it is probably a super-massive black hole. 15. Why is the lack of motion of Sgr A* important evidence in our study of the center of our galaxy? The center of our galaxy should not be in orbital motion, it is after all the center, the pivot point. Sgr A* is located in the direction that the center is believed to lie based on the orbital motions of globular clusters and material within the disk of the galaxy. Several radio sources are found in this direction, but only Sgr A* is stationary. Material on each side of it does show orbital motion. Therefore, Sgr A* appears to be located at the gravitational center of our galaxy. The radio map shows Sgr A and the Arc filaments 50 parsecs long. The image was made with the VLA radio telescope. Discussion Questions 1. How would this chapter be different if interstellar mat- ter didn't absorb starlight? It would be of how interstellar matter didn't absorb starlight, and the interstellar matter would be a lot cooler. Starlight heats interstellar dust 2. Are there any observations you could make with the Hubble Space Telescope that would allow you to better understand the nature of Sgr A*? Yes. The Hubble Space Telescope can make better observations This infrared image of the Galactic Center shows the amount of dust and other stuff blocking our view. The Galactic Center is about 8 kpc away. Proper motion also provides more evidence that Sgr A* is a black hole. Most people believe that Sgr A* is a supermassive black hole. With a radius of about 0.01 pc2 and an estimated mass of 2.6 million times the mass of the Sun, a supermassive black hole seems a probable explanation. Mass estimates are found using the proper motion of Sgr A* and stars in its vicinity2. Figure 2: Taken by the VLA, this is a 1 meter radio wave image of the galactic center. Sgr A is much more luminous in this image than in the previous image. Visible also in this image are cloud of gas and the supernova remnant. Image The inner parsecs of the galaxy, known as Sagittarius A, have baffled astronomers for years. Its contents include a cluster of young stars, a large molecular dust ring, ionized gas streamers, diffuse hot gas, a supernova-like remnant, and, the most controversial object, the possible supermassive black hole candidate, Sagittarius A* (Sgr A*)2. Blanketed by dust, observation of this region has become possible thanks to long wavelength observation such as submillimeter, gamma-ray, X-ray, infrared, and radio seen using the Very Large Array, Chandra X-Ray Observatory, and the Very Long Baseline Array among others. Study of the galactic nucleus has become quite popular, mainly due to radio point source Sgr A*. Recently released evidence may shed some light on the exact identity of Sgr A*. http://www.pas.rochester.edu/~afrank/A232/Final_Projects/ALowell/Sagittarius%20A.htm Problems 1. Make a scale sketch of our galaxy in cross section. In- clude the disk, sun, nucleus, halo, and some globular clusters. Try to draw the globular clusters to scale size. 2. Because of dust clouds, we can see only about 5 kpc into the disk of the galaxy. What percentage of the ga- lactic disk can we see? Hint: Consider the area of the entire disk and the area we can see. 3. If the fastest passenger aircraft can fly 1600 km/hr 1000 mph, how many years would it take to reach the sun? the galactic center? Hint: 1 pc = 3 X 10 13 km. Let's use the distance formula d = v t solved for time t. The time to reach the sun would be t = d / v = (1.5 x 1011 m) / (1,600,000 m/hr) = 94,000 hr = 10.7 years Thus the plane's speed is 1 AU/(10.7 yr) = 0.0935 AU/yr. To find the time to reach the galactic center, let's first find out how far the galactic center is in astronomical units. The sun is 8.5 kpc (kilo-parsecs) from the galactic center. The Appendix of our text tells us that 1 kpc = 206,265,000 AU. Therefore, the distance from the sun to the galactic center is (8.5 kpc) (206,265,000 AU/kpc) = 1.8 x 109 AU. Since the speed of the plane is 0.0935 AU/yr, the time it takes the plane to reach the galactic center would be t = d / v = (1.8 x 109 AU) / (0.0935 AU/yr) = 1.9 x 1010 years. 4. If the RR Lyrae stars in a globular cluster have apparent magnitudes of 14, how far away is the cluster? Hint: See By the Numbers 8-2. 5. If the interstellar dust makes an RR Lyrae star look 1 mag- nitude fainter than it should, by how much will we over- estimate its distance? Hint: See By the Numbers 8-2. 6. If a globular cluster is 10 minutes of arc in diameter and 8.5 kpc away, what is its diameter? Hint: Use the small- angle formula from By the Numbers 3-1. page 249 Chapter 12 The Milky Way Galaxy 7. If we assume that a globular cluster 4 minutes of arc in diameter is actually 25 pc in diameter, how far away is it? Hint: Use the small-angle formula from By the Num- bera 3-1. The angular size & theta; of the globular cluster is 4 minutes of arc = 240 seconds of arc = 240". Its actual linear size s is estimated to be 25 pc, since this is typical of many globular cluster. To find its distance d, use the small angle formula & theta; = (s/d)(206,265"/rad) solved for d. d = (s/θ)(206,265"/rad) = (25 pc / 240") (206,265"/rad) = 21,500 pc. 8. If the sun is 5 billion years old, how many times has it orbited the galaxy? To find the amount of times the sun has orbited the galaxy, lets use the amount formula, amount = (rate)(time). In this formula the amount is N, the number of times the sun has orbited the galaxy. The rate is the rate at which the sun is orbiting the galaxy; this is the reciprocal of its orbital period P. The time is the age A of the sun, which is the toal time the sun has been orbiting. The problem tells us the age A of the sun is 5 billion years. The orbital period P is given in the text as 240 million years. Putting this together we get N = (rate)(time) = (1/P)(A) = A / P = (5 x 109 yr) / (2.4 x 108) yr) = 21 revolutions 9. If the true distance to the center of our galaxy is found to be 7 kpc and the orbital velocity of the sun is 220 km/s what is the minimum mass of the galaxy? Hints: Find the orbital period of the sun, and then see By the Num- bers 8-4. By The Numbers 8-4 the total mass of two stars orbiting each other is related to the average distance a between them and their orbital period p. M + M = a3 p2 a in AU P in years.and mass in solar masses. Example A: If we observe a binary system with a period of 32 years and an average separation of 16 AU what is the total mass? Solution: The total mass equals 16 3 / 32 2, or 4 solar masses. 10. Infrared radiation from the center of our galaxy with a wavelength of a about 2 X 10 -6 m (2000 nm) comes mainly from cool stars. Use this wavelength as max and find the temperature of the stars. Hint: See By the Num- bers 6-1. By the Numbers 6-1 Black Body Radiation Wien's law 3,000,000 is divided by the temperature in degress kelvin. 5.67 X 10 -8 J/m2s degree 4. A max= 3,000,000 T Critical Inquiries for the Web 1. How is our view of other galaxies affected by our posi- tion in the Milky Way? Search the Internet or other sources for sky maps that trace the path of the Milky Way through the constellations. Which constellations does the Milky Way run through? Choose one of these constellations and another constellation that lies far from the band of the Milky Way. Consult the Web for lists of galaxies visible in these two constellations. In which one have more galaxies been found? Why is this so? Which constellations does the Milky Way run through? the Big Dipper http://www.windows.ucar.edu/tour/link=/the_universe/sky_maps.html http://www.windows.ucar.edu/tour/link=/the_universe/images/Seattle/SETMAY9_JPG_image.html Orion Nebula Andromeda, Perseus, and Cassiopeia 2. Massive stars like Eta Carinae can play a significant role in the formation of stars along a spiral arm. Look for in- formation on the Web related to Eta Carinae which is one of the most carefully studied stars in the heavens. Briefly summarize the observed behavior of this star, and discuss why it is a good candidate for contributing to self-sustaining star formation in our galaxy? http://hubblesite.org/newscenter/archive/1996/23/ Eta Carinae A huge, billowing pair of gas and dust clouds is captured in this stunning Hubble telescope picture of the super-massive star Eta Carinae. Even though Eta Carinae is more than 8,000 light-years away, features 10 billion miles across (about the diameter of our solar system) can be distinguished. Eta Carinae suffered giant outburst about 150 years ago, when it became one of the brightest stars in the southern sky. Though the star released as much visible light as a supernova explosion, it survived the outburst. Somehow, the explosion produced two lobes and a large, thin equatorial disk, all moving outward at about 1.5 million miles per hour. Estimated to be 100 times heftier than our Sun, Eta Carinae may be one of the most massive stars in our galaxy. Exploring The Sky 1. Locate Sagittarius and examine the shape of the Milky Way there and the profusion of globular clusters. Hint: To turn on Messier object labels, use labels and Setup under the View menu. Sagittarius Abbreviation: Sgr Genitive Form: Sagittarii Description: The Archer Pronunciation: SAJ' ih TAY' rih uhs Genitive Pronunciation: SAJ' ih TAY' rih eye Sky Database: Constellation Labels RA: 19h 06m 48.3s Dec: -2545'58" RA: 19h 06m 36.0s Dec: -2546'12" (Epoch 2000) Azm: 33438'28" Alt: +6630'25" Rise: 05:05 Transit: 13:16 Set: 21:27 From 9P/Tempel 1: Angular separation: 17544'05" 2. Locate the following globular clusters: M 3, M 4, M 5, M 10, M 12, M 13, M 15, M 22, M 55, M 92. Where are they located in the sky? Hint: Use Find under the Edit menu. M3 NGC 5272 Other description: Globular cluster highly resolved. Constellation: CVn Dreyer description: Very remarkable!, globular cluster, extremely bright, very large, very abruptly much brighter middle, stars of magnitude 11 and fainter; = M3. Magnitude: 6.4 RA: 13h 42m 21.8s Dec: +2821'42" RA: 13h 42m 12.0s Dec: +2823'00" (Epoch 2000) Azm: 28837'54" Alt: -2152'30" Rise: 04:13 Transit: 07:52 Set: 11:32 Size:16.2' From Sagittarius: Angular separation: 9449'26" Position angle: +29916' M4 NGC 6121 Other description: Globular cluster highly resolved. Constellation: Sco Dreyer description: Cluster, 8 or 10 bright stars line, with 5 stars, well resolved; = M4. Magnitude: 5.9 RA: 16h 23m 48.2s Dec: -2632'32" RA: 16h 23m 36.0s Dec: -2632'00" (Epoch 2000) Azm: 27844'50" Alt: +4442'56" Rise: 02:18 Transit: 10:33 Set: 18:49 Size:26.3' From M3: Angular separation: 6712'00" Position angle: +14104' 3. Compare the distribution of globular clusters with that of open clusters. Hint: Use Filters under the View menu to turn off everything but globular clusters, the Milky Way, the Galactic Equator, and Constellation Bound- aries. Use the thumbwheel at the bottom of the sky win- dow to rotate the sky. Now repeat with globular clus- ters off and open clusters on globular clusters open clusters page 250 Chapter 12 The Milky Way Galaxy Review Questions 1. Why didn't astronomers at the beginning of the 20th century recognize galaxies for what they are? Astronomers didn't have anyway of determining the distances to galaxies, and they couldn't see individual stars in them. Therefore, they didn't realize that they are huge, distant star systems, much more distant than stars in our Milky Way. Galaxies are not soli- tary beasts they collide and inter- act with each other. 90 to 99 percent of the matter in the uni- verse is invisible. We voyage out into the vast depths of the universe, out among the galaxies, space so deep it is unexplored even in fiction. Less than a century ago, astronomers did not un- derstand that they were galaxies. 2. How can a classification system aid s scientist? Their can be different kinds of galaxies to classify. We must gather some basic data concerning galaxies. We must classify the different kinds of galaxies and discover their basic properties-diameter, luminosity, and mass. The shapes of galaxies Spiral Galaxies Elliptical Galaxies Irregular Galaxies Galaxies rich in gas and dust have active star formation and contain hot bright stars. Such gal- axies tend to be bluer and contain emission nebulae. Galaxies that are poor in gas and dust contain few or none of these highly luminous stars, so those galaxies look redder and have a much more uniform look. 3. What is the difference between an E0 galaxy and an E1 galaxy? Eo galaxies are round. E1 is a elliptical galaxy. M87 is a giant elliptical galaxy classified E1. It is a number of times larger in diameter than our own galaxy and is surrounded by a swarm of over 500 globular clusters. Elliptical galaxies are round or elliptical, contain no visibe gas, dust, and lack hot bright stars. They are classified with a numbered index ranging from 1 to 7. Eos are round and E7s are highly elliptical. The index is calculated from the largest and smallest diameter of the galaxy used in the following formula and rounded to the nearest integer. 10(a - b) _______ a a is top line of galaxy b is the side of a galaxy. outline of an E6 galaxy. 4. What is the difference between an Sa and an Sb galaxy? between an SBb and an Sb? Sa galaxies are Spiral galaxies contain a disk and spiral arms. Sb galaxies are spiral. SBb galaxies are barred spiral galaxies. Sb galaxies are spiral. 5. Why can't galaxies evolve from elliptical to spiral? Why can't they evolve from spiral to elliptical? E galaxies cannot evolve into S or Irr galaxies because E galaxies have no gas or dust needed to produce the young stars and nebulae found in S and Irr galaxies. Irr galaxies apparantly don't evolve into S or E galaxies, since the presence of old stars in Irr galaxies shows that those galaxies are old and haven't evolved into another type of galaxy. an elliptical galaxy cannot become a spiral galaxy or an irregular galaxy because ellipticals contain almost no gas and dust from which to make new stars. Spiral galaxies and irregular galaxies cannot evolve into elliptical galaxies because spiral and irregular galaxies contain both young and old stars. 6. How do selection effects make it difficult to decide how common elliptical and spiral galaxies are? E and Irr galaxies are often less luminous than S galaxies and so are harder to see. E galaxies are not as luminous as they might otherwise be because they have no hot massive bright stars, only older, dimmer less massive stars. Irr galaxies are not as luminous as they otherwise might be because they are usually smaller than S galaxies. All Spiral galaxies have halos Spirals contain gas and dust and hot O and B stars Star formation is occurringin these galaxies. The amount of gas and dust in a galaxy strongly influences its ap- pearance. Galaxies rich in gas and dust have active star formation and contain hot, bright stars. Such gal- axies tend to be bluer and contain emission nebulae. Galaxies that are poor in gas and dust contain few or none of these highly luminous stars, so those galaxies look redder and have a much more uniform look. 7. Why are Cepheid variable stars good distance indica- tors? What about planetary nebulae? To find the distance to a galaxy, we must search among its stars for a familiar object whose luminosity or diameter we know. Such objects are called distance indicators. Because their period is related to their luminosity the Cepheid variable stars are reliable distance indicators. Cepheids determine the period of pulsation, and measure the average appar- ent brightness. They can then deduce the distance to the galaxy 51 million for M 100. The foundation of the distance scale rests on the Cepheid variable stars and our understanding of the luminosi- ties of the stars in the H-R diagram. Distance The distance to galaxies are so large that it is not con- venient to measure them in light-years, parsecs, or even kiloparsecs. Instead we will use the unit megaparsec MPC or 1 million pc. One Mpc equals 3.26 million ly. or approximately 2 X 10 19 miles. Planetary nebulae have proven to be very useful distance indicators, which is surprising considering how faint their central star is. However, the central stars of planetary nebulae seem faint because they are very hot and radiate most of their energy in the ultra- violet . Astronomers have been able to calibrate the bright- est planetary nebulae by studying nearby galaxies. 8. Why is it difficult to measure the Hubble constant? Modern attempts to measure H have been difficult because of the uncertainty in the distances to galaxies. The Hubble Law Although Astronomers find it difficult to measure the distance to a galaxy, they often estimate such distances using a simple relationship. red shifts implied that the galaxies had large radial velocities and were receding from us. In 1929 the American astronomer Edwin Hubble published a graph that plotted the velocity of reces- sion versus distance for a number of galaxies. Vr =Hd That is the velocity of recession V equals the distance d in millions of parsecs times the constant H. This relation between red shift and distance is known as the Hubble Law and the constant H is known as the Hubble constant. Modern attempts to measure H have been difficult because of the uncertainty in the distances to galaxies 9. How is the rotation curve method related to binary stars and Kepler's third law? Because the galaxy rotates one side moves away from us and one side moves toward us. 10. What evidence do we have that galaxies contain dark matter? Astronomers often find that the measured masses are much too large. When we studied the rotation curve of our own galaxy and concluded that it must contain large amounts of dark matter. This seems to be true of most galaxies. Measured masses of galaxies amount to 10 to 100 times more mass than we can see. The rotation curves of galaxies are strong evidence for dark matter, but astronomers have found even more evidence that most of the mass in the universe is invisible. For example in many clusters of galax- ies the velocities of the galaxies are so high the clus- ters would fly apart if they contained only the mass we see. These clusters must contain large amounts of dark matter. X-ray observations reveal more evidence of dark matter. X-ray images of galaxy clusters show that many of them are filled with very hot, low-density gas. Dark Matter in Galaxies Given the size and luminosity of a galaxy we can make a rough guess as to the amount of matter it should contain. We know how much light stars produce, and we know about how much matter there is between the stars, so it is quite possible to estimate very roughly the mass of a galaxy from its luminosity. 11. What evidence do we have that galaxies collide and merge? The best evidence is that we see galaxies in the process of colliding. Evidence of Mergers The twisted shape of NGC 7252 suggests a collision and a Hubble Space Telescope image reveals a small spiral of young stars spinning backward in the heart of the larger galaxy. This is thought to be the remains of two oppositely rotating galaxies that merged about a billion years ago. The ellipticals appear to be the product of galaxy mergers, which triggered star formation that used up the gas and dust. Most ellipticals are formed by the merger of at least two or three galaxies. 12. Why are the shells visible around some elliptical galax- ies significant? Models predict that galaxies formed of mergers could lie inside shells of stars. Faint shells of stars surround elliptical galaxy NGC 3923. As two galaxies merge they spiral around their common center of mass, which could produce shells of stars. Such faint shells are seen around some galaxies. 13. Ring galaxies often have nearby companions. What does that suggest? Ring Galaxies consist of a bright nucleus surrounded by a ring. Computer models show that they could be produced by a galaxy passing roughly per- pendicularly through the disk of a larger galaxy. Ob- servations show that many ring galaxies have nearby companions. We have seen that collisions and mergers between galaxies are commonand can produce dramatic changes in the structure of the galaxies. 14. Propose an explanation for the lack of gas, dust, and young stars in elliptical galaxies. Perhaps each E galaxy formed from a cloud of gas and dust that had little rotation. it could collapse quickly and form stars quickly, converting all the dust and gas into one generation of stars long ago. 15. How do deep images by the Hubble Space Telescope confirm our hypothesis about galaxy evolution? a Hubble Space Telescope image reveals the collision of two galaxies producing thick clouds of dust and raging star formation. Discussion Questions 1. Why do we believe that galaxy collisions are likely, but star collisions are not? Although the collision theory has some qualitative points in its favor, it has some shortcomings as well. Foremost among them is that a near collision between two stars is highly improbable. Stars are large by terrestrial standards, still minute compared to the typical distances that separate them. For example, the Sun is about 1 million kilometers in diameter, whereas the distance to the nearest (Alpha-Centauri) star system, is nearly 100 million million kilometers. Probability theory then suggests that given the number of stars, their sizes, and their typical separations, not more than a handful of such close encounters are likely throughout the entire expanse and history of the Milky Way Galaxy. Galaxy collisions are frequent, but stellar collisions are extremely rare. Current research indicates that stellar collisions are quite common. 2. Should an orbiting infrared telescope find irregular galaxies bright or faint in the far infrared? Why? What about elliptical galaxies? faint. No since Infrared radiation from the stars and other bodies is absorbed. page 270 Chapter 13 Galaxies Problems 1. If a galaxy contains a type I (classical) Cepheid with a period of 30 days and an apparent magnitude of 20, what is the distance to the galaxy? 30 20 00 60 600 2. If you find a galaxy that contains globular clusters that are 2 seconds of arc in diameter, how far away is the galaxy? Hints: Assume that a globular cluster is 25 pc in diameter, and see By the Numbers 3-1. With an angle, distance, and size involved in this problem, let's use the small angle formula θ = (s/d)(206,265"/radian). Solve this for distance d. d = (s / θ) (206,265") = (25 pc / 2") (206,265") = 2.6 Mpc = 2.6 million parsecs This is about 9 million light years. 3. If a galaxy contains a supernova that at its brightest has an apparent magnitude of 17, how far away is the gal- axy? Hints: Assume that the absolute magnitude of the supernova is -19, and see By the Numbers 8-2. 4. If we find a galaxy that is the same size and mass as our Milky Way Galaxy, what orbital velocity would a small satellite galaxy have if it orbited 50 kpc from the center of the larger galaxy? Hint: See By the Num- bers 4-1. Use the equation for the velocity of an object in a circular orbit. This formula requires the mass M of the large galaxy. The problem says M is the same as the mass of the Milky Way which or text tells us is at least 1011 solar masses. Thus, M = (1011)(2 x 1030kg) = 2 x 1041kg. Here is the formula for the velocity of an object in a circular orbit: VC = square root of (G M / r) = square root of ((6.7 x 10-11 m3/(s2 kg) (2 x 1041kg) / (50,000 x 3.09 x 1016) ) = 93,000 m/s = 93 km/s page 270 Chapter 13 Galaxies 5. Find the orbital period of the satellite galaxy described in Problem 4. Hint: See By the Numbers 8-4. By The Numbers 8-4 the total mass of two stars orbiting each other is related to the average distance a between them and their orbital period p. M + M = a3 p2 a in AU P in years.and mass in solar masses. Example A: If we observe a binary system with a period of 32 years and an average separation of 16 AU what is the total mass? Solution: The total mass equals 16 3 / 32 2, or 4 solar masses. 6. If a galaxy has a radial velocity of 2000 km/s and the Hubble constant is 70 km/s/Mpc, how far away is the galaxy? Hint: Use the Hubble Law. Hubble's law is VR = H d, where VR is the recessional velocity, H is the Hubble constant, and d is the distance to the galaxy. Solve this for d. d = VR / H = (2000 km/s) / (70 km/s / Mpc) = 28.6 Mpc. 7. If you find a galaxy that is 20 minutes of arc in diame- ter and you measure its distance to be 1 Mpc, what is its diameter? Hint: See By the Numbers 3-1. 8. We have found a galaxy in which the outer stars have orbital velocities of 150 km/s. If the radius of the galaxy is 4 kpc, what is the orbital period of the outer stars? Hints: 1 pc = 3.08 X 10 13 km, and 1 yr = 3.15 X 10 7 s. 9. A galaxy has been found that is 5 kpc in radius and whose outer stars orbit the center with a period of 200 million years. What is the mass of the galaxy? On what assumptions does this result depend? Hint: See By the Numbers 8-4. 200 x 5 1000 By The Numbers 8-4 the total mass of two stars orbiting each other is related to the average distance a between them and their orbital period p. M + M = a3 p2 a in AU P in years.and mass in solar masses. Example A: If we observe a binary system with a period of 32 years and an average separation of 16 AU what is the total mass? Solution: The total mass equals 16 3 / 32 2, or 4 solar masses. The Strange Case of Eta Carinae. The Violently Unstable. Eta Carinae is highly unstable, and prone to violent outbursts. The last of csep10.phys.utk.edu/guidry/violence/etacarinae.html Critical Inquiries for the Web 1. How far out into the universe can we see Cepheid vari- ables? Research sources on the Internet to find other galaxies whose distances have been found through ob- servation of Cepheids. List the galaxies in which Ce- pheids have been identified and the distances deter- mined from these data. Cepheid variables Cepheids http://antwrp.gsfc.nasa.gov/apod/ap960110.html Explanation: Can this blinking star tell us how fast the universe is expanding? Many astronomers also believe it may also tell us the age of the universe! The photographed "Cepheid variable" star in M100 brightens and dims over the course of days as its atmosphere expands and contracts. A longer blinking cycle means an intrinsically brighter star. Cepheids variable stars are therefore used as distance indicators. By noting exactly how long the blinking period is and exactly how bright the star appears to be, one can tell the distance to the star and hence the star's parent galaxy. This distance can then be used to match-up easily measured recessional velocity ("redshift") with distance. Once this "Hubble relation" is determined for M100, it should be the same for all galaxies - and hence tell us how fast the universe is expanding. The exact magnitude of this calibration is under dispute and so a real live debate involving the value of Hubble's constant titled The Scale of the Universe will occur in April 1996 in Washington, DC. http://www.owlnet.rice.edu/~bonnieb/Cepheids.html Types of Cepheids Type Period Range Population Type Radial/Nonradial Long-Period Variables 100-700 days I, II R Classical Cepheids 1-50 days I R W Virginis stars 2-45 days II R RR Lyrae stars 1.5-24 hours II R ? Scuti stars 1-3 hours I R, NR Cephei stars 3-7 hours I R, NR ZZ Ceti stars 100-1000 seconds I NR 2. How does the Milky Way stack up against the other gal- axies in the Local Group? Look for information on the other galaxies in our cluster, and rank the top six mem- bers in order of total mass. other galaxies in our cluster the top six members in order of total mass http://astron.berkeley.edu/~basri/astro10/lectures/lec20.html Our Galaxy is not alone in our local neighborhood of the Universe. It is gravitationally bound to many other nearby galaxies, forming what we call the Local Group. There are about 30 galaxies total in Local Group, which is analogous to a star cluster in our own galaxy, where all the stars are gravitationally bound to each other in a group. In the Local Group there are 2 massive spiral galaxies, the Milky Way and the Andromeda Galaxy (M31) which about 2 million light years away. The rest of the galaxies in the Local Group are much smaller galaxies than the spirals and are called dwarf galaxies. 3. Locate a Web page dedicated to the Messier Catalog-a list of galaxies, clusters, and nebulae that is often used as a list of targets for small telescopes. Be sure that your destination includes images of the objects. For each of the galaxies in the Messier list, determine its Hubble classification. You may be given this information at the site, but examine the images to see if the features of these galaxies conform to a particular Hubble type. http://www.seds.org/messier/index.html THE MESSIER CATALOG Messier Catalog-a list of galaxies, clusters, and nebulae http://www.ipac.caltech.edu/2mass/gallery/messiercat.html Messier 1 Crab Nebula Messier 2 globular cluster http://www.astro.northwestern.edu/labs/m100/messier.html Exploring The Sky 1. Locate the Andromeda Galaxy, also known as M 31, and its companion galaxies. Zoom in on it and estimate its angular size compared to the full moon. Hint: Use Find under the Edit menu. Andromeda Galaxy Great Nebula in Andromeda M31 NGC 224 Other description: Very elongated galaxy, dusty, bright core. Constellation: And Dreyer description: A magnificent (or otherwise interesting) object! Most extremely bright, extremely large, very moderately extended (Andromeda); = M31. Magnitude: 3.5 RA: 00h 42m 54.3s Dec: +4117'28" RA: 00h 42m 42.0s Dec: +4116'00" (Epoch 2000) Azm: 6830'55" Alt: +3925'53" Rise: 07:37 Transit: 18:51 Set: 06:10 Size:175.0' x62.0' Position Angle: 34.0 Antares SAO184415 Sky Database: Common Star Names RA: 16h 29m 35.2s Dec: -2626'24" RA: 16h 29m 23.0s Dec: -2625'54" (Epoch 2000) Azm: 22334'34" Alt: +0316'19" Rise: 06:47 Transit: 10:39 Set: 14:31 From Andromeda Galaxy: Angular separation: 13133'38" Andromeda Abbreviation: And Genitive Form: Andromedae Description: Andromeda, the Princess of Ethiopia Pronunciation: an DRAHM' ee duh Genitive Pronunciation: an DRAHM' ee dee Sky Database: Constellation Labels RA: 00h 32m 35.9s Dec: +3833'51" RA: 00h 32m 24.0s Dec: +3832'24" (Epoch 2000) Azm: 7249'15" Alt: +3931'43" Rise: 08:30 Transit: 18:41 Set: 04:56 From Andromeda Galaxy: Angular separation: 0322'00" Position angle: +21645' 2. Take a survey of galaxies and see how many are spiral and how many are elliptical. Is there any selection effect in your method? Hint: Use Filters under the View menu to turn off everything but Galaxies and Mixed Deep Sky objects. Make sure the Messier labels are switched on using the Labels and Setup under the View menu. none appeared on the screen. if any appeared would be alot. selection effect spiral elliptical 3. Locate the Sombrero Galaxy (M 104). Study the photo- graphs and discuss this galaxy's special properties. Zoom in on it and estimate its angular size compared to the moon. Sombrero Galaxy M104 NGC 4594 Other description: Edge on galaxy dusty. Constellation: Vir Dreyer description: Remarkable!, very bright, very large, extremely extended 92, very abruptly much brighter middle nucleus; = M104. Magnitude: 8.3 RA: 12h 40m 11.3s Dec: -1138'10" RA: 12h 40m 00.0s Dec: -1137'00" (Epoch 2000) Azm: 27505'43" Alt: -2025'49" Rise: 01:40 Transit: 06:50 Set: 12:01 Size: 8.0' x 5.0' From Cursor position: Angular separation: 7452'17" Position angle: +8202' 4. Study the distribution of galaxies and notice how they cluster together. Can you find the Virgo cluster? Zoom in until more galaxies appear and then scroll north to find the Coma Cluster. Zoom in on Leo to find the clus- ter there. What other clusters can you find? Leo I Leo II The Comma NGC4039 Aries Abbreviation: Ari Genitive Form: Arietis Description: The Ram Pronunciation: AY' rih eez Genitive Pronunciation: uh RY' eh tis Sky Database: Constellation Labels RA: 02h 39m 49.4s Dec: +2006'30" RA: 02h 39m 36.0s Dec: +2005'24" (Epoch 2000) Azm: 6730'03" Alt: +0650'20" Rise: 13:11 Transit: 20:48 Set: 04:29 From Coma Berenices: Angular separation: 12940'29" Position angle: +21536' Polaris North Star 5. Describe the galaxy located near the south celestial pole. page 271 Chapter 13 Galaxies Review Questions 1. What is the difference between the terms radio galaxy and active galaxy? All galaxies including our Milky Way, emit radio en- ergy from neutral hydrogen, molecules, pulsars, and so on, but some galaxies called radio galaxies emit as much as 10 million times more radio energy from a small region at their centers. Some of these galax- ies also emit powerfully at infrared, ultraviolet, and X-ray wavelengths, so the term active galaxies is also used. 2. What evidence do we have that the energy source in a double-lobed radio galaxy lies at the center of the galaxy? The geometry suggests that radio lobes are inflated by jets of excited gas emerging from the central galaxy. This has been called the double-exhaust model. Double-Lobed Radio Sources Beginning in the 1950s radio astronomers found that some sources of radio energy in the sky consisted of pairs of radio-bright regions. When optical telescopes studied the locations of these double-lobed radio sources, they revealed galaxies located between the two lobes. Apparently the galaxies were producing the radio lobes. Second, notice how the presence of synchrotron radiation and hot spots is evidence in support of the model. Third, notice is how matter falling into a cen- tral black hole can produce these jets. 3. How does the peculiar rotation of NGC 5126 help us un- derstand the origin of this active galaxy? 4. What statistical evidence suggests that Seyfert galaxies have suffered recent interactions with other galaxies? Also, about 25 percent have peculiar shapes suggesting tidal interactions with other galaxies. This statistical evidence (Window on Science 14-1) hints that Seyfert galaxies may have been triggered into activity by collisions or interactions with com- panions. Some Seyferts are expelling matter in oppo- sitely directed jets typical of matter flowing into a strong gravitational field. The cores of Seyfert galaxies contain su- permassive black holes with masses as high as millions or even billions of solar masses. En- counters with other galaxies could throw matter into the black hole and release tremendous energy from a very small region. The velocities at the center of Seyfert galaxies are roughly 10,000 km/s, about 30 times greater than velocities at the center of normal galaxies. Seyfert galaxies Although these galaxies look normal in photographs very short exposures revealed that they have unresolved nuclei. Unresolved we mean that the nucleus looks like a star with no measurable di- ameter. Seyfert galaxies are actually define by their spectra. The spectrum of the Seyfert galaxy nuclei contains broad emission lines of highly ionized atoms. Emission lines suggest a hot low-density gas, and ionized atoms suggest that the gas is very excited. Window on Science 14-1 Some scientific evidence is statistical Seyfert galaxies are three times more likely to have a nearby companion than a normal galaxy is. 5. How does the unified model explain the two kinds of Seyfert galaxies? A few thousand Seyfert galaxies are known and are divided into two categories. Type 1 Seyfert galax- ies are very luminous at X-ray and ultraviolet wave- lengths and have the typical broad emission lines. Type 2 Seyfert galaxies have narrower emission lines and are brilliant at infrared wavelengths but not at X-ray and ultraviolet wavelengths. 6. What observations are necessary to identify the presence of a supermassive black hole at the center of a galaxy? Some Seyferts are expelling matter in oppo- sitely directed jets typical of matter flowing into a strong gravitational field. All of this evidence leads modern astronomers to suspect that the cores of Seyfert galaxies contain su- permassive black holes-black holes with masses as high as millions or even billions of slar masses. En- counters with other galaxies could throw matter into the black hole and release tremendous energy from a very small region. The motion of stars and gas near the centers of galaxies is important because it can tell us the amount of mass located there. Consider the giant elliptical gal- axy M 87. It has a very small, bright nu- cleus and a visible jet of matter 1800 pc long racing out of its core. A high-resolution image shows that the core lies at the center of a spinning disk, and the jet lies along the axis of the disk. Only 60 ly from the center the gas in the disk is orbiting at 750 km/s. Substituting radius and velocity into the equation for circular velocity tells us that the central mass must be roughly 2.4 billion solar masses. 7. How does the unified model implicate collisions and mergers in triggering active galaxies? The Unified Model to find a unified model of active galaxy cores. It seems clear that the cores of active galaxies con- tain supermassive black holes surrounded by accre- tion disks that are extremely hot near the black hole but cooler farther out. This means the black hole may be hidden deep inside this narrow central well. The hot inner disk seems to be the source of the jets often seen coming out of active galaxy cores. Formation of galaxies Evidently the cores of galaxies be- come active when interactions with other galaxies throw matter into their central black holes.The impor- tance of collisions is shown by the prevalence of com- panions to Seyfert galaxies and the distorted shapes of active galaxies. 8. Why were quasars first noticed as being peculiar? The largest telescopes detect multitudes of faint points of light with peculiar emission spectra, objects called quasars also known as quasi-stellar objects or QSOs. 9. How do the large red shifts of quasars lead us to con- clude they must be very distant? A few unidentifiable emmision lines were super- imposed on a continuous spectrum. In 1963 Maarten Schmidt at Hale Observatories tried red- shifting the hydrogen balmer lines to see if they could be made to agree with the lines in 3C 273's spectrum. At a red shift of 15.8 percent three lines clicked into place. Other quasar spectra quickly yielded to this approach, revealing even larger red shifts. The red shift z is the change in wavelength. The Hubble law states that galaxies have ra- dial velocities proportional to their distances, and the distance to a quasar is equal to its radial ve- locity divided by the Hubble constant. The large red shifts of the quasars imply that they must be at great distances. The red shift of 3C 273 is 0.158 and the red shift of 3C 48 is 0.37. about 1.0 Quasars were found with red shifts much larger than that of any known galaxy. Some quasars are evidently so far away that galaxies at those distances are very difficult to detect. 10. Why do we conclude that quasars are superluminous but must be very small? The spectrum of the quasar con- tains absorption lines with the same red shift as the elliptical galaxy. Evidently the lines are produced by the thin gas in the elliptical galaxy. From this we con- clude that the quasar, though very bright is actually farther away than the distant elliptical galaxy. Bright is also very luminous. The quasars are very distant and must be super- luminous and quite small. At such great distances it is difficult to detect the host galaxy because of the glare of the quasar. 11. How do gravitational lenses provide evidence that qua- sars are distant? The gravitational lens effect A distant quasar can appear to us as multiple images if its light is deflected and focused by the mass of an intervening galaxy. The two images of quasar 0957 + 561 are produced by the gravitational lensing of the faint galaxy just above the lower image of the quasar. Red shifts show that the quasar is almost four times farther away than the galaxy. The gravitational field of a galaxy between us and the distant quasar acts as a lens, bending the light from the quasar and fo- cusing it into multiple images 12. What evidence do we have that quasars occur in distant galaxies? Quasar Distances Astronomers estimate the distances to quasars by using the Hubble law and dividing their radial veloc- ity by the Hubble constant. Know- ing the distance to a quasar depends critically on know- ing its red shift and its radial velocity. Doppler formula By the Numbers 6-2. Further evidence appeared when a new point of light appeared near quasar QSO 1059 + 730. The point of light was a supernova exploding in the galaxy that hosts the quasar. Although the galaxy is too faint to see, the appearance of the duper- nova confirms that the quasar lies in a galaxy. By blocking the glare of the quasar, astrono- mers were able to photograph the spectrum of quasar fuzz and found that it is the same as the spectrum of a normal galaxy with the same red shift as the quasar. For example 3C 273 appears to lie in a giant elliptical galaxy. Moreover some quasars have faint objects near them in the sky. The spectra of these faint objects are the same as the spectra of galaxies, and they have the same red shift as that of the quasar they accompany. This is evidence that these quasars are located in gal- axies that are part of larger clusters of galaxies. 13. How can our model quasar explain the different radia- tion we receive from quasars? This model could explain the different kinds of radiation we receive from quasars. A small percentage of quasars are strong radio sources, and the radio radiation may come from synchrotron radiation produced in the high-energy gas and mag- netic fields in the jets. 14. What evidence do we have that quasars must be trig- gered by collisions and mergers? The host galaxies appear to be involved in col- lisions and interactions with other galax- ies or they have the distorted shapes that we recognize as the result of such collisions. Because we suspect that galaxy collisions can cause the core of a galaxy to erupt, we can also suspect that quasars may have been triggered into existence in the cores of galax- ies by interactions between galaxies. 15. Why sre there few quasars at low red shifts and at high red shifts but many at red shifts of about 2? Chapter 14 page 288 Discussion Questions 1. Why do quasars, active galaxies, SS 433, and protostars have similar geometry? magnetic geometry around the star 2. By custom, astronomers refer to the unified model of AGN and not to the unified hypothesis or unified the- ory. In your opinion, which of the words seems best? unified model Identify a question, formulate a hypothesis, control and manipulate variables, devise experiments, predict outcomes. There are many types of active galaxies. Initially, when astronomers were first studying them, it was thought that the different types of AGs were fundamentally different objects. Now astronomers generally (but not universally) accept the unified model of AGs, meaning that most or all AGs are actually just different versions of the same object. Many of the apparent differences between types of AGs are due to viewing the AG at different orientations with respect to the disk, or due to observing the AG in different wavelengths of light. astronomers refer to the unified model of AGN and not to the unified hypothesis 3. Do you think that our galaxy has ever been an active gal- axy? Could it have hosted a quasar when it was young? Yes Yes. 4. If a quasar is triggered in a galaxy's core, what would it look like to people living in the outer disk of the galaxy? Could life continue in that galaxy? Begin by deciding how bright a quasar would look seen from the outer disk, considering both distance and dust. No. gravitational tidal forces in the Galaxy's core Explanation: The center of well-studied active galaxy Centaurus A is hidden from the view of optical telescopes by a cosmic jumble of stars, gas, and dust. But both radio and x-ray telescopes can trace the remarkable jet of high-energy particles streaming from the galaxy's core. With Cen A's central region at the lower right, this composite false-color image shows the radio emission in red and x-rays in blue over the inner 4,000 light-years of the jet. One of the most detailed images of its kind, the picture shows how the x-ray and radio emitting sites are related along the jet, providing a road map to understanding the energetic stream. Extracting its energy from a supermassive black hole at the galaxy's center, the jet is confined to a relatively narrow angle and seems to produce most of its x-rays (bluer colors) at the upper left, farther from the core, where the jet begins to collide with Centaurus A's denser gas. Problems 1. The total energy stored in a radio lobe is about 10 53 J. How many solar masses would have to be converted to energy to produce this energy? Hints: Use E = mc2. One solar mass equals 2 X 10 30 kg. 10 x 2 =20 2. If the jet in NGC 5128 is traveling at 5000 km/s and is 40 kpc long, how long will it take for gas to travel from the core of the galaxy to the end of the jet? Hint: 1 pc equals 3 X 10 13 km. 10 13 x 3 =30 5000 3. Cygnus A is roughly 225 Mpc away, and its jet is about 50 seconds of arc long. What is the length of the jet in parsecs? Hint: See By the Numbers 3-1. 225 50 4. Use the small-angle formula to find the linear diameter of a radio source with an angular diameter of 0.0015 second of arc and a distance of 3.25 Mpc. 0.0015 x 3.25 00075 00030 0045 =0045375 5. If the active core of a galaxy contains a black hole 10 6 solar masses, what will the orbital period be for matter orbiting the black hole at a distance of 0.33 AU? Hint: See By the Numbers 8-4. 10 6 +0.33 =1.39 By The Numbers 8-4 the total mass of two stars orbiting each other is related to the average distance a between them and their orbital period p. M + M = a3 p2 a in AU P in years.and mass in solar masses. Example A: If we observe a binary system with a period of 32 years and an average separation of 16 AU what is the total mass? Solution: The total mass equals 16 3 / 32 2, or 4 solar masses. 6. If a quasar is 1000 times more luminous than an entire galaxy, what is the absolute magnitude of such a qua- sar? Hint: The absolute magnitude of a bright galaxy is about -21. 7. If the quasar in Problem 6 were located at the center of our galaxy, what would its apparent magnitude be? Hints: See By the Numbers 8-2 and ignore dimming by dust clouds. 8. What is the radial velocity of 3C 48 if its red shift is 0.37? Hint: See By the Numbers 14-1. 0.37 By the Numbers 14-1 The Relativistic Red Shift Vr = (z + 1)2-1 where z = AA c z + 1 2 + 1 A z + 1 = 3 z + 1 2 = 9 Vr = 9 - 1 = 8 = 0.8 c 9 + 1 10 9. If the Hubble constant is 70 km/s/Mpc, how far away is the quasar in Problem 8? Hint: Use the Hubble law. 0.37 0.37 x 70 + 70 000 =1.07 = 2.59 10. The hydrogen Balmer line HB has a wavelength of 486.1 nm. It is shifted to 563.9 nm in the spectrum of 3C 273. What is the red shift of this quasar? Hint: What is ^ ? 563.9 486.1 Critical Inquiries for the Web 1. What object currently holds the distinction as the far- thest known galaxy? Search the Web for information on this distant object, and find out its red shift and distance. What is the look-back time for this object? 2. Gravitational lenses were first predicted by Einstein in 1936 but were not observed until recently. Search the Web for instances of gravitational lensing of galaxies and quasars. For a particular case, discuss how the lens effect allows astronomers to determine information about the lensing and/or lensed objects that might not have been available without the alignment. page 289 Chapter 14 Galaxies with Active Nuclei Review Questions 1. Would the night sky be dark if the universe were only 1 billion years old and were contracting instead of ex- panding? Explain your answer. Yes the sky would be dark. The stars are far away from us. 2. How can we be located at the center of the observable universe if we accept the Copernican principle? The universe has no center. No one point in the universe is any more the center than any other point. If the universe were two-dimensional and were an infinite flat sheet, then any point on the sheet is as special as any other point. No one point is the center. The cosmological principle is actually an exten- sion of the Copernican principle. Copernicus said that Earth is not in a special place: it is just one of a num- ber of planets orbiting the sun. The cosmological prin- ciple says that there are no special places in the uni- verse. 3. Why can't an open universe have a center? Why can't a closed universe have a center? The quick answer is that an edge or a center would vio- late the cosmological principle which says that every place in the universe is similar in its general properties at every other place. Then a place at the edge or center would be different, so there can be no edge and no cen- ter. 4. What evidence do we have that the universe is expand- ing? that it began with a big bang? The three primary pieces of evidence are the expansion of the universe as evidenced by the recessional velocities of all the clusters of galaxies, the observed 2.7 K background radiation which was emitted "soon" after the Big Bang, and the precise numerical abundances of the light elements H and He. The Expansion of the Universe The second fundamental observation of cosmology is that the spectra of galaxies contain red shifts that are proportional to their distances. This leads us to con- clude that the universe is expanding. The big bang The Beginning The big bang occurred long ago, and our instinct is to think of it as a historical event. The big bang isn't over, and it is happening everywhere. Because light travels at a finite speed, we see dis- tant galaxies not as they are now, but as they were when the light left them to begin its journey to Earth. 5. Why couldn't atomic nuclei exist when the universe was younger than 3 minutes? The protons, neutrons, and electrons of which our universe is made were produced during the first 4 seconds of its history. This soup of hot gas and radiation continued to cool and eventually began to form atomic nuclei. High- energy gamma rays can break up a nucleus so the for- mation of such nuclei could not occur until the uni- verse had cooled somewhat. 6. Why is it difficult to determine the present density of the universe? If the average density of the universe is equal to the critical density it will be flat. If the average density of the universe is less than the critical density the universe is negatively curved and open. If the average density is greater than the critical density, the universe is positively curved and closed. Whether the universe is open, closed, or flat de- pends on its density. It is quite difficult to mea- sure the density of the universe. 7. How does the inflationary universe theory resolve the flatness problem? the horizon problem? The exponentially great expansion of the universe which took place during inflation flattened space, just as when you blow up a ballon to a very large size--each small piece of its surface looks quite flat. If inflation took place early in the history of the universe, that means the universe remained smaller for longer than we thought (before inflation occured). The extra time allowed all parts of the universe back then to come to a common temperature and density. Consequently now the universe looks quite isotropic and homogeneous across great distances. The inlationary universe theory The key to these two problems and to others in- volving subatomic physics may lie with a theory called the inflationary universe because it predicts a sudden expansion when the universe was very young, an ex- pansion even more extreme than that predicted by the big bang theory. One of the problems is called the flatness problem. The universe seems to be balanced near the boundary between an open and a closed universe. That is, it seems nearly flat. 8. If the Hubble constant is really 100 km/s/Mpc, much of what we understand about the evolution of stars and star clusters must be wrong. Explain why? Within a few years astronomers had proposed the idea that the universe had a violent beginning and had used H to extrapolate backward to find the age of the universe. The value of H that Hubble re- ported, 530 km/s/Mpc, led them to conclude thar the universe was only half as a old as Earth. Something was wrong. In the decades that followed, astronomers figured out that Hubble's distances to galaxies were systemat- ically too small, and his value of H was too large. Modern measurements of H yield an age of the universe that is at least twice the age of Earth. We know the distance to a galaxy and its radial veloc- ity, so we can divide distance by velocity to find out how long the expansion of the universe has taken to separate the galaxies to their present distance. That is the age of the universe. The Universe is 14 billion years old. 9. Why do we conclude that the universe must have been very uniform during its first million years? The big bang created a universe of hot gas that was very uniform, as shown by the uniformity of the cosmic microwave background radiation. As the universe ex- panded the gas cooled and must have quickly formed galaxies, clusters of galaxies, filaments, and walls. Chapter 15 Cosmology page 311 10. What is the difference between hot dark matter and cold dark matter? What difference does it make to cosmology? Neutrinos travel at or nearly at the speed of light, and dark matter composed of such par- ticles is known as hot dark matter. Such fast-moving particles could not have stimulated the formation of objects as small as galaxies and clusters of galaxies, so most cosmologists believe that at least part of the dark matter is cold dark matter, which is made up of more massive particles that move more slowly. 11. What evidence do we have that the expansion of the uni- verse is accelerating? The gravity of the dark matter collecting around quantum fluctuations in space- time magnified by inflation could have pulled the mat- ter together to form the structures we see. Because ob- servations suggest the universe is flat and the expansion is accelerating, astronomers need to know how much of the density of the universe is due to dark matter. 12. What evidence do we have that the universe is flat? Astronomers believe that the remaining density to make the uni- verse flat is produced by the mass equivalent to the energy of empty space. as a mass using the equation E = mc2, is just enough to make the universe flat. Discussion Questions 1. Do you think Copernicus would have accepted the cos- mological principle? Why or why not? Yes the cosmological principle The cosmological principle is actually an exten- sion of the Copernican principle. Copernicus said that Earth is not in a special place: it is just one of a num- ber of planets orbiting the sun. The cosmological prin- ciple says that there are no special places in the uni- verse. 2. What observations would you recommend that the Hub- ble Space Telescope make to help us choose among an open, flat, or closed universe? observe the angles of the open universe flat universe closed universe greater seeing power with the lens, and mirror. 3. If we reject any model of the universe that has an edge in space because we can't comprehend such a thing, shouldn't we also reject any model of the universe that has a beginning or an ending? Are those just edges in time, or is there a difference? Just edges in time. If the universe begins then sometime in the future the universe ends. Like the way a star runs out of hydrogen gas, and the universe ends running out of energy matter. It is possible that someday that the universe is going to verly slowly or very quickly reverse direction and collapse. Problems 1. Use the data on page 296 to plot a velocity-distance di- agram, find H, and determine the approximate age of the universe. 2. If a galaxy is 8 Mpc away from us and recedes at 456 km/s, how old is the universe, assuming that gravity is not slowing the expansion? How old is the universe if it is flat? The age of the universe is related to the reciprocal of the Hubble constant H. Let's use the Hubble Law to find the reciprocal of H, which is (1/H). Start with the Hubble Law, VR = H d, and divide both sides by (H VR): (1/H) = d / VR = (8 Mpc) / (456 km/s) = (8 x 106 x 3.09 x 1016 m) / (4.56 x 105 m/s) = 5.42 x 1017 s Divide this by the number of seconds in a year, 3.16 x 107 s/yr, and we get 1.7 x 1010 years for (1/H). Our text tells us that if gravity is not slowing the expansion, then (1/H) is the age of the universe, which in this problem is 1.7 x 1010 years. Our text tells us that if the universe is flat, which requires some gravity, then the age of the universe is (2/3)(1/H), which in this problem is 1.15 x 1010 years. 2. 17.5 billion years old. 3. If the temperature of the big bang had been 10 6 K at the time of recombination, what maximum wavelength would the primordial background radiation have as seen from Earth? 4. If the average distance between galaxies is 2 Mpc and the average mass of a galaxy is 10 11 solar masses, what is the average density of the universe? Hints: The volume of a sphere is 4 tt r3. The mass of the sun is 2 X 10 33 g. 3 Recall that density = M / V, where V is any amount of volume and M is the mass in that volume. Let's consider a volume V = (2 Mpc)3; in that volume on the average we have one galaxy of mass M = 1011 MSun. With this data, let's find the average density of the universe. density = M / V = (1011 MSun) / (2 Mpc)3 = (1011 x 2 x 1030 kg) / (2 x 106 x 3.09 x 1016 m)3 = 8.47 x 10-28 kg/m3 4. 1.6 X 10 -30 gm/c 5. Figure 15-9 is based on an assumed Hubble constant of 70 km/s/Mpc. How would you change the diagram to fit a Hubble constant of 50 km/s/Mpc? The 3 lines No gravity, Open, Closed would have to be lowered. 14 9.5 Billion years ago. Figure 15-9 This figure assumes H = 70 km/s/Mpc. The expansion of the universe as a function of time. Models of the universe can be represented as curves in a graph of R, a measure of the size of the universe, and time. Open-universe models expand without end, and the corresponding curves fall in the region shaded orange. Closed models expand and then contract back to a high-density state red curve. Curves representing closed models fall in the region shaded blue. The dotted line represents a flat uni- verse, the dividing line between open and closed models. Note that the estimated age of the universe depends on the rate at which the expansion is slow- ing down. This figure assumes H = 70 km/s/Mpc. 6. Hubble's first estimate of the Hubble constant was 530 km/s/Mpc. If his distances were too small by a factor of 7, what answer should he have obtained? 6. 76 km/s/Mpc 7. What is the maximum age of the universe predicted by Hubble's first estimate of the Hubble constant? 8. If the value of the Hubble constant were found to be 60 km/s/Mpc, how old would the universe be if it were not slowed by gravity? If it were flat? 8. 16.6 billion years Critical Inquiries for the Web 1. Will the universe go on expanding forever? Search the Web for information on recent investigations that shed light on the question of the density of matter in the uni- verse. What predictions do these studies make about the fate of the universe? What kinds of observations were necessary to make these predictions? density of matter in the universe http://archive.ncsa.uiuc.edu/Cyberia/Cosmos/CosmosShape.html If the average density of matter in the universe is greater than the critical density, the force of gravity will eventually rein in expansion and cause the universe to collapse upon itself. In this case, the universe is said to be positively curved, and Omega, the ratio of the average density to the critical density, is greater than 1. Conversely, if the average density of matter in the universe is less than the critical density, gravity will lose its grip on matter and the universe will expand forever. This negatively curved universe is defined by an Omega less than one. If Omega is exactly one--that is, if the average density of the universe is equal to the critical density--then the universe will expand to a maximum density and remain there for eternity. This universe is flat; it has zero curvature. The universe's fate is intimately connected to its shape which, in turn, depends on a single number, Omega: the ratio of the average mass density of the universe to the critical value required to just maintain equilibrium. An open universe, corresponding to omega less than one, will expand forever. Matter will spread thinner and thinner. Galaxies will exhaust their gas supply for forming new stars, and old stars will eventually burn out, leaving only dust and dead stars. The universe will become quite dark and, as the temperature of the universe will approaches absolute zero, quite cold. The universe will not end, exactly, just peter out in a Big Chill. The expansion of a closed universe, with an Omega greater than one, will slow down until it reaches a maximum size, when it begins its inward collapse. Like a video of the Big Bang and expansion run backward, the universe will become denser and hotter until it ends in an infinitely hot, infinitely dense Big Crunch--perhaps providing the seed for another Big Bang. Will the universe go on expanding forever fate of the universe? 2. The steady-state theory was once a rival cosmology of the big bang. Search for Web sites that provide informa- tion on steady-state cosmology. Be careful to locate le- gitimate sites that discuss the theory, rather than sites where individuals use steady-state ideas as part of non- scientific arguments on cosmology. What were the key predictions of steady-state cosmology? How has recent evidence led to its decline? http://www.astro.ucla.edu/~wright/stdystat.htm Errors in the Steady State and Quasi-SS Models The Steady State model of the Universe was proposed in 1948 by Bondi and Gold and by Hoyle. Bondi and Gold adopted the "Perfect Cosmological Principle", and added the assumption that the Universe was the same at all times to homogeneity (the same in all places) and isotropy (the same in all directions). At the time the Steady State model was proposed, the Big Bang model was in trouble because the value of the Hubble constant was clearly bigger than the inverse of the age of the Universe. [Sound familiar?] If the Universe is the same at all times, the value of the Hubble constant must really be constant, so v = dD/dt = HD has an exponential solution and the scale factor varies like a(t) = exp(H(to-t)) Furthermore, since the radius of curvature of the Universe can not change, but must expand, the radius has to be infinite. The Steady State model has flat spatial sections like the critical density Big Bang model. Since the expansion of the Universe spreads the existing matter over a larger and larger volume, the density stays constant, the Steady State model requires continuous creation of matter. The average age of matter in the Steady State model is = 1/(3*Ho) but some galaxies are much older than the average, so the age of the globular clusters can be accomodated if the Milky Way is older than the average. The space-time diagram below shows the Steady State model: 3. Search for the Web pages of large surveys of galaxies, such as the 2dF Deep Field Survey and the Sloan Digital Sky Survey. What are they discovering about the largest and most distant objects in the universe? http://astron.berkeley.edu/~mwhite/probes-lss.html Perhaps the most obvious means for surveying the large-scale structure in the universe is to measure the position of the nearest galaxies (either in 3D -- redshift surveys -- or in the 2D projected on the sky -- angular surveys). Galaxy surveys with well defined selection criteria enable us to extract information about the clustering pattern. (Surveys during the early 1980s showed that galaxies are not distributed randomly throughout the Universe. They are found to lie in clusters, filaments, bubbles and sheet like structures.) In the early 1980s there were less than 5,000 galaxy redshifts known. We now have over 50,000 and will soon have more than 1,000,000. Recent very large scale galaxy surveys have now reached far enough out into our local volume to begin to see the end of greatness. That is they sample a large enough volume of space that the largest structures in the survey (about 100Mpc in size) are no longer of the same size of the survey itself. AAT 2dF: The Anglo-Australian Telescope 2 degree Field Galaxy Redshift Survey http://sco.stsci.edu/second_decade/Acronyms/acronyms.html 2dF Two Degree Field spectrograph 4. What is the latest news concerning observations of the cosmic microwave background radiation? NASA plans to launch a satellite to make detailed measurements, and further balloon observations are planned by a number of research teams. http://space.gsfc.nasa.gov/astro/cobe/cobe_home.html a Far Infrared Absolute Spectrophotometer (FIRAS) to compare the spectrum of the cosmic microwave background radiation with a precise blackbody, a Differential Microwave Radiometer (DMR) to map the cosmic radiation sensitively, and a Diffuse Infrared Background Experiment (DIRBE) to search for the cosmic infrared background radiation. http://cmb.physics.wisc.edu/polar/ezexp.html the remnant light from the big bang is called the cosmic microwave background radiation Exploring The Sky 1. Search for galaxy clusters. Hint: Use the View menu to turn off everything but galaxies, deep sky objects, con- stellations, and Labels. Zoom in until the field of view is 40 or smaller and the fainter galaxies appear. Search for clusters of galaxies in Ursa Major, Canes Venatici, Lynx, Virgo, and Coma Bereneces. Ursa Major Abbreviation: UMa Genitive Form: Ursae Majoris Description: The Larger She-Bear Pronunciation: UHR' suh MAY' juhr Genitive Pronunciation: UHR' see may JOH' ris Sky Database: Constellation Labels RA: 10h 40m 27.2s Dec: +5521'24" RA: 10h 40m 12.0s Dec: +5522'48" (Epoch 2000) Azm: 33602'09" Alt: +1857'13" Always above horizon. Transit: 04:51 From Canes Venatici: Angular separation: 2809'53" Position angle: +31306' Canes Venatici Abbreviation: CVn Genitive Form: Canum Venaticorum Description: The Hunting Dogs Pronunciation: KAY' neez vee NAT' ih sy Genitive Pronunciation: KAY' nuhm vee NAT' ih KOH' ruhm Sky Database: Constellation Labels RA: 13h 09m 45.9s Dec: +4107'34" RA: 13h 09m 36.0s Dec: +4109'00" (Epoch 2000) Azm: 30609'01" Alt: +2318'22" Rise: 20:08 Transit: 07:20 Set: 18:28 From SAO 258715: Angular separation: 13039'05" Position angle: +31620' Lynx Abbreviation: Lyn Genitive Form: Lyncis Description: The Wildcat Pronunciation: LINGKS Genitive Pronunciation: LIN' sis Sky Database: Constellation Labels RA: 08h 02m 05.6s Dec: +4518'31" RA: 08h 01m 48.0s Dec: +4519'12" (Epoch 2000) Azm: 35757'19" Alt: +0256'57" Always above horizon. Transit: 02:13 From Ursa Major: Angular separation: 2645'13" Position angle: +26440' Virgo Abbreviation: Vir Genitive Form: Virginis Description: The Maiden Pronunciation: VUHR' goh Genitive Pronunciation: VUHR' jih nis Sky Database: Constellation Labels RA: 13h 12m 47.1s Dec: -0344'59" RA: 13h 12m 36.0s Dec: -0343'48" (Epoch 2000) Azm: 27426'57" Alt: -0908'41" Rise: 01:37 Transit: 07:23 Set: 13:09 From Lynx: Angular separation: 8404'32" Position angle: +10125' Coma Berenices Abbreviation: Com Genitive Form: Comae Berenices Description: Berenice's Hair Pronunciation: KOH' muh BER' ee NY' seez Genitive Pronunciation: KOH' mee BER' ee NY' seez Sky Database: Constellation Labels RA: 12h 45m 46.9s Dec: +2148'26" RA: 12h 45m 36.0s Dec: +2149'48" (Epoch 2000) Azm: 29620'22" Alt: +0542'49" Rise: 23:05 Transit: 06:56 Set: 14:43 From Virgo: Angular separation: 2623'59" Position angle: +34548' 2. Locate small clusters of galaxies and compare the bright- ness of the galaxies with those in large clusters. Can you tell that the smaller clusters tend to be farther away, or is the range of galaxy luminosities too great? Hint: Click on a galaxy to find its magnitude. Andromeda Galaxy Great Nebula in Andromeda M31 NGC 224 Other description: Very elongated galaxy, dusty, bright core. Constellation: And Dreyer description: A magnificent (or otherwise interesting) object! Most extremely bright, extremely large, very moderately extended (Andromeda); = M31. Magnitude: 3.5 RA: 00h 42m 54.3s Dec: +4117'28" RA: 00h 42m 42.0s Dec: +4116'00" (Epoch 2000) Azm: 6830'55" Alt: +3925'53" Rise: 07:37 Transit: 18:51 Set: 06:10 Size:175.0' x62.0' Position Angle: 34.0 From Coma Berenices: Angular separation: 11653'59" Pinwheel Galaxy Triangulum Galaxy M33 NGC 598 Other description: Spiral galaxy structure with bright knots. Constellation: Tri Dreyer description: Remarkable! Extremely bright, extremely large, round, very gradually brighter middle nucleus; = M33. Magnitude: 5.7 RA: 01h 34m 07.0s Dec: +3040'20" RA: 01h 33m 54.0s Dec: +3039'00" (Epoch 2000) Azm: 7051'26" Alt: +2445'39" Rise: 10:56 Transit: 19:42 Set: 04:32 Size:64.0' x35.0' Position Angle: 22.0 From Andromeda Galaxy: Angular separation: 1448'09" Position angle: +13145' 3. Can you find galaxies and clusters of galaxies along the Milky Way? Turn on the Milky Way and search for gal- axy clusters within its outline. yes. Messier objects: The Andromeda Galaxy M31 and its satellites M32 and M110, as well as the Triangulum Galaxy M33. Virgo galaxy. There is a spherical component to our galaxy which contains very old stars and spherical clusters of old stars. These are often referred to as Population 2 objects. Population 1 being the objects found in the disk. The size of our galaxy is huge; light would take about 100,000 years to cross the Galaxy. galaxies and clusters of galaxies along the Milky Way? page 312 The only solutions seemed to be a uni- verse that was contracting under the influence of grav- ity or a universe in which the galaxies were rushing away from each other so rapidly that gravity could not pull them together. Chapter 15 Cosmology Part 3 The Universe of Galaxies Review Questions 1. What produced the helium now present in the sun's at- mosphere? in Jupiter's atmosphere? in the sun's core? The helium in the sun's atmosphere and in Jupiter was produced shortly after the Big Bang. Some of the helium in the sun's core was produced shortly after the Big Bang, some of the helium in the sun's core was produced there by the nuclear fusion of hydrogen. Helium was produced during the big bang. 2. What produced the iron in Earth's core and the heavier elements like gold and silver in Earth's crust? Iron in the blood were assembled inside stars. Generation after generation of stars cooked the original particles linking them together to build atoms such as carbon, nitrogen, and oxygen, and iron. calcium. The atoms heavier than iron were created by rapid nuclear reactions that can only occur when a massive star explodes. 3. What evidence do we have that disks of gas and dust are common around young stars? We have clear evidence that disks of gas and dust around young stars are common. Infrared observa- tions of T Tauri stars, for instance show that some are surrounded by gas clouds rich in dust, and spectra show that these stars are blowing away their nebulae at speeds up to 200 km/s. The presence of bipolar flows from young stars confirms that such systems contain gas and dust distributed in a disk-shaped cloud. In fact nearly all astronomers now believe that planets form naturally as a byproduct of star formation. Stars can form from the gravitational contraction of gas and dust clouds. Contraction must be triggered by compression of the gas cloud, perhaps by a nearby supernova explosion. 4. According to the solar nebula theory, why is the sun's equator nearly in the plane of Earth's orbit? As the solar nebular was collapsing due to gravity, it spun faster and flattened into a disk. The planets formed in the disk, so all of the planets orbit the sun almost in the same plane-- the plane of the disk. The sun formed at the center, with the same rotation as the disk. Therefore the sun's equator lies in what was the plane of the disk and is now the plane of the planets' orbits. Our own planetary system probably formed in such a disk-shaped cloud around the sun. When the sun became luminous enough, the remaining gas and dust were blown away into space, leaving the plan- ets orbiting the sun. This is known as the solar neb- ula theory because the planets form from the nebula around the proto-sun. 5. Why does the solar nebula theory predict that planetary systems are common? If the theory is right, then planets do form as a by-product of star formation, and thus most stars should have planetary systems. 6. Why do we think the solar system formed about 4.6 bil- lion years ago? Radioactive dating of meteorites and rocks from the moon and Mars indicate that that they are about that old. Since evidence from observations of other stars suggests that a star and its planets form together, we think our sun and the planets are about the same age as the the meteorites and rocks from the moon and Mars. It would take along time for the planets to form from the dust, and gas from the sun. 7. If you visited another planetary system, would you be surprised to find planets older than Earth? Why or why not? Yes their are planets older than Earth in another planetary system. It takes a lot of gas and dust to form planets. In a Hubble Space Telescope image has revealed a planets discovered so far tend to be massive . Figure 16-3 This Hubble Space Telescope image of the central part of the Orion Nebula reveals that many young stars there are surrounded by disks of gas and dust. 8. Why is almost every solid surface in our solar system scarred by craters? During the first half billion or billion years of the solar system, the plane of the planets' orbits was filled with planetesimals, some of which were tens to hundreds of kilometers across. As these collided with the protoplanets and protomoons with solid surfaces, they cratered those surfaces. These craters have been produced over the ages by meteorite impacts. 9. What is the difference between condensation and accretion? condensation is Condensation is the process whereby water vapour in the atmosphere is returned to its original liquid state. and accretion is Accretion (a) Collection of material together, generally to form a single body. [A84] (b) A process by which a star accumulates matter as it moves through a dense cloud of interstellar gas; or, more generally, whereby matter surrounding a star flows toward it (as in close binaries). 10. Why don't terrestrial planets have rings and large satel- lite systems like the Jovian planets? The terrestrial planets are closer to the sun and ice melts easier. All four Jovian planets have ring systems. Saturn's rings are made of ice particles. The rings of Jupiter, Uranus and Neptune are made of dark rocky particles. Terrestrial planets have no rings. 11. How does the solar nebula theory help us understand the composition of asteroids and comets? The nuclei of comets are icy bodies left over from the origin of the planets. Thus, we must conclude that at least some parts of the solar nebula were rich in ices. 12. How does the solar nebula theory explain the dramatic density difference between the terrestrial and Jovian planets? Terrestrial and Jovian planets The four inner planets are quite different from the next four outward. The planets differ dramatically in mass and density. Jupiter's family of planet's is much more massive but much less dense than Earth's family of planet's. 13. If you visited some other planetary system in the act of building planets, would you expect to see the conden- sation sequence at work, or was it unique to our solar system? Yes. The condensation sequence would be at work The sequence in which the different materials con- dense from the gas as we move away from the sun is called the condensation sequence. Table 16-3 It sug- gests that the planets forming at different distances from the sun accumulated from different kinds of materials. 14. Why do we expect to find that planets are differentiated? Once the planet formed, heat began to accumulate in its interior from the decay of short-lived radioactive elements, and this heat even- tually melted the planet and allowed it to differenti- ate. Differentiation is the separation of material ac- cording to density. When the planet melted, the heavy metals such as iron, and nickel settled to the core, while the lighter silicates floated to the surface to form a low-density crust. Planetary Differentiation The condensation process above will naturally occur at different temperatures for different elements. Things like iron, silicon, etc. will condense from the gas phase to the liquid or solid phase while still at very high temperatures (over 1000 K = more than about 1300 F), while the more volatile elements don't condense even at our room temperature (300K), where oxygen, hydrogen, helium, etc. are all still gases. Only far away from the hot center would temperatures get so low that even these elements collect on planets. Even if they don't condense to the liquid/solid form, colder gases have less kinetic energy than hotter ones (that's what temperature means, remember) and are therefore easier for a planet to capture. As the planet gathers gases, its mass increases and so does the mass-gathering ability. 15. What processes cleared the nebula away and ended planet building? Clearing the Nebula The planets in our solar system formed in a disk of gas, dust, and planetesimals around the sun 4.6 billion years ago. That solar nebula vanished while the sun was still young, and the clearing of the nebula ended planet building. Four effects helped clear the nebula. The most important was radiation pressure. When the sun be- came a luminous object, light streaming from its sur- face pushed against the particles of the solar nebula. low-mass specks of dust and in- dividual gas atoms were pushed outward and eventu- ally driven from the system. The second effect was the solar wind, the flow of ionized hydrogen and other atoms away.from the sun's upper atmosphere. The third effect for clearing the nebula was the sweeping up of space debris by the planets. Discussion Questions 1. In your opinion, if planets formed with one of the first stars to form in our galaxy, how would those planets differ from the planets in our solar system? Hint: To what population would the star belong? The older the planet would be. The planet would be the oldest planet in the solar system. To the population of Type II and a cold planet Type I or II. Earth Type: Terrestrial Temperature: Temperate Atmosphere: Type 1 (Breathable) Hydrosphere: Moderate Charr planet Type: Abandoned Wasteland Temperature: Cool Atmosphere: Type 2 (Breath Mask Suggested) 2. If the solar nebula hypothesis is correct, then there are probably more planets in the universe than stars? Do you agree? Why or why not? Yes I agree. Lots of dust in the universe so more planets can form. The Solar Nebula Hypothesis was proposed by Immanuel Kant (1755). This hypothesis states that the stars, sun, planets, moons and meteorites formed through the cooling and compression of a swirling mixture of gas and dust. Gravitational and rotational forces compressed and condensed the mass into rotating discs 4.6-5.0 Ga. The more dense terrestrial planets (MercuryVenus, Earth and Mars) cooled nearest the sun, while the less dense, more gaseous Jovian planets (Jupiter, Saturn, Uranus) stabilized farther from the sun. Our solar system began 4.6 billion years ago as a spinning cloud of gas and dust, called the solar nebula, which collapsed under its own weight to form a new star, our Sun. As the solar nebula spun and churned, dust grains stuck together to form dustballs, protoplanets and meteorites. (Painting by Don Dixon, NASA JSC photo S76-25001). Below images depict (Image and description courtesy of NASA) Orion Nebula, consisting of turbulent gas swirls and brilliant youthful stars. Orion Nebula may illustrate early stellar formation. Located 1,500 light-years from Earth, the nebula formed from collapsing interstellar gas clouds. Over 153 protoplanetary disks have been identified; these may represent embryonic solar systems. The images depicted below illustrate six views of gaseous disks circling incipient stars: four disks are seen from above; the lower two images illustrate a disk viewed edge-on in two different wavelengths. These disks represent the earliest stages of planet development. The disks' diameters range from 2-17 times that of our solar system. 1 astro-unit (AU) =1.5x10 11 m Distance sun-planet Distance between sun and FL Keys 1 light year (ly) = 9.5x10 15 m Distance between stars Distance light travels in 1 year You need to place 25,000,000,000,000,000,000,000,000,000,000 = 2.5x10 31 hydrogen nuclei (protons) side-by-side to get from the sun to alpha-Centauri-the closest star to our solar system, 4.2 ly away! (The Incredible) Speed of Light Light travels at an incredibly high speed of 3 x 108 meters per second (m/s) = 300,000 kilometers per second (km/s) = 187,000 miles per second! This is so fast that normally we can't tell that it is taking any time to get to us. We can tell that sound travels more slowly; we have all seen the light from a lightning strike before we have heard the clap of thunder, even though they take place at the same time. Similarly, you may have seen the flash of powder and smoke from a cannon before you hear the crack of the sound. What is harder to comprehend is that the light also takes some time (in this case a very short time!) to reach your eyes. The Moon is a distance of 384,000 km from the Earth, and since the distance d traveled by light is given in terms of its speed v and the time it travels t by the formula d = v x t we have that the time it takes light to get from the Moon to the Earth is t = d / v = 384,000 km / 300,000 km/s = 1.28 seconds This is why radio conversations (radio waves are a form of light with a very long wavelength compared to visible light) between the Astronauts and mission control had one-second delays in both transmission and reception. We can work out that it takes light 500 seconds or about 8 minutes to reach us from the Sun, and it takes about 11 hours to go from one side of the solar system to the other. Once outside the solar system, the nearest star is so far away that it takes 4.2 years for light from it to reach us. When we look at this star, we are looking back in time to see how it looked 4.2 years ago! Problems 1. The nearest star is about 4.2 ly away. If you looked back at the solar system from that distance, what would the maximum angular separation be between Jupiter and the sun? Hint: 1 ly equals 63,000 AU. 2. The brightest planet in our sky is Venus, which is some- times as bright as apparent magnitude -4 when it is at a distance of about 1 AU. How many times fainter would it look from a distance of 1 parsec (206,265 AU)? What would its apparent magnitude be? Hints: Remember the inverse square law Chapter 8 see Chapter 2. 2. it will look 206.26 apparent magnitude Chapter 16 page 334 3. What is the smallest-diameter crater you can identify in the photograph of Mercury on page 322? Hint: See Ap- pendix A to find the diameter of Mercury in kilometers. The smallest craters I can see on Mercury in Figure 16-7 are about 0.5 mm across in the photograph. To find out how large these craters actually are on Mercury we need to know the scale of the photograph; that is, we need to know how many kilometers there are on Mercury for each millimeter on the photograph. To find the scale just divide the diamter of Mercury in kilometers (from Table A-13) by the diamter of the photographic image of Mercury in millimeters: 4878 km / 112 mm = 43.6 km/mm. To find the size of the 0.5 mm craters, multiply this size times t he scale: (0.5 mm)(43.6 km/mm) = 22 km. Maybe you can see even smaller craters. 4. A sample of a meteorite has been analyzed and the re- sult shows that out of every 1000 nuclei of 40K origi- nally in the meteorite, only 125 have not decayed. How old is the meteorite? Hint: See Figure 16-9. 1 half life The percentage of nuclei remaining undecayed is (125/1000)(100%) = 12.5%. Figure 16-16 tells us that three half-lives have passed. The text tells us that the half-life of (40-K) is 1.3 billion years; therefore three half-lives is 3.9 billion years. This is the age of the meteorite. 4. about three times 5. In Table 16-2, which object's density differs least from its uncompressed density? Why? Table 16-2 Observed and Uncompressed Densities Planet Observed Density (g/cm3) Uncomprssed Density (g/cm3) Mercury 5.44 5.30 Venus 5.24 3.96 Earth 5.50 4.07 Mars 3.94 3.73 (Moon) 3.36 3.40 6. What composition might we expect for a planet that formed in a region of the solar nebula where the tem- perature was about 100 K? Based on Table 16-3, we would expect the planet that formed to have large amounts of methane and water ice. These are made of the common atoms H, C, and O. The planet would also have the items listed higher in the table (e.g., ammonia and water ice, troilite, feldspars, etc.), but these would be less common because most of them are made of less common atoms. 6. Large amounts of n 7. Suppose that Earth grew to its present size in 10 million years through the accretion of particles averaging 100 g each. On the average, how many particles did Earth cap- ture per second? Hint: See Appendix A to find Earth's mass. 8. If you stood on Earth during its formation as described in Problem 7 and watched a region covering 100 m2, how many impacts would you expect to see in an hour? Hints: Assume that Earth had its present radius. Thie surface area of a sphere is 4ttr2. 8. About 130 impacts 9. The velocity of the solar wind is roughly 400 km/s. How long does it take to travel from the sun to Pluto? Maximum distance from the sun 49.24 AU (73.66 X 10 8 KM) Critical Inquiries for the Web 1. How does our solar system compare with the others that have been found? Search the Internet for sites that give information about planetary systems around other stars. What kinds of planets have been detected by these searches so far? Discuss the selection effects see Win- dow on Science 13-2 that must be considered when in- terpreting these data. http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/970922f.html How common are planetary systems around other stars? Very common. The evidence is mounting that planets are quite common around other stars. Because the mass of any planet around it's parent star is much less than the star itself, it is difficult for us to observe the effects of the planets from Earth, and it really isn't possible right now to make an intelligent estimate of the percentage of other stars that have planets. However, a number of different observational studies have results that when combined imply that planets are the rule and not the exception. For example, theorists believe that planets form from a disk of material circling a star when it is young. Observations of young stars with the Hubble Space Telescope and other instruments have directly imaged such circumstellar material. Very young stars are also found to show evidence of jets coming out of the poles---jets are a very strong indicator of a disk structure. Researchers who study the precise timing of pulsars have found that some pulsars show a wobble in the period of the pulsar. That is, instead of the pulsar always having the same period, sometimes it is slightly faster, and other times slightly slower. This is strong evidence for orbital motion of the pulsar about the center of mass of a system. From careful analysis of the pulsars' period changes, orbits have been deduced that suggest planets circle the pulsars. 2. The process of protoplanetary accretion is still not well understood. Search the Web for current research in this field. From the results of your search, outline the basic steps in the formation of a protoplanet through accre- tion. What specific factors are important in these mod- els of planet building? Do these models produce plane- tary systems similar to the ones we know to exist? http://www.gps.caltech.edu/~gab/astrophysics/astrophysics.html There is an intimate connection between the early solar nebula and the interstellar medium from which it formed. The study of current star-forming environments can tell us much about how we came to be. The combination of rapidly improving observational tools and increasingly sophisticated theory has provided, for the first time, a broad outline of the physical processes associated with the assembly of Sun-like stars and their attendant planetary systems. The figure below presents an overview of the scenario developed for the formation of a single, isolated low mass star from the collapse of a molecular cloud core. http://www.owlnet.rice.edu/~spac250/matt/overview.html describe how stars and planetary systems formed. In most of these theories, a disk is created when a Giant Molecular Cloud (GMC) collapses; go here for more on both the nature of the clouds and the process of disk creation. The result is something like that in the figure above. The flaring out at the edges is just because material farther from the central object takes longer to settle down to the plane of orbit. The jets shown in the picture, by the way, are interesting topics in themselves streams of material shooting out of a young stellar disk, they can reach relativistic velocities. 3. How is radioactive dating carried out on meteorites and rocks from surfaces of various bodies in the solar sys- tem? Look for Web sites on the details of radioactive dating, and summarize the methods used to uncover the abundances of radioactive elements in a particular sample. Hint: Try looking for information on how a particular meteorite-for example, the Martian mete- orite ALH 84001-was studied, what age range was de- termined, and what radioactive elements were used to arrive at the age. the Martian meteorite ALH 84001-was studied http://www.lpi.usra.edu/lpi/meteorites/29thlpscabs.html Carbonates in Martian Meteorite ALH84001: Petrologic Evidence for an Impact Origin Many petrographic observations tend to support the theory that the carbonates and most of the shock features in ALH 84001 formed at high temperature in a single impact event (Scott et al., 1997). New observations here include the recognition: that most carbonate 'disks' in fractures are thinner at their edges than their centers; that micro-disks of magnesite (10-30m diam.) are common near carbonate disks; that the proportion of carbonate in pyroxene is correlated with its degree of fracturing; and that carbonate grains can be dispersed in linear trains within the meteorite's crush zones. explain all these carbonate textures, along with their previous observations, as having formed from carbonate melt ( plagioclase melt) formed during a single impact shock event. Exploring The Sky 1. Look at the solar system from space. Notice how thin the disk of the solar system is and how inclined the or- bits of Pluto and Mercury are. Hint: Under the View menu, choose 3D Solar System Mode, and then zoom in or out. Tip the solar system up and down to see it edge-on. Pluto was on a purple colored line. Mercury was on a purple circled line. It was very thin. 2. Look at the solar system from space and notice how small the orbits of the inner planets are compared to the orbits of the outer planets. They make two distinct groups. Hint: Use 3D Solar System Mode. very small inner planets. 3. Watch the comets orbiting around the sun. Can you lo- cate the comet C/198 M5 (Linear)? What is its orbit like? Hint: Use 3D Solar System Mode, and set the time step to 30 days (30d.) No it was 10P Tempel 2 was orbiting the sun. The object a comet would go close to the sun and away from the sun. page 335 Table 16-3 The Condensation Sequence Temperature K Condensate Planet (Estimated Temperature of Formation:K) 1500 Metal oxides Mercury (1400) 1300 Metallic Iron and nickel 1200 Silicates 1000 Feldspars Venus (900) 680 Troilite (FeS) Earth (600) Mars (450) 175 H2O ice Jovian (175) 150 Ammonia-water ice 120 Methane-water ice 65 Argon-neon ice Pluto (65) Chapter 16 The Origin of the Solar System Review Questions 1. What are the four stages in the development of a terres- trial planet? The first stage of planetary evolution is differen- tiation, the seperation of material according to den- sity. Earth now has a dense metallic core, a thick rocky mantle, and a lower-desity crust so we know it dif- ferentiated. Some of that differentiation may have oc- curred very early as the heat of formation released by in-falling matter melted the growing Earth. Some of the differentiation however, may have occurred later as radioactive decay released heat and further melted Earth allowing the denser metals to sink to the core. The second stage, cratering could not begin until a solid surface formed. The heavy bombardment of the early solar system crated Earth just as it did our moon. As the debris in the solar nebula cleared away, the rate of cratering impacts fell rapidly to its present low rate. The third stage, flooding, began as the decay of radioactive elements heated Earth's interior. Rock melted in the upper mantle where the pressure was lower than the deep interior, and some of the molten rock welled up through fissures in the crust and flooded the deeper basins. Later, as the atmosphere cooled, water fell as rain and flooded the basins to form the first oceans. Note that on Earth flooding includes both lava and water. The fourth stage, slow surface evolution, has con- tinued for at least the past 3.5 billion years or more. Earth's surface is constantly changing as sections of crust slide over each other, push up mountains, and shift continents. In addition, moving air and water erode the surface and wear away geological features. Almost all trace of the first billion years of Ea rth's geol- ogy has been destroyed by the active crust and erosion. Terrestrial planets pass through these four stages, but differences in mass, temperature, and composi- tion emphasize some stages over others and produce surprisingly different worlds. 2. Why do we expect planets to have differentiated? During the formation of planets, energy released from radioactive decay and from the infalling material caused the temperature of the planets to rise enough to melt any solid material. In a molten state, the heavier elements sunk toward the center of the planets. This is what differentiation is. Earth scientists have clear proof that Earth did differentiate. Each time an earthquake oc- curs, seismic waves travel through the interior and register on seismographs all over the world. Analysis of these waves shows that Earth's interior is divided into a metallic core, a dense rocky mantle, and a thin, low-density crust. 3. How does plate tectonics create and destroy Earth's crust? Parts of the earth's surface are continually being subducted (buried) under other surface plates. This occurs, for example, around the rim of the Pacific Ocean. New surface material is constantly upwelling from mid-oceanic rifts. The Earth's crust is divided into moving sections called plates. Where plates spread apart, lava welss up to form new crust. 4. Why do we suspect that Earth's primeeval atmosphere was rich in carbon dioxide? Our planet's first atmosphere, its primeval atmo- sphere, was once thought to contain gases from the solar nebula, such as hydrogen and methane. Modern studies, however, tell us that the planets formed hot, so gasses such as carbon dioxide, nitrogen, and water vapor would have been cooked out of the rock and metal as Earth grew. In addition, the final stages of planet building may have seen planets accreting plan- etesimals rich in easily vaporized materials such as water, ammonia, and carbon dioxide. The prime- val atmosphere must have been rich in carbon dioxide, nitrogen, and water vapor. Once Earth cooled enough, oceans began to form, and the carbon dioxide began to dissolve in the water. it reacted with dissolved compounds in the ocean water to form silicon dioxide, limestone, and other mineral sediments. The oceans transferred the carbon dioxide from the atmosphere to the seafloor and left air rich in nitrogen. 5. Why doesn't Earth have as many craters as its moon or Venus? Their hasn't been as many meteorite impacts to create craters. Meteorites striking a planet's surface deliver tremendous energy and can produce characteristic impact craters. 6. What kind of erosion is now active on Earth's moon? The main kind of erosion active on the earth's moon is erosion due to meteorites striking the moon's surface, cratering it and pulvarizing it. Meteorite impacts. Lava flooded the basins on on the Earthward side of the moon. Hardly no wind errosion. 7. Discuss the evidence and hypotheses concerning the origin of Earth's moon? The Origin of Earth's Moon The fission hypothesis proposed that our moon broke from a rapidly spinning proto-Earth. If this hap- pened after the proto-Earth differentiated, our moon would have formed from iron-poor material. However, moon rocks differ chemically from those of Earth. Also if the proto-Earth had spun fast enough to break up, the Earth-moon system should now contain much more angular momentum than it does. The condensation hypothesis suggested that Earth and its moon condensed from the same cloud of mat- ter in the solar nebula. This idea doesn't work, how- ever because Earth and its moon have different den- sities and compositions. The moon for example, is very poor in volatiles - materials such as water and sodium, which are easily vaporized. The capture hypothesis suggested that our moon formed elsewhere in the solar nebula and was later captured by Earth. However if our moon passed near enough to Earth to be captured, Earth's gravity would have ripped it to fragments. Until the mid-1980s, astronomers had no accept- able hypothesis for the origin of Earth's moon, at that point a hybrid theory offered hope. The large- impact hypothesis supposes that our moon formed when a planetesimal at least as large as Mars smashed into the proto-Earth and ejected debris into an orbit, where it formed our moon. This would explain a number of phenomena. If the collision occurred off center, it would have spun the Earth-moon system rapidly and would ex- plain the present angular momentum. Such an impact would have melted the proto-Earth, and the material falling together to form the moon would have been heated hot enough to melt. This fits our expectation that the moon formed as a sea of magma. 8. Why doesn't Earth or its moon have lobate scarps? Lobate scarps are produced when the inner part of a planet shrinks, causing its surface to buckle, much like the skin of a fruit shrivels when its inside shrink. Compared to Mercury's interior which has a high metal content, the inside of the earth and of the moon never shrank much. At first glance, we might suppose that any world with a metallic interior should have lobate scarps, other factors are also important. Earth has a fairly large metal- lic core, being a large world, it has not cooled very much, so it presumably hasn't shrunk much. Of course the geological activity on Earth's surface would erase such scarps if they did form. Earth's moon is not geo- logically active, it does not contain a significant metallic core. Although Earth's moon has lost much of its internal heat, its interior is mostly rock and didn't shrink as much as metal would have. 9. How did Earth avoid the greenhouse effect that made Venus so hot? The surface temperature on Venus is hot enough to melt lead, and we can understand that because the thick atmosphere creates a severe greenhouse effect. Sunlight filters down through the clouds and warms the surface, heat cannot escape because the carbon dioxide gas is opaque to infrared. Venus is a deadly hot desert world of volcanoes, lava flows, and impact craters lying at the bottom of a deep ocean of hot gases. Venus is only 30 percent closer to the sun. Unfortunately the surface of Venus is perpetually hidden below thick clouds. 10. What evidence do we have that plate tectonics does not occur on Venus? on Mars? on Venus The intense heat at the surface may have effected the geology of Venus by making the crust drier and more flexible so that it was unable to break into mov- ing plates as on Earth. There is no sign of plate tecton- ics on Venus, rather evidence that convection cur- rents below the crust are deforming the crust to create coronae and push up mountains such as Maxwell. The Surface of Venus There is no liquid water on Venus, so its low lands are not seafloors, and the re- maining highlands are not the well-defined conti- nents we see on Earth. Whereas Earth is dominated by plate tectonics, something different is happing on Venus. U.S. and Soviet spacecraft orbited Venus and mapped its surface by radar. Radar maps of Venus. Venus does not have plate tectonics? On Earth, plate tectonics is identifiable by the world- wide network of faults, subduction zones, volcanism, and folded mountain chains that outline the plates. Al- though some of these features are visible on Venus, they do not occur in a planetwide network that outlines plates. 11. What evidence suggests that Venus has been resurfaced within the last 300 million years? The small number of craters on the surface of Venus hints that the entire crust has been replaced within the last 300 million years or so. This may have occurred in a planetwide overturning as the old crust broke up and sank and lava flows created a new crust. Venus is a deadly hot desert world of volcanoes, lava flows, and impact craters lying at the bottom of a deep ocean of hot gases. Volcanism has helped form the high-lands. The lava that flows spread outward can create erosion. Gases. 12. Why is the atmosphere of Venus rich in carbon dioxide? Why is the atmosphere of Mars rich in carbon dioxide? The Atmosphere of Venus In composition temperature, and density, the atmo- sphere of Venus is more hotter. The air is unbreathable, very hot, and almost 100 times denser than Earth's air. U.S. and Soviet space probes have descended into the atmo- sphere and in some cases landed on the surface. In composition the atmosphere of Venus is roughly 96 percent carbon dioxide. Very small traces of deadly water vapor. 13. What evidence do we have that the climate on Mars has changed? A primary piece of evidence is that channels on the surface of Mars indicate that Mars was once warm enough to have liquid water. The present atmosphere of Mars is thin Mars may go through cli- In the past Mars had a thicker atmosphere that allowed water to flow over the surface. The escape velocity on Mars is 5 km/s. mate cycles similar to the ice ages on Earth, and that may freeze enough gas to increase the density of the at- mosphere. Polar caps are built when great dust storms on Mars scatter dust onto the polar caps. When the caps vaporize each spring, they leave the dust behind in growing layers. Because there are dif- ferent patterns of layers one on top of the other we can conclude that the climate must change occasion- ally. 14. Why do we conclude that the crust on Mars must be thick? The crust of Mars is now quite thick as shown by the mass of Olympus Mons, it was thinner in the past. 15. What evidence do we have that there has been liquid water on Mars? In the late 19th century astronomers noticed what looked like faint straight lines on Mars, lines that be- came known as canals. In 1976 two Viking spacecraft orbited Mars and sent back detailed photos that revealed two kinds of features related to water. Flood features were broad, eroded planes where sudden, massive floods appear to have scoured the surface. Some sec- tions of the crust appear to have collapsed, perhaps be- cause permafrost frozen below the surface was melted by volcanis heating and drained away in great floods. There is very little water in the Martian atmo- sphere, and the polar caps are composed of frozen car- bon dioxide dry ice with an unknown amout of frozen water trapped below. Mars was large enough to have a substantial atmosphere when it was young and may have had water falling as rain and collecting in rivers and lakes. It gradually lost much of its atmosphere and is now a cold, dry world. page 366 Chapter 17 Discussion Questions 1. If we visited a planet in another solar system and dis- covered oxygen in its atmosphere, what might we guess about its surface? Might be land, life, and water on the surface or frozen water, and ice. 2. If liquid water is rare on the surface of planets, then most terrestrial planets must have CO2 -rich atmospheres. Why? sublimation/evaporation of ices or liquids on surface requires an increase in the surface temperature such as from the greenhouse effect Mars is slightly bigger. Its volcanoes produced a CO2 and H2O atmosphere which once must have been much thicker so that the greenhouse effect could make the surface warm enough to allow water to run on the surface. We know this from observing dried riverbeds on the surface. Then what happened? The CO2 went away - through condensation and chemical reactions with the surface. This resulted in cooling of the surface and freezing of the water. Because Mars is smaller, the lithosphere thickened and there was not enough tectonic or volcanic activity to recycle the CO2 into the atmosphere. The lack of ozone meant that ultraviolet light could destroy water in Mars' atmosphere, and the light Hydrogen from the water could escape thermally while the oxygen rusted the surface. Venus is larger, like the Earth. Volcanoes produced an atmosphere of H2O and CO2 which lead to a greenhouse effect, making it still hotter. Because Venus is closer to the Sun, it was hotter than Mars and the H2O didn't condense but stayed in the atmosphere. Since H2O is a greenhouse gas, this made the planet even hotter which in turn kept the CO2 from chemically reacting with the surface. Meanwhile H2O was destroyed and lost by solar UV and thermal escape, but there was already enough CO2 in the atmosphere to keep it in a perpetual runaway greenhouse. Problems 1. If the Atlantic seafloor is spreading at 30 mm/year and is now 6400 km wide, how long ago were the continents in contact? This is a problem for one of our first formulas, d = v t, where d is the distance traveled, v is the speed of travel, and t is the travel time. We want the travel time, so solve for t: t = d / v = (6400 km) / (30 mm/yr) = (6.4 x 106m) / (3 x 10-2m/yr) = 2.1 x 108yr = 210 million years. 2. Why do small planets cool faster than large planets? Hint: Compare surface area to volume. 3. The smallest detail visible through Earth-based tele- scopes is about 1 second of arc in diameter. What size is this on Earth's moon? Hint: See By the Numbers 3-1. 4. The trenches where the seafloor slips downward are 1 km or less wide. Could Earth-based telescopes resolve such features on Earth's moon? Why are we sure such features are not present on Earth's moon? 5. How long would it take radio signals to travel from Earth to Venus and back if Venus were at its farthest point from Earth? Why are such observations impractical? 6. Repeat Problem 5 for Mercury? 6. 23 minutes 7. Imagine that we have sent a spacecraft to land on Mer- cury, and it has transmitted radio signals to us at a wave- length of 10 cm. If we see Mercury at its greatest angu- lar distance west of the sun, to what wavelength must we tune our radio telescope to detect the signals? Hints: See Data File Four to find Mercury's orbital velocity, and then see By the Numbers 6-2. This is a Doppler shift problem. The Doppler formula is VR = (Δλ / lambda;0) c We need to find how much the wavelength changes Delta;& lambda;. Delta; lambda; = VR lambda;0 / c = (47.9 km/s) (10 cm) / (3 x 108m/s) = 0.0016 cm Therefore, we must tune our radio telescope to receive a wavelength of 10.0016 cm. By the Numbers 6-2 The Doppler Formula Astronomers can measure radial velocity by using the Doppler effect. Vr = AA C A The radial velocity divided by the speed of light, c is equal to AA divided by A. In astron- omy radial velocities are almost always given in kilometers per second so we will express c as 300,000 km/s. For example suppose the laboratory wave- length of a certain spectral line is 600.00 nm and the line is observed in a star's spectrum at a wave- length og 600.10 nm. Then AA is + 0.10 nm and the velocity is 0.10/600 multiplied by the speed of light. The radial velocity equals 50 km/s. Be- cause AA is positive we know the star is receding from us. 8. The smallest feature visible through an Earth-based telescope has an angular diameter of about 1 second of arc. If a crater on Mars is just visible when Mars is near- est to Earth, what is its linear diameter? Hint: See By the Numbers 3-1. 8. about 380 km 9. What is the maximum angular diameter of Phobos as seen from Earth? What surface features should we be able to see using Earth-based telescopes? Hint: See By the Numbers 3-1. page 367 10. Phobos orbits Mars at a distance of 9380 km from the center of the planet and has a period of 0.3189 days. Calculate the mass of Mars. Hints: Convert to AU and years, and then see By the Numbers 4-1. This is a problem using our generalized Kepler's third law. Mars is "A," Phobos is "B," and the Sun is "S." (MA + MB)/ MS = (a / 1 AU)3 / (P / 1 year)2 = (9380 km / 1.5 x 108km)3 / (0.3189 d / 365.25 d)2 = 3.2 x 10-7 Therefore, the mass of Mars plus Phobos is this last answer times the mass of the sun. (MA + MB) = (3.2 x 10-7) ( MS) = 6.4 x 1023kg. Since the mass of Phobos is so small, this last number is essentially the mass of Mars alone. 10. 6.4 X 10 23 kg Critical Inquiries for the Web 1. Who decides how planetary features are named? Sur- face features on Venus are mostly named after female figures from history and mythology, whereas figures from the arts and music are used to name features on Mercury. Look for Information on planetary nomencla- ture, aund summarize the way different types of features on Venus are assigned names. http://curious.astro.cornell.edu/question.php?number=372 Who named the planets and who decides what to name them? Who decides what to name the planets? And who named them? The planet names are derived from Roman and Greek mythology, except for the name Earth which is Germanic and Old English in origin. who's in charge of naming solar system objects that are discovered now? Since its organization in 1919, the International Astronomical Union (IAU) has been in charge of naming all celestial objects. When an astronomer discovers an object, or wants to name a surface feature, they can submit a suggestion to the IAU, and the IAU either approves i      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~t or suggests a different name. Since we don't think there are any undiscovered planets, the IAU focuses on the naming of moons, surface features, asteroids, and comets and has websites about naming conventions for each. 2. "Martians" have fascinated humans for the last century or more. There are many online sources that chronicle the representation of life on Mars throughout history and in literature. Read about the Martians as presented by a particular literary work or nonfiction account, and dis- cuss to what extent it is or is not based on realistic views of the nature of Mars both in terms of our current understanding and the views of that period. http://rsd.gsfc.nasa.gov/marslife/marslif6.htm METEORITE YIELDS EVIDENCE OF PRIMITIVE LIFE ON EARLY MARS A NASA research team of scientists at the Johnson Space Center and at Stanford University has found evidence that strongly suggests primitive life may have existed on Mars more than 3.6 billion years ago. The NASA-funded team found the first organic molecules thought to be of Martian origin; several mineral features characteristic of biological activity; and possible microscopic fossils of primitive, bacteria-like organisms inside of an ancient Martian rock that fell to Earth as a meteorite. This array of indirect evidence of past life will be reported in the Aug. 16 issue of the journal Science, presenting the investigation to the scientific community at large to reach a future consensus that will either confirm or deny the team's conclusion. Because the water was saturated with carbon dioxide from the Martian atmosphere, carbonate minerals were deposited in the fractures. http://www-curator.jsc.nasa.gov/curator/antmet/marsmets/SearchForLife/SearchForLife.htm ALH84001 arrived on Earth 13,000 years ago (15, 21) and appears to be essentially free of terrestrial weathering (8). ALH84001 does not have the carbon isotopic compositions typically associated with weathered meteorites (12, 15), and detailed mineralogical studies (8) show that ALH84001 has not been significantly affected by terrestrial weathering processes. ALH84001 is somewhat friable and breaks relatively easily along preexisting fractures. It is these fracture surfaces that display the carbonate globules. We analyzed freshly broken fracture surfaces on small chips of ALH84001 for polycyclic aromatic hydrocarbons (PAHs) using a microprobe two-step laser mass spectrometer http://spot.colorado.edu/~marscase/cfm/c6papers/rshaid.html Mars and drain the water reservoirs in the ice caps by actually building the imaginary channels. 3. What would it be like to walk on the lunar surface? Apollo astronauts visited six different locations on our moon, exploring a variety of lunar terrain. Describe the horizons and general relief of the landing locations of the different missions by exploring Web sites that pro- vide lunar surface photography from the missions. What differences do you see between images from landings in highlands and maria? It is less gravity and air on the lunar surface on the Moon. More floating. The maria are nearly free of craters. The ancient highlands are heavily cratered. Exploring The Sky 1. Locate Mercury and Venus and Zoom in until you can see their phases. Explain their phases by discussing their locations in their orbits. Mercury Rise: 6:16 AM on 6/6/91 Transit: 1:54 PM on 6/6/91 Set: 9:32 PM on 6/6/91 RA: 05h 52m 36.4s Dec: +2514'24" Azm: 5644'53" Alt: +0026'05" Phase: 87.478%, Apparent magnitude: -1.13 Heliocentric ecliptical coordinates: l: 12938'42.9" b: +0654'21.5" r: 0.328115 Geometric geocentric ecliptical coordinates: l: +8820'32" b: +0149'40" r: 1.236890 Mean geometric ecliptical coordinates: l: +8820'00" b: +0149'38" r: 1.237016 True equatorial coordinates: RA: 05h 52m 36s Dec: +2514'29" Physical Data Apparent angular diameter: 5.43 Venus Rise: 4:50 AM on 6/6/91 Transit: 12:04 PM on 6/6/91 Set: 7:18 PM on 6/6/91 RA: 04h 03m 52.4s Dec: +1955'19" Azm: 7715'19" Alt: +1608'45" (with refraction: +1612'06") Phase: 97.335%, Apparent magnitude: -3.89 Heliocentric ecliptical coordinates: l: 4406'15.2" b: -0152'13.5" r: 0.723202 Geometric geocentric ecliptical coordinates: l: +6251'51" b: -0048'32" r: 1.672217 Mean geometric ecliptical coordinates: l: +6251'28" b: -0048'33" r: 1.672155 True equatorial coordinates: RA: 04h 03m 52s Dec: +1955'22" Physical Data Apparent angular diameter: 9.98 From Mercury: Angular separation: 2536'31" Position angle: +26301' 2. Repeat the activity above for Mars. Mars Set: 1:41 AM on 6/6/91 Rise: 12:58 PM on 6/6/91 Transit: 7:18 PM on 6/6/91 RA: 11h 19m 35.4s Dec: +0522'28" Azm: 34033'02" Alt: -4301'36" Phase: 88.799%, Apparent magnitude: 0.54 Heliocentric ecliptical coordinates: l: 20743'16.5" b: +0042'33.3" r: 1.607007 Geometric geocentric ecliptical coordinates: l: +16836'48" b: +0056'52" r: 1.202650 Mean geometric ecliptical coordinates: l: +16836'36" b: +0056'52" r: 1.202713 True equatorial coordinates: RA: 11h 19m 35s Dec: +0522'33" Physical Data Dec Earth: 25.65, Dec Sun: 25.65 Position angle: 20.37 Longitude of central meridian: 281.95 Defect of illumination: 0.87, Position angle: 113.09 Apparent angular diameter: 7.78 From Mercury: Angular separation: 8015'05" Position angle: +9156' 3. Watch Venus and Mars circle the sky and discuss the way their phases change with their positions in their orbits. Mars moves out away from Venus for a short while, and then comes closer to Venus. up and down. page 367 Chapter 17 The Earthlike Planets Review Questions 1. Why is Jupiter so much richer in hydrogen and helium than Earth? Since Jupiter formed far from the sun compared to the Earth, many substances condensed under those cold conditions. Thus Jupiter grew massive from condensation and accretion. In this way Jupiter gained some hydrogen and helium. As a massive planet, Jupiter than became even more massive as it gravitationally attracted and held hydrogen and helium from the solar nebular. Density is mass divided by volume, and the density of Jupiter is only a third greater than that of water. Data File Seven For comparison, Earth is over 4 times more dense than Jupiter. Theoretical models of Jupiter based on this density conclude that it is com- posed mostly of hydrogen and helium. 2. How can Jupiter have a liquid interior and not have liquid surface? Although Jupiter is mostly liquid hydrogen there is no ocean surface. The base of the atmosphere is so hot and the pressure is so high that there is no real distinction between liquid and gas. We would see the gas density increasing around us until we were sinking through liquid, we would never splash into a liquid surface. 3. How does the dynamo effect account for the magnetic- fields of Jupiter and Saturn? Magnetic fields are produced by electric currents. Jupiter and Saturn are massive enough to squeeze their hydrogen into a liquid metallic state, which conducts electric currents very easily. The rotation of the planets and their interior convection drive large internal electric currents which produce the planets' magnetic fields. This is the dynamo effect. This mass of conducting liquid, stirred by convection currents and spun by the planet's rapid rotation, drives the dynamo effect and generates a powerful magnetic field. Jupiter's field is over 10 times stronger than Earth's. One consequence of this magnetic field is aurora. liquid hydrogen liquid metallic hydrogen a material that is a very good conductor of electricit. 4. Why are the belts and zones on Saturn less distinct than those on Jupiter? On Saturn In a Jovian planet, the light-colored zones form in high- pressure regions where rising gas cools and condenses to form icy crystals of ammonia, which we see as bright clouds. Saturn is twice as far from the sun as Jupiter, so sunlight is weaker and the atmosphere is colder. The rising gas currents don't have to rise as high to reach temperatures cold enough to form clouds. Because the clouds form deeper in the hazy atmosphere, they are not as brightly illuminated by sunlight and look dimmer. A layer of methane haze above the clouds makes the belts and zones look even less distinct. Jupiter's atmosphere is dominated by belt-zone circulation, which circles the planet parallel to its equator. The colors of the belts and zones are believed to arise from molecules in the clouds produced by sunlight or by lightning in- teracting with ammonia and other compounds. 5. Why do we conclude that neither Jupiter's ring nor Sat- urn's rings can be left over from the formation of the planets? The small particles in the rings would have been swept away from the planets by the solar wind long ago if those particles had been there since the formation of the solar system. We can be sure that the ring particles are not old. Another reason the ring particles can't be old is that the intense radiation around Jupiter will grind the dust specks down to nothing in a century or so The rings we see today can't be material left over from the forma- tion of Jupiter. The rock's gravity cannot hold the dust speck. And the billions of dust specks in the ring can't pull themselves together to make a new moon because of tidal forces inside the Roche limit. Saturn's rings are made of bright ice chunks. The particles in Jupiter's ring are very dark and reddish. The ring is rocky rather than icy. 6. How can a moon produce a gap in a planetary ring system? Some gaps in the rings, such as Cassini's division, are caused by resonances with moons. A particle in Cassini's division orbits Saturn twice in the time the moon Mimas takes to orbit once. On every other orbit, the particle feels a gravitational tug from Mimas and is eventually pulled out of its circular orbit into an elliptical orbit. Such an orbit is dangerous because the particle crosses the orbits of other particles, which results in collisions. In this way, resonances can remove particles from a gap. Jupiter has at least 16 known moons. The large family of moons may be mostly captured asteroids. Asteroids and comets. 7. Explain how geological activity on Jupiter's moons varies with distance from the planet? The activity we see in the Galilean moons must be driven by energy flowing outward. The violently active volcanism of Io is apparently caused by tidal heating. I o is too small to have remianed hot from its formation its orbit is slightly elliptical as it moves closer to and then farther from Jupiter, the planetary gravitational field flexes the moon with tides, and friction heats its interior. and the Galilean moons affected by tidal heating. 8. What makes Saturn's F ring and the rings of Uranus and Neptune so narrow? Because of collisions among ring particles planetary rings should spread outward. The sharp outer edge of the A ring and the narrow F ring are confined by shepherd satellites that gravitationally usher straying particles back into the rings. Eleven narrow rings have been found on Uranus. Planetary rings should gradually spread into thin sheets, so the narrowness of these rings suggests that they are confined by shephered moons. Voyager 2 photos show two moons less than 25 km in diameter shepherding the E ring and the other rings are presumably confined by moons too small to detect. Why are the rings of Uranus so narrow? Unlike the rings of Jupiter and Saturn, the rings of Ura- nus are quite narrow, like hoops of wire. Two small moons have been found orbiting just inside and outside of the E ring. If a ring particle drifts away from the ring, the corre- sponding moon's gravity will boost it back into the ring. Moons too small to detect are thought to shepherd the other rings. The rings of Uranus resemble Saturn's narrow F ring. The rings of Uranus are tenuous and dark. They total less mass than the thin material in Cassini's division. The rings of Uranus are composed of very dark material as black as lumps of coal. Saturn's rings are very thin and nearly vanish when seen edge-on. Although ripples in the rings caused by waves may be hundreda of meters high the sheet of particles may be only a dozen or so meters thick. 9. Why is the atmospheric activity of Uranus less than that of Saturn and Neptune? The material was less dense and where orbital motions were slower. Uranus never grew massive enough to capture large amounts of gas from the nebula as did Saturn. Uranus is rich in water and ice rather than in hydrogen and helium. Uranus and Neptune are ice planets. Saturn is a gas giant planet. 10. Why do we suspect that Enceladus has been geologically active more recently than some other moons? 11. What are the seasons on Uranus like? Winter Spring 12. Why are Uranus and Neptune blue? The methane in the atmosphere of both planets preferentially absorbs red light and reflect blue light. When the sun's white light hits the planets, we see the reflected light which is more blue than red. The light we receive from Nep- tune is sunlight that is scattered from various layers of Neptune. Because sunlight con- tains a distribution of photons of all visible wavelengths, it looks white to us sunlight entering Neptune's at- mosphere must pass through hydrogen gas that contains a small amount of methane, which is a good absorber of longer wavelengths. As a result red photons are more likely to be absorbed than blue photons, and that makes the light bluer. 13. What evidence do we have that Triton has been geolog- ically active recently? the dark smudges visible in the southern polar cap. photos reveal tha these are deposits pro- duced when liquid nitrogen in the crust warmed by the sun, erupts through vents and spews up to 8 km high into the atmosphere. Methane in the gas is con- verted by sunlight into dark deposits that fall to the surface, leaving the black smudges. These geysers make Triton one of only three in our solar sys- tem observed to be active. 14. How do astronomers account for the origin of Pluto? It is thought that icy planetesimals formed outside the orbit of Neptune, when the solar system was forming. Many of these planetesimals were destroyed in collisions or were ejected into the distant parts of the solar system by gravitationals interactions with the Jovian planets. Pluto was perhaps one of the largest of these planetesimals. Somehow it survived and captured another smaller one, its moon Charon. An old theory proposed that Pluto was an es- caped moon of Neptune. Pluto is much like Neptune's moon Triton. 15. What evidence do we have that catastrophic impacts have occurred in the solar system's past? The high inclination of Pluto and the orbit of Charon suggest a past collision or, at least a close gravitational interaction with a passing body. And the Pluto-Charon system is not the only evidence for major impacts in the past. The peculiar orbits of Neptune's moons Triton and Nereid, the fractured and cratered state of the Jovian satellites, and the peculiar rotation of Uranus all hint that impacts and encounters with large planetesimals or the heads of comets have been important in the history of these worlds. page 396 Chapter 18 Discussion Questions 1. Some astronomers argue that Jupiter and Saturn are un- usual, while other astronomers argue that all solar sys- tems should contain one or two such giant planets. What do you think? Support your argument with evidence. No doubt humans have struggled with the question of whether we are alone in the universe since the beginning of consciousness. Today, armed with evidence that planets do indeed orbit other stars, astronomers wonder more specifically: What are those planets like? Of the 100 billion stars in our Milky Way galaxy, how many harbor planets? Among those planets, how many constitute arid deserts or frigid hydrogen balls? Do some planets contain lush forests or fertile oceans teaming with life? For the first time in history, astronomers can now address these questions concretely. During the past two and a half years, researchers have detected eight planets orbiting sunlike stars. In October 1995 Michel Mayor and Didier Queloz of Geneva Observatory in Switzerland reported finding the first planet. Observing the star 51 Pegasi in the constellation Pegasus, they noticed a telltale wobble, a cyclical shifting of its light toward the blue and red ends of the spectrum. The timing of this Doppler shift suggests that the star wobbles because of a closely orbiting planet, which revolves around the star fully every 4.2 days--at a whopping speed of 482,000 kilometers (299,000 miles) an hour, more than four times faster than Earth orbits the sun. Another survey of 107 sunlike stars, performed by our team at San Francisco State University and the University of California at Berkeley, has turned up six more planets. Of those, one planet circling the star 16 Cygni B was independently discovered by astronomers William D. Cochran and Artie P. Hatzes of the University of Texas McDonald Observatory on Mount Locke in western Texas. Detection of an eighth planet was reported in April 1997, when nine-member team led by Robert W. Noyes of Harvard University detected a planet orbiting the star Rho Coronae Borealis. A ninth large object, which orbits the star known by its catalogue number HD114762, has also been observed--an object first detected in 1989 by astronomer David W. Latham of the Harvard-Smithsonian Center for Astrophysics and his collaborators. But this bulky companion has a mass more than 10 times that of Jupiter--large, though not unlike another large object discovered around the star 70 Virginis, a similar object with a mass 6.8 times that of Jupiter. The objects orbiting both HD114762 and 70 Virginis are so large that most astronomers are not sure whether to consider them big planets or small brown dwarfs, entities whose masses lie between those of a planet and a star. Detecting Extrasolar Planets Finding extrasolar planets has taken a long time because detecting them from Earth, even using current technology, is extremely difficult. Unlike stars, which are fueled by nuclear reactions, planets faintly reflect light and emit thermal infrared radiation. In our solar system, for example, the sun outshines its planets about one billion times in visible light and one million times in the infrared. Because of the distant planets' faintness, astronomers have had to devise special methods to locate them. The current leading approach is the Doppler planet-detection technique, which involves analyzing wobbles in a star's motion. Here's how it works. An orbiting planet exerts a gravitational force on its host star, a force that yanks the star around in a circular or oval path--which mirrors in miniature the planet's orbit. Like two twirling dancers tugging each other in circles, the star's wobble reveals the presence of orbiting planets, even though we cannot see them directly. 2. Why don't the terrestrial planets have rings? If you were to search for a ring among the terrestrial planets, where would you look first? Can terrestrial planets have rings? It turns out that all of the planets, Earth included, did have rings at one time. The thing is, these rings were unstable and the material was either lost to space or collected into the satellites of these planets. The difference between the terrestrial and giant planets is the giant planets have the gravity to capture and hold onto a large satellite system, and these satellite systems are the source of the ring material. For a ring to be stable, it must be held tightly by the planet's gravity, and the planet must also exert tidal forces on the particles in the ring. Tidal forces result from the fact that the pull of gravity is inversely related to distance, so the farther away an object is, the less force it experiences from the object it is orbiting. Therefore, a planet pulls a little bit more on the inner side of its moons than on the outer side. Close enough to a planet, this might cause a moon to break apart, and also keeps the bits of material that form a ring from collecting together into a moon. However, it's possible that Mars might develop a ring in the future. Its moon, Phobos, is close enough to the planet that it feels the effects of the planet's tidal forces, and eventually it may break apart and form a ring. Earth, Mercury, Venus and Pluto will probably never again have observable rings, although if you dumped tons and tons and tons of sand near the planet, it would probably form a ring. Problems 1. What is the maximum angular diameter of Jupiter as seen from Earth? Repeat this calculation for Saturn and Pluto. Hints: See Data Files Seven, Eight, and Eleven, and also By the Numbers 3-1. 2. What is the angular diameter of Jupiter as seen from Callisto? from Io? Hint: See By the Numbers 3-1. 3. Measure the photograph in Data File Eight and calcu- late the oblateness of Saturn. 4. If we observe light reflected from Saturn's rings, we should see a red shift at one edge of the rings and a blue shift at the other edge. If we observe a spectral line and see a difference in wavelength of 0.056 nm, and the unshifted wavelength observed in the laboratory is 500 nm, what is the orbital velocity of particles at the outer edge of the rings? Hint: See By the Numbers 6-2. We need to use the Doppler shift formula to find the radial (line-of-sight) velocity VR of particles at the outer edge of the ring. The problem says the shift in wavelength from blue shift to red shift is 0.056 nm. Therefore the change in wavelength compared to the no-shift value of 500 nm is half of 0.056 nm, which is 0.028 nm. VR = (Delta;lambda; / lambda;0) c = (0.028 nm / 500 nm) (3 x 108 m/s) = 1.68 x 104 m/s = 16.8 km/s 4. 16.8 km/s 5. One way to recognize a distant planet is by its motion along its orbit. If Uranus circles the sun in 84 years, how many seconds of arc will it move in 24 hours? Hint: Ig- nore the motion of Earth. 6. If the E ring is 50 km wide and the orbital velocity of Uranus is 6.81 km/s, how long a blink should we ex- pect to see when the ring crosses in front of the stars? The duration of the blink is just the time it takes for the ring to move its width, 50 km. T his problem requires the use of the distance equals speed times time formula: d = v t. We need to solve for the length of time t. t = d / v = (50 km) / (6.81 km/s) = 7.3 s 6. 7.3 s 7. What is the angular diameter of Pluto as seen from the surface of Charon? Hint: See Figure 18-22. Figure 18-22 The circular orbit of Charon is seen here at an angle. The orbit is only a few times bigger than Earth and is tipped 118 degrees to the plane of Pluto's orbit. distance from Charon to Pluto is 19,640 km 8. If Pluto has a surface temperature of 50 K, at what wave- length will it radiate the most energy? Hint: See By the Numbers 6-1. By the Numbers 6-1 Black Body Radiation Wien's law 3,000,000 is divided by the temperature in degress kelvin. 5.67 X 10 -8 J/m2s degree 4. A max= 3,000,000 T 8. 60.000 nm 60 microns 9. How long did it take radio commands to travel from Earth to Voyager 2 as it passed Neptune? This problem also uses the d = v t formula; once again we need to solve for the length of time t. The distance d to Neptune can be found in our textbook. The speed of radio waves is the speed of light c. t = d / v = (4.5 x 1012 m) / (3 x 108 m/s) = 15,000 s = 4.2 hr 10. Use the orbital radius and orbital period of Charon to calculate the mass of the Pluto-Charon system. Hints: Express the orbital radius in meters and the period in seconds. Then see By the Numbers 4-1. 10. 1.5 X 10 22 kg Critical Inquiries for the Web 1. If you lived on the surface of Pluto and looked into the sky to observe Charon, what phases would you see? Hint: Be sure to consider your location on the planet when answering this question. At other times in its orbit, Charon will depict crescent and gibbous phases repeating every 153 hours or so, never achieving full phase or new moon. http://quest.arc.nasa.gov/hst/journals/journal21.html The week before last, we asked: Pick two spots on Pluto (other than the poles) that are on opposite hemispheres. From these two vantage points, describe the phases of the moon, Charon, that you would see. (Hint: there are two periods of time to worry about, one long and one short.) 1.Let's pick two spots on Pluto. To make life easy, both locations are on the equator, one at 0 degrees longitude (Pluto's equivalent of Greenwich, England), and one at 180 degrees longitude (Pluto's equivalent of the international date line). The trick here is knowing that the length of Pluto's day is exactly the same as Charon's orbital period. That means Pluto always presents the same face toward Charon and Charon always presents the same face toward Pluto. This is just what our own Moon does. What is different, is that standing on Pluto, Charon does not appear to move relative to the horizon. That means you never see a moonrise from Pluto's surface. This also means that over half the planet, you can never see Charon. The longitude system is defined such that 0 degrees longitude is in the center of the hemisphere that can see Charon. So, from 180 degrees you wouldn't see any phases at all. The rest of the answer will deal just with the side of Pluto where you can see Charon. 2.The quickest change in the phase of Charon happens during one orbit of Charon around Pluto. This orbit takes about 6.4 days. Let's consider the time when the Sun is directly over the equator. This time of "year" coincides with the start of spring. At "new" moon, Charon would be directly in front of the Sun and we'd see an eclipse. Then as Charon moves in it's orbit it would progress through the same phases that our own moon exhibits every month, taking only 6.4 days to complete the cycle. 3.Another longer term cycle also affects the phases and that is Pluto's orbit around the sun. Since Pluto's rotation axis is tipped on its side, it has very pronounced seasons. Unlike our Moon, Charon also undergoes the same exact seasons. The only time you'd see a completely full moon would be at the start of spring or fall. At the start of summer or winter, just over half the surface would be illuminated at its fullest, either the north or south polar regions depending on the season. To describe the phase of Charon from Pluto requires a knowledge of 1.Charon's orbit around Pluto 2.Pluto's orbit around the Sun 3.the orientation of Pluto in orbit Remember, that "phase" describes the amount of illumination from the Sun, so we need to determine the relative positions of the Sun and Charon as seen from a point on the surface of Pluto. Pluto's orbital is now relatively well known, although future improvements are likely since Pluto has now been observed since its discovery in 1930 for less than one third of its orbit around the sun (one Pluto year equals 248 earth years). Its orientation has been determined from the knowledge of the orbit of its moon, Charon, which has a period of 153 hours 18 minutes (6.38722 days) around Pluto. Charon is believed to be in synchronous rotation around Pluto, meaning that Pluto must also rotate about itself in the same period, 6.38722 days. Further, Charon's orbit is believed to mark the equatorial plane of Pluto, i.e. the spin axis of Pluto is parallel to that of Charon. This direction can be described by the location of Pluto's 'pole' position, or the positive spin axis which is at 311.63 degrees right ascension and 4.18 degrees declination rightascension and declination are astronomical equivalents of longitude and latitude in the equatorial co-ordinate system. But exactly where is this direction? Unless you can visualize RA and Dec values into position in the sky, here is a clue. Between 1985 and 1990, we know that earth was in the plane containing the orbital plane of Charon (and hence Pluto). We need to know to see how the phase of Charon would appear from Pluto. Since the orientation of the spin axes does not change as Pluto and Charon move in their respective orbits, it is clear that at some point in future, roughly a quarter of the orbital period from about 1985, or, around 2050 (=1987+248/4), the equatorial plane would be normal (90 degrees) with the direction to the Sun. At that point, from any point on Pluto, the phase of Charon would be half moon, continuously, from either the north or the south hemispheres of Pluto, and in fact, the entire northern hemisphere of Pluto will be in continuous sunlight and the southern hemisphere will be in continuous darkness (ignoring for the moment the orbital inclination of Pluto, which is actually substantial, 17 degrees). This situation will change when Pluto changes its orbital position enough so that its spin axis is sufficiently away from the direction to the sun from Pluto, until half an orbital period later, when it will recur, except that the dark and lit hemispheres of Pluto will be switched. In between, roughly half an orbital period from 1985-1990, the earth will pass again through the Charon orbital plane, recreating the mutual eclipse events. At that time, Charon will undergo a complete change between new moon and full moon every 153 hours. Of course, Charon and solar eclipses will also occur occasionally, At other times in its orbit, Charon will depict crescent and gibbous phases repeating every 153 hours or so, never achieving full phase or new moon. 2. What factors caused Voyager 2 to see a bland atmosphere when it encountered Uranus in 1986? Given these cir- cumstances, would images of the Uranian atmosphere taken by a space probe arriving at Uranus in 2006 be similar to those taken in 1986, or would there be signif- icant differences? Their would be significant differences. 3. Imagine what you'd think if you had been the first per- son ever to see Saturn through a telescope. When Gali- leo first observed Saturn in 1610, he did not recognize that it was a ringed planet. It was many years later be- fore the strange apparition of Saturn was finally attrib- uted to a ring structure. Search the Web for information on historical observations of Saturn, summarize the ob- servations of Galileo and others, and determine who was first to recognize what he saw as a ring. Why do you suppose it took so long to understand that Saturn is a ringed planet? http://www.seds.org/nineplanets/nineplanets/saturn.html Saturn has been known since prehistoric times. Galileo was the first to observe it with a telescope in 1610; he noted its odd appearance but was confused by it. Early observations of Saturn were complicated by the fact that the Earth passes through the plane of Saturn's rings every few years as Saturn moves in its orbit. A low resolution image of Saturn therefore changes drastically. It was not until 1659 that Christiaan Huygens correctly inferred the geometry of the rings. Saturn's rings remained unique in the known solar system until 1977 when very faint rings were discovered around Uranus (and shortly thereafter around Jupiter and Neptune). Exploring The Sky 1. Zoom in on Jupiter and observe the orbital motion of its moons. Hint: Under Tools choose Time Skip and then Tracking Setup. Lock on Jupiter. Then use the Time Skip buttons to make the moons move around their orbits. 2 Pallas object or moon Distance from Earth: 1.791765 astronomical units. Distance from Sun: 2.136902 astronomical units. Heliocentric: l:127.8175 b:-26.3676 r:2.1369 Magnitude: 8.3 Magnitude: 8.3 RA: 06h 25m 57.5s Dec: -0843'26" RA: 06h 23m 04.6s Dec: -0841'17" (Epoch 2000) Azm: 32759'06" Alt: -5509'29" Rise: 12:57 Transit: 18:29 Set: 00:06 From Cursor position: Angular separation: 12803'08" 3 Juno Distance from Earth: 2.609270 astronomical units. Distance from Sun: 1.992490 astronomical units. Heliocentric: l:71.7824 b:-12.8254 r:1.9925 Magnitude: 9.7 Magnitude: 9.7 RA: 06h 12m 50.6s Dec: +1338'11" RA: 06h 09m 17.8s Dec: +1339'11" (Epoch 2000) Azm: 10046'40" Alt: +3343'09" Rise: 02:19 Transit: 09:08 Set: 15:56 From 2 Pallas: Angular separation: 2236'05" Position angle: +35116' 2. Calculate the mass of Jupiter from your own observa- tions of the orbital period and orbital radius of Jupiter's moons. Hint: See By the Numbers 4-1. 3. Repeat Activities 1and 2 above for Saturn. 4 Vesta Distance from Earth: 1.433816 astronomical units. Distance from Sun: 2.315847 astronomical units. Heliocentric: l:-23.6152 b:-5.6685 r:2.3158 Magnitude: 6.5 Magnitude: 6.5 RA: 23h 48m 14.4s Dec: -1117'01" RA: 23h 45m 00.9s Dec: -1137'53" (Epoch 2000) Azm: 22417'19" Alt: +2743'27" Rise: 21:16 Transit: 02:40 Set: 08:08 From 3 Juno: Angular separation: 9831'30" 3 Juno Distance from Earth: 2.609270 astronomical units. Distance from Sun: 1.992490 astronomical units. Heliocentric: l:71.7824 b:-12.8254 r:1.9925 Magnitude: 9.7 Magnitude: 9.7 RA: 06h 12m 50.6s Dec: +1338'11" RA: 06h 09m 17.8s Dec: +1339'11" (Epoch 2000) Azm: 10046'40" Alt: +3343'09" Rise: 02:19 Transit: 09:08 Set: 15:56 From 4 Vesta: Angular separation: 9831'30" Saturn Rise: 3:17 AM on 8/7/62 Transit: 10:39 AM on 8/7/62 Set: 6:02 PM on 8/7/62 RA: 06h 44m 58.8s Dec: +2220'42" Azm: 8729'58" Alt: +3325'12" (with refraction: +3326'41") Phase: 99.897%, Apparent magnitude: -0.31 Heliocentric ecliptical coordinates: l: 9642'42.3" b: -0044'43.9" r: 9.031343 Geometric geocentric ecliptical coordinates: l: +10023'49" b: -0041'02" r: 9.844161 Mean geometric ecliptical coordinates: l: +10023'42" b: -0041'03" r: 9.844139 True equatorial coordinates: RA: 06h 44m 59s Dec: +2220'42" Physical Data Latitude of the Earth B: -25.564 Latitude of the Sun Bd: -26.105 Position angle: 354.167 Semi-major axis of rings: 38.13" Semi-minor axis of rings: 16.45" Apparent equatorial diameter: 16.8 Apparent polar diameter: 13.7 From 3 Juno: Angular separation: 1134'46" Position angle: +3945' page 397 Chapter 18 Worlds of the Outer Solar System The Four Bases A purple color for Adenine c Cytosine G green for Guanine T dark brown for Thymine Review Questions 1. What do Widmanstatten patterns tell us about the history of iron meteorites? When iron meteorites are sliced open, polished, and etched with nitric acid, they reveal regular bands called Widmanstatten patterns. The patterns arise from crystals of nickel-iron alloys that have grown very large, indicating that the meteorite cooled from a molten state no faster than a few de- grees per million years. Explaining how iron mete- orites could have cooled so slowly will be a major step in analyzing their history. 2. What do chondrules tell us about the history of chon- drites? Stony meteorites called chondrites have chemical compositions that resemble a cooled lump of matter from the sun with the volatile gases removed. Al- though there are many kinds of chondrites, most con- tain chondrules, rounded bits of glassy rock ranging from microscopic to as big as a pea. The origin of chondrules is unknown, they appear to have formed as droplets of molten rock that cooled and hardened rapidly. 3. Why are there no chondrules in achondritic meteorites? Stony meteorites of the third type are called achon- drites because they contain no chondrules. They also lack volatiles and appear to have been subjected to in- tense heat that melted chondrules and drove off vola- tiles. 4. Why do astronomers refer to carbonaceous chodrites as unmodified? The carbonaceous chondrites generally contain both chondrules and volatile compounds, including significant amounts of carbon. Heating would have modified and driven off these fragile compounds. The carbonaceous chondrites are, along with certain kinds of chondrites, among the least modified bodies in our solar system. The meteorites carry hints about the origin of our solar system., they also carry an older secret. Among the smallest grains in meteorites are specks of miner- als whose abundance of isotopes brands them as star dust-grains of interstellar matter that predate our solar system. 5. How do observations of meteor showers reveal one of the sources of meteoroids? We can find evidence of the ori- gin of meteors through one of the most pleasant observations in as- tronomy. We can observe a meteor shower, a display of meteors that are clearly related in a common ori- gin. For example, the Perseid meteor shower occurs each year in August, and on that night we might see as many as 60 meteors an hour. On any clear, moonless night of the year we might see 5 to 15 meteors per hour. 6. How can most meteors be cometary if all meteorites are asteroidal? Cometary fragments are very fragile, easily vaporized. They easily burn up in the Earth's atmosphere, so they easily form meteors and don't usually reach the ground. Asteroid fragments are more substantial, rockier, and less easily vaporized. They make it to the ground as meteorites more easily than cometary fragments. Meteors in a shower are debris left behind as a comet's icy nucleus vaporizes. The rocky and metallic bits of matter spread along the comet's orbit. If Earth passes through such material, we see a meteor shower. Studies of meteors show that most of the meteors we see on any given night whether or not there is a shower are produced by tiny bits of debris from comets. 7. Why do we think the asteroids were never part of a planet? The total mass of the asteroids is too small to have formed a planet. Furthermore if they had formed a planet at one time, it is hard to imagine what would have later provided the great amount of energy needed to break apart the supposed planet to form the asteroids. Thousands of asteroids are known, most are quite small. Asteroids are distant objects too small to study in de- tail with Earth-based telescopes. The vast majority of meteorites must have originated in the asteroids which are tiny rocky worlds. The irregular shape of asteroids. 8. What evidence do we have that the asteroids are fragmented? The NEAR spacecraft for example, has orbited the asteroid Eros, which is a 34-km-long fragment of rock. The asteroids are not only irregular in shape also are marked by craters, the marks of violent collisions in their past. Asteroids are clearly rocky and have low densities, and Mathilde asteroid is a collection of fragments only loosely held together by their mutual gravity. Asteroids have a history filled with collisions. 9. What evidence do we have that some asteroids have had active surfaces? Some asteroids are heavily cratered. Apparently, only the largest asteroids have enough gravity to overcome the strength of the rock they are made from and pull themselves into spherical shape. 11. How is the composition of meteorites related to the for- mation and evolution of asteroids? Meteorites are basalts typical lava flows. Chemical and crystalline structure show once was a part of a molten lava flow on Vesta. The compositions of the meteorites. They are fragments of planetesimals the largest of which developed molten cores, differentiated may have suffered lava flows on their surfaces, and then cooled slowly.The largest asteroids we see today may be nearly unbroken planetesimals. The rest are just fragments produced by 4.6 billion years of cllisions. Asteroid Vesta diameter 525 km it's surface is patchy, and spec- tra of these regions suggest solidified lavas. a giant crater 87 percent of Vesta's diameter. 12. What is the difference between a type I tail and a type II tail? A type I tail is made of ionized gas released by the comet. It is long, narrow, and blue-white. Since it is affected by magnetic fields, it is sometimes bent or kinked by the Sun's magnetic field and solar wind. A type II tail is made of dust released by the comet. It is broader, sometimes shorter, and yellow-white. It may gently curve. A type I tail or ion tail is produced by ionized gas carried away from the nucleus by the solar wind. The spectrum of an ion tail is an emission spectrum. The atoms are ionized by the ultraviolet light in sunlight. The wisps and kinks in ion tails are produced by the magnetic field embedded in the solar wind. A type II tail or dust tail is produced by dust from the vaporizing ices of the nucleus. The dust is pushed gently outward by the pressure of sunlight, and it reflects an absorption spectrum, the spectrum of sunlight. The dust is not affected by the magnetic field of the solar wind, so dust tails are more uniform than ion tails. Dust tails are often curved because the dust particles follow their individual orbits around the sun once they leave the nucleus. 13. What evidence do we have that cometary nuclei are rich in ices? Spectra of ion tails reveal atoms and ions such as H2O, CO2, CO, H, OH, O, S, C, and so on. These are released by the vaporizing ices or produced by the breakdown of those molecules. Some gases, such as hydrogen cyanide HCN, must be formed by chemical reactions. The nucleus of a comet is a small fragile lump of porous rock containg ices of water, carbon dioxide, ammonia, and so on. Comet nuclei can be 10 to 100 km in diameter. The coma of a comet is the cloud of gas and dust that surrounds the nucleus. It can be over 100,000 km in diameter bigger than the sun. 14. Why do short-period comets tend to have orbits near the plane of the solar system? About 100 or so of the 600 well-studied comets have orbits with periods less than 200 years. These short-period comets follow orbits that lie within 30 degrees of the plane of the solar system, and most revolve around the sun counterclockwise the same direc- tion the planets orbit. Comet Halley with a period of 76 years is a short-period comet. Comets cannot survive long before the heat of the sun drives away their ices and reduces them to inac- tive bodies of rock and dust. 100 to 1000 orbits around the sun. 15. How did the bodies in the Kuiper belt and the Oort cloud form? It is thought that in the early days of the solar system, great numbers of icy planetesimals formed from the orbit of Saturn out to several hundred astronomical units. Many of these planetesimals were swept up by the Jovian protoplanets; the planetesimals are now part of the Jovian planets. Many others of these icy planetesimals were ejected to the distant reaches of the solar system by close encounters with the Jovian planets. These planetesimals now form the Oort cloud. Some of the icy planetesimals remain where they were formed long ago; these form the Kuiper belt. Comet tails point generally away from the sun, their precise direction depends on the flow of the solar wind and the orbital motion of the nucleus. Discussion Questions 1. Futurists suggest we may someday mine the asteroids for materials to build and supply space colonies. What kinds of materials could we get from asteroids? Hint: What are S-, M-, and C-type asteroids made of? M-type, S-type and C-type. M-type is a iron asteroid while S-type is a combination of stony and iron. C-type asteroids are made of carbon chondrite and are the "chondrite" asteroids described above. Comets are merely an asteroid with a tail. 2. If cometary nuclei were heated by internal radioactive decay rather than by solar heat, how would comets dif- fer from what we observe? comets are icy cold rocky material. comets might be made up of radioactive decay with uranium metal, and other radioactive decay particles. 3. From what you know now, do you think the government should spend money to locate near-Earth asteroids? How serious is the risk? Yes since their might be such an near-Earth asteroid more likely to crash into planet Earth someday in the future. Problems 1. Large meteorites are hardly slowed by Earth's atmo- sphere. Assuming the atmosphere is 100 km thick and that a large meteorite falls perpendicular to the surface, how long does it take to reach the ground? Hint: About how fast do meteoroids travel? 2. What is the orbital velocity of a meteoroid whose aver- age distance from the sun is 2 AU? Hint: Find the or- bital period from Kepler's third law. See Table 4-1 . 2. 21.1 km/s page 415 Chapter 19 Meteorites, Asteroids, and Comets 3. If a single asteroid 1 km in diameter were fragmented into meteoroids 1 m in diameter, how many would it yield? Hint: Volume of sphere = 4 ttr 3. 3 This problem requires us to think and not simply plug into a pre-existing formula. The volume VA of the asteroid is to be broken up into N meteoriods, each with volume VM. Therefore, N = VA / VM = (4 π RA3/3) / (4 π RM3/3) = (RA/RM)3 = (1000 m / 1 m)3 = 109 = one billion pieces. We have assumed that each volume is spherical. 4. What is the orbital period of a typical asteroid? Hint: Use Kepler's third law. See Table 4-1. 4. 5.2 years 5. If half a million asteroids each 1 km in diameter were assembled into one body, how large would it be? Hint: Volume of sphere = 4 tt r 3. 3 6. What is the maximum angular diameter of Ceres as seen from Earth? Could Earth-based telescopes detect sur- face features? Could the Hubble Space Telescope? Hint: See By the Numbers 3-1. 6. 0.77 seconds of arc 7. If the velocity of the solar wind is about 400 km/s and the visible tail of a comet is 10 8 km long, how long does it take an atom to travel from the nucleus to the end of the visible tail? Here is another problem for the distance formula d = v t. We need to solve for the time t. t = d / v = (108 km) / (400 km/s) = 250,000 s = 2.9 days 8. If you saw Comet Halley when it was 0.7 AU from Earth and it had a visible tail 5 long, how long was the tail in kilometers? Suppose that the tail was not perpendic- ular to your line of sight. Is your answer too large or too small? Hint: See By the Numbers 3-1. 8. 9.2 X 10 6 km 9. What is the orbital period of a cometary nucleus in the Oort cloud? What is its orbital velocity? Hints: Use Kepler's third law. The circumference of a circular or- bit = 2 tt r. 10. The mass of an average comet's nucleus is about 1 0 12 kg. If the Oort cloud contains 200 X 10 9 cometary nuclei, what is the mass of the cloud in Earth masses? Hint: Mass of Earth = 6 X 10 24 kg. 10. 0.033 Earth Masses This is similar to problem 3, however in this problem we are working with masses not volumes or radii. The total mass M is equal to the mass m of each nucleus times the number N of cometary nuclei. M = m N = (1012 kg) (200 x 109) = 2 x 1023 kg. To convert the answer to units of "Earth masses", multiply the answer by (ME)/(6 x 1024 kg). M = (2 x 1023 kg) (ME)/(6 x 1024 kg) = 0.033 ME This answer is proobably too low for the mass M of the Oort cloud, since the mass of an average cometary nucleus is probably more that 1012 kg. Critical Inquiries for the Web 1. Some nights are better for looking for meteors than oth- ers see Table 19-1. We know these showers are associ- ated with comets, but how are these associations made? There are several sites on the Internet that provide in- formation on meteor showers, including historical data and information on parent comets. Pick a shower whose parent comet is known, and summarize how we came to know that the meteors and the comet are related. http://liftoff.msfc.nasa.gov/academy/space/solarsystem/meteors/Showers.html Many meteor showers can be predicted, as they repeat every year when the earth passes through the path of a comet. The bits of debris left behind by the comets, most no larger than a grain of sand, create a spectacular light show as they enter the earth's atmosphere. comet Tempel-Tuttle taken on February 19, 1998. A tail can be seen extending in the anti-solar direction. The comet was 0.98 AU from the sun and 1.22 AU from the Earth. http://www.lowell.edu/users/farnham/tt/index.html http://www.lowell.edu/users/farnham/tt/shower.html Meteors (or shooting stars) are very different from comets, although the two can be related. A Comet is a ball of ice and dirt, orbiting the Sun (usually millions of miles from Earth). As the ices in the nucleus are heated and vaporized by the Sun, gas escapes, taking dust particles along with it. The dust and gas are pushed away from the Sun by the solar wind and radiation pressure, producing the comet's tail (which always points away from the Sun, by the way). Large particles (the size of a grain of sand or larger) take a long time to be pushed around, so they remain in nearly the same orbit as the comet for months or even years, forming a large "cloud" of dust. A Meteor on the other hand, is a grain of dust or rock (see where this is going) that burns up as it enters the Earth's atmosphere. The brief trail that is seen is the material that has burned off and is still glowing as it cools. Most meteors are only about the size of a grain of sand, and burn up completely in the atmosphere. Once in a while, however, a large meteor will survive and hit the ground (at which time it is called a meteorite). A Meteor Shower takes place when the Earth crosses one of the "clouds" of dust that is in the comet's orbital plane. The duration and intensity of the shower is determined by the size and density of the cloud of dust. If the cloud is widely spread out, then the shower could be seen as a few meteors per hour for several days. On the other hand, if the cloud is very small and dense, then the shower will consist of thousands of meteors raining down in only a few minutes (which is classified as a Meteor Storm ). meteor showers, including historical data and information on parent comets 2. The chances are small that you will be killed by an aster- oid impact, but if there are objects out there that astron- omers are not aware of whose orbits intersect Earth, we could be in for a surprise one day. Look for information on the LONEOS project and other investigations into near-Earth asteroids. How many such objects have been discovered? What is the record for closest known pas- sage of an asteroid to Earth? http://asteroid.lowell.edu/asteroid/loneos/loneos_disc.html http://asteroid.lowell.edu/asteroid/loneos/loneos1.html The Lowell Observatory Near-Earth-Object Search (LONEOS) is an ongoing search for asteroids and comets that can approach our planet. These bodies, collectively called near-Earth objects (NEOs), have bombarded Earth throughout history, sometimes with devastating consequences. A mass extinction of more than half the species on Earth, including the dinosaurs, about 65 million years ago was likely precipitated by the impact of a 15 kilometer (km) asteroid or comet in theYucatan peninsula. There are thought to be roughly 1,000 near-Earth asteroids (NEAs) larger than 1 km. As of 2001 March, about 450 have been found. One kilometer is considered the threshold size for an asteroid that would have globally catastrophic consequences if it hit Earth. Exploring The Sky 1. Of the five brightest asteroids, which has the most in- clined orbit? Hint: Use Filters in the View menu to turn off everything, but the sun, stars, ecliptic, and minor planets. You can use Tracking Setup under Time Skip in the Tools menu to lock onto an object and follow it along the ecliptic as time passes. start left lower corner go up toward center and down to lower right corner while following. 2. Do asteroids go through retrograde motion? Hint: Use Filters in the View menu to turn off the stars and turn on the Equatorial Grid. See Activity 1 above. yes 3. Of the five brightest asteroids, which has the most el- liptical orbit? Hint: Use 3D Solar System Mode in the View menu and watch objects orbit the sun. 9P/Tempel 1 Distance from Earth: 5.357381 astronomical units. Distance from Sun: 4.447667 astronomical units. Heliocentric: l:87.5737 b:3.3968 r:4.4477 Magnitude: 25.3 Magnitude: 25.3 RA: 06h 12m 59.6s Dec: +2613'32" RA: 06h 12m 46.4s Dec: +2613'36" (Epoch 2000) Azm: 23401'18" Alt: +6935'59" Rise: 03:46 Transit: 11:25 Set: 19:05 From Cursor position: Angular separation: 9637'19" Position angle: +6331' 4. Of the comets shown, which has the smallest orbit? Hint: Use 3D Solar System Mode in the View menu and watch objects orbit the sun. 9P/Tempel 1 Distance from Earth: 5.357381 astronomical units. Distance from Sun: 4.447667 astronomical units. Heliocentric: l:87.5737 b:3.3968 r:4.4477 Magnitude: 25.3 Magnitude: 25.3 RA: 06h 12m 59.6s Dec: +2613'32" RA: 06h 12m 46.4s Dec: +2613'36" (Epoch 2000) Azm: 23401'18" Alt: +6935'59" Rise: 03:46 Transit: 11:25 Set: 19:05 From 10P/Tempel 2: page 416 Chapter 19 Meteorites, Asteroids, and Comets Chapter 20 Life on Other Worlds page 437 Review Questions 1. If life is based on information, what is that information? Life is based on information that contains the directions for the processes of duplication and preservation. If the information handed down by a life form was always the same, every generation would be identical to ever preceding generation. A life form of this type would soon become extinct. Since the information handed down is always the same, the life form cannot adapt to changes in environment. When changes come that are unfavorable, the organism will no longer be able to survive. 2. What would happen to a life form if the information handed down to offspring was always the same? How would that endanger the future of the life form? 3. How does the DNA molecule produce a copy of itself? How is DNA reproduced? Every organism must produce copies of itself in order to pass on genetic information to its young before it dies...so must the cells in that organism, as well. How does the cell make a copy of itself and how is DNA copied as a part of that process? What is the three-dimensional structure of DNA? The components which make up DNA come together in a specific way to form the DNA molecule. How do these components relate to one another to form DNA? The DNA molecule reproduces itself by splitting down the middle to produce two separate strands. Sugars and bases quickly combine to form the other halves of each of the separate strands, thereby forming two DNA molecules where only one had existed. Any mutation that leaves the organism with a clear advantage over members of its species will have an easier time existing and reproducing. This increases that organism's chance to pass the mutation on to its offspring and for the mutation to slightly alter the species. The example cited in the text is of a rabbit with slightly stronger teeth, which would allow the rabbit to forage on a larger variety of plants. Another example might include a mutation that allows an animal to better blend with its surroundings. In prey animals, it would make it less likely to be killed by a predator, and in predators, it would make it easier for them to get closer to and catch their prey. Other examples might be improved sight, hearing, sense of smell, the ability to better tolerate heat or cold in temperate climates, the ability to store larger amounts of food or water in body tissue, increased mobility, etc. 4. Give an example of natural selection acting on new DNA patterns to select the most advantageous characteristics? The oldest fossils known are of marine life. These fossils date back 500 to 600 million years, while the earliest fossils of land creatures are 400 million years old. Studies show that microscopic marine fossils may date back as far as 3.5 billion years. 5. Why do we believe that life on Earth began in the sea? 6. Why do we think that liquid water is necessary for the origin of life? The presence of liquid water is important for the evolution and sustaining of life because it provides the medium to transport nutrients and wastes within the organism. Chemical evolution of the build-up of molecules of certain types because some reactions are more likely to occur than others, and some molecules are more stable than others. In this way certain molecules will be more abundant than other molecules. Chemical evolution then is the combining of two or more parts to form something different from any of the combined parts. Biological evolution occurs when a molecule can reproduce a c opy of itself. Biological evolution is the splitting of a molecule to form two molecules the same as the original. 7. What is the difference between chemical evolution and biological evolution? Chemical is a liquid Biological 8. What was the significance of the Miller experiment? The Miller experiment showed that complex organic compounds form easily when basic chemicals and energy are present. Complex organic compounds such as amino acids are necessary for life to begin and for life to be sustained. Simply put, intelligence increases our options for survival. Intelligence allows an organism to see alternative ways for survival. An intelligent organism can learn from its mistakes. An organism that can learn is more likely to avoid fatal situations and hence is more likely to reporduce. It is important to note that evolution does not claim that what was learned will be passed on, only the ability for the offspring to learn. The intelligence is passed on, not the specific knowledge. 9. How does intelligence make a creature more likely to survive? 10. Why are upper-main-sequence stars unlikely sites for intelligent civilizations? Science believes that it takes billions of years of somewhat stable conditions for life to develop. Upper main sequence stars don't live that long. So conditions around an upper main sequence star are not stable long enough for life to develop. The most serious problem with upper-main-sequence stars is the amount of time that they remain on the main-sequence. Life takes several billion years to develop. Stars on the upper main sequence have lifetimes that are less than a billion years. Therefore the energy output of these stars is not stable for a long enough period of time to allow life to develop. 11. Why do we suspect that travel between stars is nearly impossible? Even if we could travel at nearly the speed of light, the travel times to most stars is so long that the trip would be nearly impossible. Stars are at greater distances from each other. Travel between stars is nearly impossible because the distances are too great and because nature has imposed a very real speed limit that we cannot exceed. Nothing can travel faster than the speed of light, and human travel cannot be expected to exceed even a small fraction of the speed of light. Therefore, the time it would take to travel even to the nearest star, Proxima Centauri (about four light years away) would take much more than four years. Traveling at the incredible speed of 3000 km/sec, it would take 400 years —many generations of space travelers— to reach the nearest star. 12. How does the stability of technological civilizations af- fect the probability that we can communicate with them? The technological stability of a life form is important in determining the length of time over which it is capable of communicating. A life form that can communicate but that possesses this ability for only a few years has little chance of being detected. The signal these life forms send out must reach another life form with advanced enough technology to receive and process the signal. The length of time that the signal is sent increases the odds that it will reach a communicative life form. 13. What is the water hole, and why would it be a good place to look for other civilizations? The water hole is a part of the spectrum of radio wavelengths between an emission line of hydrogen at a wavelength of 21 cm and an emission line of OH at a wavelength of 18 cm. It is called the water hole because H and OH make water, and because extraterrestrial civilizations might "gather at the water hole" to communicate use wavelengths between 18 and 21 cm. The water hole is a band of frequencies in the radio portion of the electromagnetic spectrum that lie between the 21-cm line of neutral hydrogen and the 18-cm line of OH. It is called the water hole because H and OH can combine to form H2O, water. It is a good place to look for radio signals from other civilizations because it is not cluttered by natural radio emission from other sources and it is close to the 21-cm radiation of neutral hydrogen. Additionally, water is very important to the existence of life, and this gap in the spectrum between H and OH could easily be identified by other possible life forms with technology. Discussion Questions 1. What would you change in the Arecibo message if hu- manity lived on Mars instead of Earth? Explanation: What are these Earthlings trying to tell us? The above message was broadcast from Earth towards the globular star cluster M13 in 1974. During the dedication of the Arecibo Observatory - still the largest radio telescope in the world - a string of 1's and 0's representing the above diagram was sent. This attempt at extraterrestrial communication was mostly ceremonial - humanity regularly broadcasts radio and television signals out into space accidentally. Even were this message received, M13 is so far away we would have to wait almost 50,000 years to hear an answer. The above message gives a few simple facts about humanity and its knowledge: from left to right are numbers from one to ten, atoms including hydrogen and carbon, some interesting molecules, DNA, a human with description, basics of our Solar System, and basics of the sending telescope. Several searches for extraterrestrial intelligence are currently underway, including Project Phoenix and one where you can use your own home computer. 2. What do you think it would mean if decades of careful searches for radio signals for extraterrestrial intelligence turned up nothing? It was just an attempt to see if we would pick up any radio signals. Perhaps if there is other life they maynot be involved in radio signals or are much more advanced or unadvanced then we are. They may even be transmitting signals in some other unknown form. Problems 1. A single human encloses about 1.5m of DNA con- taining 4.5 billion base pairs. What is the spacing be- tween these base pairs in nanometers? That is, how far apart are the rungs on the DNA ladder? 2. If we represent the history of the Earth by a line 1 m long, how long a segment would represent the 400 million years since life moved onto the land? How long a seg- ment would represent the 3-million-year history of hu- man life? This is a problem in which we must convert times into distances on a one meter long line. The conversion factor is (1 m)/(4.5 billion years) since 1 m represents the age of the earth, which is 4.5 billion years. In the first part we want the length corresponding to 400 million years. (400 million yr)(1 m)/(4.5 billion years) = 0.09 m = 9 cm In the second part we want the length corresponding to only 3 million years. (3 million yr)(1 m)/(4.5 billion years) = 6.7 x 10-4 m = 0.67 mm 2. 8.9 cm 0.67 mm 3. If a human generation the time from birth to childbear- ing, is 20 years, how many generations have passed in the last million years? This question asks how many 20 year generations divide into one million years. This is just a division problem. (1,000,000 yr)/(20 yr) = 50,000 generations 4. If a star must remain on the main sequence for at least 5 billion years for life to evolve to intelligence, how massive could a star be and still harbor intelligent life on one of its planets? See By the Numbers 9-1. 4. about 1.3 solar masses Chapter 20 5. If there are about 1.4 X 10 -4 stars like the sun per cubic light-year how many lie within 100 light-years of Earth? The Volume of a sphere is 4 ttr 3 3 6. Mathematician Karl Gauss suggested planting forests and fields in a gigantic geometric proof to signal to pos- sible Martians that intelligent life exists on Earth. If Mar- tians had telescopes that could resolve details no smaller than 1 second of arc, how large would the smallest ele- ment of Gauss's proof have to be? See By the Numbers 3-1. 6. 380 km 7. If we detected radio signals with an average wavelength of 20 cm and suspected that they came from a civiliza- tion on a distant planet, roughly how much of a change in wavelength should we expect to see because of the orbital motion of the distant planet? See By the Numbers 6-2. 8. Calculate the number of communicative civilizations per galaxy from your own estimates of the factors in Table 20-1 . Critical Inquiries for the Web 1. The popular movie Contact focused interest on the SETI program by profiling the work of a radio astronomer dedicated to the search for extraterrestrial intelligence. Visit Web sites that give information about the movie. SETI programs, and radio astronomy, and discuss how realistic the movie was in capturing how such research is done. 2. Where outside the solar system would you look for hab- itable planets? NASA has increasingly focused its inter- est on this question. Look for information online about programs dedicated to detecting which planets might support life as we know it. What criteria are used to choose targets for the planned searches? What methods will be used to carry out the searches? page 437 chapter 20. Answers for horizons Chapter 1 2. 1.3 s 4. 1.8 X 10 7 km Chapter 2 2. 2800 4. The sun is 400,000 times brighter than the moon. Chapter 3 2. new, full 4. 14 days 6. 690 arc second 8. The eclipse Chapter 5 2. 2.9 cm, radio 4. Each of the keck telescopes 6. 5.8 seconds of arc 8. 13mm 10. 45 cm. Chapter 6 2. 150 nm 4. 2 4 = 16 6. 250 nm 8. B.F.M.K. 10. 0.6 nm Chapter 7 2. 730 km 4. 3.6 times brighter 6. 2.5 billion megatons of TNT 8. 3.9 X 10 9 s= 124 years. Chapter 8 m M d(pc) 7 7 10 11 1 1000 1 -2 40 4 2 25 5. B2 to B4 8. Using Figure 8-7, M d = 400 pc 08V star from figure 8 10. a, c, c, c, d 12. 4.23 X 10 11 m, 6.7 X 10 16 masses. 14. A 4-solar mass main sequence nosity of approximately 12 A 9 solar mass main sequence of approximately 2187 solar 7 solar mass main sequence of 907 solar luminosities. Chapter 9 8. 2000 nm = 2 micrometers 10. 5.6 implies a distance Chapter 10 2. 1.7 ly 4. 32,000 years 6. 1.3 X 10 6 times 8. 980 km/s 10. about 310 million Chapter 11 4. 3 m, 1.1 X 10 - 25 m 6. 8.1 minutes 8. 3.0 X 10 8 m/s, photons Chapter 12 2. 16% 4. 5000 pc 6. 25 pc 8. 20 times 10. 1500 K Chapter 13 2. 2.6 X 10 6 pc 2.6 Mpc 4. 93 km/s 6. 28.6 Mpc 8. 165 million years Chapter 14 2. 7.6 million years 4. 0.024 pc = 4900 AU 6. -28.5 9. 0.3c 10. AA = 77.8 nm, z A2 1.5 parsecs X 3 light years/parsec = 4.5 ly TABLE 1-1. Large Numbers 1 10 0 one 10 10 1 ten 100 10 2 hundred 1,000 10 3 thousand 1,000,000 10 5 million thousand thousand 1,000,000,000 10 9 billion thousand million 1,000,000,000,000 10 12 trillion million million Astronomy Chapter 11 The Formation of the Stars Review Questions 1. What factors resist the contraction of a cloud of inter- stellar matter? If the internal temperature is high enough, pressure pushes outward just enough to balance gravit. 2. Explain four different ways a giant molecular cloud can be triggered to contract. Gravity makes clouds of gas contract. Thermal energy in the gas is present as mo- tion among the atoms and molecules. The second factor that gravity must overcome to make the cloud contract. The third factor is rotation. Everything in the uni- verse rotates to some extent. As the gas cloud begins to contract, it spins more and more rapidly as it con- serves angular momentum. Fourth trigger is the spiral pattern of our Milky Way Galaxy. The spiral arms mabe shock waves. 3. What evidence do we have that (a) star formation is a continuing process? (b) protostars really exist? (c) the Orion region is actively forming stars? Evidence is of observations of star formation. Through the observations at infrared wavelengths to search for hidden protostars. How observations give us insight into the process by which stars form. 4. How does a contracting protostar convert gravitational energy into thermal energy? Whenever a cloud of gas contracts, gravitational en- ergy is converted into thermal energy, and the gas grows hotter. 5. How does the geometry of bipolar flows and Herbig- Haro objects support our hypothesis that protostars are surrounded by rotating disks? Matter flowing into a protostar swirls through a thick disk and, by a process believed to involve magnetic fields, ejects high-energy jets in opposite directions. Observation of these bipolar flows is evidence that protostars are surrounded by disks because only disks could focus the flows into jets. 6. How does the CNO cycle differ from the proton-proton chain? How is it similar? The CNO cycle a hydrogen fusion process that uses carbon, nitrogen, and oxygen as steppingstones. The CNO cycle and the proton-proton chain both fuse hydrogen to make helium, their sensitivity to temperature is dramatic. 7. How does the extreme temperature sensitivity of the CNO cycle affect the structure of stars? The CNO cycle is so critically sensitive to tem- perature that it produces almost no energy at all at temperatures below 16 million Kelvin. The proton- proton chain can make energy at these lower temper- atures, the proton-proton chain makes slightly more energy, the CNO cycle produces a flood of energy. Stars with higher-temperature cores make nearly all of their energy using the CNO cycle while stars with cooler cores make their energy using the proton- proton chain. 8. How does energy get from the core of a star, where it is generated, to the surface, where it is radiated into space? Conduction is an unimportant means of energy flow. Stars can exist only as long as energy can move from their cores to their sur- faces. In the material of which stars are made, energy, can move by conduction, radiation, or convection. 9. Describe the principle of hydrostatic equilibrium as it relates to the internal structure of a star. The gravity-pressure balance that supports the sun is a fundamental part of stellar structure known as the law of hydrostatic equilibrium. It says that, in a stable star like the sun, the weight of the material press- ing downward on a layer must be balanced by the pressure of the gas in that layer. Hydro implies that we are discussing a fluid-the gases of the star. Static implies that the fluid is stable-neither expanding nor contracting. The law of hydrostatic equilibrium can prove to us that the interior of the sun must be very hot. Near the sun's surface, there is little weight pressing down on the gas, so the pressure must be low, implying a low temperature. As we go deeper into the sun, the weight becomes larger, so the pressure, and therefore the temperature, must also increase. 10. How does the pressure-temperature thermostat control the nuclear reactions inside stars? Once nuclear fusion begins, energy flows outward, heats the layers of gas, and stops the contraction. The relationship between pressure and temperature. In a star, the nuclear reactions generate just enough energy to balance the inward pull of gravity. Suppose the nuclear reactions began making too little energy. Then the star would contract slightly, increasing the central temperature and den- sity and increasing the nuclear energy generation. The stability of a star depends on this relation be- tween pressure and temperature. Discussion Questions 1. Ancient astronomers, philosophers, and poets assumed that the stars were eternal and unchanging. Is there any observation they could have made or any line of rea- soning that could have led them to conclude that stars don't live forever? 2. How does hydrostatic equilibrium relate to hot-air-bal- looning? Hot air ballooning is when heat from a gas keeps the balloon up in the sky. Problems 1. If the Horsehead Nebula shown in figure 11-1 is about 1 pc in overall diameter and is 500 pc from Earth, how large does it look in the sky in seconds of arc? in degress? 2. If a giant molecular cloud is 50 pc in diameter and a shock wave can sweep through it in 2 million years, how fast is the shock wave going in kilometers per second? 3. If a giant molecular cloud has a mass of 10 35 kg, and it converts 1 percent of its mass into stars during a single encounter with a shock wave, how many star can it make? Assume the stars each contain 1 solar mass? 4. If a protostellar disk is 200 AU in radius, and the disk plus the forming star contain 2 solar masses, what is the orbital velocity at the outer edge of the disk in kilome- ters per second? 5. If a contracting protostar is five times the radius of the sun and has a temperature of only 2000 K, how lumi- nous will it be? see chapter 9 6. The gas in a bipolar flow can travel as fast as 100 km/s. If the length of the jet is 1 ly, how long does it take for a blob of gas to travel from the protostar to the end of the jet? 7. If a T Tauri star is the same temperature as the sun but is ten timess more luminous, what is its radius? see chapter 9. 8. Circle all of the 1 H and 4 He nuclei in Figure 11-9 and explain how the CNO cycle can be summarized by 4 3 H > 4 He + energy. 9. How much energy is produced when the CNO cycle converts 1 km of mass into energy? Is your answer dif- ferent if the mass is fused by the proton-proton chain? 10. If the Orion Nebula is 8 pc in diameter and has a den- sity of about 600 hydrogen atoms/cm3. what is its total mass? Hint The Volume of a sphere is 4 tt R3. 3 11, The hottest star in the Orion Nebula has a surface tem- perature of 40,000 K. At what wavelength does it radi- ate the most energy? See Chapter 7. 234 and 235 Part 2 The Stars Problems 1. The thickness of the plastic in plastic bags is about 0.001 mm. How many wavelengths of red light is this? 2. Measure the actual wavelength of the wave in Fig-ure 3-1. In what portion of the electromagnetic spec-trum would it belong? 3. Compare the light-gathering powers of a 5-m telescope and a 0.5-m telescope. 5-m telescope 0.5-m telescope 4. How does the light-gathering power of one of the Keck telescopes compare with that of the human eye? (Hint: Assume that the pupil of your eye can open to about 0.8 cm.) 5. What is the resolving power of a 25-cm telescope? What do two stars 1.5 seconds of arc apart look like through this telescope? 25-cm telescope 1.5 seconds of arc apart 6. Most of Galileos telescopes were only about 2 cm in di-ameter. Should he have been able to resolve the two stars mentioned in Problem 5? 7. How does the resolving power of the 5-m telescope compare with that of the Hubble Space Telescope? Why does the Hubble Space Telescope outperform the 5-m telescope? 8. If we build a telescope with a focal length of 1.3 m, what focal length should the eyepiece have to give a magnification of 100 times? 9. Astronauts observing Critical Inquiries for the Web 1. How do professional astronomers go about making ob-servations at major astronomical facilities? Visit sev-eral observatory Web sites to determine the process an astronomer would go through to secure observing time and make observations at the facility. 2. NASA is in the process of completing a fleet of four space-based Great Observatories. (The Hubble Space Telescope is one; what are the others?) Examine the cur-rent state of these missions by visiting their home pages on the Internet. What advantages would these facilities have over ground-based observatories? Exploring TheSky 1. Astronomical telescopes using equatorial mountings must be aligned precisely with the north celestial pole. Locate Polaris and determine how far it is from the north celestial pole. (Hint: Use Reference Lines under the View menu and check Grid under Equitorial. Be sure the spac-ing is set to auto/fine. Then locate the Little Dipper and zoom in on Polaris.) not far from center north top. west. Polaris The North Star SAO 308 GSC 4628:237, HIP 11767, PPM 431, HD 8890, B+88 8 Flamsteed-Bayer: 1-Alpha Ursae Minoris Spectral: F7:Ib-IIv SB **** Data from Hipparcos Catalog **** Proper motion (mas/yr): RA = 44.22, Dec = -11.74 Magnitudes Bt: 2.756, Vt: 2.067 Parallax: 7.560 mas, 132.2751 pc Distance: 431.42 light-years, 27283753.74 astronomical units Magnitude: 1.97 RA: 02h 37m 31.673s Dec: +8917'09.199" RA: 02h 31m 49.084s Dec: +8915'50.794" (Epoch 2000) Azm: 0102'32" Alt: +4728'12" Always above horizon. Transit: 20:46 Position error: 0.60 mas From Cursor position: Angular separation: 0011'28" Position angle: +10700' Go to the Brooks/Cole Astronomy Resource Center (www.brookscole. com/astronomy) for critical thinking exercises, articles, and addi-tional readings from InfoTrac College Edition, Brooks/Coles online student library. 1. The thickness of the plastic in plastic bags is about 0.001 mm. How many wavelengths of red light is this? 1. How do professional astronomers go about making ob-servations The Electromagnetic Spectrum A spectrum is an array of electromagnetic radiation in order of wavelength. We are most familiar with the spectrum of visible light, which we see in rainbows. The colors of the spectrum differ in wavelength, with red having the longest wavelength and violet the short- est, as shown in the visible spectrum at the top of Fig- ure 3-2. The average wavelength of visible light is about 0.0005 mm. We could put 50 light waves end to end Microwave transmissions, used for radar and long-distance telephone communi-cations, for instance, have wavelengths from a few centimeters down to about 1 mm. 3-2 Optical Telescopes Astronomers build optical telescopes to gather light and focus it into sharp images. This requires sophisti-cated optical and mechanical designs, and it leads astronomers to build gigantic telescopes on the tops of high mountains. To begin, we need to understand the terminology of telescopes, but it is more impor-tant to understand how different kinds of telescopes work and why some are better than others. Two Kinds of Telescopes Astronomical telescopes focus light into an image in one of two ways, as shown in Figure 3-3. A lens bends (refracts) the light as it passes through the glass and brings it to a focus to form a small inverted image. A mirror a concave piece of glass with a reflective surface forms an image by reflecting the light. In either case, the focal length is the distance from the lens or mirror to the image formed of a distant light source, such as a star (Figure 3-4). Short-focal-length lenses and mirrors must be strongly curved, and long-focal-length lenses and mirrors are less strongly curved. Grinding the proper shape on a lens or mirror is a deli-cate, time-consuming, and expensive process. Light can be focused in two ways, by a lens or by a mirror. Consequently there are two kinds of astro-nomical telescopes. Refracting telescopes use a large lens to gather and focus the light, whereas reflecting telescopes use a concave mirror (Figure 3-5). This limitation on the amount of information in an image is related to the limitation on the accuracy of a measurement. All measurements have some built-in uncertainty (Window on Science 3-1), and scien-tists must learn to work within those limitations. The third and least important power of a telescope is magnifying power, the ability to make the image bigger. Since the amount of detail we can see is lim-ited by the seeing conditions and the resolving power, very high magnification does not necessarily show us more detail. Also, we can change the magnification by changing the eyepiece, but we cannot alter the tele-scopes light-gathering or resolving power. Compare an astronomers telescope with a biolo-gists microscope. A microscope is designed primar- ily to magnify and thus show us things too small to see. An astronomers telescope solves a different prob-lem. Its primary function is to gather light and show us things too faint to see. Nearly all major observatories are located far from major cities and usually on high mountains. Astrono-mers avoid cities because light pollution, the bright-ening of the night sky by light scattered from artificial outdoor lighting, can make it impossible to see faint objects (Figure 3-10). In fact, many residents of cities are unfamiliar with the beauty of the night sky be-cause they can see only the brightest stars. Astrono-mers prefer to place their telescopes on carefully se-lected high mountains. The air there is thin and more transparent, but, most important, astronomers select Have you ever seen a movie in which the hero magnifies a newspaper photo and reads some tiny detail? It isnt really possi-ble, because newspaper photos are made up of tiny dots of ink, and no detail smaller than a single dot will be visible no matter how much you magnify the photo. In fact, all images are made up of elements of some sort, and that means there is a limit to the amount of detail you can see in an image. In an astronomical image, the resolution is often set by seeing. It is foolish to attempt to see a detail in the image that is smaller than the resolution. This limitation is true of all measure-ments in science. A zoologist might be try-ing to measure the length of a live snake, or a sociologist might be trying to measure the attitudes of people toward drunk driv-ing, but both face limits to the resolution of their measurements. The zoologist might specify that the snake was 43.28932 cm long, and the sociologist might say that 98.2491 percent of people oppose drunk driving, but a critic might point out that it isnt possible to make these measurements that accurately. The resolution of the tech-niques does not justify the accuracy implied. Science is based on measurement, and whenever we make a measurement we should ask ourselves how accurate that measurement can be. The accuracy of the measurement is limited by the resolution of the measurement technique, just as the amount of detail in a photograph is limited by the resolution of the photo. Window on Science 3-1 Resolving Power and the Accuracy of a Measurement Figure 3-10 This composite of satellite images shows the entire Earth at night. Light pollution caused by outdoor lighting is growing rapidly, drowning out the fainter stars, and interfering with our view of the night sky. Of course, light lost into space also wastes electrical power. Many astronomers work with city govern-ments to enact laws that improve lighting on the ground and reduce light scattered into the night sky. (Defense Meteorological Satellite Program/NASA) mountains where the air flows smoothly and is not turbulent. This produces the best seeing. Building an observatory on top of a high mountain far from civi-lization is difficult and expensive, but the dark sky and steady seeing make it worth the effort (Figure 3-11). When we compare telescopes, we should consider their powers. By the Numbers 3-1 shows how to cal-culate the powers of a telescope. This will be useful if you decide to buy a telescope of your own. Buying a Telescope Thinking about how we should shop for a new tele-scope will not only help us if we decide to buy one but will also illustrate some important points about astronomical telescopes. Assuming we have a fixed budget, we should buy the highest-quality optics and the largest-diameter telescope we can afford. Of the two things that limit what we see, optical quality is under our control. We cant make the atmosphere less turbulent, but we should buy good optics. If we buy a telescope from a toy store and it has plastic lenses, we shouldnt expect to see very much. Also, we want to maximize the light-gathering power of our telescope, so we want to pur-chase the largest-diameter telescope we can afford. Figure 3-11 The two 8-meter Gemini telescopes are located atop high moun-tains. One is at 8895 ft in the Chilean Andes, and the other is atop a 13,600-ft volcano in Hawaiian altitude where the air is hardly thick enough for breathing but is ideal for astronomy. Here the northern Gemini dome is seen against star trails circling the north celestial pole. (Gemini Observatory/AURA/NOAO/NSF) 72 Part 1 The Astronomers Sky By the Numbers 3-1 The Powers of a Telescope Light-gathering power is proportional to the area of the telescope objective. A lens or mirror with a large area gathers a large amount of light. The area of a circular lens or mirror of diameter D is (D/2)2 . To compare the relative light-gathering powers (LGP) of two telescopes A and B, we can calculate the ratio of the areas of their objectives, which re-duces to the ratio of their diameters (D) squared. 2 Example A: Suppose we compare a 4-cm tele-scope with a 24-cm telescope. How much more light will the large telescope gather? Solution: 2 6 2 36 times more light Example B: Our eye acts like a telescope with a diameter of about 0.8 cm, the diameter of the pupil. How much more light can we gather if we use a 24-cm telescope? Solution: 2 (30)2 900 times more light The resolving power of a telescope is the angu-lar distance between two stars that are just barely visible through the telescope as two separate im-ages. The resolving power a, in seconds of arc, equals 11.6 divided by the diameter of the tele-scope in centimeters: a  Example C: What is the resolving power of a 10-cm telescope? Solution: a  1.16 seconds of arc If the lenses are of good quality and if the seeing is good, we should be able to distinguish as sepa-rate points of light any pair of stars farther apart than 1.16 seconds of arc. If the stars are any closer together, diffraction fringes blur the stars together into a single image (see Figure 5-8). The magnification M of a telescope is the ratio of the focal length of the objective lens or mirror F o divided by the focal length of the eyepiece F e : M  Example D: What is the magnification of a tele-scope whose focal length is 80 cm used with an eye-piece whose focal length is 0.5 cm? Solution: The magnification is 80 divided by 0.5, or 160 times. F o  F e 11.6  10 11.6  D 24  0.8 LGP 24  LGP eye 24  4 LGP 24  LGP 4 D A  D B LGP A  LGP B Given a fixed budget, that means we should buy a re-flecting telescope rather than a refracting telescope. Not only will we get more diameter per dollar, but our telescope will not suffer from chromatic aberration. We can safely ignore magnification. Department stores and camera stores may advertise telescopes by quoting their magnification, but it is not an important number. What we can see is fixed by light-gathering power, optical quality, and Earths atmosphere. Be-sides, we can change the magnification by changing eyepieces. Other things being equal, we should choose a tele-scope with a solid mounting that will hold the tele-scope steady and allow us to point at objects easily. Computer-controlled pointing systems are available for a price on many small telescopes. A good telescope on a poor mounting is almost useless. We might be buying a telescope to put in our back-yard, but we must think about the same issues astron-omers consider when they design giant telescopes to go on mountaintops. In fact, some of the newest tele-scopes solve these traditional problems in new ways. New-Generation Telescopes For most of the 20th century, astronomers faced a se-rious limitation on the size of astronomical telescopes. Traditional telescope mirrors were made thick to avoid sagging that would distort the reflecting surface, but those thick mirrors were heavy. The 5-m (200-inch) mirror on Mount Palomar weighs 14.5 tons. These tra-ditional telescopes were big, heavy, and expensive. Modern astronomers have solved these problems in a number of ways. Study Modern Astronomical Telescopes on pages 00 and 00 and notice three im-portant advances in telescope design made possible by high-speed computers. First, astronomers can now build simpler, stronger telescope mountings and de-pend on computers to move the telescope and follow the westward motion of the stars as Earth rotates. Second, computer control of the shape of tele-scope mirrors allows the use of thin, lightweight mir-rors either floppy mirrors or segmented mirrors. Lowering the weight of the mirror lowers the weight of the rest of the telescope and makes it stronger and less expensive. Thinner mirrors also cool faster at nightfall and suffer less distortion from uneven ex-pansion and contraction. The third advance is the way astronomers use high-speed computers to reduce seeing distortion caused by Earths atmosphere. Only a few decades ago, many astronomers argued that it wasnt worth building more large telescopes on Earths surface be-cause of the limitations set by seeing. Now a number of new giant telescopes have been built, and more are in development. An international collaboration of astronomers built the Gemini telescopes with 8.1-m thin mirrors (Figure 3-12a). One is located in the northern hemi-sphere and one in the southern hemisphere to cover the entire sky. The European Southern Observatory has built the Very Large Telescope (VLT) high in the remote Andes of northern Chile (Figure 3-12b). The VLT consists of four telescopes with computer-controlled mirrors 8.2 m in diameter and only 17.5 cm (6.9 in.) thick. The four telescopes can work singly or can combine their light to work as one large telescope. Italian and American astronomers are building the Large Binocular Telescope, which carries a pair of 8.4-m mirrors on a single mounting. Around the world astronomers are drawing plans for large telescopes. R E V I E W Critical Inquiry Why do astronomers build observatories at the tops of mountains? Astronomers have joked that the hardest part of build-ing a new observatory is constructing the road to the top of the mountain. It certainly isn't easy to build a large, delicate telescope at the top of a high mountain, but it is worth the effort. A telescope on top of a high mountain is above the thickest part of Earths atmo-sphere. There is less air to dim the light, and there is less water vapor to absorb infrared radiation. Even more important, the thin air on a mountaintop causes less disturbance to the image, and thus the seeing is better. A large telescope with modern, high-quality optics is capable of detecting very small details. It is the seeing that limits the detail that astronomers can detect. It really is worth the trouble to build telescopes atop high mountains where the air is thin, dry, and steady. Astronomers not only build telescopes on mountain-tops; they also build gigantic telescopes many meters in diameter. What are the problems and advantages in building such giant telescopes? meter attached to a telescope. Today, however, most such measurements are made on CCD images. Be-cause the CCD image is easily digitized, brightness and color can be measured to high precision. The Spectrograph To analyze light in detail, we need to spread the light out according to wavelength into a spectrum, a task performed by a spectrograph. We can understand how this works if we reproduce an experiment performed by Isaac Newton in 1666. Boring a hole in his window shutter, Newton admitted a thin beam of sunlight into his darkened bedroom. When he placed a prism in the beam, the sunlight spread into a beautiful spectrum on the far wall. From this Newton concluded that white light was made of a mixture of all the colors. Newton didnt think in terms of wavelength, but we can use that modern concept to see that the light passing through the prism is bent at an angle that de-pends on the wavelength. Violet (short wavelength) bends most, and red (long wavelength) least. Thus, the white light entering the prism is spread into a spectrum (Figure 3-15a). A typical prism spectrograph contains more than one prism to spread the light far-ther and lenses to guide the light into the prism and to focus the light onto a photographic plate. Nearly all modern spectrographs use a grating in place of a prism. A grating is a piece of glass with thousands of microscopic parallel lines scribed onto its surface. Different wavelengths of light reflect from the grating at slightly different angles, so white light is spread into a spectrum and can be recorded, often by a CCD camera. Because astronomers need to measure the precise location of features in a spectrum, most spectrographs can add a comparison spectrum above and below the spectrum of the observed object (Figure 3-15b). The features in this comparison spectrum have known wavelengths carefully measured in the laboratory, and thus the astronomer can use the comparison spectrum as a precise calibration of the wavelengths in the spec-trum being studied. Because astronomers understand how light in-teracts with matter, a spectrum carries a tremendous amount of information (as we will see in the next chapter), and that makes a spectrograph the astrono-mers most powerful instrument. An astronomer re-cently remarked, We dont know anything about an object til we get a spectrum, and that is only a slight exaggeration. Chapter 3 Astronomical Tools 77 Figure 3-14 (a) This digital image has been reproduced as a negative, with the sky light and bright objects dark. The arrow points to a faint galaxy so distant the light took 10 billion years to reach us. (Hyron Spinrad) (b) This digital image of the galaxy NGC1232 has been given false colors to represent different levels of brightness. (CARA) a b R E V I E W Critical Inquiry What is the difference between light going through a lens and light passing through a prism? A refracting telescope producing chromatic aberration and a prism dispersing light into a spectrum are two ex-amples of the same thing, but one is bad and one is good. When light passes through the curved surfaces of a lens, different wavelengths are bent by slightly differ-ent amounts, and the different colors of light come to focus at different focal lengths. This produces the color fringes in an image called chromatic aberration, and thats bad. But the surfaces of a prism are made to be precisely flat, so all of the light enters the prism at the same angle, and any given wavelength is bent by the So far, our discussion has been limited to visual wave-lengths. Now it is time to consider the rest of the elec-tromagnetic spectrum. 3-4 Radio Telescopes Operation of a Radio Telescope A radio telescope usually consists of four parts: a dish reflector, an antenna, an amplifier, and a recorder (Fig-ure 3-16). The components, working together, make it possible for astronomers to detect radio radiation from celestial objects. The dish reflector of a radio telescope, like the mirror of a reflecting telescope, collects and focuses radiation. Because radio waves are much longer than light waves, the dish need not be as smooth as a mir-ror. In some radio telescopes, the reflector may not 78 Part 1 The Astronomers Sky Figure 3-15 (a) A prism bends light by an angle that depends on the wavelength of the light. Short wavelengths bend most and long wavelengths least. Thus, white light passing through a prism is spread into a spec-trum. (b) In this negative image, the bright spectrum of the star is the dark streak running from left to right in the middle of the image. The gaps in the stars spectrum are produced by different gases in the star. Above and below the stellar spectrum are the lines of a comparison spectrum, which astronomers use to calibrate the spectrum of the star and measure wavelengths to high precision. (Caltech) White light Prism Ultraviolet Short wavelengths Infrared Long wavelengths Visible light spectrum Comparison spectrum Stellar spectrum Comparison spectrum b a same amount wherever it meets the prism. Thus, white light is dispersed into a spectrum. We could call the dispersion of light by a prism controlled chromatic aberration, and that's good. CCDs have been very good even be dish-shaped, or the telescope may contain no reflector at all. Though a radio telescopes dish may be many me-ters in diameter, the antenna may be as small as your hand. Like the antenna on a TV set, its only function is to absorb the radio energy and direct it along a cable to an amplifier. After amplification, the signal goes to some kind of recording instrument. Most radio obser-vatories record data directly into computer memory. However it is recorded, an observation with a radio telescope measures the amount of radio energy com-ing from a specific point on the sky. Because humans cant see radio waves, astrono-mers must convert them into something perceptible. One way is to measure the strength of the radio signal at various places in the sky and draw a map in which contours mark areas of uniform radio intensity. We might compare such a map to a seating diagram for a baseball stadium in which the contours mark areas in which the seats have the same price (Figure 3-17a). Contour maps are very common in radio astronomy and are often reproduced using false colors (Figure 3-17b). Computer Amplifier Dish reflector Cable Antenna Figure 3-16 In most radio telescopes, a dish reflector concentrates the radio signal on the antenna. The signal is then amplified and recorded. For all but the shortest radio waves, wire mesh is an adequate reflector (above). (Courtesy Seth Shostak/SETI Institute) Figure 3-17 (a) A contour map of a baseball stadium shows regions of similar admission prices. The most expensive seats are those behind home plate. (b) A false-color radio map of Tychos supernova remnant, the expanding shell of gas produced by the explosion of a star in 1572. The radio contour map has been color-coded to show intensity. Red is the strongest radio intensity and violet the weakest. (Courtesy NRAO) Seat prices in a baseball stadium Astronomy College books You may mail your order to us at : Specialty Books, Inc. 6000 Poston Rd Athens, Ohio 45701 Prod # Name Style Price Qty Total Price DISCOVERING ASTRONOMY FLEX-TEXT NEW $68.95 DISCOVERING ASTRONOMY ACTIVITIES MANUAL & KIT NEW $47.95 DISCOVERING ASTRONOMY NEW $6.25 Purchase Subtotal: $123.15 Specialty Books FAQ 4b. How much does shipping cost? Service Cost Average Shipping time in business days Shipping Options for Domestic US Orders UPS Ground $6.00 2-6 days DISCOVERING ASTRONOMY Order By: AST302S/G ISBN: Used for: AST302 Class ID#: AST302 Required DISCOVERING ASTRONOMY ACTIVITIES MANUAL & KIT Order By: SHAWL ISBN: 0787255599 Ed: 4 Used for: AST302 Class ID#: AST302 Required DISCOVERING ASTRONOMY FLEX-TEXT Order By: SHAWL ISBN: 0787264873 Ed: 4 Used for: AST302 Class ID#: AST302 Required University of Texas Astronomy Course Prod # Name Style Price Qty Total Price 0-534-38185-5 HORIZONS (W/CD ) NEW $90.25 Purchase Subtotal: $90.25 Inquiry Inquiry 2-4 The following are examples of inductive or deductive reasoning, or situations in which induction or deduction can be used. For each case, which type of reasoning is used? Explain your answer. (a) A detective who finds a blood-stained weapon next to a victim, blood of the same type on the clothes of a suspect, scratches on the suspect's face, and skin under the fingernails of the victim would use what process in concluding who commit- ted the crime? (b) Economist Abelieves the health of the U.S. economy is dependent only on the money supply allowed by the Federal Reserve. What reasoning process is used in attempting to validate the idea? (c) A social scientist believes that women make better scientists than men. What reasoning process is used to try to validate the idea? (d) A person suggests that UFOs prove the existence of extraterrestrial intelligence. What reasoning process might be used to study this question? Inquiry 2-5 Making conclusions based on the properties of objects at different distances requires making an important assumption. What might that be? Inquiry 2-6 What would you conclude about galaxies if you were told there was evidence that the most distant galaxies show explosions taking place in their cores, whereas the nearby galaxies do not? Inquiry 2-7 Express the estimated size of the universe in kilometers in scientific notation. The universe is 10 billion ly light-years across. 100,000,000,000,000,000,000,000 km. containing billions of galaxies in the universe. Inquiry 2-8 What are three arguments that would persuade others that the Hollow Earth theory is wrong? Inquiry 2-9 What arguments might a Hollow Earther use to explain away each of the arguments you thought of in Inquiry 2-8? Inquiry 2-10 Can you find a statement in this chapter that if quoted out of context, would support a conclusion opposite to that intended by the authors? Inquiry 2-11 There is only one way to be sure that a quotation has not been made out of context. What is it? Inquiry 3-1 Assume that a child's hand and arm, being smaller than an adult's, are still in the same proportion to body size. Are the angles given above still correct for the child? Explain. angular size = 57.3degrees true size distance angular size of Moon = 57.3 X 3475 km = 385,000 km = 0.52 =31.2 minutes of arc. =1872 seconds of arc. Inquiry 3-2 If the Sun's true size is 1.39 million km in diameter and its distance from Earth is 149.6 million km, what is its angular size (angular diam- eter) in degrees? How does it compare to the angular diameter of the Moon? Inquiry 3-3 Suppose you measured your bedroom five times and got the following values: (a) 12 ft 8 in., (b) 12 ft 9 in., (c) 12 ft 7 in., (d) 12 ft 6 in., and (e) 13 ft. What would be the best value you could give for the length of your room? Inquiry 3-4 Show that your measurement would be systematically too small by calculating how many inches across the room is if you measure four yards when in fact you had measured four meters. Inquiry 3-5 What kind of errors, random or systematic, are shown in each part of Figure 3-11? In the diagram showing precision without accuracy, are the random and systematic errors of equal importance? Inquiry 5-14 Uranus's orbital period is 84 years. To the nearest tenth of an AU, what is its average distance from the Sun? Hint: 20 X 20 X 20 = 8000. Inquiry 16-1 What principle or law would you use to compute the energy radiated by the Sun, given an observation of the amount of energy falling on Earth? Inquiry 16-2 The conclusion that gravitational contraction is not the source of the Sun's energy depends on what assumptions about the constancy of the Sun's energy output over time? Explain. Inquiry 16-4 Because the photon is massless and chargeless show that this reaction conserves atomic mass number and charge. Inquiry 16-5 Nuclear reactions must conserve charge and total atomic mass number. On the basis of such considerations, which of the following reactions is impossible and why? (a) 2 H + 1H > 3H + e + + y 1 1 1 (b) 2H + 1H > 3H + e+ + y 1 1 1 Inquiry 16-11 Why would it be of great value to detect neutrinos that have traveled from the center of the Sun? Hint: Do we receive any of the photons that are created in the center of the Sun? Inquiry 16-20 Figure 16-19a illustrates photospheric granulation with the slit of the spec- trograph across a number of the bright and dark areas. Figure 16-19b shows one spectral line at a wavelength of 8542 A. Wavelength increasing to the right as shown, indicates that light from the bright areas is blueshifted whereas that from the dark areas is redshifted. How should this observa- tion be interpreted? A pattern of light and dark areas known as granulation. The bright areas are slightly hotter than the darker ones. Inquiry 17-1 The densities and temperatures of many interstellar clouds are quite low. Why might these facts have led astronomers to believe that the formation of molecules in such clouds Inquiry 17-2 Using what you know about spectra, how do you think spectral lines formed in molecular clouds might appear different from the lines formed in less dense diffuse clouds? Explain your reasoning. Inquiry 17-3 Dark clouds and hot, luminous massive therefore young main-sequence stars are associated with each other far more often than chance would predict. What conclusion do you draw from this fact? Inquiry 17-4 What are six mechanisms that astronomers believe can initiate the collapse of a cloud to form a star? Explain how each causes collapse to occur. Inquiry 17-6 Stars of type O and B are often seen in regions where star formation is thought to be taking place. How do we know that such OB associations must be young? Inquiry 17-7 Observationally it has been found that the youngest OB associations are also the smallest in size. What conclusion can you draw from this fact? Inquiry 17-9 What is one reason why astrono- mers are interested in studying the infrared portion of the spectrum? Inquiry 17-11 How would you expect intense mass loss in T Tauri stars to make its presence observable? Inquiry 18-1 With its internal-energy source gone, what will happen to the core of the star? Inquiry 18-2 You will notice that at point 3 in Figure 18-3 the star has a slightly larger luminosity and nearly the same temperature as at its ZAMS location at point 1. What does this imply about the star's diameter? Inquiry 18-5 The three helium nuclei together have a greater mass than the resulting carbon nucleus. What has happened to the rest of the mass? Inquiry 18-6 What happens to the luminosity of a star whose diameter and temperature vary regularly? Inquiry 18-7 What would you expect to happen to the rate at which a star evolves if the star were to lose a substantial amount of its mass? Inquiry 18-9 What happens to the evolution of each star due to the transfer of mass between them? Inquiry 18-18 Describe what specific information astronomers had to have to determine the mass of Sirius B. Inquiry 18-19 Does the companion of Sirius B obey the mass-luminosity relationship discussed in Chapter 15? Suggest a reason for this. Inquiry 18-20 To be detected the gravitational redshift of a white dwarf must be distinguished from any velocity redshift Doppler shift due to the motion of the star in space. How this might be done in the case of Sirius B? Hint: Sirius B is a member of a binary system. Inquiry 18-21 Statistics on the proportion of stars of various masses and the number of white dwarfs that are seen today indicate that most stars end up as white dwarfs, including all stars up to at least several times the Sun's mass. What must happen during the evolution of these more massive stars for them to end up as white dwarfs? Inquiry 18-22 How is the age of a cluster related to the position of its turnoff point? Inquiry 18-23 The turnoff point provides a method for estimating the age of a cluster using a plot of its Hertzsprung-Russell diagram. What crucial assumption is being made about the forma- tion of the cluster for this to be a valid method? By the Numbers 6-2 The Doppler Formula Astronomers can measure radial velocity by using the Doppler effect. _V_ = ___ c the radial velocity divided by the speed of light. c is equal to ^ ^ divided by ORBITS MERCURY VENUS EARTH MARS Orbit eccentricity 0.206 0.007 0.017 0.093 Orbit inclination() 7 3 0 2 ROTATION MERCURY VENUS EARTH MARS Rotation period 58.65 243 days 23.9 24.6 days retrograde hours hours orange red sun 2 large dark spots on left five on far right 1 near middle sunspots are cooler regions on the photosphere that are strongly associated with solar magnetism, which varies with a 22-year cycle. Solar flares and promi- nences are other phenomena observed to be associ- ated with solar magnetic activity. The solar wind is made up of charged particles emitted by the Sun through coronal holes. The observable outer regions of the Sun include the visible photosphere, the chromosphere , and the transition region into the corona. Convection is observable in the Sun as granulation in the solar photosphere. The straight line relation from Hubble's dia- gram can be written as the simple equation Vr =Hd Vr equals the distance d in millions of parsecs time the constant H. This relation between red shift and distance is known as the Hubble law, and the constant H is known as the Hubble constant. Figure 15-20 The mass-luminosity relationship for main- sequence. On left handside of figure is Solar luminosities. 10 6 10 4 10 2 1 10 -2 10 -4 10 -5 The line starts out in lower left corner and rises up to the upper right corner. 0.1 0.2 0.3 0.5 1 2 3 5 10 20 30 50 100 numbers are underneath the figure. Mass solar masses Space Astronomy Sample Chapter 3 Astronomical Tools Review Questions 1. Why would you not plot sound waves in the electromagnetic spectrum? Sound waves are a mechanical disturbance that travels through the air from source to ear. Sound requires a medium; so on the moon, where there is no aire, there can be no sound. 2. If you had unlimited funds to build a large telescope, what type would you choose, a refractor or a reflector? Why? should buy a reflecting telescope rather than a refracting telescope. Get more diameter per dollar, and telescope will not suffer from a chromatic aberration. a refractor I would build a refractor telescope since refracting telescopes use a large lens to gather and focus the light. a reflector a reflecting telescopes use a concave mirror. 3. Why do nocturnal animals usually have large pupils in their own eyes? How is that related to astronomical telescopes? The larger the lens is the more light is collected. 4. Why do optical astronomers sometimes put their tele- scopes at the tops of mountains, while radio astrono- mers sometimes put their telescopes in deep valleys? Astronomers avoid cities because light pollution in cities building on mountains is better since it is farther away.. 5. Optical and radio astronomers both try to build large telescopes but for different reasons. How do these goals differ? Building large telescopes because a telescope can help our eyes in three important ways - the three powers of a telescope. The most important of these depends on the diameter of the telescope. a telescope that can gather large amounts of light to produce a bright image. 6. What are the advantages of making a telescope mirror thin? What problems does this cause? Thinner mirros also cool faster at nightfall and suffer less distortion from uneven expansion and contraction. 7. Small telescopes are often advertised as "200 power" or "magnifies 200 times." As someone knowledgeable about astronomical telescopes, how would you improve such advertisements? 8. An astronomer recently said, "Some people think I should give up photographic plates." Why might she change to something else? They can detect both bright and faint objects in a single exposure, are much more sensitive than a photographic plate, and can be read directl into computer memory. 9. What purpose do the colors in a false-color image or false- color radio map serve? Astronomers also manipulate images to produce false-color images in which the colors represent different levels of intensity and are not related to the true colors of the object. 10. How is chromatic aberration related to a prism spectro- graph? Analyze light in detail we need to spread the light out according to wavelength into a spectrum, a task performed by a spectrograph. When placing a prism in a thin beam of light that is streaming into a darkened room a beautiful spectrum appears on the far wall. 11. Why would radio astronomers build identical radio telescopes in many different places around the world? radio astronomers can combine two or more radio telescopes to improve the resolving power. A linkup of radio telescopes is called a radio interferometer. 12. Why do radio telescopes have poor resolving power? Because the radio telescopes are small in size diameter. The only way to improve the resolving power is to build a bigger telescope. 13. Why must telescopes observing in the far infrared be cooled to low temperature? If a telescope observes at far-infrared wavelengths, then it must be cooled. Infrared radiation is emitted by heated objects, and if the telescope is warm it will emit many times more infrared radiation than that coming from a distant object. 14. What might we detect with an X-ray telescope that we could not detect with an infrared telescope? X-ray Specifics? Well for starters how about Diablo Canyon N35 12.48 W120 52.21 #2204big These high resolution photos show every detail of the place, every fence, wire, barricade, spent fuel pools, every rock. PG&E and SCE in California. http://www2.californiacoastline.org/ 7. What does the word apparent mean in apparent visual magnitude? 8. In what ways is the celestial sphere a scientific model? 9. Why do astronomers use the word on to describe angles on the sky rather than angles in the sky? 10. If Earth did not rotate, could we define the celestial poles and celestial equator? 11. Where would ou go on Earth if you wanted to be able to see both the north celestial pole and the south celes- tial pole at the same time? 12. Where would you go on Earth to place a celestial pole at your zenith? 13. Explain how to make a simple astronomical observa- tion that would determine your latitude. 14. Why does the number of circumpolar constellations de- pend on the latitude of the observer? 15. How could we detect Earth's precession by examing star charts from ancient Egypt? Discussion Questions 1. All cultures on Earth named constellations. Why do you suppose this was such a common practice? 2. If you were lost at sea, you could find your approximate latitude by measuring the altitude of Polaris. But Polaris isn't exactly at the celestial pole. What else would you need to know to measure your latitude more accurately? Problems 1. If light from one star is 40 times more intense than light from another star, what is their difference in magnitudes? 2. If two stars differ by 8.6 magnitudes, what is their in- tensity ratio? 3. Star A has a magnitude of 2.5; Star B, 5.5; and Star C, 9.5. Which is brightest? Which are visible to the unaided eye? Which pair of stars has an intensity ratio of 16? Chapter 2 The Sky page 19 4. By what factor is sunlight more intense than moonlight? see figure 2-6 5. If you are at a latitude of 35 degrees north of Earth's equator, what is the angular distance from the northern horizon up to the north celestial pole? from the south- ern horizon down to the south celestial pole? Critical Inquiries for the Web 1. We've all heard of the "North Star," but is there a "South Star" for southern hemisphere observers? As Earth pre- cesses on its axis, Polaris will cease to be our pole star and others will eventually hold that title. Search for star maps available online and for information on preces- sion, and determine what bright southern hemisphere stars might be future pole stars for people living south of the equator. 2. Would the stars of a familiar constellation like Pegasus or Orion look similar to our view from Earth if we lived on planets orbiting other stars? Find a table of distance data for the bright stars in a familiar constellation and construct a diagram that visualizes the positions of these stars in space similar to Figure 2-3. 3. Most cultures around the world developed their own constellations. Search the Web to find constellation mythology from other cultures. Native Americans have a rich mythology of the sky, as do Polynesian, Asian, and African peoples. Exploring The Sky 1. Describe the apparent rotation of the sky in the north, south, east, and west sky from your location. (Hint Use Site Information under the Data menu to set location and date. Use Look under the Orientation menu and use the time skip arrows in the toolbar to go forward in time.) 2. What constellations are visible in your evening sky? (Hint Use Reference Lines and Labels under the View menu to turn on the constellations and their labels. Use Site Information under the Data menu to set your loca- tion, date, and time.) 3. What would the sky look like if you could not see stars fainter than second magnitude? (Hint Use Filters under the View menu to change the magnitude limits.) 4. Locate the constellation Orion as seen from your loca- tion, from Earth's north pole, and from Earth's south pole. Part 1 The Sky page 20 Chapter 2 Review Questions 1. What is the difference between the daily and annual motions of the sun? 2. If Earth did not rotate, could we still define the eclip- tic? Why or why not? 3. What would our seasons be like if Earth were tipped 35 degrees instead 23.5 0? What would they be like if Earth's axis were perpendicular to light-year (ly) Review Questions 1. What is the largest dimension you have personal knowl- edge of? Have you run a mile? hiked 10 miles? run a marathon? 2. Why are astronomical units more convenient than miles or kilometers for measuring some astronomical distances? The orbits of the planets are not perfect circles apparent for Mercury Its orbit carries it as close to the sun as 0.307 AU and as far away as 0.467 AU. 3. In what ways is our planet changing? Our planet is experiencing different weather, beach erosion, volcanoe eruptions, and sometimes a meteor lands on the deserts on the planet. 4. What is the difference between our solar system, our gal- axy, and the universe? The Solar System consists of 9 planets 5. Why do all stars, except for the sun, look like points of light as seen from Earth? Its the brightness of the stars light. 6. Why are light-years-rather than miles or kilometers- more convenient units for measuring some astronomi- cal distances? 7. Why is it difficult to see planets orbiting other stars? 8. Why can't we measure the diameters of stars from the size of the star images on photographs? What does the diameter of a star image really tell us about the star? 9. How long does it take light to cross the diameter of our solar system? of our galaxy? 10. What are the largest known structures in the universe? 11. How many planets inhabited by intelligent life do you think the universe contains? Explain your answer? Problems 1. If 1 mile equals 1,609 km and the moon is 2160 miles in diameter, what is its diameter in kilometers? in meters? 2. If sunlight takes 8 minutes to reach Earth, how long does moonlight take? 3. How many suns would it take, laid edge to edge, to reach the nearest star? 4. How many kilometers are there in a light-minute? (Hint: The speed of light 3 x 10 5 km/s) 5. How many galaxies like our own, laid edge to edge, would it take to reach the nearest large galaxy (which is 2 x 10 6 ly away)? Exploring The Sky 1. Locate and center one example of each of three differ- ent types of objects: a. A planet, such as Saturn. Find its rising and setting time. Such objects have distances measured in astro- nomical units (AU). How to proceed: Decide on the object you want to lo- cate Then find and center the object by pressing the Find button on the Object Toolbar. The second method is to press the F key. The third is to click Edit, then Find. Once you have the Object Information window, press the center button. b. A star. All stars in The Sky belong to our Milky Way Galaxy. Give the star's name, its magnitude, and its distance in light-years. How to proceed: Click on any star, which brings up an Object Information window containg a variety of information about the star. c. A galaxy; give its name and/or its designation. How to proceed: To show galaxies, click on the Galax- ies button in the Object Toolbar, then click on any galaxy. Distances to galaxies are on the order of mil- lions and billions of light years. 2. Look at the solar system from beyond Pluto by clicking on View and then on 3D Solar System Mode. Tip the solar sydtem edge on and then face on. Zoom in to see the inner planets. Under Tools, set the Time Skip Incre- ment to 1 day and then go forward in time to watch the planets move. 3. Identify some of the brightest constellations located along the Milky Way. (Hint: See View, Reference Lines.) Chapter 1 8 Part 1 The Sky Review Questions 1. Why have astronomers added modern constellations to the sky? 2. What is the difference between an asterism and a con- stellation? Give some examples. A group of stars with a local name, that are not officially constellations, are called asterisms. The big dipper is an asterism. The constellations are lines connecting from star to star. A constellation is group of stars like a dot-to-dot puzzle. If you join the dots--stars, that is--and use lots of imagination, the picture would look like an object, animal, or person. For example, Orion is a group of stars that the Greeks thought looked like a giant hunter with a sword attached to his belt. 3. What characteristic do stars in a constellation or aster- ism share? They are stars bright stars, and dim stars. A human grouping of stars in the night sky into a recognizable pattern. 4. Do people from other cultures on Earth see the same stars, constellations, and asterisms that you see? No. The Earth is always moving. 5. How does the Greek-letter designation of a star give us a clue to its brightness? 6. Give two features of the magnitude scale that might seem confusing. What is the origin of theits orbit? 4. Why are the seasons reversed in the southern hemi- sphere? The reason that the Sun's position in the sky is north of the celestial equator for six months, and then south of the celestial equator for six months is that the Earth's axis of rotation is TILTED compared to it's orbit around the sun. If the Earth's axis was not tilted, it would form a right angle (90º) with our orbit around the sun.  But the Earth's axis is tilted 23½º.  This tilt is solely responsible for the different position of the sun in our sky during the year, and consequently, it is solely responsible for the Earth's seasons. 5. Do the phases of the moon look the same from every place on Earth, or is the moon full at different times in different locations? We see different views of the Universe from where we live as Earth makes its yearly trip around the solar system. 6. What phase would Earth be in if you were on the moon when the moon was full? at first quarter? at waning crescent? 7. How does the moon slow Earth's rotation, and how does Earth change the moon's orbit? The energy needed to cause the Moon to move away from Earth is at the expense of the energy involved in Earths rotation. Earth is slowing down in its rotation. 8. Why have most people seen a total lunar eclipse, while few have seen a total solar eclipse? At any particular moment, one face of the Earth is beneath the Moon. Assuming that Earths population is uniformly distributed, one-half of Earths population is able to witness a total lunar eclipse. The path of totality of a total solar eclipse is quite narrow, so that only a small fraction of Earths population are in a position to witness it. 9. Why isn't there an eclipse at every new moon and at every full moon? 10. Why is the moon red during a total lunar eclipse? 11. Why should the eccentricity of Earth's orbit make win- ter in the northern hemisphere different from winter in the southern hemisphere? 12. How could small changes in the inclination of Earth's axis affect world climate? Discussion Questions 1. Do planets orbiting other stars have ecliptics? Could they have seasons? 2. Why would it be difficult to see prominences if you were on the moon during a total lunar eclipse? Problems 1. If Earth is about 4.6 billion (4.6 x 10 9) years old, how many precessional cycles have occurred? 2. Identify the phases of the moon if on March 21 the moon were located at (a) the vernal equinox, (b) the au- tumnal equinox, (c) the summer solstice, (d) the winter solstice. 3. Identify the phases of the moon if at sunset the moon were (a) near the eastern horizon, (b) high in the south, (c) in the southeast, (d) in the southwest. 4. About how many days must elapse between first-quarter moon and third-quarter moon? 5. Draw a diagram showing Earth, the moon, and shadows during (a) a total solar eclipse, (b) a total lunar eclipse, (c) a partial lunar eclipse, (d) an annular eclipse. 6. Phobos, one of the moons of Mars, is 20 km in diameter and orbits 5982 km above the surface of the planet. What is the angular diameter of Phobos as seen from Mars? (Hint See By the Numbers 3-1). 7. A total eclipse of the sun was visible from Canada on July 10, 1972. When did this eclipse occur next? From what part of Earth was it total? 8. When will the eclipse described in Problem 7 next be total as seen from Canada? Critical Inquiries for the Web 1. There are many Web sites on astrology, but nearly all accept the old superstition as real. Can you find Web sites that analyze astrology logically? Begin at the Web site for the Committee for the Scientific Investigation of Claims of the Paranormal. 2. Find photographs of recent lunar and solar eclipses. Where did the observers have to go to see the eclipse? 3. What is the latest news concerning the Milankovitch hypothesis? Is there new evidence to test the theory? Exploring The Sky 1. How long is daylight on the day of the summer solstice and on the day of the winter solstice for your location? Use Site Information under the Data menu to set your lo- cation and the date. Find the time of sunrise and sunset. 2. How many days must elapse between new moons? Use the Moon Phase Calendar under the Tools menu to lo- cate new moons. 3. Where would you have to go to see a total solar eclipse in the next year or two? Use the Eclipse Finder under the Tools menu to select a total solar eclipse. Check Show Path of Totality to see a map of Earth. 4. View the total solar eclipse you located in Activity 3. In the Eclipse Finder, click View to see the sun and moon from your present location. Then use Site Information under the Data menu to change your location to a spot in the path of totality. Set the Time Step to 5 minutes and click the arrows to move the moon across the sun. 5. Locate and observe an annular and a total lunar eclipse. Part 1 The Sky page 40 Chapter 3 Review Questions 1. Why did Greek astronomers conclude that the heavens were made up of perfect crystalline spheres moving at constant speeds? 2. Why did classical astronomers conclude that Earth had to be motionless? 3. How did the Ptolemaic model explain retrograde motion? 4. In what ways were the models of Ptolemy and Coperni- cus similar? 5. Why did the Copernican hypothesis win gradual accep- tance? 6. Why is it difficult for scientists to replace an old para- digm with a new paradigm? 7. Why did Tycho Brahe expect the new star of 1572 to show parallax? Why was the lack of parallax evidence against the Ptolemaic model? 8. How was Tycho's model of the universe similar to the Ptolemaic model? How did it resemble the Copernican model? 9. Explain how Kepler's laws contradict uniform circular motion? 10. What is the difference between a hypothesis, a theory, and a law? 11. How did the Alfonsine Tables, the Prutenic Tables, and the Rudolphine Tables differ? 12. Review Galileo's telescope discoveries, and explain why they supported the Copernican model and contra- dicted the Ptolemaic model. 13. Galileo was condemned by the Inquisition, but Kepler, also a Copernican, was not. Why not? 14. Why did Newton conclude that gravitation had to be universal? 15. Explain why we might describe the orbital motion of the moon with the statement, "The moon is falling." Discussion Questions 1. Historian of science Thomas Kuhn has said that De Revo- lutionibus was a revolution-making book, but not a rev- olutionary book? How was it classical? 2. Why might Tycho Brahe have hesitated to hire Kepler? Why do you suppose he appointed Kepler his scientific heir? 3. How does the modern controversy over creationism and evolution reflect two ways of knowing about the physi- cal world? Problems 1. If you lived on Mars, which planets would describe ret- rograde loops? Which would never be visible as cres- cent phases? 2. Galileo's telescope showed him that Venus has a large angular diameter (61 seconds of arc) when it is a cres- cent and a small angular diameter (10 seconds of arc) when it is nearly full. Use the small-angle formula to find the ratio of its maximum distance to its minimum distance. Is this ratio compatible with the Ptolemaic universe shown on page 45? 3. Galileo's telescopes were not of high quality by modern stand      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~ards. He was able to see the rings of Saturn, but he never reported seeing features on Mars. Use the small- angle formula to find the angular diameter of Mars when it is closest to Earth. How does that compare with the maximum diameter of Saturn's rings? 4. If a planet had an average distance from the sun of 10 AU what would its orbital period be? 5. If a space probe were sent into an orbit around the sun that brought it as close as 0.5 AU to the sun and as far away as 5.5 AU, what would its orbital period be? 6. Pluto orbits the sun with a period of 247.7 years. What is its average distance from the sun? 7. Calculate the circular velocity of Venus and Saturn around the sun. Hint: The mass of the sun is 2 x 10 30kg. 8. The circular velocity of Earth around the sun is about 30 km/s. Are the arrows for Venus and Saturn correct in figure 4-4b? Hint: see problem 7. 9. What is the orbital velocity of an Earth satellite 42,000 km from Earth? How long does it take to circle its orbit once? Critical Inquiries for the Web 1. The trial of Galileo is an important event in the history of science. We now know, and the Church now recog- page 64 chapter 4 nizes, that Galileo's view was correct, but what were the arguments on both sides of the issue as it was un- folding? Research the Internet for documents chroni- cling the trial. Galileo's observations and publications, and the position of the Church. Use this information to outline a case for and against Galileo in the context of the times in which the trial occurred. 2. It's hard to imagine that an observatory could exist be- fore the invention of the telescope, but Tycho Brahe's observatory at Hveen was a great astronomical center of its day. Search the Web sites on Tycho and his instru- ments and describe what an observing session at Hveen might have involved. Exploring The Sky 1. Go to Stonehenge in southern England and watch the sun rise on the morning of the summer solstice. Where does it rise on the morning of the winter solstice? 2. Observe Mars going through its retrograde motion. Hint Use Reference Lines under the View menu to turn on the ecliptic. Be sure you are in Free Rotation under the Orientation menu. Locate Mars and use the time skip arrows to watch it move.) 3. Compare the size of the retrograde loops made by Mars, Jupiter, and Saturn. 4. Can you recognize the effects of Kepler's second law in the orbital motion of any of the planets? Hint Use 3D Solar System Mode under the View Menu. 5. Can you recognize the effects of Kepler's third law in the orbital motion of the planets? Chapter 4 page 65 Review Questions 1. Why would you not plot sound waves in the electro- magnetic spectrum? 2. If you had unlimited funds to build a large telescope, which type would you choose, a refractor or a reflector? Why? 3. Why do nocturnal animals usually have large pupils in their eyes? How is that related to astronomical telescopes? 4. Why do optical astronomers sometimes put their tele- scopes at the tops of mountains, while radio astrono- mers sometimes put their telescopes in deep valleys? 5. Optical and radio astronomers both try to build large telescopes but for different reasons. How do these goals differ? 6. What are the advantages of making a telescope mirror thin? What problems does this cause? 7. Small telescopes are often advertised as "200 power" or "magnifies 200 times." As someone knowledgeable about astronomical telescopes, how would you improve such advertisements? 8. An astronomer recently said, "Some people think I should give up photographic plates." Why might she change to something else? 9. What purpose do the colors in a false-color image or false- color radio map serve? 10. How is chromatic aberration related to a prism spectro- graph? 11. Why would radio astronomers build identical radio telescopes in many different places around the world? 12. Why do radio telescopes have poor resolving power? 13. Why must telescopes observing in the far infrared be cooled to low temperatures? 14. What might we detect with an X-ray telescope that we could not detect with an infrared telescope? 15. If the Hubble Space Telescope observes at visual wave- lengths, why must it observe from space? Discussion Questions 1. Why does the wavelength response of the human eye match so well the visual window of Earth's atmosphere? 2. Basic research in chemistry, physics, biology, and similar sciences is supported in part b industry. How is as- tronomy different? Who funds the major observatories? page 88 chapter 5 Astronomical Tools Problems 1. The thickness of the plastic in plastic bags is about 0.001 mm. How many wavelengths of red light is this? 2. Measure the actual wavelength of the wave in Figure 5-1. In what portion of the electromagnetic spectrum would it belong? 3. Compare the light-gathering powers of a 5-m telescope and a 0.5-m telescope. 4. How does the light-gathering power of the largest tele- scope in the world compare with that of the human eye? Hint: Assume that the pupil of your eye can open to about 0.8 cm. 5. What is the resolving power of a 25-cm telescope? What do two stars 1.5 seconds of arc apart look like through this telescope? 6. Most of Galileo's telescopes were only about 2 cm in di- ameter. Should he have been able to resolve the two stars mentioned in Problem 5? 7. How does the resolving power of the 5-m telescope compare with that of the Hubble Space Telescope? Why does the Hubble Space Telescope outperforms the 5-m telescope? 8. If we build a telescope with a focal length of 1.3 m, what focal length should the eyepiece have to give a magnification of 100 times? 9. Astronauts observing from a space station need a tele- scope with a light-gathering power 15,000 times that of the human eye, capable of resolving detail as small as 0.1 second of arc, and having a magnifying power of 250. Design a telescope to meet their needs. Could you test your design by observing stars from Earth? 10. A spy satellite orbiting 400 km above Earth is suppos- edly capable of counting individual people in a crowd. What minimum-diameter telescope must the satellite carry? Hint Use the small-angle formula. Critical Inquiries for the Web 1. How do professional astronomers go about making ob- servations at major astronomical facilities? Visit sev- eral observatory Web sites to determine the process an astronomer would go through to secure observing time and make observations at the facility. 2. NASA is in the process of completing a fleet of four space-based "Great Observatories." (The Hubble Space Telescope is one: what are the others?) Examine the cur- rent state of these missions by visiting their home pages on the Internet. What advantages would these facilities have over ground-based observatories? Exploring The Sky 1. Astronomical telescopes using equatorial mountings must be aligned precisely with the north celestial pole. Locate Polaris and determine how far it is from the north celestial pole. Hint: Use Reference Lines under the View menu, and check Grid under Equitorial. Be sure the spacing is set to auto/fine. Then locate the Little Dipper and zoom in on Polaris.) page 89 Chapter 5 Astronomical Tools Review Questions 1. Why might we say that atoms are mostly empty space? 2. What is the difference between an isotope and an ion? 3. Why is the binding energy of an electron related to the size of its orbit? 4. Explain why ionized calcium can form absorption lines but ionized hydrogen cannot? 5. Describe two ways an atom can become excited? 6. Why do different atoms have different lines in their spectra? 7. Why does the amount of black body radiation emitted depend on the temperature of the object? 8. Why do hot stars look bluer than cool stars? 9. What kind of spectrum does a neon sign produce? 10. Why are Balmer lines strong in the spectra of medium- temperature stars and weak in the spectra of hot and cool stars? 11. Why are titanium oxide features visible in the spectra of only the coolest stars? 12. Explain the similarities among Table 6-1, Figure 6-7c, Figure 6-8, and Figure 6-9. 13. Why does the doppler effect detect only radial velocity? 14. How can the Doppler effect explain shifts in both light and sound? 15. Explain why the presence of spectral lines of a given ele- ment in the solar spectrum tells us that element is pres- ent in the sun, but the absence of the lines would not mean the element was absent from the sun? Chapter 6 Atoms and Starlight page 107 Discussion Questions 1. In what ways is our model of an atom a scientific model? In what ways is it incorrect? 2. Can you think of classification systems we use to sim- plify what would otherwise be complex measurements? Consider foods, movies, cars, grades, and clothes. Problems 1. Human body temperature is about 310 K (98.6 F). At what wavelength do humans radiate the most energy? What kind of radiation do we emit? 2. If a star has a surface temperature of 20,000 K, at what wavelength will it radiate the most energy? 3. Infrared observations of a star show that it is most in- tense at a wavelength of 2000 nm. What is the tempera- ture of the star's surface? 4. If we double the temperature of a black body, by what factor will the total energy radiated per second per square meter increase? 5. If one star has a temperature of 6000 K and another star has a temperature of 7000 K, how much more energy per second will the hotter star radiate from each square meter of its surface? 6. Transition A produces light with a wavelength of 500 nm. Transition B involves twice as much energy as A. What wavelength light does it produce? 7. Determine the temperatures of the following stars based on their spectra. Use Figures 6-8 and 6-9. a. medium-strength Balmer lines, strong helium lines b. medium-strength Balmer lines, weak ionized-calcium lines c. strong TiO bands d. very weak Balmer lines, strong ionized-calcium lines 8. To which spectral classes do the stars in Problem 7 belong? 9. In a laboratory, the Balmer beta line has a wavelength of 486.1 nm. If the line appear in a star's spectrum at 486.3 nm, what is the star's radial velocity? Is it ap- proaching or receding? 10. The highest-velocity stars an astronomer might observe have velocities of about 400 km/s. What change in wave- length would this cause in the Balmer gamma line? Hint: Wavelengths are given on page 99. Critical Inquiries for the Web 1. The name for the element helium has astronomical roots. Search the Internet for information on the discovery of helium. How and when was it discovered, and how did it get its name? Why do you suppose it took so long for helium to be recognized? Janssen obtained the first evidence of helium during the solar eclipse of 1868 when he detected a new line in the solar spectrum. 2. How was the model of the atom presented in the text you read developed? Search the Web for Information on historical models of the atom, and compile a time line of important developments leading to our current under- standing. What evidence exists that supports our model? Exploring The Sky 1. Locate the following stars, click on them, and deter- mine their spectral types: Antares in Scorpius, Betel- geuse in Orion, Aldebaran in Taurus, Sirius in Canis Major, Rigel in Orion. 2. How are spectral types correlated with the colors of stars? Hint: Locate Orion and choose Spectral Colors under the View menu.) page 100 Chapter 6 Table 6-2 The Most Abundant Elements in the Sun Element Percentage by Number of Atoms Percentage by Mass Hydrogen 91.0 70.9 Helium 8.9 27.4 Carbon 0.03 0.3 Nitrogen 0.008 0.1 Oxygen 0.07 0.8 Neon 0.01 0.2 Magnesium 0.003 0.06 Silicon 0.003 0.07 Sulfur 0.002 0.04 Iron 0.003 0.3 page 100 Chapter 6 Part 2 The Stars Review Questions 1. Why can't we see deeper than the photosphere? 2. What evidence do we have that granulation is caused by convection? 3. How are granules and supergranules related? How do they differ? 4. How can a filtergram reveal structure in the chromo- sphere? 5. What evidence do we have that the corona has a very high temperature? 6. What heats the chromosphere and corona to high tem- perature? 7. How are astronomers able to explore the layers of the sun below the photosphere? 8. What evidence do we have that sunspots are magnetic? 9. How does the Babcock model explain the sunspot cycle? 10. What does the spectrum of a prominence tell us? What does it shape tell us? 11. How can solar flares affect Earth? 12. Why does nuclear fusion require high temperatures? 13. Why does nuclear fusion in the sun occur only near the center? 14. How can astronomers detect neutrinos from the sun? 15. How can neutrino oscillation explain the solar neutrino problem? Discussion Questions 1. What energy sources on Earth cannot be thought of as stored sunlight? 2. What would the spectrum of an auroral display look like? Why? 3. What observations would you make if you were ordered to set up a system that could warn astronauts in orbit of dangerous solar flares? Such a system exists. Problems 1. The radius of the sun is 0.7 million km. What percent- age of the radius is taken up by the chromosphere? 2. The smallest detail visible with ground-based solar telescopes is about 1 second of arc. How large a region does this represent on the sun? Hint: Use the small- angle formula. 3. What is the angular diameter of a star like the sun lo- cated 5 ly from Earth? Is the Hubble Space Telescope able to resolve detail on the surface of such a star? 4. If a sunspot has a temperature of 4200 K and the solar surface has a temperature of 5800 K, how many times brighter is the surface compared with the sunspot? Hint: Use the Stefan-Boltzmann Law, By the Numbers 6-1 . 5. A solar flare can release 10 25 J. How many megatons of TNT would be equivalent? Hint: A 1-megaton bomb produces about 4 X 10 15 J. 6. The United States consumes about 2.5 X 10 19 J of en- ergy in all forms in a year. How many years could we run the United States on the energy released by the solar flare in Problem 7? 7. Neglecting energy absorbed or reflected by our atmo- sphere, the solar energy hitting 1 square meter of Earth's surface is 1360 J/s the solar constant. How long does it take a baseball diamond 90 ft on a side to receive 1 megaton of solar energy? Hint: See Problem 5. 8. How much energy is produced when the sun converts 1 kg of mass into energy? 9. How much energy is produced when the sun converts 1 kg of hydrogen into helium? Hint: How does this problem differ from Problem 8? 10. A 1-megaton nuclear weapon produces about 4 X 10 15 J of energy. How much mass must vanish when a 5-megaton weapon explodes? Critical Inquiries for the Web 1. Do disturbances in one layer of the solar atmosphere produce effects in other layers? We have seen that fil- tergrams are useful in identifying the layers of the solar atmosphere and the structures within them. Visit a Web site that provides daily solar images, choose today's date or one near it, and examine the sun in several wave- lengths to explore the relation between disturbances in various layers. 2. Page 123 shows the auroral displays caused by interac- tion of particles from the sun with our magnetosphere and ultimately our atmosphere. Explore the Web to find out how auroral activity is affected as solar activity rises and falls through the solar cycle. What changes in auroral visibility occur during this cycle? In what other ways can the increased activity associated with a solar maximum affect Earth? 3. What can you find on the Web about Earth-based efforts to generate energy through nuclear fusion? How do nu- clear fusion power experiments attempt to trigger and control nuclear fusion? So-called cold fusion has been largely abandoned as a false trail. How did it resemble nuclear fusion? Exploring The Sky 1. Locate the six photos of the sun provided in The Sky and attempt to draw in the sun's equator in each photo. Hint: In the sun's information box, choose More Infor- mation and then Multimedia. What features are visible in these images that help us recognize the orientation of the sun's equator? page 129 Chapter 7 The Sun - Our Star Review Questions 1. Why are Earth-based parallax measurements limited to the nearest stars? 2. Why was the Hipparcos satellite able to make more ac- curate parallax measurements than are ground-based telescopes? 3. What do the words absolute and visual mean in the de- finition of absolute visual magnitude? 4. What does luminosity measure that is different from what absolute visual magnitude measures? 5. Why does the luminosity of a star depend on both its radius and its temperature? 6. How can we be sure that giant stars really are larger than main-sequence stars? 7. Why do we conclude that white dwarfs must be very small? 8. What observations would we make to classify a star ac- cording to its luminosity? Why does that method work? 9. Why does the orbital period of a binary star depend on its mass? 10. What observations would you make to study an eclips- ing binary star? 11. Why don't we know the inclination of a spectroscopic binary? How do we know the inclination of an eclips- ing binary? 12. How do the masses of stars along the main sequence il- lustrate the mass-luminosity relation? 13. Why is it difficult to find out how common the most lu- minous stars are? The least luminous stars? 14. What is the most common kind of star? 15. If you look only at the brightest stars in the night sky, what kind of star are you likely to be observing? Why? Discussion Questions 1. If someone asked you to compile a list of the nearest stars to the sun based on your own observations, what measurements would you make, and how would you analyze them to detect nearby stars? 2. The sun is sometimes described as an average star. Is that true? What is the average star really like? Problems 1. If a star has a parallax of 0.050 second of arc, what is its distance in pc? in ly? in AU? 2. If you place a screen of area 1 m2 at a distance of 2.8 m from a 100-watt lightbulb, the light flux falling on the screen will be 1 J/s. To what distance must you move the screen to make the flux striking it equal 0.01 J/s? This assumes the lightbulb emits all of its energy as light. 3. If a star has a parallax of 0.016 second of arc and an ap- parent magnitude of 6, how far away is it, and what is its absolute magnitude? 4. Complete the following table. P m Mv d(pc) (sec of arc) ___ 7 10 _______ 11 ___ 1000 _______ ___ -2 ____ 0.025 4 ___ _____ 0.040 5. The unaided human eye can see stars no fainter than those with an apparent magnitude of 6. If you can see a bright firefly blinking up to 0.5 km away, what is the absolute magnitude of the firefly? Hint: Convert the distance to parsecs and use the formula in By the Num- bers 8-2? 6. If a main-sequence star has a luminosity of 400 L, what is its spectral type? Hint: See Figure 8-7. 7. If a star is 10 times the radius of the sun and half as hot, what will its luminosity be? Hint: See By the Num- bers 8-3. 8. An 08 V star has an apparent magnitude of +1. Use the method of spectroscopic parallax to find the distance to the star. Why might this distance be inaccurate? 9. Find the luminosity and spectral type of a 5-M main- sequence star. 10. In the following table, which star is brightest in appar- ent magnitude? most luminous in absolute magnitude? largest? least dense? farthest away? Star Spectral Type m a G2 V 5 b B1 V 8 c G2 Ib 10 d M5 III 19 e White dwarf 15 11. If two stars orbit each other with a period of 6 years and a separation of 4 AU, what is their total mass? Hint: See By the Numbers 8-4. 12. If the eclipsing binary in Figure 8-17 has a period of 32 days, an orbital velocity of 153 km/s, and an orbit that is nearly edge-on, what is the circumferences of the orbit? the radius of the orbit? the mass of the system? 13. If the orbital velocity of the eclipsing binary in Fig- ure 8-17 is 153 km/s and the smaller star becomes com- pletely eclipsed in 2.5 hours, what is its diameter? 14. What is the luminosity of a 4-solar mass star? of a 9-solar- mass star? of a 7-solar-mass star? page 153 Chapter 8 The Properties of Stars Critical Inquiries for the Web 1. Hertsprung-Russell diagrams allow astronomers to rep- resent the distribution of stellar properties. Examine H-R diagrams at various Web sites, and determine which types of stars are the most and least numerous in our galax. 2. What if Algol were oriented such that its stars did not eclipse each other as seen from Earth? Use the Internet to find information about Algol and binary stars in gen- eral, and discuss how astronomers could still deter- mine that it is a multiple-star system. Exploring The Sky 1. Locate the following stars and determine their apparent magnitude, parallax, distance in parsecs and in light- years, and spectral classification: Sirius, Aldebaran, Vega, Deneb, Betelgeuse, Antares, and Altair. Hint To center on an object, use Find under the Edit menu and type the object's name followed by a period. 2. Use the spectral type and parallax of the stars above to estimate their distance from Earth. Compare with distances given in The Sk. 3. Take a survey of the stars. Center on Orion and adjust the field until it is about 100 wide. Click on the ten brightest stars and record the spectral types. Now zoom in until only a few dozen stars are in the frame. Click on the 10 faintest stars and record their spectral types. Is there a difference between the brightest and faintest stars? Is the result what you expected? Are certain kinds of stars missing? 4. Repeat exercise 3 for a region centered on the Big Dipper. page 154 Chapter 8 Review Questions 1. What evidence do we have that the spaces between the stars are not empty? spaces between the stars are not empty 2. What evidence do we have that the interstellar medium contains both gas and dust? 3. Why would an emission nebula near a hot star look red, but a reflection nebula near its star looks blue? 4. Why do astronomers rely heavily on infrared observa- tions to study star formation? 5. What observational evidence do we have that star for- mation is a continuous process? 6. How are Herbig-Haro objects related to star formation? 7. How do the proton-proton chain and the CNO cycle re- semble each other? How do they differ? 8. Why does the CNO cycle require a higher temperature than the proton-proton chain? 9. How does the pressure-temperature thermostat control the nuclear reactions inside stars? 10. Step by step, explain how energy flows from the sun's core to Earth? 11. Why is there a mass-luminosity relation? page 178 Chapter 9 12. Why is there a lower limit to the mass of a main-sequence star? 13. Why does a star's life expectancy depend on its mass? 14. What evidence do we have that star formation is hap- pening right now in the Orion Nebula? Discussion Questions 1. When we see distant streetlights through smog, they look dimmer and redder than they do normally. But when we see the same streetlights through fog or falling snow, they look dimmer but not redder. Use your knowledge of the interstellar medium to discuss the rel- ative sizes of the particles in smog, fog, and snow com- pared with the wavelength of light? 2. If planets form as a natural by-product of star forma- tion, which do you think are more common-stars or planets? Problems 1. The interstellar medium dims starlight by about 1.9 mag- nitudes/1000 pc. What fraction of photons survive a trip of 1000 pc? Hint: See By the Numbers 2-1. 2. A small Bok globule has a diameter of 20 seconds of arc. If the nebula is 1000 pc from Earth, what is the diame- ter of the globule? 3. If a giant molecular cloud has a diameter of 30 pc and drifts relative to neighboring clouds at 20 km/s, how long will it take to travel its own diameter? 4. If the dust cocoon around a protostar emits radiation most strongly at a wavelength of 30 microns, what is the temperature of the dust? Hint: See By the Num- bers 6-1. 5. The gas in a bipolar flow can travel as fast as 300 km/s. How long would it take to travel 1 light-year? 6. Circle all 1H and 4H nuclei in Figure 9-9. Explain how both the proton-proton chain and the CNO cycle can be summarized as 4 1H > 4HE + energy? 7. In the model shown in Figure 9-14, how much of the sun's mass is hotter than 12,000,000 K? 8. If a brown dwarf has a surface temperature of 1500 K, at what wavelength will it emit the most radiation? Hint: See By the Numbers 6-1. 9. What is the life expectancy of a 16-solar-mass star? 10. If the 06 V star in the Orion Nebula is magnitude 5.4, how far away is the nebula? Hint: Use spectroscopic parallax. 11. The hottest star in the Orion Nebula has a surface tem- perature of 40,000 K. At what wavelength does it radi- ate the most energy? Hint: See By the Numbers 6-1. page 178 Chapter 9 Critical Inquiries for the Web 1. Use the Web to supply additional details concerning the evolution of protostars and T Tauri stars. 2. Astronomers have been searching for brown dwarfs for years. How is the search going? About how many have been found so far? How do astronomers distinguish brown dwarfs from low-mass red dwarfs? 3. What is BM Orionis? W here is it located and what does it do? Why might it be interesting to observe with even a small telescope? Exploring The Sky 1. The following nebulae are all star-formation regions: M42, M20, M8, M17. What kind of nebulae are they? Hint: To center on an object use Find under the Edit menu. Choose Messier Objects, and pick from the list. 2. Locate M8 in the sky, zoom in, and identify other nebu- lae in the region. Study the photo of NGC 6559. page 179 Chapter 9 The Formation and Structure of Stars Review Questions 1. Why does helium fusion require a higher temperature than hydrogen fusion? 2. How can the contraction of an inert helium core trigger the ignition of a hydrogen-fusion shell? 3. Why does the expansion of a star's envelope make it cooler and more luminous? 4. Why is degenerate matter so difficult to compress? 5. How does the presence of degenerate matter in a star trigger the helium flash? 6. How can star clusters confirm our theories of stellar evolution? 7. Why don't red dwarfs become giant stars? 8. What causes an aging giant star to produce a planetary nebula? 9. Why can't a white dwarf contract as it cools? What is its fate? 10. Why can't a white dwarf have a mass greater than 1.4 solar masses? 11. How can a star of as much as 8 solar masses form a white dwarf when it dies? 12. How can we understand the Algol paradox? 13. How can the inward collapse of the core of a massive star produce an outward explosion? 14. What is the difference between type I and type II super- novae? 15. What is the difference between a supernova explosion and a nova explosion Discussion Questions 1. How do we know the helium flash occurs if it cannot be observed? Can we accept something as real if we can never observe it? 2. False-color radio images and time exposure photographs of astronomical images show us aspects of nature we can never see with our unaided eyes. Can you think of common images in newspapers or on television that re- veal phenomena we cannot see? Problems 1. About how long would a 0.4-M star spend on the main sequence? Hint: See By the Numbers 9-1. 2. The Ring Nebula in Lyrae is a planetary nebula with an angular diameter of 72 seconds of arc and a distance of 5000 ly. What is its linear diameter? Hint: See By the Numbers 3-1. 3. If the Ring Nebula is expanding at a velocity of 15 km/s, typical of planetary nebulae, how old is it? 4. Suppose a planetary nebula is 1 pc in radius. If the Doppler shifts in its spectrum show it is expanding at 30 km/s, how old is it? Hints: 1 pc equals 3 X 10 13 km, and 1 year equals 3.15 X 10 7 seconds. 5. If a star the size of the sun expands to form a giant 20 times larger in radius, by what factor will its average density decrease? Hint: The volume of a sphere is (4ttr3.) 3 6. If a star the size of the sun collapses to form a white dwarf the size of Earth, by what factor will its density page 202 increase? Hints: The volume of a sphere is (4)ttr3. See 3 Appendix A for the radii of the sun and Earth. 7. The Crab Nebula is now 1.35 pc in radius and is ex- panding at 1400 km/s. About when did the supernova occur? Hint: 1 pc equals 3 X 10 13 km. 8. If the Cygnus Loop is 40 pc in diameter and is 20,000 years old, with what average velocity has it been ex- panding? Hints: 1 pc equals 3 X 10 13 km, and 1 year equals 3.15 X 10 7 seconds. 9. Observations show that the gas ejected from SN1987A is moving at about 10,000 km/s. How long will it take to travel one astronomical unit? one parsec? Hints: 1 AU equals 1.5 X 10 8 km, and 1 pc equals 3 X 10 13 km. 10. If the stars at the turnoff point in a star cluster have masses of about 4 M, how old is the cluster? Critical Inquiries for the W eb 1. As seen in on pages 190 and 191 there is an incredible diversity of appearance for planetary nebulae. Browse the Web for images and information on these dying stars, and discuss why there is a range of shapes of planetary nebulae that we see. 2. Naked-eye supernovae in our galaxy are rare, but astron- omers have noted supernovae in other galaxies for years. Look for summaries of observations of recent supernovae. Are similar numbers of type I and type II being seen? Compare the number of supernovae seen during the last few years with that of two decades ago. Why are we find- ing so many more supernovae in recent years than in the past? Exploring The Sky 1. Locate the planetary nebulae M 57, M 97, and M 27. How does their shape distinguish them from the star formation nebulae such as M 42 and M 8? Hint: To find an object use Find under Edit. Choose Messier Objects, and pick from the list. 2. The Crab Nebulae is M 1. Locate it, zoom in, measure its angular size in seconds of arc, and compute its diame- ter, assuming it is about 6000 ly from Earth. 3. Locate the supernova remnant called the Cygnus Loop just south of e Cygni. How big is this object in angular diameter compared to the full moon? Hint: Under the View menu choose Labels and Setup. 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