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.

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

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