Chapter 7 The Birth and Evolution of Planetary Systems hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY Ideas about the origins of the Sun, the Moon, and Earth objects in the Solar System. It is important for the stu- hn hk io il sy SY are older than written history. Greek and Roman mythol- dents to realize that you can determine many of the prop- ogy, as well as creation myths of the Bible, represent some erties of a planet by knowing its mass, size, and distance hn hk io il sy SY of humanity’s earliest attempts to explain how the heav- from the Sun. hn hk io il sy SY ens and Earth ­were created. Thousands of years and Although planetary scientists are confident they have hn hk io il sy SY the scientific revolution have ultimately debunked many the big picture of Solar System formation correct, the ancient creation stories, but our new theories of solar sys- details are still in question. Questions remain concerning hn hk io il sy SY tem formation are still relatively in their infancy. Jupiter’s formation. It might not have been able to form hn hk io il sy SY The nebular model of solar system formation was first via accretion but may have formed similarly to the Sun. proposed by Immanuel Kant in 1755. It has undergone To complicate matters further, Uranus and Neptune significant revision in the last 250 years, but the details of appear to be too large to have formed at their current posi- the pro­cess remain elusive. In broad strokes, our current tions in the Solar System. Additionally, more than one understanding suggests that the Solar System formed hundred large extrasolar planets have been discovered from a collapsing cloud of interstellar gas and dust. Most in odd orbits about other stars, hinting that perhaps of the infalling material fell to the center of the cloud, our Solar System is not representative of solar systems where it became the proto-­Sun, but a significant fraction throughout the galaxy. As you teach this chapter, be sure of the cloud’s mass was flattened into a disk-­shaped neb- to remind your students that the story being told is still ula surrounding the proto-­Sun. Because of the higher incomplete and probably even partially incorrect. It is, temperatures in the inner nebula, only relatively rare however, a great demonstration of science in action, illus- refractory materials (rock and metal) condensed from the trating how our understanding of the universe changes nebula. However, in the outer nebula, abundant volatile and grows as technology provides us with better observa- materials also condensed as ices. Rock and metal particles tions and more data. It also is very exciting to be able to in the inner nebula accreted into small protoplanetary teach your students about other worlds and the fact that seeds that gave birth to terrestrial planets. In the outer even more are being found as you read this and speak to nebula, ices dominated the larger protoplanetary seeds them. that became the Jovian planets. All of the protoplanets captured hydrogen and helium atmospheres from the Discussion Points nebula; but when nuclear fusion ignited the Sun, powerful solar winds cleared the nebula of loose material and • Based on the amount of material near the Sun, how stripped the small terrestrial planets of their atmospheres. likely is it that the Sun has siblings that formed from Meanwhile, because they ­were larger and farther from the the same parental molecular cloud? Discuss with stu- Sun, the Jovian planets retained their atmospheres despite dents the possibility of stellar siblings of any kind. buffeting from the solar winds. After the Sun cleared the • Have students use the equation for spin angular nebula, protoplanets on crossing orbits swept up any momentum to calculate the values for the Sun, Jupiter, remaining material and collided with each other, ulti- Saturn, Neptune, Earth, Pluto, and a typical molecular mately growing to the full-­sized planets we see today. cloud. Discuss these in the context of conservation of Eventually the volcanically active terrestrial planets angular momentum. (aided by comet impacts that delivered volatile material • Ask students, if they had to choose just one technique from the outer Solar System) outgassed secondary atmo- among the leading choices for finding extrasolar plan- spheres. As implied from the above statements, the envi- ets that might be as habitable as Earth, which one they ronment created by the Sun dictated a great deal of the would pick. What makes that method advantageous properties for each of the planets, as well as the other over the others? —-1 —0 —+1 1

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• Discuss with students the significance of finding as many extrasolar planets as we have so far. Consider the Demonstrations and Activities discovery of Earth-­like planets in habitable zones. Demonstration: Ranking Task for Formation What cultural impact does that have? What questions of Planetary Systems does it provoke? If there ­were life, would we be able to communicate with it? Note to Instructors hn hk io il sy SY Ranking tasks are excellent means of helping students hn hk io il sy SY Exploration think about the progression of events, whether it is in hn hk io il sy SY space, as in sizes of objects, or in time, as in which came Exploration 1: Using the Transit Method hn hk io il sy SY first, second, and so on. This ranking task asks students to Detect Exoplanets to think about the pro­cess of the formation of a star and hn hk io il sy SY Note to Instructors its planetary system. Rather than memorizing the steps, students should visualize the birth of a star and planets hn hk io il sy SY This alternate Exploration is about searching for extra­ in steps that begin from a giant, cool, rotating molecu- solar planets (exoplanets) using transits. It pairs nicely hn hk io il sy SY lar cloud and end with fully formed planets orbiting a with the Exploration in the textbook since Doppler shift hn hk io il sy SY genuine star. This is listed as a demonstration rather is the way astronomers have found most of the known than an activity because students should be able to dis- hn hk io il sy SY extrasolar planets to date, but transits—­specifically cuss, explain, and explore what they think is the correct with the Kepler mission—­are how astronomers are cur- order. rently searching for new extrasolar planets. Transits hold the best potential for finding Earth-­like extrasolar Learning Goals planets. • Chapter learning goal(s) addressed: Summarize the Learning Goals role that gravity, energy, and angular momentum play in the formation of stars and planets. Describe the • Explain thoroughly how transits can be used to dis- modern theory of planetary system formation. cover extrasolar planets. • State what mea­sur­able pa­ram­e­ters transit observations Required Materials can yield. • Cards or strips of paper containing the steps Required Materials • Same steps attached to strips of magnetic sheeting • Computer with Internet access Instructions Pre- Post-­Assessment Question Make enough copies of the list of the stages in star​/planet formation (included with the worksheets for this chap- Assuming the mass of the parent star is known, what ter) to hand out to teams of 3–­5 students. Cut the sheet information about an extrasolar planet can be inferred into strips, shuffle the strips, and place sets of strips into from a mea­sured light curve? envelopes. Students should work in teams to put the stages a. Orbital period into the correct order, according to the current theory b. Planet radius of planet formation. If events occur nearly simultane- c. Orbital radius ously, then those stages should be put into the same pile. d. All of the above can be inferred. After students have had enough time to discuss the order Answer: d of the stages and placed the strips in order from first to last, call on a team to read off their order, or have a rep- Post-­Exploration Debriefing resentative of a team come to the board and place the After completing this activity, have students visit NASA’s magnetic strips in the order his or her team found. Kepler website to research the latest discoveries at htt­ p:// Correct order according to the text (*coeval events): kepler.nasa.gov/Mission/discoveries. 1. Cloud of interstellar gas starts to collapse under the force of its own self-­gravity. Instructions 2. *Gravitational potential energy of collapsing inter- Complete instructions are included on the student stellar gas cloud is converted into heat and radiative -1— worksheet. energy. 0— +1—

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3. *Cloud of interstellar gas rotates faster and faster • Investigate the claim that a Jupiter-­like planet could as it collapses because of conservation of angular not exist at the location of 51 Pegasi b. momentum. • Chapter learning goal(s) addressed: Describe how 4. *Inner parts of flattening cloud begin to fall freely astronomers find planets around other stars and what inward, raining down on growing object at the center. those discoveries tell us about our own and other solar 5. Material makes its final inward plunge, landing on a systems. thin, rotating accretion disk. hn hk io il sy SY 6. Motions push smaller grains of material back and Required Materials hn hk io il sy SY forth past larger grains; smaller grains stick to larger • Scientific calculator hn hk io il sy SY grains. hn hk io il sy SY 7. Planetesimals form that are massive enough to have Instructions gravity that begins to attract nearby bodies. Detailed instructions for this activity are found on the hn hk io il sy SY 8. Planetesimals continue to accrete material until they student worksheet. become large enough to be called planets. hn hk io il sy SY hn hk io il sy SY Activity 2: Surveying the Known Extrasolar hn hk io il sy SY Activity 1: 51 Pegasi: The Discovery Planets of a New Planet hn hk io il sy SY Sekill L vel: Medium Sekill L vel: Medium Note to Instructors Note to Instructors This activity has students compare the characteristics of This activity walks students through a review of the known extrasolar planets to those of the planets in our graphed 1995 discovery data of the first extrasolar planet Solar System by using an interactive table and histograms orbiting a Sun-­like star and interpretation of the radial of all known extrasolar planets. velocity data for 51 Pegasi that led to that discovery. Stu- When introducing this activity, it is worth emphasizing dents determine the planet’s orbital period and the radial that students are using real, cutting-­edge astronomical velocity amplitude of the parent star and then use these data to answer scientific questions; students are imple- data with equations derived from Kepler’s and Newton’s menting the scientific method in much the same way a laws in order to find the radius of the orbit and a lower professional scientist would. limit on the mass of the planet. Students then compare 51 Pegasi b to planets in our Solar System to gain an appre- Learning Goals ciation for how bizarre many of the known extrasolar planets are when compared to what we think of as conven- • Survey the characteristics of known extrasolar tional for planets. planets. The current working theory for how a gaseous planet got • Compare the extrasolar planets to those in our Solar so close to its star is also addressed as students work through System. the same calculations that astronomers used immediately • Use histograms and data tables to analyze the after the discovery. This activity gives students direct discoveries. insight as to how science works when new and unexpected • Chapter learning goal(s) addressed: Describe how discoveries are made. astronomers find planets around other stars and what those discoveries tell us about our own and other solar Learning Goals systems. • Experience the basic steps involved in the pro­cess of Required Materials discovering a planet orbiting another star. • Apply Kepler’s and Newton’s laws to find orbital and • Computer with Internet access physical characteristics of an exoplanet. • Compare an extrasolar planet to the more familiar Instructions planets of our Solar System. Detailed instructions for performing this activity are in­­ • Summarize the current theory on the formation of the cluded on the student worksheet. You may wish to pro- planet orbiting 51 Pegasi. vide students with the scatterplot and two histograms • State how we know the motion of a star. shown on page 4 for the more difficult comparisons. —-1 —0 —+1

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Known exoplanets as of October 5, 2012 2.0 Summary S elf-­Test 1. d, b, f, a, e, c 1.5 2. a, b, h, c, e, d, f, g 3. (c) Since angular momentum L = mvr, we see that hn hk io il sy SY if the distance r is halved, the speed v must be doubled. hn hk io il sy SY 1.0 4. (b) Since the inner material was hotter than the hn hk io il sy SY outer Solar System, inner planets ­were not able to hn hk io il sy SY Semi-major axis (AU) 0.5 capture the large amounts of gas that are present in the Jovians. hn hk io il sy SY 5. (b) It is hard to guarantee that every star has planets, but 0.0 it is also hard to argue that only a few stars have them. hn hk io il sy SY 0.2 0.4 0.6 0.8 1.0 6. (e) While not all methods are very efficient, they have hn hk io il sy SY Mass (M ) Jupiter all been used. hn hk io il sy SY 7. (b) The distance at which planets form from their hn hk io il sy SY central star controls their composition. 8. (a) Large planets close to the star tug the stars by a Known exoplanets as of October 5, 2012 160 large amount with short periods, making them the easiest to detect. 140 9. (c) The disk instability theory proposes that a disk could break up into large planet-­like components, 120 which happens very rapidly, as opposed to the slow 100 pro­cess of accretion. 10. (b) Some books or instructors suggest that if a theory 80 is wrong, it has to be thrown out. Strict adherence to 60 that disproven theory has to be stopped, but the the- Temperature (K) Temperature ory can be modified in a consistent and realistic way 40 to include new observations.

20

0 10–3 10–2 10–1 100 101 102 True/False and Multiple Choice Mass (M ) Jupiter 11. True: Aside from chemical bonds that hold together rocks or chunks of metal, almost everything in the universe that is held together is done so by gravity. 12. True: Gravity is what causes the cloud to collapse and Known exoplanets as of October 5, 2012 what allows larger planetesimals to grow. Angular 500 momentum spins up the protostellar disk as it con- tracts, which increases the rate of collisions of small 400 particles, allowing them to grow into planetesimals. 13. True: Volatile materials turn into liquids and vapors 300 at moderate to high temperatures. 14. True: The same can be said for all stars, too! 15. True: The difference between microlensing and tran- Number 200 sits is that in the latter, the bright object is dimmed; while in the former, the object becomes brighter. 100 16. (a) This is an expression of the conservation of angu- lar momentum. 0 17. (b) In the early Solar System, collisions between par- -1— 0510 15 20 25 30 35 ticles and clumps cause them to stick together and Mass (M ) 0— Jupiter grow in size. +1—

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18. (d) One of the hallmarks of good science is demon- level. Another way to think about this is that con- strated when the same conclusions are reached from served quantities are like a bank, because if you bor- many different avenues. row from the bank, you have to pay it back. Consider 19. (a) As shown in Math Tools 7.1, spin angular momen- ice skaters, who have a fixed amount of energy in their tum scales as L ∝ R2/P so if the radius is cut in half, bodies. If they skate very quickly (i.e., a lot of energy the period must drop by a factor of 22 = 4. of motion), they must get that energy from some- 20. (d) Angular momentum, whether spin or orbital, where, in this case, they convert some of their own hn hk io il sy SY depends on all three factors listed. biochemical energy. An object can always gain or lose hn hk io il sy SY 21. (a) Spectroscopic radial-­velocity mea­sure­ments pick mass, energy, momentum, or angular momentum, hn hk io il sy SY up the Doppler shift of a star being tugged by its orbit- but it has to come from somewhere. hn hk io il sy SY ing planet, much like a very small chihuahua tugs on 30. Spin is a conserved quantity, but it is mea­sured by the its owner when being walked. product of how fast you rotate and how far away you hn hk io il sy SY 22. (b) Volatiles are easily evaporated or broken up with are from your axis of rotation. Thus, when an ice heat, and it was too hot in the inner Solar System for skater spins slowly with her arms extended, she has a hn hk io il sy SY these compounds to be present. fixed amount of spin. As she brings her arms in toward hn hk io il sy SY 23. (d) Remember that the “primary” atmosphere is the her body, that distance gets smaller, so the rate of hn hk io il sy SY one the planet was formed with; the terrestrial plan- rotation has to increase to keep the product of the two hn hk io il sy SY ets all lost theirs soon after formation. constant. Thus, the skater spins much faster. 24. (d) Asteroids and comets are made almost exclusively 31. Just as a skater spins rapidly as she pulls her arms in of the pristine material out of which the Solar System toward her body, so would the Sun that formed at the formed. center of the presolar nebula in early theories of solar 25. (c) Jupiter-­like planets cannot form close to their stars, system formation. These early models did not con- according to our theories of planet formation; there- sider that angular momentum was transferred from fore, they must have migrated inward to be located the collapsing star to the accretion disk and so pre- close to their stars, as we currently find many of them dicted that the star that formed would spin so rapidly to be. it would shred itself apart. 32. An accretion disk is the thin, rotating disk that forms as a gas cloud collapses on itself. It is out of this disk Thinking about the Concepts that the central star and planets form. These disks are also found in a wide variety of astrophysical environ- 26. The hydrogen and most helium atoms that make up ments, as we shall later learn. Broadly speaking, this our Sun and Solar System all came from the Big Bang, disk allows material to spiral into the star by convert- while everything ­else was formed in previous genera- ing its orbital energy into heat. When material spins tions of stars. in a disk around a star, there is friction between the 27. Stellar astronomers looking at young stellar objects different particles, which heats up that material and have noticed that many of them are located within allows it to give off energy. As particles give off that dark, dusty disks, as shown in Figure 7.2. Planetary energy, they spiral deeper down the disk toward the scientists looking at our own Solar System noticed star, eventually landing on the star. that the planets all lie in a disk orbiting in the same 33. Small grains of dust moving about the Sun in similar direction, and by studying meteorites, they found evi- orbits randomly collide with each other at gentle dence of larger objects being built up from smaller speeds. These small grains stick to each other electro- ones. This suggests that our Solar System formed statically to form increasingly larger seeds. As the from a disk of gas and dust. The two findings suggest a seeds grow, gravity becomes more important. Rather common origin for solar systems. than awaiting random collisions, the largest seeds 28. A protoplanetary disk is the spinning disk of gas begin to gravitationally attract nearby particles, and and dust out of which planets form around the cen- their growth rate increases exponentially. Eventually, tral star. The inner part will be hotter because it is the largest seeds reach planetesimal sizes. When plan- closer to the central star and because the inner mate- etesimals collide, the resulting accretion leaves planet-­ rial has gained energy when moving from the outer sized objects. regions in. 34. As I blow “dust bunnies” toward each other (provided 29. “Conservation” of a quantity means that the sum total they don’t fly apart), they tend to become tangled —-1 of that quantity within a system must keep a fixed with each other and grow in size. While on Earth they —0 —+1

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will never grow enough in size to pull on each other in pler shifts of emission lines from the star due to an a meaningful way, this could certainly happen in orbiting planet dragging the star around in space, outer space. The Earth’s gravity and friction with which requires the system’s orbital plane be close to the ground keep the bunnies from forming self-­ edge-­on; (2) transit method: we watch a star regularly gravitating collections. change brightness due to a planet passing in front of 35. Rocky materials are refractory and condense at very the star and eclipsing it; this is only possible if one star hn hk io il sy SY high temperatures. As such, we expect that rocky passes in front of the other, which means the orbital hn hk io il sy SY materials will solidify in all regions of the nebula, plane must be even closer to edge-­on than the first hn hk io il sy SY including near the proto-­Sun. Volatiles condense only method; (3) microlensing method: a planet passes in at very low temperatures. Only the regions of the front of a distant star, and its gravity bends some of hn hk io il sy SY solar nebula far from the proto-­Sun are cool enough the star light toward Earth, making the star brighten for these materials to solidify. This picture is consis- for a short period of time; this requires an exact hn hk io il sy SY tent with modern observations of our Solar System. alignment of the two stars, which is very rare; and hn hk io il sy SY Furthermore, barring other issues (such as migration (4) direct imaging: we take an image that resolves hn hk io il sy SY of giant planets inward owing to complex gravita- the planet from its central star, which is only possi- hn hk io il sy SY tional interactions), we expect the same to be true of ble if the stars are near enough to Earth to be hn hk io il sy SY other planetary systems. resolved. 36. The giant planets had several advantages over the 39. Stars are very bright and far away, while planets terrestrial planets to enhance the growth of their appear very close to their stars. Thus, it is difficult to atmospheres. (1) Giant planet seeds formed from mask out the light of the star and still see close enough more abundant volatile materials, whereas terrestrial to it to see the reflected light of a planet. planet seeds formed from the much less abundant 40. We have only detected planets around a few hundred refractory materials. (2) Giant planets had a much stars, which is a very small sample of the stars in the larger area from which to accumulate raw materials, Milky Way, and not all of them look like our Solar whereas terrestrial planets could draw material only System. In par­tic­u­lar, our technology for detecting from a narrow zone in the immediate vicinity of the extrasolar planets is biased toward detecting giant forming proto-­Sun. (3) Accreting in the outer nebula planets in close orbits about their parent stars. We are left giant planets less vulnerable to the powerful solar only now finding Earth-­like planets in Earth-­like winds emanating from the proto-­Sun; the terrestrial orbits. Based on this sample and our current techno- planets ­were scoured clean due to their proximity to logical limitations, it is too early to make any general the proto-­Sun. Because of their ability to quickly conclusions about whether our Solar System is grow very large and the lower temperature of the gas, unusual. the Jovian planets captured and retained significant amounts of the primordial hydrogen and helium from the solar nebula. Meanwhile, the terrestrial Applying the Concepts planets ­were too small and too exposed to the young Sun to capture or retain significant atmospheres 41. Setup: The conversion from m/s to mph is m km mi 3,600 s from the solar nebula where the gases ­were hotter. 1 ×=2.3mph 37. Debris from the formation of the Solar System exists s 1,000 m 1.6km hr in the form of asteroids and comets (some of which Solve: In Figure 7.20, the maximum radial velocity is still occasionally cross planetary orbits). Jupiter keeps about 35 m/s, or 35 × 2.3 = 80.5 mph, which is a fac- the outer regions of the terrestrial planet zone gravita- 67,000 tor of = 832 smaller than the Earth’s orbit tionally stirred, thereby preventing the asteroids in 80.5 the asteroid belt from accreting into another terres- around the Sun. trial planet. The Oort Cloud was formed when the Review: As expected, a star moves much less than the giant planets ejected comets to the fringes of the Solar planet orbiting it. We see ­here that astronomers are System. Finally, the Kuiper Belt is composed of icy indeed detecting very small radial motions in a star to planetesimals that ­were too sparsely distributed for detect the planet orbiting it. accretion into large worlds. 42. Setup: The table gives us values of planetary masses 38. The four methods that astronomers currently use to in terms of Earth’s mass, so for this question we must -1— search for extrasolar planets are (1) spectroscopic sum up the planet masses and then compare them to 0— radial-­velocity method: we watch the periodic Dop- Jupiter and Earth. +1—

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Solve: (a) The sum of the eight planets in our Solar Sys- about its axis, which is why it has so little spin angular

tem is Mp = (0.055 + 0.815 + 1.00 + 0.107 + 317.83 + momentum. 95.16 + 14.54 + 17.15) M ⊕ = 446.66 M⊕. (b) To find Jupi- 45. Setup: For this problem, we will use the equation of ter’s percentage of that mass, divide its mass by the total: orbital angular momentum Lorb = mvr. However, instead M 317. 83ᮍ of computing L for each planet, let’s directly compare J ==⇒ (c) The same 0..7116 71 16%. the values of both by constructing a ratio. This has MP 446. 66ᮍ Mᮍᮍ1.00M two added advantages: (1) it gets rid of the constants hn hk io il sy SY for Earth yields ==0.0022⇒ 0.22% in our formula, and (2) we can use what­ever units we hn hk io il sy SY MP 446.66Mᮍ want as long as we are consistent, that is, masses in hn hk io il sy SY Reviewts : I i no surprise that the Solar System’s plan- terms of Earth mass, speeds in km/s, and distance in hn hk io il sy SY etary mass is dominated by Jupiter, given how large astronomical units. and massive it is compared to everything ­else. We can       simply see from inspection that almost all the mass is LJJm v JJr 1331./km s hn hk io il sy SY Solve: =       =×318 ×=52. 727 in Jupiter and Saturn. Lᮍᮍ m   vᮍᮍ  r  29./8 km s hn hk io il sy SY 43. Setup: For this problem, we will use the equation of LJJ m   v JJ  r  1331./km s hn hk io il sy SY 4πmR 2Solve: =       =×318 ×=52. 727 spin angular momentum L = and orbitalL m v r 29./8 km s hn hk io il sy SY spin 5P ᮍᮍ   ᮍᮍ   angular momentum Lorb = mvr. Remember to use con- Review: Given Jupiter’s huge mass and greater dis- hn hk io il sy SY sistent units (i.e., time in seconds, distance in meters). tance from the Sun, it is no wonder that it has so much Solve: For Earth’s spin angular momentum, more angular momentum around the Sun than the P = 24 h = 86,400 s, so Earth. 45π ⋅×.971024 kg ⋅×(.6 378 1062m) 46. Setup: Conservation of angular momentum states L = = 7.001× 033 kg ms2/ spin 58⋅ 6,400 s that the cloud’s angular momentum just before it col- 45π ⋅×.971024 kg ⋅×(.6 378 1062m) lapses must equal its angular momentum after it has L = = 7.001× 033 kg ms2/ spin 58⋅ 6,400 s become the Sun. Expressing this as an equation, we By comparison, using 1 AU = 1.5 × 1011 m 2 2 4πmRcloudcloud 4πmR 24 3 11 L = mvr = 5.97 × 10 kg ⋅ 29.8 × 10 5m/s ⋅ 1. × 10 have LLcloud ==or . orb 5P 5P = 2.7 × 1040 kg m2/s which is about 3.8 million times cloud  Solve: Since a large percentage of the cloud becomes more than the spin angular momentum. the Sun, we will assume the mass does not change. Review: Recall that angular momentum depends on Solving above, we have how fast an object is spinning and how far away that 2 2 object is from its spin axis. Thus, it stands to reason  R   14. ×109 m  P =  P = 10668yr =×19..6100− yr = 6 s. that all the angular momentum of a planet is in its    cloud  16  2  Rcloud  10 m orbit, since the planet’s distance from the Sun is so 9 2 R  14. ×10 m  668− much greater than its size.P =   Pcloud =   10 yr =×19..6100yr = 6 s.  R   1016 m  44. Setup: For this problem, we clouwidll use the equation Review: Recall that angular momentum depends on 4πmR 2 of spin angular momentum Lspin = ,however how fast an object is spinning and how far that object 5P is away from its spin axis. Thus it stands to reason that instead of computing the spin for each planet, let’s a huge cloud that collapses into a star would spin at an directly compare the spins of both by constructing a almost mind-­numbing rate, if there ­were no means by ratio. This has two added advantages: (1) it gets rid of which that cloud could transfer its angular momen- the constants in our formula, and (2) we can use what­ tum away from the central object. Thank goodness for ever units we want as long as we are consistent, that is, accretion disks. masses in terms of Earth mass and periods in days. 2 47. Setup: To compute the density of Vesta, we will assume L  m   R   Pᮍ  that1 Vesta is a sphere. The volume of a sphere is Solve:0VenusV=  enus   Venus  =×.815 0.9492 ×=0.003   4 33π mass Lᮍᮍ m   Rᮍ   PVenus  243πrd= . Density is ρ = . 2 36 volume      P  LVenusVm enus RVenus ᮍ 2 1 Solve:0=       =×.815 0.949 ×=0.003 L m R P 243 Solve: (a) The density of Vesta is ᮍᮍ   ᮍ   Ve nus  20 Review: Although Venus is practically Earth’s twin 2.71× 0kg ρ = = 3,464 kg/m 3. —-1 in terms of mass and size, Venus rotates very slowly π × 53 (/6)(5.3 10 m) —0 —+1

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(b) Vesta’s density is significantly higher than that of 51. Setup: This question asks for distance from a star water or rock. This implies that Vesta’s composition knowing the orbital period, so we must use Kepler’s must include a significant component of metal in third law. Since the star’s mass is that of the Sun, this addition to rock. is simply P2 = A3; however, we are given the period in Review: This confirms what we have found in mete- days, so we must convert to years. yr ors that have landed on Earth: they are mostly com- Solve: (a) P =×3.525 day =×9.65 10−3 yr, hn hk io il sy SY posed of metals. 365.25 day hn hk io il sy SY 48. Setup: The Sun’s brightness will drop in propor- so solving for distance we find the planet is = 2⁄3 = −3 2⁄3 = hn hk io il sy SY tion to how much area of the Sun is blocked. This A P 0 (9.65 × 1 ) 0.045 AU from HD 209458. ignores limb darkening, which will be discussed in (b) Mercury is located at 0.387 AU from the Sun. hn hk io il sy SY chapter 14. That is 8.5 times farther from the Sun than Osiris is Solve: The Sun’s radius is 7 × 108 m, and Jupiter’s radius from HD 209458! Osiris is very likely to be tidally hn hk io il sy SY is 7.2 × 107 m. Since area is proportional to radius locked to its parent star, with one side of the planet hn hk io il sy SY 2  7.21× 07  experiencing temperatures much higher than any hn hk io il sy SY squared, Jupiter’s area is   ≅ 0.01 as much  71× 08  planet in our Solar System. Because Osiris is a gas giant, hn hk io il sy SY the thick atmosphere of the planet probably transports as the Sun. So the Sun’s light would drop by this the heat efficiently to the dark side of the planet. hn hk io il sy SY amount, or about 1 percent. Review: Do we expect Osiris to be very close to its Review: Aliens should easily be able to detect this star? Compare its period to that of Mercury (about drop with even crude instruments. However, this drop one-­half year), and you find that it must indeed have a only happens once every 11.8 years, so they will have very small distance to have such a short period. to watch the Sun very carefully and for a long time to 52. Setup: The dimming of the light from HD 209458 is detect it. proportional to the cross-­sectional area of Osiris. 49. Setup: Doppler shifting tells us that the relative shift Expressed as an equation, we can write this as of a line away from its center equals the relative speed A D2 of the moving object compared to the speed of the Dimming ==Osiris Osiris where A represents the A D2 wave. In this case, the wave is light, so our formula star star ⌬λ v cross-­sectional area of the object and D represents the can be written = . object’s diameter. λ c 0 Solve: (a) Rewriting our dimming equation to solve Solve: Solve our Doppler shift equation for for the diameter of Osiris and substituting in appro- v 1ms/ ==×=65× ⌬λλ== =×−6 priate numbers yields DDOsiris star Dimming 1.710km0.017 2.210km. 0 575 nm 8 19..210 nm c 31× 0 ms/ 65 DDOsiris ==star Dimming 1.71×=0km0.017 2.21× 0km. (b) Taking the ratio Review: This shows us that to observe Doppler shifts of the diameter of Osiris to Jupiter, we find that in visible light, large velocities are required. D 2.21× 0k5 m Osiris = =1.585. Osiris is approximately 50. Setup: Doppler shifting tells us that the relative shift 5 DJ 1.41× 0km of a line away from its center equals the relative speed 59 percent larger than Jupiter. of the moving object compared to the speed of the Review: From Problem 51, we know that Osiris is wave. In this case, the wave is light, so our formula extremely close to its host star, and from this problem ⌬λ v can be written = . and in this problem we are we know it is larger than Jupiter, which shows us that a λ0 c very large planet is orbiting extremely close to its star. looking for the shift Δ l away from a central wave- Certainly this begs the question, how can a giant planet be right next to its host star if our planet forma- length l0 = 500 nm. Solve: Solve our Doppler-­shift equation for tion scenario requires that giant planets only form far from their stars? v 0.09 m/s ⌬λλ==500 nm =×1.510n−7 m. 53. Setup: Following Math Tools 7.3, we can estimate 0 c 31× 0m8 /s the percentage reduction of light during a transit as 2 R planet Review: This is much smaller than the size of sub- . In this problem, we are given units in terms −15­ 2 atomic particles (10 m) and thus shows how nearly Rstar -1— hopeless it is to ever be able to detect such a gravita- of a solar radius (7 × 108 m) and Earth’s radius 0— tional signal with radial velocities. (6.4 × 106 m). +1—

577-53231_ch01_4P.indd 8 3/9/13 8:16 AM Chapter 7 The Birth and Evolution of Planetary Systems ◆ 9

Solve: Given the values in the problem, the per­ mum mass, the semimajor axis, period, and eccentric- centage decrease in light during the transit is ity of its orbit, its density, and radius if available, and (4.5⋅×6.410)62 whether it is Jovian or terrestrial. =×1.410−3 or 0.14%. (1.1⋅×710)82 58. (a) At the time of this writing, Kepler has confirmed Review: Transits of planets should produce very 105 planets and has 2,740 candidates. Follow-­up obser- small drops in brightness given how tiny the planet is vations consist of direct imaging of the star to rule out compared to the star, as we have found ­here. Note that eclipsing binaries and spectroscopy to determine the hn hk io il sy SY in Math Tools 7.3, we see that Kepler-­11C produces a spectral type of the star, to rule out eclipsing binaries, hn hk io il sy SY drop of only 0.0008 or 0.08%, which confirms that and to try to determine the planet’s orbital details hn hk io il sy SY through radial-­velocity mea­sure­ments. (b) Answers this planet is larger than Kepler-­11C. hn hk io il sy SY 54. Setup: Volume scales as size V ∝ R3 and density will vary. 59. Answers will vary. The answer will include why it is ρ = M / V. hn hk io il sy SY Solve: (a) Since volume scales with size cubed, the better to have many people looking at data, rather hn hk io il sy SY ratio of the volumes will be the cube of the ratios of than one person, and a copy of any stars observed for 3  1.7 the citizen-­science project. hn hk io il sy SY the sizes, i.e., = 4.9 or the new planet has  1  60. Gaia will detect the wobble of stars being tugged by hn hk io il sy SY 4.9 times more volume. If the density is the same, then their planets, as well as transits. The science goals hn hk io il sy SY M ∝ V so the new planet has 4.9 times more mass as include confirmation of suspected planets, a more com- well. plete census of nearby faint stars, mea­sure­ment of the Review: Note that part (b) is based on an assumption orbital planes of planets, mass estimates, and creation of that may not be true. However, if we can use the orbital a database of targets for upcoming planet-­finding mis- details of the planet to find its mass, we can combine sions. The launch, on a Soyuz-­STB/Fregat rocket from these data to find the density. Kourou, French Guiana, is delayed until October 2013. 55. Setup: We are comparing to Jupiter, so note that 27 7 MJ = 1.9 × 10 kg and Rj = 7.15 × 10 m Solve: 27 27 Exploration (a) M = 2.33 MJ = 2.33 ⋅ 1.9 × 10 kg = 4.43 × 10 kg. 7 8 1. Earth’s orbital eccentricity is very small, meaning the (b) R = 1.43 RJ = 1.43 ⋅ 7.14 × 10 m = 1.02 × 10 m. 4 4 orbit is close to circular. (c) ==ππ38×=32× 43 VR (1.0210m)4.4410m. 2. Mercury has the most eccentric orbit, differing from a 3 3 circle by about 20 percent. M 4.43× 1027 kg ρ == = 3 3. In general, our planets have mostly round orbits. (d) 24 3 0.997kg/m , which is V 4.44× 10 m 4. Mercury has the largest orbital inclination. just slightly less than the density of water. This planet 5. The inclination of orbits is defined relative to ours. must be gaseous. 6. The Solar System is a flat disk, like a pancake. Review: We expect that planets this large and mas- 7. Venus rotates clockwise, i.e., opposite all other planets. sive will all be gas giants, and the density we found 8. The rotations of the Solar System bodies suggest that confirms this expectation. they all formed together at the same time from the same body. Using the Web 9. The Solar System is revolving counterclockwise and mostly rotating in that same direction. 56. Answers will vary. The answer will include what meth- 10. Our Solar System today is ordered the same as a flat ods are used to detect planets from the ground and spinning disk of gas, which, we believe, happens when from space, whether planets have been found in each of gas clouds collapse. two projects, and if any ­were, what types of planets they 11. If there ­were insufficient gravity in the cloud, only a are. A future project will also be presented, including star, and no planets, might form. when it will begin and what method(s) it will use. 12. If angular momentum is not conserved, the system 57. Answers will vary. Part (a) will show a graph showing might not flatten to a disk, and only a star, and no the distances of exoplanets from its central star, not- planets, might form. ing the masses and sizes of each planet, and compar- 13. A giant impact by a planet-­sized body might have ing them to our solar system. (b) will give a recently affected Venus such that its original rotation (in a —-1 discovered and published planet, including its mini- CCW sense) was halted and slowly reversed (CW). —0 —+1

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EXPLORATION: Using the Transit Method to Detect Exoplanets Go to the StudySpace at wwnorton​.com/studyspace and open the “Exoplanet Transit Simulator” from the list of Inter- active Simulations for Chapter 7. hn hk io il sy SY On the light curve graph, move the red line back and forth to see the correlation between the graph and a planet’s hn hk io il sy SY motion. hn hk io il sy SY hn hk io il sy SY 1. What is happening when the planet’s brightness drops? hn hk io il sy SY hn hk io il sy SY 2. Why ­doesn’t the brightness drop instantly? hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY

Vary the radius of the planet (in the “Planet Properties” box) and observe the effect this has on the light curve.

3. Can a planet’s radius be inferred from the light curve data? Explain why or why not, supporting your answer.

4. Can a planet’s orbital period be inferred from a light curve? Explain why or why not, supporting your answer.

5. Assume the parent star’s mass is known. How can Kepler’s laws be used to find the planet’s orbital radius?

Planetary systems are oriented randomly with respect to our point of view, so they aren’t always viewed edge-­on. Play with the “Inclination” slider (in the “System Orientation and Phase” box) and see what happens to the light curve when a planet is not edge-­on.

6. Can transits be used to find all extrasolar planets? Explain why or why not, supporting your answer.

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577-53231_ch01_4P.indd 12 3/9/13 8:16 AM DEMONSTRATION: Ranking Task for Formation of Planetary Systems

Cloud of interstellar gas starts to collapse under the force of its own self-­gravity. hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY Gravitational potential energy of collapsing interstellar gas cloud is converted into heat and radiative energy. hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY Cloud of interstellar gas rotates faster and faster as it hn hk io il sy SY collapses because of conservation of angular momentum.

Inner parts of flattening cloud begin to fall freely inward, raining down on growing object at the center.

Material makes its final inward plunge, landing on a thin, rotating accretion disk.

Motions push smaller grains of material back and forth past larger grains; smaller grains stick to larger grains.

Planetesimals form that are massive enough to have gravity that begins to attract nearby bodies.

Planetesimals continue to accrete material until they become large enough to be called planets.

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ACTIVITY: 51 Pegasi: The Discovery of a New Planet In just the past few years, astronomers have announced discoveries of hundreds of planets orbiting nearby stars. These discoveries seem to finally answer the question of whether or not our solar system is unique. We should note, however, that when astronomers state that they have discovered a new planet, what they are really saying is that their data can hn hk io il sy SY best be interpreted as a planet orbiting a star. One cannot “prove” that these other planets exist; one can only state that, hn hk io il sy SY until the hypothesis is disproved, a planet orbiting the star best explains the observations. We can mea­sure only indi- hn hk io il sy SY rectly the influence each one has on its parent star as the star and planet orbit their common center of mass. The planet hn hk io il sy SY makes the star “wobble,” so we can use the Doppler method to detect it. That is the method we explore ­here. We enter this realm of discovery by working with actual discovery data from observations of the star 51 Pegasi made at the hn hk io il sy SY Lick Observatory in California. These data are the mea­sure­ments of the Doppler shift of the wavelengths of the absorp- hn hk io il sy SY tion lines seen in the spectra of 51 Peg. hn hk io il sy SY hn hk io il sy SY Observations hn hk io il sy SY Take a look at the graph shown ­here. It shows the mea­sured radial velocities as a function of time recorded in days. As you can see, the radial velocities are sometimes positive (the light is redshifted) and sometimesnegative (the light is blueshifted), indicating that sometimes the star is receding (redshifted) from us and sometimes approaching us (blueshifted) when viewed from our frame of reference, Earth. This wobble of the star was the first indication that the star 51 Pegasi had an invisible companion.

100 1 cycle 2 cycles etc.

50

0 Velocity (m/s) Velocity –50

–100 51015202530 Time (days) Figure 1 Discovery data for the planet orbiting the star 51 Pegasi.

The above plot represents the observed radial velocities over a period of about 33 days. The data ­were obtained by mea­ sur­ing the Doppler shift for the star using the formula

λλshiftr− est vc=× radial λ (Eqn. 1) rest That is, we simply find out what the shift in the wavelength is compared to the rest wavelength and multiply that num- ber by the speed of light. For example, let’s say we mea­sured a spectral line of hydrogen that should be at a wave- length of 656.3 nm (10−­9 m), and find it at a wavelength of 656.3001 nm. Then, from Eqn. 1 and a value for the speed of light of 3 × 108 m/s, we get  656.. 3001− 656 3 ()31××088  =×(/3101ms)(× .. 51×≈04−7 )/6 ms —-1  656.3  —0 —+1

577-53231_ch01_4P.indd 15 3/9/13 8:16 AM Here, c is the speed of light and v is the velocity of the object being observed. Note that the difference in the shifted wave- length and the rest wavelength is extremely small for these planetary detections and that the ratio of the change in the wavelength to the rest wavelength means the units come out to be what­ever units in which the speed of light is expressed.

Procedure and Questions hn hk io il sy SY 1. A period is defined as one complete cycle, that is, where the radial velocities return to the same position on the hn hk io il sy SY curve but at a later time. How many cycles did the star go through during the 33 or so days of observations? hn hk io il sy SY Number of cycles = ______hn hk io il sy SY 2. What is the period, P, in days of one complete cycle? (Number of days for these observations divided by number of hn hk io il sy SY cycles.) = hn hk io il sy SY Period ______days hn hk io il sy SY 3. What is P in years? (Hint: divide the period in days by the number of days in a year; the answer will be a decimal hn hk io il sy SY number smaller than 1.) hn hk io il sy SY P = ______years 4. What is the uncertainty in your determination of the period? That is, by how many days or fractions of a day could your value be wrong? (This is a number that you decide. There is no set “rule” for this, but the uncertainty has to be less than a day ­here.) Uncertainty = ______days

1 5. What is the amplitude, K? To find this, take 2 of the value of the full range of the velocities. For example, if the 1 1 velocities went from +12.4 m/s to – 13.8 m/s, 2 of the full range (26.2 m/s) would be 2 of 26.2 m/s = 13.1 m/s.

K = ______m/s

6. How accurate is your determination of this value? Uncertainty = ______m/s

We make some assumptions in order to simplify the equations we have to use for determining the mass of the planet. In this case, we use Jupiter as our reference planet. The equation we use is 1 3 M planet  PK   =   ×   (Eqn. 2) M  12  13 Jupiter P should be expressed as a fraction of a year, and K in m/s. Twelve years is the approximate orbital period for Jupiter and 13 m/s is the magnitude of the “wobble” of the Sun caused by Jupiter’s gravitational pull. The answer we get is the

ratio of the mass of the planet (Mplanet) to the mass of Jupiter (MJupiter), since that is so much easier to envision than a number times 10 raised to the power of 27. 7. Put in your values for P and K and calculate the mass of this new planet in terms of the mass of Jupiter. That is, your cal- culations will give the mass of the planet as some factor times the mass of Jupiter. Astronomers like doing these kinds of direct ratios so that the large numbers basically cancel each other, as do all units like m/s, and seconds, and years.

Mplanet = ______MJupiter

We assume that the parent star is 1 solar mass and that the planet is much, much less massive than the star (the case with our solar system). We can then calculate the distance this planet is away from its star, in astronomical units, using Kepler’s third law: -1— 3 a 1 = or 3 = 2 or aP=()2 3 0— 2 1 a P (Eqn. 3) +1— P

577-53231_ch01_4P.indd 16 3/9/13 8:16 AM 8. Again, P is expressed as a fraction of a year, and a represents the AU’s. Solve for a: a = ______AU

9. Do a search for the actual values for the following quantities and list them in column 2, giving your references in the space provided. List your results in column 3.

Characteristic Published Value My Value hn hk io il sy SY Mass hn hk io il sy SY Period (P) in days hn hk io il sy SY Amplitude (K) in m/s hn hk io il sy SY Distance from star (AU) References: hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY How close ­were you for all of these values? (Be sure to list all quantities.) hn hk io il sy SY 10. Compare the orbit of this planet to those in our Solar System by referring to this rough scale model. Mercury is hn hk io il sy SY 0.4 AU from the Sun; Venus, 0.7 AU; Earth, 1.0 AU; Mars, 1.5 AU; Jupiter, about 5 AU. Where would the new planet fit in if it­we re in our Solar System?

MSun VE M J 11. Science is based on the ability to predict outcomes. However, nothing prepared astronomers for the characteris- tics of this “new” Solar System. Consider the mass of this planet as well as its distance from its star. Why was the discovery such a surprise when compared to our Solar System?

12. If this actually is a planet, is it possibly hospitable to life? Comment on what it would be like on this planet.

13. Summarize the current working theory on how the planet orbiting 51 Pegasi came to rest so close to its planet. Use what­ever information sources are available to you.

14. In your own words, summarize how astronomers determine the velocity of a star and how they know when the star is coming toward us and when it is going away from us based on the spectrum of the star.

One of the original pessimistic views held by many astronomers after this discovery was that this could not be a gas- eous planet so close to its star because the planet would not have been able to hold onto its atmosphere for billions of years. We can test that hypothesis with a few simple equations. First, let’s just give this planet the mass and radius of Jupiter. We know that Jupiter held onto its atmosphere from when and where it formed. The question becomes: would Jupiter have been able to hold onto its atmosphere if it migrated to the inner part of our solar system to ~0.05 Au?

15. If we graph the approximate temperature versus distance from the Sun, we can extrapolate to the temperature at the distance that this planet is away from its star. The fit is definitely a power function (see Eqn. 4); the temperature is proportional to the inverse distance squared. What would be the approximate temperature in the inner part of the planetary system, 0.05 AU, where this planet is located? T = ______K —-1 —0 —+1

577-53231_ch01_4P.indd 17 3/9/13 8:17 AM 389 (Eqn. 4) TK()= D(AU)

Temperature of the Solar System vs distance from the Sun 700

600 hn hk io il sy SY 500 hn hk io il sy SY hn hk io il sy SY 400 hn hk io il sy SY 300 Temperature (K) Temperature hn hk io il sy SY 200 hn hk io il sy SY 100 hn hk io il sy SY 0 036912 15 18 21 24 27 30 hn hk io il sy SY Distance from Sun (AU) hn hk io il sy SY Figure 2. Temperature of the Solar System vs. distance from the Sun. Temperatures ­were approximated using the Stefan-­ Boltzmann law, a surface temperature of the Sun of 5700 K, solar radius = 7.0 × 10 8, 1 AU = 1.5 × 10 11.

We need to find out if this temperature is great enough to evaporate the planet’s atmosphere, and that involves deter- mining the velocity of the gas in its atmosphere versus the escape velocity of the planet.

2GM Escape velocity of Jupiter: v =≈60,/ 000 ms (Eqn. 5) escape r The condition for a planet to hold onto its atmosphere for billions of years is that the gas velocity must be less than 1 one-­sixth of the escape velocity of the planet: vv< . (Eqn. 6) gase6 scape temperature The velocity of the gas can be calculated by vm(/s)=157 (Eqn. 7) gas molecule,smass 16. Assume that the planet’s atmosphere is primarily molecular hydrogen with a molecular mass of 2. Then answer this question: If the planet orbiting 51 Pegasi formed at a distance where it had a substantial molecular hydrogen atmo- sphere, would the planet retain that atmosphere at its current distance of 0.05 AU.? Show all calculations ­here.

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ACTIVITY: Surveying the Known Extrasolar Planets Go to the Extrasolar Planets Encyclopaedia website at ­http://​exoplanet​.eu​/catalog hn hk io il sy SY Part 1: The Data Table hn hk io il sy SY The planets at the Extrasolar Planets Encyclopaedia can be sorted by pa­ram­e­ters by clicking on a column heading. The hn hk io il sy SY mass and radius of a planet are expressed as comparisons to Jupiter. The period is expressed in Earth days, and the dis- hn hk io il sy SY tance from its star (a) is expressed in astronomical units. Sort the data appropriately to answer the questions below. 1. How many planetary systems, planets, and multiple planet systems does this cata­log include? hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY 2. What was the least massive planet discovered most recently? hn hk io il sy SY 3. What is the mass of that planet in terms of Jupiter’s mass? hn hk io il sy SY 4. The distance of that planet from its star is AU, and it takes days or years for 1 orbit. 5. List the name, mass, period, orbital distance (a), and eccentricity (e) of the planet that was most recently discov- ered (and has values for all of these characteristics) by clicking on Col. 10, Discovery.

6. Find the planet HD 180314b, which orbits star HD 180314, by clicking on the star’s name. Note the masses of the planet and its star. Also find the orbital period in years and the distance the planet is from its star. Does the relation- ship between the period and orbital distance of this exoplanet obey Kepler’s third law, which assumes the star is Sun-­like and contains 99.99% of the mass of the planetary system? Click on the planet name to find out more about the parent star to understand any discrepancies and summarize ­here. Show any calculations ­here.

7. For the next question, you will need to set criteria for determining whether a planet is Earth-­like in terms of mass and Earth-­like in terms of orbital distance. What numbers are you going to use?

One of the goals of extrasolar planet hunters is to find other Earths. Since the masses of the planets in this database are expressed in terms of Jupiter, you need the fact that Earth’s mass is 0.003 that of Jupiter’s. Be sure to use the criteria you set above; you are free to change your numbers, but if you do, make sure you note that.

8. How many extrasolar planets are Earth-­like in terms of mass? How many extrasolar planets are Earth-­like in terms of orbital distance (Col. 5)? How many extrasolar planets are Earth-­like in terms of both of these pa­ram­e­ters? 9. How many of the known extrasolar planets orbit their parent star closer than Mercury orbits our Sun (0.39 AU)? What percent of the known extrasolar planets is this?

10. How many of the known extrasolar planets have masses that are 1 MJupiter or more? What percent of the known extrasolar planets is this? (Hint: Figure out how many are listed on each screen and then multiply by how many screens it takes for the list. A rough estimate is fine.)

Part 2: Histograms For this part, you need to click on “Diagrams” in the top menu. Also choose “Histogram plot” shown just under the top menu. Select the category you want for the x-axis. (Note: You can get a better estimate for these answers by —-1 clicking on log scale for the x-axis and remembering that 100 equals 1.) —0 —+1

577-53231_ch01_4P.indd 19 3/9/13 8:17 AM 11. Approximately how many of the extrasolar planets have masses that are 10 MJupiter or less? What percent of the known extrasolar planets is this?

Part 3: Scatter Plots Change your option to “Scatter plot.” hn hk io il sy SY hn hk io il sy SY 12. Is the typical extrasolar planet similar to Earth? To answer this question, set the x-axis Planetary Mass max value to 0.01. Earth’s mass is roughly 0.003 that of Jupiter’s. Set y-axis Semi-­Major Axis min to 0.4 and max to 2. hn hk io il sy SY Explain using your analysis of the scatter plot. It is best to use log scales for both axes and remember that 100 hn hk io il sy SY equals 1 and 10−­2 equals 0.01. Compare your answer ­here to your answer in question 7. hn hk io il sy SY hn hk io il sy SY 13. Choose x-axis “Year of discovery” and Y axis “Distance to a host star” and plot. What does this scatter plot tell you hn hk io il sy SY about our rate of discoveries (be sure to unclick log scale)? What is the distance at which the number of discoveries hn hk io il sy SY drops off significantly (state both in parsecs, pc, and light years)? Extrapolate the information into the future. ­Predict what would happen to our rate of discoveries if we ­were able to accurately detect planets past 1000 pc hn hk io il sy SY (3,260 light-years).

14. Pick two of the characteristics of the exoplanets from the many options listed and create a scatter plot from those data. State what you graphed and summarize what you learned from that graph.

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hn hk io il sy SY The stars are so far away that we see them only as pin- matter was poorly understood at the time. In 1901 Annie hn hk io il sy SY points of light, even through telescopes. Apparent bright- Jump Cannon made some progress by noting that the 22 hn hk io il sy SY ness and color are the only semiquantifiable mea­sures of known stellar spectral classes could be combined into 7 hn hk io il sy SY stars that our eyes can perceive directly. It is not at all and reordered as a temperature sequence. With the dis- surprising that even as recently as the turn of the 20th coveries of the electron by J. J. Thompson in 1898 and the hn hk io il sy SY century, astronomers knew almost as little about stars as atomic nucleus by Ernest Rutherford in 1911, the general hn hk io il sy SY the ancient Greeks of 2,500 years ago. Yet with some structure of the atom was determined. Work in the bur- clever human ingenuity, combined with ever-­more-­ geoning field of quantum mechanics in the 1920s led to a hn hk io il sy SY sophisticated technology, 21st-­century astronomers rou- more complete understanding of how light interacts with hn hk io il sy SY tinely mea­sure fundamental properties of stars like atoms, finally providing the Rosetta stone astronomers hn hk io il sy SY distance, size, luminosity, surface temperature, composi- needed to fully interpret stellar spectra. Today we can tion, and mass. Today we plot these properties against quickly determine the surface temperature of a star by each other using the Hertzsprung-­Russell diagram (also noting the relative strengths of par­tic­u­lar spectral lines. known as the H‑R diagram) to develop an understanding We have also learned how to uncover compositional dif- of stars that would have astonished astronomers of just a ferences from stellar spectra, leading to the discovery of century ago. Two important breakthroughs in the previ- two stellar populations: metal-­rich and metal-­poor stars. ous century set the stage for our modern understanding of Our ability to interpret stellar spectra has matured to the stars: high-­precision parallax mea­sure­ments and the mat- point that we can use subtle gradations in line strength to uration of the field of spectroscopy. subdivide Annie Cannon’s seven spectral classes into Mea­sur­ing distances across the Solar System with paral- nine subclasses each. We can now literally distinguish lax is easy to do. Even the most distant planets show mea­ two separate subclasses of stars whose surface tempera- sur­able motion across the sky over a few weeks. However, tures may vary by as little as 100 Kelvin! the stars are another matter entirely. The nearest bright In the early 1900s, the astronomical toolbox expanded star, Alpha Centauri, lies 4.4 light-­years from the Sun. That dramatically. The most important tool for stellar astrono- is nearly 7,000 times farther than Pluto! A parallax-­second mers joined the toolbox in 1910, when Ejnar Hertzsprung (or parsec) is defined to be the distance at which Earth’s and Henry Norris Russell in­de­pen­dently discovered a maximum angular separation from the Sun is 1 arcsecond relationship between a star’s luminosity and its surface tem- (1/3,600°). This angle is so small that we would have to perature. The relationship is best illustrated by a diagram travel 3.26 light-­years from the Sun before the radius of that bears the names of its discoverers: the Hertzsprung-­ Earth’s orbit would be contained in such a narrow angle of Russell diagram. Plotting a star’s absolute magnitude or sky. Yet the nearest star resides 30 percent farther still. At stellar luminosity on the vertical axis against its surface Alpha Centauri’s distance of 4.4 light-­years (1.3 parsecs), temperature (in reverse), color, or spectral classification the entire diameter of Earth’s orbit subtends an angle of on the horizontal axis reveals that 90 percent of stars fall only 1.54 arcseconds (1/2,340°). Shifting to the perspec- along a narrow band from the upper left to the lower right tive of an Earthbound observer, it is Alpha Centauri that of the diagram. This band is called themain sequence, and displays that maximum parallax shift of 1.54 arcseconds stars spend approximately 90 percent of their lives in this when viewed from opposite ends of Earth’s orbit. This shift stage. The remaining stars generally fall into smaller clus- is so tiny that it was not mea­sur­able until the middle of the ters on the diagram. Stars in their finalred giant or red 19th century! Modern instruments now permit us to mea­ supergiant stages are large and luminous with cool surface sure stellar distances out to ~50 parsecs (our immediate temperatures, and they plot to the upper right of the main stellar neighborhood) using parallax. sequence. White dwarf stars are small and faint with very By the late 19th century, astronomers ­were routinely hot surface temperatures, and they plot to the lower left of observing stellar spectra. Unfortunately, interpreting the main sequence. The H‑R diagram is so fundamental to spectra was difficult because the interaction of light with stellar astronomy that it is used to show the relationships —-1 —0 —+1 21

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between stars, to determine the ages of star clusters, to • Explain why each element absorbs radiation at exactly plot the life track of a single star, and to identify all the the same wavelengths as it emits. physical properties of an individual star. In short, just • State the relationship between absorption line strength about everything we know about stars can be expressed and temperature, and give an example. with the H‑R diagram. • Chapter learning goal(s) addressed: Determine the composition of stars from their spectra. Classify hn hk io il sy SY stars, and or­ga­nize this information on an H‑R Discussion Points hn hk io il sy SY diagram. hn hk io il sy SY • Discuss the utility of the spectral classification of stars. hn hk io il sy SY Have students look for information about the new Required Materials spectral classes defined in the 21st century such as “L” • Computer with Internet access hn hk io il sy SY and “T” dwarfs. • Discuss the Sun in terms of its relation to other stars Pre-­Assessment Question hn hk io il sy SY concerning size, temperature, mass, luminosity, and its hn hk io il sy SY spectrum. How many other elements have the same emission spec- hn hk io il sy SY • In thinking about binary star systems, discuss what trum as hydrogen? hn hk io il sy SY effects could be observable and what information a. 0 could be inferred from those observations if an extra- b. All the alkali metals terrestrial observer could see the orbital plane of Earth c. All the gases around the Sun edge-on. Answer: a • Discuss the possibility of finding planets in the habit- able zones of stars and the difficulties that could Instructions be encountered when considering different types of stars. Students follow the guidelines given within the Spectrum • Discuss the history of the stellar classification system, Explorer pages. including its original incarnation and its rearrange- ment once temperature was realized to be the impor- Post-­Exploration Questions tant quantity. 1. What does the observation that each element has its own unique spectral signature tell us about the energy levels in the atoms of the different elements? Explorations 2. Reproduce the diagram of the energy levels of the Exploration 1: Spectrum Explorer hydrogen atom and include the first four energy levels. Note to Instructors a. How many absorption lines are possible? This alternate Exploration has students using the “Spec- b. How many emission lines are possible? trum Explorer” Interactive Simulation found on the c. Are the energy differences in each of the transitions StudySpace to examine emission and absorption line in absorption and emission the same? spectra. The primary lesson of this activity has important d. Will the wavelengths be the same? ramifications for understanding the science of astronomy. 3. The riginalo spectral sequence went alphabetically. Students benefit from seeing firsthand that elements have Why did it have to be changed? unique spectra and that this makes it possible for astrono- mers to infer a wealth of information about a distant object using only the light it emits. Questions 6 through 9 Exploration 2: Using Wien’s Law to Estimate ask students to probe the reasons for both O and M the Surface Temperatures of Stars stars—­stars on the opposite ends of the temperature scale—­having weak hydrogen lines. Review of key parts Note to Instructors of Chapter 5 may be necessary. Stars are not perfect blackbody radiators by the very fact that we can see them. The fit of a blackbody curve to a Learning Goals star’s spectrum, then, is, at best, an estimate—­sometimes • Explain how we know that each element has its own good enough, but most of the time just rough. There is a -1— unique spectral signature. lot of stellar astrophysics involved that is well beyond an 0— +1—

577-53231_ch02_4P.indd 22 3/9/13 8:17 AM Chapter 13 Taking the Mea­sure of Stars ◆ 23

introductory text. Even so, with a brief introduction and connection between the units and the equivalent in practice fitting curves to the spectra of stars covering a nanometers. range of spectral types, students can get a good under- We have estimated an uncertainty for the “best fit” standing of what is involved. They then should be able to blackbody temperature by manipulating the temperature find the surface temperatures of a sample of stars to within of a star until we get one that is “too hot” and one that is 20 percent or so. “too cool” (but still not unreasonable fits). For each spectral type, we then subtract the coolest possible temperature hn hk io il sy SY Learning Goals allowed from the hottest possible temperature allowed hn hk io il sy SY and divide by 2 to get our uncertainty. • Infer surface temperatures of stars from their peak hn hk io il sy SY The Java applet also shows how the peak wavelength wavelength using Wien’s law. hn hk io il sy SY for O and B stars is well into the UV part of the spec- • State the shortcomings of relying entirely on Wien’s trum. When working with the M5V spectrum, the broad law for surface temperatures of stars. hn hk io il sy SY swaths of absorption by molecules shows how inexact • Chapter learning goal(s) addressed: Infer the tempera- this method is for such cool stars. Emphasize, however, hn hk io il sy SY tures and sizes of stars from their colors. that as inexact as this science is, we still get an estimate hn hk io il sy SY of the star’s temperature to within a few hundred or a hn hk io il sy SY Required Materials thousand degrees, and that’s not bad, considering our hn hk io il sy SY • Computer access thermometer. • Projection system

Pre- Post-­Assessment Questions Exploration 3: Details of the H‑R Diagram • How close to perfect blackbodies are stars? Note to Instructors • What are some of the difficulties in fitting a blackbody This exploration is a slightly different version from the curve to stellar spectra? one in the textbook, asking the student more analytical questions. It uses the simulation “H-R Diagram Explorer,” Instructions found on the StudySpace. Stellar spectra contain many bumps and wiggles that correspond to absorption lines because of a variety of ele- Learning Goals ments present in their atmospheres. Our efforts in infer- • Demonstrate an understanding of the interrelationship ring a star’s surface temperature start with the shape of among a star’s surface temperature, radius, and its continuous spectrum and an estimate of the peak luminosity. wavelength. Perhaps the hardest part of this is getting a • State what the classifications of stars between brightest feel for how to determine the shape of the star’s contin- and nearest imply about their distances and actual uum. The shape is hard to figure out because of the many numbers. absorption lines that affect the spectra of stars, pulling out light at certain wavelengths. A‑stars have what’s Required Materials known as the Balmer jump, where light from the star short-­ward of around 400 nm is almost totally absorbed. • Computer with Internet access That radiative energy has to get out somehow, and the photons jostle around until their energies are equivalent Pre-­Exploration Question to wavelengths where the opacity isn’t so great, and they Locate the following sections for the H‑R diagram shown escape. ­here: It is important that the students read the instructions carefully, as they need to know how to select a region of Stars that are hot and have high luminosity _____ the spectrum and thus how to zoom in on the part that Stars that are hot and have low luminosity _____ they are interested in. Also emphasize that the y‑axis scale Stars that are cool and have low luminosity _____ will change depending on how much of the spectrum is Stars that are cool and have high luminosity _____ being viewed. Have students leave the fiducial wavelength Region of stars that have the largest radii _____ at 5000.0 angstroms (Å). Since the x‑axis uses a scale that Region of stars that have the smallest radii _____ might be hard for students to understand, emphasize the The location of the Sun on this H‑R diagram _____ —-1 —0 —+1

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sented by steps on the ladder. As with all quantum phe- ABnomena, a macroscopic visual aid is useful in conveying to students the unintuitive nature of the quantum world.

Learning Goals • State what is meant by the “discrete energy levels” of hn hk io il sy SY an atom. hn hk io il sy SY • Explain the pro­cesses by which energy is conserved hn hk io il sy SY during transitions of an electron. • Chapter learning goal(s) addressed: Summarize how hn hk io il sy SY the energy levels of an atom determine the wavelength of the light that the atom emits and absorbs (Chapter 5). hn hk io il sy SY hn hk io il sy SY 1 R Required Materials hn hk io il sy SY • Short stepladder (four steps is sufficient), or a lecture hn hk io il sy SY ______y -axis title: hall with stairs hn hk io il sy SY • Rainbow assortment of colored plastic balls—­ VIBGYOR—like those typically used in ball pits • Bucket or mitt to catch the balls

CD Pre- Post-­Assessment Questions

O5 B0 B5 A0 F0 G0 K5 M5 • What do we mean when we talk about electron x-axis title: ______“jumps” or transitions in an atom? • What is meant by the “discrete energy levels” of an atom? Instructions • Can electrons exist in between those energy levels? • How is energy conserved during transitions of an elec- Complete instructions are given within the student work- tron? What transpires? sheet for this exploration. If you assigned the activity that had the students calculate the distances, temperatures, Instructions brightness, luminosities, and radii of the four stars in Cyg- 1. Stand next to the stepladder and explain to students nus, this Exploration provides a review. that electrons in atoms cannot have just any energy, just as you ­can’t stand in between steps. Electrons can Post-­Exploration Questions have only certain energies, just as you must stand on 1. How can a star be cool, say, ~3500 K, and have a very specific steps. Having discrete levels of energy means high luminosity? that energy levels for electrons in atoms are “quan- 2. What is the characteristic of a star that is both extremely tized.” Review energy conservation and the fact that hot and has low luminosity? because the energy spacing between quantized levels 3. The Sun, of course, is both the nearest star and the is set for each element, photons must have just the brightest star. Is it represented on the H‑R diagram in right energy to be absorbed. this simulation? 2. Hand out balls of different colors. Decide for yourself beforehand what colors represent what amounts of energy. This is important later as you “transition” to Demonstrations and Activities different levels of the stepladder. Demonstration: Atomic Spectrum 3. Start by standing on ground level next to the steplad- der (“atom”). Explain that you will represent an elec- Note to Instructors tron and the stepladder represents the atom and its This is a variation on the demonstration used for Chapter energy levels that you can inhabit. State that you are 5. It provides a different macroscopic visual before further starting at the ground state energy level. discussion about emission and absorption spectra, most 4. Have a few students gently toss balls to you. If a ball likely necessary to refresh students’ memories. Balls of has the right color for you to transition to the first -1— various colors represent different energy photons while energy level of the “atom,” grab it and step up. Ask, 0— you represent the electron at various energy levels repre- “What happened?” +1—

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5. Have a few more students gently toss balls to you. If Learning Goals a ball has the right “energy” for the next transition • Apply Wien’s law to real spectra and estimate the sur- (or higher transitions), “absorb” it and move up to face temperatures of 14 stars. the appropriate level. Explain what has just • Classify stars by their surface temperatures. occurred. 6. Now, toss back a ball and move down one or more Required Materials energy levels, as appropriate, depending on the • Computer with Internet access hn hk io il sy SY color of the ball. Ask students, “What happens • Printer hn hk io il sy SY when an electron goes (when I go) down in energy • Calculator hn hk io il sy SY levels?” hn hk io il sy SY 7. At least once, absorb a single photon that transitions Instructions you up by two energy levels and then move back down Detailed instructions for performing this activity are hn hk io il sy SY two, one step at a time, emitting two different pho- contained in the student worksheet, which students can hn hk io il sy SY tons. Ask students, “Is energy still conserved if an fill out and turn in. electron emits (I “emit”) two photons?” Suggestion: Print the spectra and attach to index cards, hn hk io il sy SY 8. At some point, work your way up to the “top” of your with enough sets to use with teams. Students then place hn hk io il sy SY step’s “energy levels.” When a student tosses a ball of the spectra in order of temperature, reinforcing the shift hn hk io il sy SY sufficient energy to you, jump off the ladder and run not only in the peak wavelength but also in the strengths of away, representing ionization. Explain. the hydrogen absorption lines. Students can then discuss 9. When you return to the “atom,” toss out a ball of the the proper fit of a thermal radiation curve. The student right color in order to recombine with the atom and instructions suggest copying each image into a document return to a previous energy level. Explain what has just editor for more rapid reviewing. occurred. Depending on the speed of the college or university’s Ethernet system, the number of students accessing the data simultaneously, and the SDSS computers, finding the data for the individual spectra can take time, especially if stu- Activity 1: Spectral Classification of Stars dents need to access each spectrum multiple times. Skill Level: Easy Note to Instructors Activity 2: Finding Distances to Stars This activity uses real astronomical data from the Sloan Using Parallax Mea­sure­ments Digital Sky Survey (SDSS). Students look at the spectra of Skill Level: Intermediate several stars, fit them with a curve to find the continuum, find the peak wavelength, calculate the star’s temperature Note to Instructors using Wien’s law, and then classify the stars based on the As simple as the equation is for determining the distances estimated surface temperature. to stars via their mea­sured parallaxes, students enjoy dis- This activity is a subset of a larger activity posted on covering the relationship themselves through simple the SDSS website (­http://​www​.sdss​.org). The full experimentation and mea­sur­ing. This activity asks stu- activity has students classify stars by both temperature dents to predict what they think the relationship is, based and hydrogen absorption line strength and poses a upon their blinking alternate eyes or maybe holding their series of questions about why the stars with the stron- thumbs in front of their faces, and then test that predic- gest hydrogen lines are neither the hottest nor the tion. Even students who are familiar with the concept coolest. may be surprised. If this is the first time your students have looked at The idea of how those results ­were obtained are then real spectra data, you will want to give them a “tour” of applied to the mea­sured parallaxes of stars, including one of the spectra. Specifically, you will need to show the use of arc seconds and parsecs. The desirability of a them what noise looks like, what absorption lines look longer baseline versus the feasibility of actually mea­sur­ like (both narrow and broad), and how to see the under- ing from a greater distance, such as from Pluto, is lying blackbody curve through the many bumps and explored. The activity ends with ranking of distances squiggles of a spectrum. This activity has students based on actual parallax values and calculation of dis- begin on a Web page that walks them through this very tances in parsecs given the actual parallaxes of another —-1 tour. set of stars. —0 —+1

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Learning Goals Instructions While working in pairs with a meter stick and toothpick, Full instructions guiding the students through the predic­ students will: tions, experimentation, consideration of their results, and extension to the stars are given on the student worksheet. • Demonstrate mea­sured parallax by noting scale of The activity ends with a short writing exercise. Students “jumps” at different distances from eyes. can come very close to the actual y = 1/x graph, as noted in • Derive the relationship between the distance of the hn hk io il sy SY the sample shown here that came from actual student data. toothpick and the corresponding “jumps.” hn hk io il sy SY Using any kind of projector, reproduce a mea­sur­ing • Approximate the limit to the detection of the “jumps.” hn hk io il sy SY grid such as the one shown ­here to give the students refer- • Apply this knowledge to the parallax of stars. ence points for their mea­sure­ments. hn hk io il sy SY • Chapter goal(s) addressed: Use the brightness of nearby stars and their distances from Earth to deter- hn hk io il sy SY mine their luminosity. hn hk io il sy SY Activity 3: Deriving Stellar Properties for Four hn hk io il sy SY Required Materials Stars in Cygnus hn hk io il sy SY • Meter sticks or long bamboo skewers Skill Level: Advanced (Scientific Notation, • Toothpick (anything very thin) hn hk io il sy SY Ratios, Estimating Uncertainties) • Protractor • Overhead or pre­sen­ta­tion slide with finely spaced grid Note to Instructors (sample provided ­here) Students seem to enjoy working problems when they real- ize that they are using real data and references that profes- Deriving the distance—parallax relationship sional astronomers use. This activity uses SIMBAD (SIMBAD database, operated at CDS, Strasbourg, 14 France), which has the complete details on not only stars 12 but also galaxies, nebulae, observations at different wave- lengths, and more. If earlier in the course you discussed 10 celestial coordinates, then you can draw student attention to the listed equatorial coordinates. 8 id marks “jumped” id marks The demonstration of Wien’s law is graphically displayed using an applet at ­http://​www​.jb​.man​.ac​.uk​/distance​/life​ 6 /sample​/java​/spectype​/specplot​.htm and given as an 4 Exploration for this chapter is useful for seeing how a con- tinuum blackbody spectrum should be fit to the data. It is Number of fine gr 2 also important to make sure that students thoroughly understand Wien’s law and how extremely hot stars have 0102030405060708090100 blackbody spectra that peak far into the ultraviolet, and Distance from eyes (cm) extremely cool stars have spectra that peak far into the infrared. For these extremes, only a maximum peak wave- Measuring grid for deriving the distance versus parallax relationship length (minimum surface temperature) can be estimated for the hot stars and only a minimum peak wavelength (maximum surface temperature) can be estimated for the cool stars. Some students are uncomfortable with not being able to get the “exactly right” answer, but often in astronomy we just do the best we can with the observa- tions we have. Some students are also unable to figure out their uncer- tainties in the peak wavelength. We have found it helpful to give them lots of practice work on similar stars before attempting this activity. Here is another opportunity for students to work in -1— teams, sharing computations, or even set up a spreadsheet 0— to streamline their work. +1—

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Learning Goals 12. False: As blackbodies, hotter stars are bluer. • Use Wien’s law to estimate the surface temperatures of 13. False: A star with no carbon absorption could have no four stars, including uncertainties. carbon, or the atoms could not be in the right energy • Demonstrate proficient use of SIMBAD in locating state to absorb light. observational data for stars. 14. True: Conservation of momentum dictates that the • Compare surface temperatures derived using Wien’s more massive star moves more slowly. law versus published values. 15. True: Once one knows the mass of a star, then its size, hn hk io il sy SY • Given the published values for observed properties of temperature, and luminosity are known, but provided hn hk io il sy SY these four stars, find the distances, luminosities, and radii. one also knows its evolutionary state. hn hk io il sy SY 16. (d) Since parallax depends on the distance of the hn hk io il sy SY Required Materials planet from the Sun, a larger orbital distance would make the pc longer and would allow us to see parallax • Computer access to the Internet hn hk io il sy SY motion of more distant stars. hn hk io il sy SY Instructions 17. (a) More distant stars have smaller parallax angles. 18. (d) A radian is about 57.3 degrees while an arcmin hn hk io il sy SY Full instructions are provided on the student worksheet. and arcsec are fractions of a degree. hn hk io il sy SY 2 19. (a) Brightness b changes as L/d , so the fact that star hn hk io il sy SY Summary S elf-­Test A appears twice as bright but is twice as far away means that L ∝ b × d2 = 2 × 22 = 8 using proportional 1. (b) Since the two stars are at the same distance, the reasoning. only reason one would appear brighter is if it is more 20. (b) The high temperatures of O stars mean that most luminous. atoms are either ionized or the electrons are in such 2. (a) Since a star acts like a blackbody, we know that a high-­energy states that they will not absorb many bluer intrinsic color means the star’s surface is hotter. photons to produce absorption lines. 3. (c) Since the stars are at the same distance and appear 21. (b) Helium has four times the mass of hydrogen. the same brightness, they must have the same lumi- 22. (a) L ∝ R2 T4, so if one star is hotter, it must be smaller nosity. The fact that one is red means it is cooler; to have the same luminosity as the cooler one. therefore, it must be larger to have the same luminos- 23. (b) Capella will be made of yellow (G‑type) and red ity as the blue star. (M‑type) stars, and since both pairs are close to each 4. (a) For two identical stars, the depth of the absorp- other, the brighter G‑type stars will dominate. There- tion line indicates the amount of the element present. fore, the color will appear yellow. 5. (b) Star A is moving three times faster than star B; 24. (d) While the H‑R diagram shows the temperatures, therefore, it must be less massive. colors, and luminosities of stars, we combine these 6. (b) If stars are at the same distance, then brightness with the paths that stars take along the diagram to indicates luminosity. Since star A is brighter than B understand how they evolve. but at the same temperature, it must be larger; that is, 25. (a) Most stars are main-­sequence stars; since this stage A is a giant, and B is a dwarf or main-­sequence star. accounts for about 90 percent of any star’s lifetime, we 7. (d) If stars are at the same distance, then brightness see most stars during their longest-­lived phases. indicates luminosity. Since star A is brighter than B but the same color, then it must be larger. We can rule out either star being a red giant because of the blue color, leaving only answer (d). Thinking about the Concepts 8. (c) This response lists the correct axes. 26. The sky is essentially a two-­dimensional sphere with 9. (b) and (d). The lower right of the H‑R diagram is low distances mea­sured in angles rather than in linear temperature and low luminosity. units. Because astronomers mea­sure parallax as the 10. (a), (c), and (e). Since both stars are main-­sequence, angular shift of foreground objects in the sky, they more massive implies hotter, larger, and more luminous. prefer to define a distance unit that can be obtained directly from parallax mea­sure­ments. Therefore, the parsec is the distance unit of choice among most True/False and Multiple Choice astronomers because distances in parsecs can quickly 11. False: Brightness depends on distance as well as be converted to parallax shifts in arcseconds and vice —-1 luminosity. versa. —0 —+1

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27. If we can mea­sure parallax down to 2 milli­arcseconds, 32. Every blackbody emits light at all wavelengths; the then nearby stars (parallaxes much larger than this) question is just how much light is emitted. A star with a can be mea­sured fairly accurately (around 10 percent 2,500 K blackbody emits mostly in the red and infrared error), but as the star becomes more distant, the paral- but still gives off a mea­sur­able amount of blue light. lax decreases and the relative error goes up. If a star has 33. Betelgeuse is cooler than Rigel. Nothing can be said a parallax as small as 4 milli­arcseconds, then the dis- about their luminosity or size without knowing their hn hk io il sy SY tance is uncertain by up to 50 percent. distance. hn hk io il sy SY 28. As parallax angles become smaller, they approach the 34. The riginalo classification of stars was based on the hn hk io il sy SY precision limits of our instruments. For example, con- strength of hydrogen absorption lines in their spectra, sider an instrument that can mea­sure any parallax with A‑class stars exhibiting the strongest absorption. hn hk io il sy SY with a precision of ± 1 milliarcsecond. For an object Hydrogen absorption lines are strongest at tempera- with a 50 milli­arcsecond parallax, the uncertainty tures near 10,000 K, but stellar surface temperatures hn hk io il sy SY is only 1 milliarcsecond out of 50 milliarcseconds span the range from 3,000 K to 30,000 K. At surface hn hk io il sy SY or 2 percent. On the other hand, an object with a temperatures well below 10,000 K (F, G, K, and M hn hk io il sy SY 2 milliarcsecond parallax has an uncertainty of 1 milli­ stars), much of the hydrogen in a star’s upper atmo- hn hk io il sy SY arcsecond out of 2 milliarcseconds, or 50 percent! sphere remains unexcited, so fewer electron transi- hn hk io il sy SY Therefore, due to the limited precision of our instru- tions are observed in the stellar spectrum. At ments, uncertainties in distance grow as parallax temperatures well above 10,000 K (O and B stars), angles get smaller. most of the hydrogen in a star’s upper atmosphere is 29. Size and mass are distance-­dependent quantities. Size ionized, so no electrons remain in the hydrogen to depends on distance because objects that are farther produce spectral lines. When spectral classes are away look smaller. Mass depends on distance because arranged as a temperature sequence, the O and B we need to know the size of a binary companion’s spectral classes move ahead of the A‑class stars orbit to accurately determine the star’s mass. Tem- despite their weaker hydrogen lines. As a final note, perature, color, spectral type, and composition are all the missing letters in the OBAFGKM sequence are a distance-­independent quantities that can be deter- result of merging the original 22 spectral classes into mined from a star’s spectrum. Because stellar spectra the 7 larger spectral classes that we have today. show the same lines whether viewed from up close or 35. We can directly mea­sure stellar masses only by using from halfway across the galaxy, the mea­sured proper- Kepler’s third law, which requires us to observe the ties of a star derived from its spectrum are not influ- orbit of a companion. To accurately mea­sure the enced by the star’s distance. semimajor axis of the companion’s orbit, we need to 30. Parallax is not dependent on the brightness of a star, know how its orbit is inclined to our line of sight. so dimming (or “extinction”) does not affect this Only eclipsing or visual binaries allow us to mea­sure method. However, spectroscopic parallax compares both the orbital period of a companion star and its the observed brightness to the theoretical luminosity semimajor axis directly. Note that we can use Kepler’s of a star. If the star appears dimmer than it really is, third law with spectroscopic binaries as well; but then we will compute a distance that is larger than the because we do not know the orbital inclination of the star’s actual value. companion star, we can determine only a minimum 31. (a) The color of a star is a direct mea­sure of its tem- mass for the system, not the exact mass. perature. Blue stars have hot surface temperatures, 36. Stars are all governed by the same laws of physics, and and red stars have cooler surface temperatures. their evolution depends almost exclusively on their Because the golden star lies closer to the redder side of mass (which controls their central temperature, pres- the spectrum than the sapphire blue star, we conclude sure, ­etc.), so we assume that if two stars have similar that the sapphire blue star has the higher temperature composition and mass, they will act and evolve the of the pair. (b) Blue stars emit more light per unit area same way since there is no physical mechanism to than yellow stars. Therefore, if the blue star was the change this. same size as or larger than the golden star, it would be 37. We mea­sure the masses of stars using two different the brighter star in the pair. Because our observations approaches, both of which require some untested demonstrate that the golden star is actually the assumptions and significant uncertainties. The first brighter of the two, we must conclude that it has a approach is to determine a star’s spectral class from its -1— much larger surface area than the blue star to appear spectrum. If the star is on the main sequence, then 0— more luminous. its mass can be estimated directly from knowledge of +1—

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its spectral class. Another approach is to estimate the Solve: The horizontal axis (temperature) goes from distance to the star in order to compute its luminos- above 40,000 K to 3,000 K with uneven spacing; that ity. From its luminosity and color, a star’s position on is, the distance from 10,000 K to 20,000 K is much the H‑R diagram can be ascertained, and its mass can larger than from 20,000 K to 30,000 K; therefore, it is be determined from its placement on the diagram. logarithmic. For nearby stars, a third (and much more reliable) Review: Note that in this par­tic­u­lar H‑R diagram, it approach becomes possible if we can detect the star’s is logarithmic but you may sometimes see tempera- hn hk io il sy SY gravitational “wobble” due to the influence of planets ture in linear plots as well. hn hk io il sy SY or an unseen companion. If we can see a star wobble 43. Setup: Figure 13.20 shows mass versus (a) luminos- hn hk io il sy SY under the influence of gravity, then we can use ity, (b) radius, and (c) temperature for a variety of hn hk io il sy SY Kepler’s third law to determine the mass of the star stars. Note that for (a) and (b) they are in solar units, directly. so one can directly compare to the Sun. For tempera- hn hk io il sy SY 38. There ­were very few massive elements in the early ture, we have to recall that the Sun’s surface tempera- universe, and they ­were created by the first generation ture is about 5,770 K. hn hk io il sy SY of stars and spread throughout gas to be incorporated Solve: For a star 10 times the mass of the Sun, we look hn hk io il sy SY

in future generations. along the vertical line at 10 M and interpolate hn hk io il sy SY 39. Spectral identification uses the lines of H and He between points if need be. For luminosity, the star hn hk io il sy SY extensively, so we could do much of the identification will be about 1,000 times more luminous than the with just these two elements. However, note that the Sun. For radius, it will be about 7 times larger. For identification of stars relates to their temperatures, temperature, it will be around 2 × 104 K, which is which one can also identify with a spectrum (or col- roughly 3.5 times hotter than the surface of the Sun. ors) regardless of whether any absorption features are Review: These plots show us that as mass of a main-­ present. sequence star increases, L, R, and T all increase as 40. Binary stars can transit each other just like planets, so well. But in par­tic­u­lar, luminosity is a strong function any mission that is looking for regular time variability of mass, while the other two grow much more slowly is going to discover binary stars (and pulsating vari- with increasing mass. ables as well). 44. Setup: The angular size of an object q is related to its dθ s size s and distance d, by s ==or d 57.3 57 3.  θ Applying the Concepts Solve: Knowing that the size s between our eyes is 6cm == 41. Setup: Quite often in astronomy, graphs need to 6 cm, d 57.3  688cm. cover a huge range of pa­ram­e­ters, which is easier to 05. visualize using logarithms as they spread out the Review: 6.88 m is about 21 feet. A 6-­cm object will ­whole range of values so that one can identify the appear to be half a degree in size from about 21 feet positions of points with both small and large values away. Try it! equally well. To find the difference in luminosity 45. Setup: Parallax uses the parallax formula d = 1/p between the least and most luminous stars, read off where d is in parsecs and p is in arcseconds. To convert the values on the y-­axis. parsecs to light-­years, recall 1 parsec = 3.26 light-­​​​years. Solve: The most luminous star is shown with about Solve: The distance to Sirius is d = 1/0.379" = 2.64 6 −3 32. 6 ly 10 L while the faintest are about 2 × 10 L for a or 26. 4 pc ×=86. ly. total difference of about 109 or a billion. pc Review: It is remarkable that stars that have a small Review: Remember that a parallax of 1 arcsecond range of masses and temperatures can have such a equals a distance of 1 parsec. Sirius has a distance huge range of luminosities; we will learn why this close to one-­third of an arcsecond, so its distance happens in future chapters. should be about 3 parsecs, as we found. 42. Setup: To check whether an axis is logarithmic or lin- 46. Setup: Remember that luminosity L ∝ R2 T4 and that ear, look at how the axis values increase. If they if two stars are the same spectral type, then they have increase by powers of 10, it is logarithmic. If the spac- the same temperature. ing between increasing values decreases (i.e., there is Solve: Both stars are the same spectral type (A) and more space from 1 to 2 than from 2 to 3, ­etc.) then it is therefore have the same temperature. If the —-1 logarithmic as well. Pup appears 6,800 times fainter and is the same —0 —+1

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­distance as the Dog Star (which they are since they Review: When we talk about “nearby” stars, we are are a binary pair), then the Pup is 6,800 times less really including those up to many hundreds of light-­ luminous. Since luminosity L ∝ R2 T4 we can use the years from Earth. above information to infer that the Pup Star must 49. Setup: Light travels 1 light-­year in 1 year, so the time be RL∝≈/(T 4416/,800)/11==/,6 800 18/ 3 it takes for light to reach us from a star is the same as times the size, or 83 times smaller than the Dog the star’s distance in light-­years. Using the parallax hn hk io il sy SY Star. distance formula d= 1/p, we can compute the dis- hn hk io il sy SY Review: If two stars are the same temperature and tance to Proxima Centauri in parsecs and then con- hn hk io il sy SY one is very faint, then it must be very small. Why? All vert it to light-­years knowing 1 pc = 3.26 ly. Solve: hn hk io il sy SY blackbodies with a temperature T give off the same 3.26 ly amount of energy per tiny patch of area. So if a star d = 1/0.772" = 1.30 pc, or 1.30 pc ×=4.22 ly. hn hk io il sy SY has small luminosity, it has small area, that is, small pc Therefore it takes 4.22 years for light from Proxima hn hk io il sy SY size! 47. Setup: For binary stars, we know the ratio of the Centauri to reach us. hn hk io il sy SY masses is equal to the inverse ratio of the orbital Review: Remember that a parallax of 1 arcsecond hn hk io il sy SY equals a distance of 1 parsec. Proxima Centauri has a M1 v2 velocities, that is, = , and Kepler’s law for two parallax close to an arcsecond, so its distance should hn hk io il sy SY M v 2 1 be about 1 parsec, as we found. a3 stars is P2 = . 50. Setup: Parallax uses the parallax formula d = 1/p MM12+ where d is in pc and p is in arcseconds. Solve: Using the parallax formula, the distance is Solve: (a) For binary stars, we know the ratio of the 131.1 parsecs. The uncertainty means the distance masses is equal to the inverse ratio of the orbital could be as low as 0.00763 – 0.00164 = 0.00599 M v velocities, that is, 1 = 2 , so if the companion arcsec or as high as 0.00927 arcseconds, yielding M2 v1 distances of 108 and 167 parsecs. The uncertainty is has 2.35 times higher velocity, it has 2.35 times therefore half this difference, or about 30 parsecs. lower mass. Its mass is then about equal to that of Review: Remember that a parallax of 1 arcsecond the Sun, since Sirius A has a mass 2.35 times larger equals a distance of 1 parsec. Betelgeuse has a paral- than the Sun. (b) Using a total mass of 1.00 + lax close to one-­hundredth arcsecond, so its distance 2.35 = 3.35 solar masses and a period of 50 years, we should be about 100 parsecs, as we found. The uncer- tainty is about 4 times smaller than the parallax, so aM=+3 ()MP2 =⋅3 33. 55022 = 0AU find 12 . the corresponding uncertainty in distance should be Review: In our Solar System, a period of 50 years about one-­fourth the actual value: 30 parsecs is about would place a planet between Saturn and Uranus one-­fourth of 135.9 parsecs (more or less), again con- (about 14 AU). Since the Sirius system has more mass, firming our calculation. the period of a given object will be shorter than ­were 51. Setup: Parallax uses the parallax formula d = 1/p it orbiting the Sun. It therefore makes sense that the where d is in pc and p is in arcsec. Luminosity, bright- companion must be more than 14 AU away to have a ness, and distance are related by L ∝ bd2. period of 50 years. Solve: Rigel is at a distance of d = 1/0.00412”= 243 pc 48. Setup: To solve this problem, we begin by noting that based on its parallax. Rigel is twice as far as Betelgeuse; the apparent brightness (b) of a star is proportional to so to appear the same brightness, Rigel must be about its luminosity (L) and inversely proportional to its four times more luminous. Betelgeuse is red and so much L distance squared (d2) as b ∝ . cooler (perhaps half the temperature of Rigel), so it d2 must be larger than Rigel to give off so much energy. Solve: Taking a ratio of the brightnesses of Review: To check whether Betelgeuse is larger than Polaris and Sirius and solving for distance yields Rigel, using our luminosity, size, and temperature relationship that L ∝ R2T 4, we find Betelgeuse should dPolaris 2,350L =×23 = 49.6. Polaris is 49.6 times be about RL∝≈/(T 4414/)/(12/) ==42 times d 22L Sirius  larger. farther from us than Sirius is. Knowing from 52. Setup: We know that the brightness of an object -1— ­problem 45 that Sirius is 8.6 ly from the Sun, we decreases as the square of the distance according to 0— compute that Polaris must be 426 ly from the Sun 1 b ∝ . We can form a ratio of the brightness of the +1— (49.6 × 8.6 ly =­426 ly). d2

577-53231_ch02_4P.indd 30 3/9/13 8:18 AM Chapter 13 Taking the Mea­sure of Stars ◆ 31

Sun to the brightness of a distant solar-­mass star, 57. As the inclination increases, the orbit goes from look- b ing circular to linear. The simulation shows an elliptical dd=×Sun yielding a distance star Sun . orbit until the inclination is exactly 90 degrees. As sep- bstar Solve: (a) Using the numbers in the problem, we find aration increases, the dips during an eclipse become narrower, and vice versa as the stars approach closer. If d =×11AU .61×=0413 ×106 AU . (b) Converting star star 2 grows three times larger and cools to 4,000 K, this distance to light-­years yields eclipses disappear at an inclination around 77 degrees. hn hk io il sy SY 149.61× 06 km 1 ly 41××06 AU × = 63..2 ly. This shows that highly inclined systems are the easiest hn hk io il sy SY 1 AU 94. 61× 012 km to detect by looking for eclipsing light curves. hn hk io il sy SY Review: This is a shockingly small number, which 58. Answers will vary with time. The student will report hn hk io il sy SY tells us that most of the stars that we see in the sky are the number of confirmed planets and how many of significantly brighter than our Sun. these are eclipsing binaries. From the eclipsing binary hn hk io il sy SY 53. Setup: The solar neighborhood has an average den- cata­log, a few stars will be listed, including their incli- sity of one star every 360 cubic light-­years. If there are nation, and the depth of their light curves. hn hk io il sy SY 100 billion stars, then the total volume is the product 59. Answers will vary. My favorite constellation is Orion. hn hk io il sy SY 13 3 of these, or about 3.6 × 10 ly . To find the radius, we Two stars in par­tic­u­lar have very different colors, Betel- hn hk io il sy SY 4 geuse (red) and Rigel (bluish white). The three brightest must use the volume formula Vr= π 3 and solve for hn hk io il sy SY stars are Rigel (T = 12,100 K, L = 117,000 L , d = 773 ly), = 3 π . 3  rV34/ Betelgeuse (T = 3,100 K, L = 120,000 L , d = 640 ly), Solve: We find a radius of about 20,000 ly.  and Bellatrix (T = 22,000 K, L = 6,400 L , d = 243 ly). Review: Is this a reasonable first guess for the size  60. An O5 is hard to fit because it is unclear whether one is of our galaxy? It is about five times too small, but as fitting the peak (which is noisy with absorption lines) we will learn later, our galaxy is not a sphere but a or the longer-­wavelength tail. The best fit to the peak is disk. about 20,000, while the longer-­wavelength data require 54. Setup: Wien’s law tells us the star emits most of its higher than 30,000 (the limit of the applet). Fitting 29.n×106 m radiation at λ = . to the longer-­wavelength tail only, the B5 is well fit T around 29,000 K; the A5 around 11,000 K; F5 around Solve: At 10,000 K, Wien’s law gives 7,200 K; G5 around 5,900 K; K5 around 4,300 K; and 2.91× 0n6 m λ = = 290 nm. This is in the near-­UV. M5 around 2,300 K. From Table 13.2, the tempera- T tures for these stars are O5 42,000 K; B5 15,200 K; This star will appear very blue/white to us. A5 8,300 K; F5 6,650 K; G5 5,560 K; K5 4,410 K; M5 Review: Although the majority of the energy comes 3,170 K. The fits would be a little better if we didn’t out in the near-­UV, this star is still emitting copious have air absorbing at some wavelengths, but that is a amounts of radiation in the visible spectrum, which is small effect. why we can see a blackbody that peaks outside the visible range of light. Exploration 55. Setup: Using the relationship between brightness b, luminosity L and distance d, L = 4πd2b, we can solve 1. As I move to the left of the H‑R diagram, the star’s for the luminosity of the Sun, knowing we are 1 AU or radius becomes smaller. 1.5 × 1011 m away and b = 1,470 W/m2. 2. As I move to the right of the H‑R diagram, the star’s Solve: This yields a luminosity of 4 × 1026 W. radius becomes larger. Review: This is very close to the actual value of 3. As I move up the H‑R diagram, the radius becomes 3.8 × 1026 W. larger. 4. As I move down the H‑R diagram, the radius becomes Using the W eb smaller. 5. To move the star into the white dwarf area of the H‑R 56. As of October 2012, Gaia is still under develop- diagram, the temperature must be very high and the ment and is set to launch in 2013. As listed on the luminosity very low. mission page, the satellite will “precisely chart 6. In general, the nearest stars are cooler than the Sun. [the] positions, distances, movements and changes 7. In general, the nearest stars are less luminous than the in brightness” of about 1 billion stars. The L2 orbit Sun. —-1 was chosen to provide an unobstructed view of the 8. In general, the brightest stars are hotter than the Sun, —0 sky. but there are still many that are cooler. —+1

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9. Except for one straggler, all the stars are more lumi- have lower temperatures and luminosities than our nous than the Sun. Sun. This shows us that the nearest stars are not 10. The rightestb stars all have high luminosities and a always the brightest stars, meaning that space is more range of temperatures. The nearest stars almost all filled with low-­luminosity stars than high ones. hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY

-1— 0— +1—

577-53231_ch02_4P.indd 32 3/9/13 8:18 AM Name______Date ______Section ______

EXPLORATION: Spectrum Explorer Open the “Spectrum Explorer” interactive simulation on the StudySpace at wwnorton.com/studyspace. Select ­“Emission” under “Spectrum.” Check the box beside two options under “Elements,” view the spectra, and then uncheck the boxes. hn hk io il sy SY hn hk io il sy SY 1. Do this for all options and compare the emission spectra you see. In general, how do the spectra differ? Do any of hn hk io il sy SY the elements shown have the same spectra? hn hk io il sy SY

hn hk io il sy SY 2. If none of them have identical spectra, do any of them share lines that occur at exactly the same wavelength? If so, hn hk io il sy SY state which ones. hn hk io il sy SY hn hk io il sy SY 3. Can emission lines be used to uniquely identify the chemical compositions of stars? Explain, and give a more famil- hn hk io il sy SY iar example of how we use unique patterns to identify things.

Select “Absorption” under “Spectrum” and check each option under “Elements.” Note the spectral type and luminosity class of the “star” that is being examined, which is important because not all stars show the same absorption lines. Compare the absorption spectrum of each element to its corresponding emission spectrum by toggling between those two spectrum options.

4. Do elements emit at the same wavelengths that they absorb? Does your answer make sense? Explain.

With the “Absorption” spectrum still selected, check only the box beside “Hydrogen” from “Elements.” Move the “Spectral Type” slider to adjust the temperature of the source.

5. Review Figure 5.16 in your textbook that shows the electron transitions required to make hydrogen emission lines. For the absorption lines you are looking at, from what energy level are the electrons transitioning?

6. Electrons generally occupy energy levels determined by the temperature of their environment. What energy level would you expect most electrons to occupy in a “low-­temperature” star, such as an “M” star?

7. Why don’t “M”-­type stars show strong hydrogen absorption lines?

8. What energy level would you expect most electrons to occupy in a “high-­temperature” star, such as an “O”star?

9. Why don’t “O”-­type stars show strong hydrogen absorption lines?

10. Using the slider, find the temperature at which hydrogen is most efficient at absorbing visible wavelength light. This will be when the absorption lines are the darkest or blackest. Record this value: ______—-1 —0 —+1

577-53231_ch02_4P.indd 33 3/9/13 8:18 AM 11. Before astronomers knew how to mea­sure the temperatures of stars, stars ­were originally classified by the strength of their hydrogen absorption lines. a. Which spectral type shows the strongest hydrogen absorption lines?

b. For which spectral types do the hydrogen lines disappear? hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY c. Does this mean those stars have no hydrogen in their atmospheres? Explain. hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY

-1— 0— +1—

577-53231_ch02_4P.indd 34 3/9/13 8:18 AM Name______Date ______Section ______

ExPloraTion: using Wien’s law to Estimate the Surface Temperatures of Stars Th e applet at htt p:// www .jb .man .ac .uk /distance /life /sample /java /spectype/specplot .htm is useful for seeing how a continuum blackbody spectrum should be fi t to the data. Go to that website hn hk io il sy SY and start by reviewing the instructions above the Java window. hn hk io il sy SY Once the applet is loaded, select the G5V star from the menu on hn hk io il sy SY the bott om left of the applet. Try fi tt ing the continuum. Remember hn hk io il sy SY that since the absorption lines remove light, the continuum fi t should be mostly above the data. You should fi nd that around the hn hk io il sy SY peak of the curve, and certainly on the short-wavelength side of it, hn hk io il sy SY a perfect fi t is impossible, as Figure 1 at right shows. Demonstrate the fl exibility of this program by picking any of the hn hk io il sy SY spectral types and manipulating the temperature via the slider or hn hk io il sy SY fi lling in a value aft er Temperature / K. Figure 2 shows one such hn hk io il sy SY analysis and comments.

Figure 1. Demonstration of fi tt ing a curve to a G5V star.

Figure 2a. Figure 2b. Figure 2c. Figure 2a. Figure 2a shows a fi t that is “too hot”; Figure 2b, “too cold.” Even overall the “just right” line is not an exact fi t, but we see that the peak of the spectrum occurs around 400 nm (0.4 × 104 angstroms, Figure 2c). Compared to a “by eye” fi t, the program helps us fi nd a fairly exact temperature, 7180 K, and gives us a range that we can use to state our uncertainties in that temperature: surface temperature = 7180 ± 850 K.

Fill in the following table using the Wien’s law applet, estimate the uncertainty in temperature, and comment.

Spectral Temperature Uncertainty in Comments concerning Difficulties in Figuring out Type of “Best Fit” the the Best Blackbody Temperature Curve Temperature

O5V

B5V

A5V

K5V

M5V —-1 —0 —+1

577-53231_ch02_4P.indd 35 3/9/13 8:18 AM hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY

-1— 0— +1—

577-53231_ch02_4P.indd 36 3/9/13 8:18 AM Name______Date ______Section ______

EXPLORATION: Details of the H‑R Diagram Visit the StudySpace and click on the simulation “H-R Diagram Explorer.” The Cursor Properties sliders control tempera­ ture and luminosity. hn hk io il sy SY Work with the sliders a bit, and then move the X to make it as hot and as luminous as you can while still keeping it on hn hk io il sy SY the main sequence (luminosity ~105 solar, T ~ 40,000 K). This would represent one of the stars in Orion’s belt or the hn hk io il sy SY Trapezium in the Orion nebula. hn hk io il sy SY

1. What is the radius of this star compared to the Sun? hn hk io il sy SY hn hk io il sy SY 2. Does that seem surprising? Comment briefly. hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY Exploring the relationship between radius, luminosity, and temperature some more: You can replicate Betelgeuse’s size by keeping the X at the same luminosity but decreasing the temperature to ~3500 K, a published value for this star’s surface temperature. Then, increase this star’s radius to ~1200 times that of the Sun, a published radius for this star.

3. Keeping the surface temperature and radius at the published values, what is a more accurate value for Betelgeuse’s luminosity?

Under Plotted Stars, click on nearest stars (124) and then brightest stars (149).

4. As far as the H‑R diagram goes, how do these populations differ? What do the data say about the nearest stars?

5. We are looking at the plot of luminosity and temperature, intrinsic properties of these stars. Many of the brightest stars are also the most luminous and some are very large. What does this say about their distances?

6. Do these data eliminate the possibility that there are no high-­luminosity, extremely large stars nearby?

7. Do these data eliminate the possibility that there are no low-­luminosity, Sun-­like stars far away?

8. Comment on your answers to the previous two questions.

—-1 —0 —+1

577-53231_ch02_4P.indd 37 3/9/13 8:18 AM hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY

-1— 0— +1—

577-53231_ch02_4P.indd 38 3/9/13 8:18 AM Name______Date ______Section ______

AC TIviTY: Spectral Classification of Stars Begin by visiting the Sloan Digital Survey website at ­http://​skyserver​.sdss​.org​/public​/en​/proj​/basic​/spectraltypes​.

This introductory information is important to review, as it relates our classification of stars into better-­known classifica- hn hk io il sy SY tion schemes of biology, geology, and chemistry. Be sure you have reviewed Wien’s law for finding the surface tempera- hn hk io il sy SY ture of stars and understand what an absorption line is. hn hk io il sy SY hn hk io il sy SY Then click on “Spectra of Stars: A Brief Review” and read through the information on that page to learn how to inter- pret a real spectrum. Notice that while the length unit of nanometers is now the most commonly used unit in astron- hn hk io il sy SY −­10 omy, the wavelengths of these spectra are given in angstroms (Å). An Å is equal to 10 meters. There are 10 Å per hn hk io il sy SY nanometer. hn hk io il sy SY 1. Define continuum peak, absorption line, noise, and thermal radiation. hn hk io il sy SY hn hk io il sy SY

2. What events occurring in the atoms result in the absorption lines?

When you have finished reading the information, click the “Next” button on the bottom of the page.

The “Exploring Spectra” page provides links to the spectra for 14 different stars. Click on each spectrum to view a larger image; open up a new document editor and copy each of the spectra into the document in order. This will make it more con­ve­nient and quicker to answer the following questions for these 14 different stars.

For each spectrum, imagine a smooth curve that fits over the entire spectrum (review Chapter 5). Find the wavelength at which each star’s spectrum peaks and enter it in the table below labeled “Spectral Classifications.” You will fill in the other columns as this activity progresses.

Recall Wien’s law: the peak wavelength of blackbody radiation in Å, lpeak, is inversely proportional to its temperature, T, in Kelvin. (We have reverted to Å because that is the unit used for the Sloan Survey spectra.) 29,, 000 000  λ = Α peak T 3. Which is hotter, a star that peaks at 5,000 Å or a star that peaks at 6,000 Å? How do you know?

4. Use this relationship to find the temperatures of the 14 stars. Also note the color of each star as seen on each page. Enter your values in Table 1, “Characteristics and data for 14 SDSS stars based on examination of their spectra.”

—-1 —0 —+1

577-53231_ch02_4P.indd 39 3/9/13 8:18 AM Table 1. Characteristics and data for 14 SDSS stars based on examination of their spectra

Star Star Color Peak Wavelength Surface Temperature Spectral Type [Å] [K] 1 2 3 hn hk io il sy SY 4 hn hk io il sy SY 5 hn hk io il sy SY 6 hn hk io il sy SY 7 hn hk io il sy SY 8 9 hn hk io il sy SY 10 hn hk io il sy SY 11 hn hk io il sy SY 12 hn hk io il sy SY 13 14

5. Do you notice a relationship between a star’s color as seen on each page and the appearance of its spectrum? Describe.

6. Use the information in Table 2 to find the spectral type for each of the 14 stars. Enter the type of each star into Table 1.

Table 2. The spectral types associated with each surface temperature range for stars

Spectral Type Temperature Spectral Type Temperature Range [K] Range [K] O 33,000 F 6000−7500 B 10,000−33,000 G 5200−6000 A 7500−10,000 K 3700−5200 M < 3700

-1— 0— +1—

577-53231_ch02_4P.indd 40 3/9/13 8:18 AM Name______Date ______Section ______

AC TIviTY: Finding Distances to Stars Using Parallax Mea­sure­ments One of the most difficult problems in astronomy is determining the distances to objects in the sky. There are four basic methods of determining distances: radar, parallax, standard candles (there are many), and Hubble’s law. Each of these methods is most useful at certain distances, with radar being useful nearby (for example, the Moon), and Hubble’s hn hk io il sy SY law being useful at the most distant scales. In this activity, we investigate the use of mea­sured stellar parallax to hn hk io il sy SY determine distances. hn hk io il sy SY hn hk io il sy SY Even when observed with the largest telescopes, stars are still just points of light. Although we may be able to tell a lot about a star through the flux we mea­sure, these observations do not give us a reference scale to use to mea­ hn hk io il sy SY sure their distances. We need to rely on a method that you are actually already familiar with: the parallax of an hn hk io il sy SY object. hn hk io il sy SY You can see the parallax effect by holding your thumb out at arm’s length. View your thumb relative to a distant back- hn hk io il sy SY ground while you alternate opening and closing each eye. Does your thumb seem to jump back and forth relative to this hn hk io il sy SY background? This is because the centers of your eyes are 7 to 8 centimeters apart from each other, so each eye has a slightly different angled point of view. This is why we see in three dimensions.

y = x y = 1 x–

y y

0 0 x x

y = x + 3 y = sqrt(x)

y y

0 0 x x Figure 1. Graphed possible relationship between distance of an object and its parallax.

First, we will make our predictions about the relationship between distance from our eyes at which the toothpick is held and number of fine grid marks that the toothpick appears to jump between. Which relationship, as graphed ­here, would you predict is the correct one?

_____ y = x _____ y = 1/x _____ y = −x + constant _____ y = sqrt(x)

Now let’s test how the parallax of an object varies with distance from the baseline mea­sure­ment of our eyes. 1. Person A takes the meter stick and places the pencil vertically at the 50 cm mark, centering the pencil on the meter stick. The other person, B (that would be you), places the “zero” end of the meter stick against her or his chin, hold- ing it out horizontally. Person B then alternates opening and closing each eye, noting how the pencil moves against —-1 the fine grid marks being projected at the front of the room. —0 —+1

577-53231_ch02_4P.indd 41 3/9/13 8:18 AM 2. Person A now moves the pen half of the original distance (to 25 cm). When you alternate opening and closing each eye, does the pen appear to move more or less than before? Quantify how much more or less by noting the number of grid marks the tip of the toothpick moves. 3. Now, have your lab partner move the pen twice the original distance to you, to the end of the meter stick, 100 cm away. When you alternate opening and closing each eye, does the pen appear to move more or less than before? Quantify how much more or less by noting the number of scale marks the tip of the toothpick moves, filling in the hn hk io il sy SY “Distance (cm)” and “No. of Grid Marks” columns in the table below. hn hk io il sy SY Deriving the distance—parallax relationship hn hk io il sy SY Distance No. of (cm) Grid hn hk io il sy SY Marks 50 hn hk io il sy SY 25 hn hk io il sy SY 100 hn hk io il sy SY ____ hn hk io il sy SY “jumped” id marks hn hk io il sy SY ______

____

Number of fine gr ____

____ 0102030405060708090100 ____ Distance from eyes (cm)

Make at least five more mea­sure­ments and fill in this table with your results; the more data you have, the more reliable your results. Graph the number of fine grid marks the toothpick jumped versus the distance from your eyes.

4. State the relationship between the distance from the baseline of your eyes and the number of fine grid marks the toothpick jumped. Show your logic ­here. Was your earlier prediction correct?

Let’s take a look at the approximate relationship between distance and parallax from a different viewpoint by exam- ining the angles shown in Figure 2.

5. The distance d2 is twice the distance d1. Does it qualitatively appear that angle a2 is one-­half of a1? _____ Use a protractor and mea­sure the angles a1 and a2. Quantitatively, is angle a2 one-half­ that of a1? State your findings.

x a1

d1

x a d2 2 Figure 2. Sketch showing the change in parallax angle with distance object is away.

-1— When the distance is large enough that the parallax angle is very small, which is certainly true for all stars except 0— the Sun, the parallax angle is proportional to the inverse of the distance (1/d). Conversely, if we can mea­sure the paral- +1— lax angle, we know that the distance to the object is proportional to the inverse of that angle. Note: Although the

577-53231_ch02_4P.indd 42 3/9/13 8:18 AM baseline for the mea­sure­ment of parallax angles for stars is the diameter of the Earth’s orbit, we use the value of 1 AU, the radius of the Earth’s orbit, as the stellar parallax angle.

6. You and your partner(s) can find your parallax limits by taking the toothpick farther and farther away from the “observer” until the parallax becomes undetectable. hn hk io il sy SY hn hk io il sy SY a. One of you should take the role of “observer” while the other walks straight away, holding the toothpick out at hn hk io il sy SY arm’s length. Mea­sure how far away you ­were from each other when the pencil stopped moving relative to hn hk io il sy SY objects at “infinity,” or at least extremely far away. Approximately how far was that for you? hn hk io il sy SY b. Switch roles and repeat the previous step. hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY c. Wh at does this step say about the usefulness of parallax at different distances? For example, how meaningful is it to hn hk io il sy SY speak of the parallax of a toothpick located at your school if you are trying to mea­sure its parallax from another state?

To mea­sure a parallax, we need a baseline: the distance between observation points. In the activity above, the baseline is equal to the distance between the center of your eyes. In astronomy, to mea­sure the parallaxes of stars and thus calculate their distances, we use a baseline of the diameter of the Earth’s orbit and then define the parallax angle as half of the total shift. The infinite background is usually stars much farther away than the star or object in question.

From ground-­based telescopes, we can mea­sure accurate parallaxes of thousands of stars; unfortunately, most stars are simply too far to get an accurate parallax, and we must resort to other methods to determine their distances. One method we have already succeeded in is putting a telescope into orbit and thus above the smearing effect of our atmo- sphere. The Hipparcos satellite succeeded in mea­sur­ing parallaxes of about 1 million stars!

7. The ongerl the baseline, the more objects move relative to the background—­the more distant objects can be and still have an accurate parallax mea­sured. Which of the following scenarios would be the most desirable? _____ Which of the following scenarios do you think would be the most feasible? _____ a. A satellite orbiting high above the Earth (above the atmosphere) b. A telescope on the Moon c. A satellite orbiting the Sun at Jupiter’s distance (5.2 AU’s) d. A telescope sitting on Pluto (Pluto is 40 AU’s from the Sun)

Explain your reasoning for the “most desirable.”

Explain your reasoning for the “most feasible.”

8. Because the stars are so far away, the distance at which we can mea­sure a parallax is directly proportional to B the length of the baseline. Use the following small angle formula: Baseline = a × Distance or B = a × D Distance Show mathematically that the distance is directly pro- —-1 portional to the baseline used using a simple algebraic —0 manipulation. —+1

577-53231_ch02_4P.indd 43 3/9/13 8:18 AM 9. Let’s say we put a telescope on Pluto, ~40 AU’s from the Sun, beaming back information to us. How much farther will we be able to mea­sure accurate parallaxes compared to our work ­here on Earth? Explain your answer.

10. At your telescope, you mea­sure the parallax of the star Regulus in the constellation of Leo to be one-­half of the par- allax of the star Denebola, also in Leo. What do you immediately know about their relative distances from us? hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY 11. We can now investigate more thoroughly why parsecs and arcseconds not only work but are con­ve­nient for astron- omy. The Earth in its orbit wouldsubtend a maximum angle of 1 arcsecond from the Sun from a distance of 1 parsec hn hk io il sy SY away. We can reverse that and state that a star 1 parsec away has a parallax angle of 1 arcsecond. A star that is 2 par- secs away has a parallax angle of 1⁄2 arcsecond, and so on. Fill in the missing quantities in the following table. Paral- hn hk io il sy SY laxes are in arcseconds; distances will be in parsecs. hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY Rank: closest = 1, farthest = 4 Calculating distances to stars hn hk io il sy SY Rank Star Name Parallax Star Name Parallax Distance Antares 0.024 Arcturus 0.090 Ross 780 0.213 Procyon 0.288 Regulus 0.045 Hadar 0.006 Betelgeuse 0.009 Altair 0.194

12. In your own words, briefly explain how we calculate a star’s distance in parsecs by mea­sur­ing its parallax angle in arcseconds. Give an example.

-1— 0— +1—

577-53231_ch02_4P.indd 44 3/9/13 8:18 AM Name______Date ______Section ______

AC TIviTY: Deriving Stellar Properties for Four Stars in Cygnus A. Analysis of Four Stars in the Constellation of Cygnus

For this assignment, you are going to develop your skills in estimating peak wavelengths of the spectra of four stars in hn hk io il sy SY the constellation of Cygnus and in calculating their surface temperatures. Using the larger images of the spectra of hn hk io il sy SY these four stars that are given at the end of this activity, provide the missing information in Table 1, using Wien’s law and the best curve you can find. Use the information provided for 41 Cygni as a guide. hn hk io il sy SY hn hk io il sy SY 1a. Estimate the peak of the continuum and the uncertainty for the peak for each star using the spectra provided toward the end of this document; fill in columns 2 and 3 of Table 1. hn hk io il sy SY 1b. Using Wien’s law, calculate the surface temperature of the star, state an uncertainty, and fill in columns 4 and 5 of hn hk io il sy SY Table 1. 2. Use the “Spectral Type Characteristics,” Table 5, determine each star’s spectral type, assuming it is a luminosity hn hk io il sy SY class V star. State a “hottest” and “coolest” spectral type possible based upon thepeak wavelength uncertainty. This hn hk io il sy SY will require adding and subtracting the uncertainty and using Wien’s law and Table 5. hn hk io il sy SY

Table 1. Wien’s Law results for three stars in Cygnus (assuming all stars are Luminosity Class V) Star ID 1a. Peak Peak 1b. Surface Temp. 2. Spectral “Hottest” “Coolest” Wavelength Uncertainty Temp. Uncertainty Type Type Type 41 Cygni 410 nm +/− 25 nm 7250 K +/− 450 K F0 A7 F2 52 Cygni 69 Cygni xi Cygni

B. Using SIMBAD to Find Published Information on Stars SIMBAD is the database that professional astronomers use to find published, reliable information on stars, galaxies, nebulae, and other celestial objects. We will use it ­here to find the spectral types, apparent visual magnitudes, and par- allaxes (they are listed in milliarcsecond units) for these four stars. • W e b s i t e : h tt p : //​s i m b a d ​. u ​‑ s t r a s b g ​. f r ​/s i m b a d • Click on Queries—­basic search • Type 41 Cygni in the basic search form; click SIMBAD search • We want the information for the parallax, spectral type, and fluxes through the V filter, so look for: • Parallaxes mas: • Spectral type: • Fluxes (): find the flux (given as the spectral type) through the V filter

There are notes on each of these categories after Table 2. After finding the necessary quantities for each of the four stars, be sure to read the notes before moving to the next steps. Especially important is remembering that you must divide the parallax in milliarcseconds (mas) by 1,000 to put the value into seconds of arc (arcseconds).

—-1 —0 —+1

577-53231_ch02_4P.indd 45 3/9/13 8:18 AM C. Confirming YourV alues and Deriving the Rest of the Stellar Properties for the Stars’ Temperatures, Apparent Magnitudes, and Parallaxes 3. Knowing the spectral type and surface temperature of a star, when combined with its apparent magnitude and mea­sured parallax shift, leads directly to finding other fundamental properties of a star. Use SIMBAD to find the values of spectral type, apparent magnitude, and parallax for these four stars and fill in Table 2 below. hn hk io il sy SY Table 2. Listing of SIMBAD’s values for all four stars hn hk io il sy SY Star ID Spectral Type Surface Temp. App. V Mag. Parallax (mas) Parallax hn hk io il sy SY (K) (arcsec) hn hk io il sy SY 41 Cygni F5Iab 7400 4.016 4.24 236 52 Cygni hn hk io il sy SY 69 Cygni hn hk io il sy SY xi Cygni hn hk io il sy SY hn hk io il sy SY Note on Spectral Type hn hk io il sy SY The spectral types may have additional identifications besides what we have covered, but we are concerned just with the main spectral types—­A1, F0, M4, G2, and so on—­and the luminosity classes—­V, IV, III, II, and I. How close did your values for spectral types using Wien’s law come to the published values? ­Were your values a ­whole spectral type off? ­Were they a few subtypes off? Remember that the difference between an A7 star and an F2 star is five spectral sub- types, not a ­whole spectral type. ___ Very close! ___ Somewhat close ___ Somewhat off ___ Way off!

Note on stellar surface temperatures Now we have not only the actual spectral types but also the spectral classes available to us. Using the “Spectral Type Characteristics,” Table 5, determine each star’s temperature based on its full classification. You may need to use some interpolation if the exact spectral type is not given.

Note on apparent magnitude SIMBAD lists the apparent magnitudes through a number of different filters: B, V, R, I, J, H, and K. We want just the V (visual) magnitude that results from mea­sur­ing the flux through a filter centered at a wavelength of approximately 550 nm.

Note on mea­sured parallax shift The numbers for the parallax shift are given in milliarcseconds (1/1000 or 0.001 arcseconds). We need to convert them to arcseconds in order to find the distances in parsecs. To do that, you must divide the number given in milliarcseconds (mas) by 1000, or simply move the decimal over to the left three places. Example: Sirius has a parallax of 379.21 mas. That is 0.37921 arcseconds.

Note on finding the distance from the mea­sured parallax Once we have a parallax value, calculating the distances to stars is easy. The distance (in units of parsecs) is simply the inverse of the parallax: 1 dpc = (Eqn. 1) parallax(sarc ec)

Note on calculating the absolute magnitude If the parallax is known, we can use the star’s mea­sured distance to determine its absolute magnitude by the following equation:

-1— Mv = mv − 5log10(dpc) + 5 (Eqn. 2) 0— +1—

577-53231_ch02_4P.indd 46 3/9/13 8:18 AM Apparent magnitude is given by a lowercase “m”; absolute magnitude, by an uppercase “M.” To indicate the filter through which the magnitude is mea­sured, a subscript is added. For example, if the light from the star is mea­sured

through the V‑band (“visible”) filter, the apparent magnitude is labeled as m“ v.” If no subscript is given, we assume the mea­sure­ment was through a V‑filter.

Table 3. Basic astronomical data for the Sun Solar Property Value hn hk io il sy SY Mass (M) 1.989 x 1030 kg hn hk io il sy SY Radius (R) 6.96 x 108 meters hn hk io il sy SY Surface temperature (T) 5770 K hn hk io il sy SY Luminosity (L) 3.84 x 1026 Watts Spectral type and luminosity G2 V hn hk io il sy SY class hn hk io il sy SY Apparent magnitude (m ) −­26.7 v hn hk io il sy SY Absolute magnitude (M ) +4.74 v hn hk io il sy SY hn hk io il sy SY Note on finding the luminosity of the star as compared to the Sun Astronomers often use the Sun as a comparison star, which puts the quantities into a form that is much easier to grasp. Because we know the physics of the Sun, we know its properties to high accuracy. Table 3 above gives the fundamental properties for the Sun. It is easier to calculate the luminosity of a star relative to the Sun because by using a ratio, the physical constants (“big G,” σ, π, ­etc.) drop out of the equations. We can relate the ratio of the luminosity of stars to the Sun with their absolute magnitudes as shown in Equation 3.

L ()Msun −Mstar star =10 25. (Eqn. 3) Lsun

Note on calculating the radius of the star as compared to the Sun The luminosity of a star depends on two factors: its radius and its temperature. Assuming a spherical star, the tempera- ture of a star can be calculated by the Stefan-­Boltzmann equation:

L = 4πR2σT 4 (Eqn. 4)

where the Stefan-­Boltzmann constant is σ = 5.67 × 10−8 W/m 2/K4. 24     Lstar Rstar Tstar =     (Eqn. 5) Lsun  Rsun   Tsun 

Equations 4 and 5 would be alternate ways of calculating the luminosity of a star directly or in units of the Sun’s lumi- nosity, but we would have to have an accurate, in­de­pen­dent mea­sure­ment of the star’s radius, which we rarely do. We have direct mea­sure­ments of the radii of roughly only a dozen stars, and that is it! We can now rearrange Eqn. 5 to solve for the ratio of the radius of the star with respect to the radius of the Sun, as noted in Eqn. 6. When using ratios such as this, the constants all con­ve­niently cancel.

2   Rstar Tsun Lstar =   (Eqn. 6) Rsun  Tstar  Lsun

4. Use the table given at the end of this activity, “Spectral Class versus Surface Temperature for Supergiants (I), Giants (III), and Main Sequence (V) Stars,” to get each star’s surface temperature. Then, using the appropriate —-1 equations, complete Table 4. —0 —+1

577-53231_ch02_4P.indd 47 3/9/13 8:18 AM Table 4. Calculations for four of the stars in Cygnus

2 1 L ()Msun −Mstar R  T  L Star 2.5 star sun star Star Name dpc = Mv = mv − 5log10(dpc) + 5 = 10 =   parallax(arcsec) Lsun Rsun  Tstar  Lsun

2 1 (.4742+ .)85  5770 41 Cygni = 236 4.016 − 5log(236) + 5 = −2.85 1025. = 103   1000= 19 0.00424  7400 hn hk io il sy SY 52 Cygni hn hk io il sy SY 69 Cygni hn hk io il sy SY xi Cygni hn hk io il sy SY 5. Pick the star with the largest radius. How far into the solar system would it reach if the Sun ­were replaced by this hn hk io il sy SY star? Show your calculations, and mark the extent of the star on the following scale. hn hk io il sy SY hn hk io il sy SY Sun MVEM J hn hk io il sy SY hn hk io il sy SY How to do this: • Convert the star’s radius from solar radius to meters. • Convert the star’s radius from meters to AU’s. • Plot semicircle on above scale.

6. Summarize in a well-­written paragraph the pro­cess we use to determine the basic properties of stars. Include such things as which observations are needed, what formulae are used, and a short personal critique of the steps used.

Peak wavelength 41 Cygni Uncertainty Uncertainty 52 Cygni 1.0 1.0 0.9 0.9 0.8 0.8 0.7 0.7 0.6 0.6 0.5 0.5 0.4 0.4 Normalized flux 0.3 0.3 Normalized flux 0.2 0.2 0.1 0.1 0 0 350 390 430 470 510 550 590 630 670 710 750 350 390 430 470 510 550 590 630 670 710 750 Wavelength (nm) Wavelength (nm)

69 Cygni xi Cygni 1.0 1.0 0.9 0.9 0.8 0.8 0.7 0.7 0.6 0.6 0.5 0.5 0.4 0.4 0.3 0.3 Normalized flux Normalized flux 0.2 0.2 0.1 0.1 0 0 350 390 430 470 510 550 590 630 670 710 750 350 390 430 470 510 550 590 630 670 710 750 -1— Wavelength (nm) Wavelength (nm) 0— Figure 1. Sample fit for 41 Cygni and spectra for 52 Cygni, 69 Cygni, and xi Cygni +1—

577-53231_ch02_4P.indd 48 3/9/13 8:18 AM Table 5: Spectral type characteristics

Main Sequence Stars (V) Giants (III) Supergiants (I) Spectral Temp. (K) Spectral Type Temp. (K) Spectral Type Temp. (K) Type O5V 39800 O9V 35480 O8III 31620 hn hk io il sy SY B0V 28180 B0III 25790 B0I 36000 hn hk io il sy SY B1V 22390 B1-­2III 19950 B2I 21800 hn hk io il sy SY B3V 19060 B3III 17370 B4I 17600 hn hk io il sy SY B5-­7V 14130 B5III 14790 B5I 14400 B8V 11750 B9III 11090 B8I 11200 hn hk io il sy SY A0V 9550 A0III 9570 A0I 9720 hn hk io il sy SY A2V 8910 A2III 9200 hn hk io il sy SY A3V 8790 A4III 8830 hn hk io il sy SY A5V 8490 A5III 8450 F0V 7210 F0III 7590 F0I 7690 hn hk io il sy SY F2V 6780 F2.5III 7060 F2.5I 7160 F5V 6530 F5III 6530 F5I 6640 F8V 6040 G0III 5610 F8I 6100 G0V 5810 G2III 5460 G0I 5510 G2V 5640 G4III 5310 G2.5I 5280 G5V 5590 G5III 5160 G5I 5050 G8V 5340 G8III 5010 G8I 4590 K0V 5190 K0III 4850 K0I 4420 K2V 4890 K1.5III 4610 K2I 4260 K3.5V 4540 K3III 4370 K4I 3990 K5V 4190 K5III 4010 K7V 4000 K7III 3910 M0V 3800 M0III 3820 M2I 3450 M2V 3540 M2.5III 3620

M4V 3110 M3III 3520 M5V 2950 M5III 3420

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577-53231_ch02_4P.indd 49 3/9/13 8:18 AM hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY hn hk io il sy SY

-1— 0— +1—

577-53231_ch02_4P.indd 50 3/9/13 8:18 AM Credits

hn hk io il sy SY A ctivities, Demonstrations, and Chapter 13, Activity 3: Activity inspired by Research-­ hn hk io il sy SY Explorations Based Science Education for Undergraduates (RBSEU), a hn hk io il sy SY collaborative program between the University of Alaska Chapter 7, Activity 1: Discovery data for the planet orbit- hn hk io il sy SY Anchorage and Indiana University. RBSEU is funded ing the star 51 Pegasi. Plot produced by Ana Larson from through a grant from the National Science Foundation. original data supplied by Geoff Marcy and Paul Butler, hn hk io il sy SY private communication. Chapter 13, Activity 3: Scale model of solar system by hn hk io il sy SY Ana Larson. hn hk io il sy SY Chapter 7, Activity 1: Scale model of solar system and Temperature of the Solar System vs. distance from the hn hk io il sy SY Chapter 13, Activity 3: Four spectra, Research-­Based Sci- Sun by Ana Larson. hn hk io il sy SY ence Education for Undergraduates (RBSEU), a collab- orative program between the University of Alaska Chapter 7, Activity 2: Known exoplanet histograms and Anchorage and Indiana University. RBSEU is funded scatterplot by Ana Larson. through a grant from the National Science Foundation. Chapter 13, Exploration 2: Images of applet at ­http://​ Chapter 13, Activity 3: Table adapted from The Stellar www​.jb​.man​.ac​.uk​/distance​/life​/sample​/java​/spectype​ Spectral Flux Library by Pickles 1998, Space Telescope /specplot​.htm. Dr. T. J. ­O’Brien, University of Manchester. Science Institute, ­http://​www​.stsci​.edu​/hst​/HST​_overview​ /documents​/synphot​/AppA​_Catalogs5​.html​#65839. Chapter 13, Exploration 3: H‑R diagram by Ana Larson. Retrieved 07​/14​/2012. Some linear interpolation of tem- peratures has been made for student ease in determining Chapter 13, Activity 2: Figures by Ana Larson. stellar surface temperatures.

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