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Spring Semester 2010 Exam 3 – Answers a Planetary Nebula Is A

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Spring Semester 2010 Exam 3 – Answers a Planetary Nebula Is A ASTR 1120H – Spring Semester 2010 Exam 3 – Answers PART I (70 points total) – 14 short answer questions (5 points each): 1. What is a planetary nebula? How long can a planetary nebula be observed? A planetary nebula is a luminous shell of gas ejected from an old, low- mass star. It can be observed for approximately 50,000 years; after that it its gases will have ceased to glow and it will simply fade from view. 2. What is a white dwarf star? How does a white dwarf change over time? A white dwarf star is a low-mass star that has exhausted all its nuclear fuel and contracted to a size roughly equal to the size of the Earth. It is supported against further contraction by electron degeneracy pressure. As a white dwarf ages, though, it will radiate, thereby cooling and getting less luminous. 3. Name at least three differences between Type Ia supernovae and Type II supernovae. Type Ia SN are intrinsically brighter than Type II SN. Type II SN show hydrogen emission lines; Type Ia SN don't. Type II SN are thought to arise from core collapse of a single, massive star. Type Ia SN are thought to arise from accretion from a binary companion onto a white dwarf star, pushing the white dwarf beyond the Chandrasekhar limit and causing a nuclear detonation that destroys the white dwarf. 4. Why was SN 1987A so important? SN 1987A was the first naked-eye SN in almost 400 years. It was the first SN for which a precursor star could be identified. And it is the only SN from which neutrinos have been detected. 1 5. Who discovered pulsars? What are these objects? Jocelyn Bell discovered pulsars in 1967. Pulsars are pulsating radio sources thought to be rapidly rotating neutron stars. 6. How do pulsars emit pulses? The radio pulses we see come from beams of radiation emanating from the neutron star's magnetic poles as the beams sweep past the Earth. These beams are produced by charged particles spiraling around the neutron star's magnetic field lines. 7. How does the Crab Nebula and its associated pulsar provide a convincing case that pulsars are physically related to supernova remnants? The total synchrotron radiation observed coming from the Crab Nebula is equal to the amount of rotational energy lost by the Crab pulsar as a result of its slowing down. 8. Name the three fundamental physical parameters that are affected by motion, according to the special theory of relativity. How are they affected? Mass is increased. Length is contracted. Time is dilated (increased). 9. Name four experimental tests of General Relativity. The gravitational bending of light, as demonstrated during the total solar eclipse of 1919. The precession of the perihelion of Mercury. Gravitational time dilation, which causes gravitational redshift. Gravitational waves (or radiatio), as demonstrated, for example, by the binary pulsar, for which Taylor and Hulse won the 1993 Nobel Prize in Physics. 2 10. What are the data that convince astronomers that Cyg X-1 is a stellar-sized black hole? Detection of x-rays, which could arise from an accretion disk around a black hole. Identification of the system (Cyg X-1 and HDE 226868) as a single-lined spectroscopic binary, which is necessary for accretion to take place. Rapid variability of the x-rays, indicating a maximum size to Cyg X-1 of about 3000 km, i.e., a compact object. Solution of Newton's version of Kepler's 3rd law indicating that the mass of Cyg X-1 is larger than the maximum mass of a neutron star. 11. How did Harlow Shapley determine the size and extent of the Milky Way Galaxy and our location in it? Shapley applied Henrietta Leavitt's Period-Luminosity Relation for Cepheid variables to the RR Lyrae variables in Globular Clusters to determine the distances to the clusters. The result was a spherically symmetric distribution of the clusters, with a diameter of about 100,000 LY. He then argued that the center of the distribution of clusters must be the center of the Galaxy. He also showed that the center of the Galaxy was about 25,000 LY from the solar system. 12. How have astronomers determined that the Milky Way Galaxy has a spiral arm structure? Observations of the 21 cm spin-flip transition in hydrogen have allowed astronomers to map the distribution of atomic hydrogen in the Galaxy. This distribution, as well as those of hot, massive main-sequence stars and H II regions (= emission nebulae), is concentrated in numerous arched lanes, that we call spiral arms. 3 13. Why do astronomers think that the Milky Way Galaxy contains dark matter? The galactic rotation curve shows that the orbital speed of material in the Galaxy does not fall off with distance. The calculated mass necessary to produce such a rotation curve far exceeds the sum of all of the visible matter in the Galaxy, hence, that excess mass is unseen, which is why it is called dark matter. 14. What is the evidence that there is a supermassive black hole at the center of the Milky Way Galaxy? Using high-resolution observatories, numerous stars have been observed to orbit the center of the Galaxy, i.e., Sagittarius A*, very close to Sag A*. By applying Newton's version of Kepler's 3rd law, the mass of Sag A* is calculated to be millions of solar masses. The most plausible explanation for such a large mass in such a confined region is a supermassive black hole. PART II (30 points) – 6 mathematical problems (5 points each): 1. The distance to the LMC is about 50,000 pc. The absolute magnitude of SN1987A at its brightest was about –15 (negative 15). What was its apparent magnitude when it was at its brightest? The naked-eye limit is about apparent magnitude = 6. According to your calculations should SN1987A have been visible to the naked-eye? m – M = 5 log d – 5 ⇒ m = 5 log d – 5 + M = 5 log (50,000) – 5 + (–15) = 3.49. Since 3.49 is less than 6, SN 1987A would have been (and was!) visible to the naked-eye. 4 2. A pulsar’s period is 6 × 10-1 sec and it is increasing at a rate of 3 × 10-12 s/s. Calculate the approximate age of this pulsar. Could Galileo (who lived from 1564-1642) have seen the supernova that created this pulsar (assuming it was bright enough)? t = P/2R = (6 × 10-1 sec)/2(3 × 10-12 s/s) = 1 × 1011 s = 3175 years. Galileo could only have seen SN that occurred 368 to 446 years ago, so he wouldn't have been able to see this one. 3. If a black hole were to have the same size as the Earth (RE = 6,378 km), how massive would it be? 2 RSch = (2GM)/c ) = 3 km (M/Mo) ⇒ (M/Mo) = RSch / 3 km = 6,378 km / 3 km = 2126 ⇒ M = 2126 Mo 4. According to the Hawking, black holes may evaporate. (a) If the “temperature” of a black hole is inversely proportional to its mass, i.e., T ∝ 1/M, how is the “luminosity” of the black hole dependent on its mass? (Hints: use the formulae for Rs and for L that we discussed many times in class.) (b) What about the time it would take to “evaporate”; how does that depend on the black hole’s mass? (Hint: Recall that the lifetime of a star, or in this case a black hole, is dependent on its fuel supply, i.e., its energy, and on the rate at which it loses energy.) 2 (a) RSch = (2GM)/c ) and T ∝ 1/M, so L = 4πR2σ T4 ∝ R2 T4 ∝ M2 M-4 ∝ M-2 (b) t = E/L = Mc2 / L ∝ M / M-2 ∝ M3 5 5. A gas cloud located in the spiral arm of a distant galaxy is observed to have an orbital velocity of 400 km/s = 4 × 105 m/s. If the cloud is 20,000 pc = 6.2 × 1020 m from the center of the galaxy and is moving in a circular orbit, what is the mass of the galaxy 30 contained within the cloud’s orbit? The Sun’s mass is 1 Mo = 2 × 10 kg. M = (Rv2)/G = (6.2 × 1020 m)(4 × 105 m/s)2)/(6.67 × 10-11 m3 / (kg s2)) 42 11 = 1.5 × 10 kg = 7.5 × 10 Mo 6. If the supermassive black hole at the center of the Milky Way galaxy (Sagittarius A*) 6 7 has a mass of 3.7 × 10 Mo, then its Schwarzschild radius is about 10 km = 0.07 AU. If that black hole could be seen, what angular size would it have as seen from the star SO-16, which comes as close as 45 AU to Sagittarius A*? Is this bigger or smaller than the angular size of the Moon as seen from Earth (about 0.5°)? D = (αd)/(206,265) ⇒ α = 206,265 D / d = 206,265 (2 × 0.07 AU) / 45 AU = 642″ = 0.18° This is smaller than the angular size of the Moon as seen from the Earth. 6 .
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