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Physics 160 Fall 2013 Midterm Exam Physics 160: Stellar Astrophysics Midterm Exam 31 October 2013 Score (out of 128) 118 Name: S O L U T I O N S Student ID #: INSTRUCTIONS – READ ME! 1. You have 80 minutes to complete the exam. 2. Complete all of the questions in Part 1: Short Answer Questions, but do not spend too long on this section. 3. Complete as many questions as you can in Part 2: Long Answer Questions; you do not need to do all of them to get full credit. 4. All answers should be written on these pages in the space provided. 5. For all questions, clearly circle/box your final answer(s). 6. Be sure to show your work, as partial credit will be awarded. Extra sheets for work are available after the last question. 7. Equations and unit conversions are provided on the last three pages for reference (you might consider tearing these out for ease of use). 8. You may use a non-­‐graphing calculator for computation. DO NOT TURN OVER THIS PAGE UNTIL INSTRUCTED TO DO SO PUT AWAY ALL BOOKS, NOTES, PHONES, COMPUTERS AND STUDY AIDS Good luck! Page 1 of 16 Physics 160 Fall 2013 Part 1: Short-­‐Answer Questions
Midterm Exam Answer all of the questions. You do not need to show your work. Do not spend too long on this section (1) [18 pts] Order of Magnitude Guesstimation: Estimate the following quantities in the units indicated to the nearest power of 10 (3 pts each). Note: answers within a factor of 10 were given partial credit (a) Diameter of the Galaxy’s disk (in parsec): 2x104 (b) Mass density of stars in the vicinity of the Sun (in Msun/pc3): 0.14 (c) Total number of stars in the Galaxy: 2x1011 (d) The mass of the supermassive black hole in the center of our Galaxy (in solar masses): 4x106 (e) Resolution of the eye (in arcseconds): 60 (f) Fraction of the Sun’s mass lost due to fusion over 1010 yr (in %): L/c2 = 4e26W/(3e8m/s)2 ≈ 4e9 kg/s x πe7 s/yr x 1e10 yr ≈ 1e27 kg ≈ 10-­‐3 Msun so 0.1% Page 2 of 16 Physics 160 Fall 2013 Midterm Exam (2) [20 pts] Were you paying attention? Short Answers Answer each of the following (4 pts each); you do not need to show your work. Question (b) refers to letters written on the chalkboards. (a) What is the current Sidereal Time? At 2:30pm (4:00pm) on 10/31, it will be 16:21 (17:51). You can estimate by remembering that in Week 1 LST midnite was 0:00, so 2:30pm would have been 14:30. However, 4 weeks have passed, which is roughly a month, or 2 hours of sidereal motion. (b) Which of the letters on the boards corresponds to an azimuth of 90º? It will be the one pointing due East – A, B or C was accepted (c) What is the current phase of the Moon? If you choose crescent or gibbous, be sure to indicate whether it is wanning or waxing. New or Wanning Crescent. A hint from class: our originally scheduled night lab on Friday 10/18 was a lunar eclipse, and hence a full moon; 2 weeks later is ½ of the lunar cycle (d) What is the fastest proper motion measured for a star in arcseconds/year, and what star is it? 11 arcseconds/year for Barnard’s Star. You can estimate this by assuming stars move around 50 km/s and nearby stars are rough 5 pc away, so µ = Vtan/4.74d = 2 arcseconds/yr would have been a good guess (e) Around what element or proton number do you cease gaining energy through fusion? Iron, or Z = 26; this is the peak of the binding energy curve Page 3 of 16 Physics 160 Fall 2013 Midterm Exam Part 2: Long-­‐Answer Questions Answer as many of the following questions as you can. Full credit for this part is 90 points; any additional points you earn above this will be added as 30% bonus points (for a maximum of 120 points). (1) [Total: 40 pt] Let’s meet Vega Vega (α Lyr) is the brightest star in the constellation Lyra, an A0 V star which by definition has an apparent visual magnitude V = 0. It also has a parallax π = 0.130 arcseconds, a proper motion of µ = 0.35 arcseconds/year and a surface temperature of 9600 K. Its Balmer Hα line (n = 2-­‐>3) is observed to be at 656.2 nm. (a) [5 pts] What is the distance to Vega in pc? Distance (pc) = 1/π (arcseconds) = 1/0.130 = 7.7 pc (b) [5 pts] What is the absolute visual magnitude of Vega? M – m = -­‐5log10(d/10 pc) => MV = V -­‐ 5log10(d/10 pc) = 0 – 5log10(7.7 pc/10 pc) = 0.57 mag Page 4 of 16 Physics 160 Fall 2013 Midterm Exam (c) [5 pts] What is the tangential speed of Vega in km/s? Vtan (km/s) = 4.74 µ (“/yr) d(pc) = 4.74 x 0.35 x 7.7 = 13 km/s (d) [5 pts] What is the radial velocity of Vega in km/s? Be sure to indicate whether it is moving toward us or away from us. Vrad = c (Δλ/λ0) = (3 x 105 km/s) x (656.2 nm – 656.3 nm)/(656.3 nm) = -­‐46 km/s The line is at a shorter wavelength (toward the blue), so it is blueshifted and therefore moving toward us. Page 5 of 16 Physics 160 Fall 2013 Midterm Exam (e) [10 pts] A0 stars have a bolometric correction Mbol – MV = 0.15. Using the Sun’s absolute bolometric magnitude as a guide, calculate the luminosity of Vega in units of solar luminosities, and its radius in units of Solar radii. Is Vega a dwarf (Main Sequence) or giant star? Mbol = 0.57+0.15 = 0.72 Mbol for the Sun is 4.74, and using the standard definition of magnitude: Mbol(Vega) – Mbol(Sun) = -­‐2.5log10(LVega/LSun) => LVega = LSun x 10-­‐0.4(0.72-­‐4.74) = 41 LSun L = 4πR2σT4 => R = (L/4πσT4)1/2 Or in solar units: (R/R¤) = (L/L¤)1/2(T/T¤)-­‐2 = (41)1/2(9600/5800)-­‐2 R = 2.3 R¤ This is a dwarf star. (f) [10 pts] What is the apparent angular size (diameter) of Vega in arcseconds as viewed from Earth? Can this be resolved with the Keck telescopes (resolution ≈ 0.05 arcseconds) θ (“) = diameter (AU) / distance (pc) = 2 x radius (AU) / distance (pc) = 2 x (2.3 R¤) x (6.96x108 m/R¤) x (AU/1.50x1011 m) / 7.7 pc = 2.8x10-­‐3 arcseconds or 2.8 milliarcseconds This could not be resolved with Keck. Page 6 of 16 Physics 160 Fall 2013 Midterm Exam (2) [Total: 50 pt] Taking Jupiter’s Temperature. (a) [5 pts] What is the radiant flux received at Jupiter (orbital radius 5.2 AU) from the Sun in W/m2? Flux = Luminosity/Area = L/4πd2 = 3.84x1026 W / 4π x (5.2 x 1.50x1011 m)2 = 50.2 W/m2 (b) [10 pts] What fraction of the Sun’s luminosity is absorbed by Jupiter, assuming it reflects 20% of incident sunlight? Jupiter has a radius about 0.1 times that of the Sun. Power = Flux x Area = 0.8 x 50.2 W/m2 x πRJ2 = 40.2 W/m2 x π(6.96x107 m)2 = = 6.11x1017 W Power absorbed/Luminosity = 6.11x1017 W/3.84x1026 W = 1.59x10-­‐9 (c) [10 pts] Jupiter has a stable effective temperature of about 125 K. What fraction of the radiant energy coming from Jupiter’s is thermally emitted versus reflected from the Sun? Thermal emission = 4πRJ2σTJ4 Reflected emission = L/4πd2 x 0.2 x πRJ2 Ratio = 4πRJ2σTJ4 x 4πd2 / 0.2LπRJ2 = 80π d2 σTJ4 / L = 80π (5.2 x 1.50x1011 m)2 (5.67x10-­‐8 W/m2-­‐K4) (125 K)4 / (3.84x1026 W) = 0.97 Almost the same! Page 7 of 16 Physics 160 Fall 2013 Midterm Exam (d) [15 pts] Derive a general equation relating the equilibrium surface temperature Tp of a planet of radius Rp to the surface temperature T* of a star of radius R* separated by a distance d. Assume that both objects radiate as uniform blackbodies at these temperatures, and that the planet reflects a fraction A of the incident radiation back into space (this value is known as the albedo). Thermal equilibrium implies that the energy per second absorbed by the planet is equal to the energy per second emitted. The powers in and out are: Power in = σT*4 x (R*/d)2 x πRp2 x (1-­‐A) Power out = 4πRp2σTp4 Power out = Power in => 4πRp2σTp4 = σT*4 x (R*/d)2 x πRp2 x (1-­‐A) => Tp = 0.71 T*(R*/d)1/2 (1-­‐A)1/4 Note that this is independent of the planet’s radius! Page 8 of 16 Physics 160 Fall 2013 Midterm Exam (e) [10 pt] Suppose Jupiter is actually orbiting Vega, which has T* = 9600 K and R* = 2.4 Rsun. What would be the equilibrium temperature of its photosphere? Assume again that it reflects 20% of incident light. Compare this temperature with the freezing and boiling points of water, and comment on the potential habitability (possible presence of life) of moons around a Vegan Jupiter. We can either use the result from (d), or note from (c) that the fraction TJ4/L ~ TJ4/R*2T*4, so both imply scaling TJ with T*R*1/2 will preserve a ratio of unity Tp = 125 K (9600 K/5800 K)(2.4/1.0)1/2 = 320 K This is right between the freezing (273 K) and boiling (373 K) points of water, suggesting that moons around Vegan Jupiter could have liquid water on their surfaces if they have sufficient atmospheric pressure. Liquid water is a critical ingredient for life on Earth, so it’s presence on other worlds is seen as a requirement for habitability. Page 9 of 16 Physics 160 Fall 2013 Midterm Exam (3) [Total: 50 pts] Nubulium Prof. Burgasser has proposed a new element called Nubulium, which is a Hydrogen-­‐like atom that has a low ionization potential, χNu = 3 eV. He estimates that Nubulium has a cosmic abundance 10-­‐8 that of Hydrogen, and that the Sun is a good place to look for it. (a) [5 pts] At what wavelengths and in what parts of the electromagnetic spectrum (UV, visible, infrared, etc.) would we expect to see absorption lines from Nubulium for the n=1-­‐>2 (Lyman α) and n=2-­‐>3 (Balmer α) transitions? En = -­‐χNu/n2 => E1 = -­‐3 eV, E2 = -­‐0.75 eV, E3 = -­‐0.33 eV ΔE = hc/λ => λ = hc/ΔE => 1-­‐>2 transition occurs at 1.24 eV µm/2.25 eV = 552 nm – visible => 2-­‐>3 transition occurs at 1.24 eV µm/0.42 eV = 2.95 µm – infrared (b) [10 pts] Assume the Sun’s atmosphere contains Nubulium. What is the ratio of neutral Nubulium in the n = 2 to n = 1 states? This makes use of the Boltzmann Equation: N2/N1 = g2/g1 e-­‐(ΔE21)/kT = (2x22)/(2x12) e-­‐[2.25 eV/(8.62x10-­‐5 eV/K x 5800 K)] = 4e-­‐4.5 = 0.044 Page 10 of 16 Physics 160 Fall 2013 Midterm Exam (c) [15 pts] If the cosmic fraction of Nubulium to Hydrogen atoms is 10-­‐8 (about the same as Lithium to Hydrogen), what are the number densities of Hydrogen and Nubulium atoms in the Sun’s photosphere (τ = 2/3)? For simplicity, assume the Sun’s atmosphere is pure Hydrogen, has a mean opacity κ = 0.1 m2/kg, and the surface pressure P = 0 at τ = 0. nNu = 10-­‐8 nH = 10-­‐8 P/kT To solve for the pressure, use the relation we derived in the homework: dP/dτ = g/κ => Pph = 3/2 GM/R2 1/κ = 3/2 (6.67x10-­‐11 N m2/kg2)(1.99x1030 kg)/(6.99x108 m)2 /0.1 m2/kg = 4.07 x 103 N/m2 => nH = (4.07x103 N/m2)/[(1.38x10-­‐23 J/K)(5800 K)] = 5.08x1022 m-­‐3 => nNu = 10-­‐8nH = 5.08x1014 m-­‐3 Page 11 of 16 Physics 160 Fall 2013 Midterm Exam (d) [15 pts] What is the fraction of ionized to neutral Nubulium atoms in the Sun’s photosphere? To simplify the calculation, assume that the partition function for ionized Nubulium atoms Z+ = 1, that all neutral Nubulium atoms are in the n=1 ground state, and that all of the free electrons in the gas come from ionized Nubulium atoms (so N+ ≈ Ne). This is the Saha equation: N+Ne/N0 ≈ N+2/No = 2 Z+/Zo (2πmekT/h2)3/2 e-­‐ χ/kT Z+ = 1, Z0 = g1 e-­‐ΔE11/kT = 2, so 2 Z+/Z0 = 1 the numerical factor: 2πmekT/h2 = 2π(mec2)kT/(hc)2 = 2π(0.511x106 eV)(0.5 eV)/(1.24 eV µm)2 = 1.04x106 µm-­‐2 x (106 µm/m)2 = 1.04x1018 m-­‐2 therefore N+2/N0 = (1.04x1018 m-­‐2)3/2 e-­‐(3 eV/0.5 eV) = 2.63x1024 m-­‐3 We want the fraction N+/N = x, and N+2/N0 = (N+/N)2/[(N-­‐N+)/N2] = N x2/(1-­‐x) = 2.63x1024 m-­‐3 => x2/(1-­‐x) = 5.18x109 based on part (c) Since 0 < x < 1, x must be very close to 1, so we can approximate x = 1-­‐(1-­‐x) = 1-­‐(1-­‐[N+/N]) ≈ 1-­‐(1/5.18x109) = 1-­‐1.9x10-­‐10 ≈ 1 Nearly all of the Nubulium is ionized! Page 12 of 16 Physics 160 Fall 2013 Midterm Exam (e) [5 pts] Based on the results from (d), is Prof. Burgasser’s prediction that we would see Nubulium absorption lines in the Solar spectrum a reasonable one? Why or why not? What might be a better type of star to look for this hypothesized element? Since only a tiny fraction of Nubulium is neutral (the equivalent abundance would be 10-­‐18 of Hydrogen, which is less than every other atom), we would see no Nubulium absorption lines in a Solar spectrum even if it was there. We need to look in a cooler class of star, such as an M dwarf, to detect any bound-­‐bound transitions.
Page 13 of 16 Physics 160 Fall 2013 Midterm Exam (4) [Total: 40 pts] A stellar model Assume a star of mass M and radius R has a density that varies linearly with radius; i.e.: ρ = ρc
!
r"
1−
R where ρc is the central radius density. (a) [10 pts] Using the equation of mass conservation, write down an expression for M(r), the mass of the star interior to radius r, as a function of the total mass M and radius R ∫ dM = ∫ 4πr2ρ(r)dr = 4πρc ∫ (r2-­‐r3/R)dr integrate from M = o to M = M(r) and from r = 0 to r = r M(r) – M(0) = 4πρc (1/3 r3 -­‐ 1/4 r4/R) M(r) = 4π/3 ρc r3 (1-­‐3/4 r/R) At the maximum radius: M(R) = M = π/3 ρcR3 <-­‐-­‐-­‐¼ the value you’d get for a uniform sphere! => M(r) = M (r/R)3 (4-­‐3r/R) Page 14 of 16 Physics 160 Fall 2013 Midterm Exam (b) [5 pts] Determine a relation between the average density <ρ> of the star to the central density ρc which does not depend explicitly on M or R. <ρ> = 3/4π M(R)/R3 substituting in above <p> = 3/4π [4π/3 ρc R3 (1-­‐3/4) ] / R3 => <p> = 1/4 ρc (c) [10 pts] Derive an expression for the core density pressure in terms of G, M and R. Use the equation of hydrostatic equilibrium: ∫ dP = -­‐ ∫ g(r)ρ(r) dr = -­‐G ∫ [M(r)ρ(r)/r2] dr = -­‐1/4 GM 4πρc ∫ [(r/R)3 (1-­‐3/4 r/R) (1-­‐r/R)]/r2 dr <-­‐-­‐ substitute x = r/R = -­‐ 4π GM/R <ρ> ∫ [x (1-­‐3/4 x) (1-­‐x)] dx Note that I have substituted in <ρ> and pulled 1/R out of the integral. Expanding the terms in the integral and evaluating P = P(0) = Pc to P(R) = 0 and x = 0 to 1: P(R)-­‐P(0) = -­‐Pc = -­‐4π GM/R <ρ> ∫ [3/4x3 – 7/4x2 +x] dx => Pc = 4π GM/R <ρ> [3/16 – 7/12 + 1/2] = 5/48 4π GM/R [3/4π M/R3] = 5/16 GM2/R4 NOTE: THIS PROBLEM WAS WORDED INCORRECTLY IN THE EXAM, SO IT WAS NOT INCLUDED IN THE GRADING Page 15 of 16 Physics 160 Fall 2013 Midterm Exam (d) [10 pts] Using the Sun’s mass and radius, evaluate ρc and Pc. Then, assuming this to be a fully ionized, pure hydrogen, ideal gas, find Tc. ρc = 4 <ρ> = 4 x 3/4π x 1.99x1030 kg/(6.96x108 m)3 = 5640 kg/m3 Pc = 5/16 x (6.67x10-­‐11 N m2/kg2) x (1.99x1030 kg)2 / (6.96x108 m)4 = 3.52x1014 N/m2 = 3.52 Gbar For fully ionized hydrogen, µ = 0.5 Tc = PcµmH/ρck = (3.52x1014 N/m2)(0.5)(1.67x10-­‐27 kg)/[(5640 kg/m3)(1.38x10-­‐23 J/K)] = 3.77x106 K NOTE: BECAUSE THE PREVIOUS PROBLEM WAS WORDED INCORRECTLY IN THE EXAM, THIS ONE WAS ALSO NOT INCLUDED IN THE GRADING (e) [5 pts] Is this a good model for the Sun? Why or why not? This is not a good model because the inferred core temperature is too low to produce the level of fusion needed to keep the Sun as bright as it is. The Sun must be more centrally condensed than this model, as this would raise the pressure and result in a higher core temperature. Page 16 of 16