Physics 160: Stellar Astrophysics. Midterm Exam. 27 October 2011 INSTRUCTIONS READ ME!

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1 Physics 160: Stellar Astrophysics 27 October 2011 Name: S O L U T I O N S Student ID #: INSTRUCTIONS READ ME! 1. There are 4 questions on the exam; complete at least 3 of them. 2. You have 80 minutes to complete the exam. 3. All answers should be written on these pages in the space provided. 4. For all questions, clearly circle your final answer(s). 5. Be sure to show your work, as partial credit will be awarded. Extra workspace is provided after question Equations and unit conversions are provided on the last three pages for reference. 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 11

2 (1) Regulus (α Leo) is the brightest star in the constellation Leo, a B2 star with an apparent visual magnitude V = 1.35, a parallax π = 0.042, a proper motion of µ = 0.25 /year and a surface temperature of 20,000 K. Its weak Hα line is observed to be at nm. (a) What is the distance to Regulus? [5 pts] Distance (pc) = 1/π (arcseconds) = 1/0.042 = 24 pc (b) What is the absolute visual magnitude of Regulus? [5 pts] M m = - 5log10(d/10 pc) => MV = V- 5log10(d/10 pc) = log10(24 pc/10 pc) = mag (c) How far could Regulus be in parsecs and still be visible to our eyes? (assume we cannot see anything fainter than V = 6) [5 pts] using same equation: MV V = = = - 5log10(d/10pc) => d = (10 pc) x 10 (6.53/5) = 202 pc (d) What is the tangential speed of Regulus? [5 pts] Vtan (km/s) = 4.74 µ ( /yr) d(pc) = 4.74 x 0.25 x 24 = 28 km/s Page 2 of 11

3 (e) What is the radial velocity of Regulus? Is it moving toward us or away from us? [5 pts] Vrad = c (Δλ/λ0) = (3 x 10 5 km/s) x (656.5 nm nm)/(656.3 nm) = 91 km/s The line is at a longer wavelength (toward the red), so it is redshifted and therefore moving away from us. (f) Regulus has a luminosity of 1500 L. What is its radius in units of Solar radii? Is Regulus a dwarf (Main Sequence) or giant star? [5 pts] L = 4πR 2 σt 4 => R = (L/4πσT 4 ) 1/2 Or in solar units: (R/R ) = (L/L ) 1/2 (T/T ) - 2 = (1500) 1/2 (20,000/5800) - 2 This is a dwarf star. R = 3.3 R (g) What is the apparent angular size (diameter) of Regulus in arcseconds as viewed from Earth? [5 pts] θ ( ) = diameter (AU) / distance (pc) = 2 x radius (AU) / distance (pc) = 2 x (3.3 R ) x (6.955x10 8 m/r ) x (AU/1.496x10 11 m) / 24 pc = 1.3x10-3 arcseconds or 1.3 milliarcseconds Page 3 of 11

4 (h) At what wavelength does the blackbody continuum of Regulus peak in intensity? What part of the electromagnetic spectrum is this? [5 pts] Use Wein s Law: λpeakt = 2898 µm K => λpeak = (2898 µm K)/(20,000 K) = 0.14 µm = 140 nm This is in the UV part of the electromagnetic spectrum (i) Assuming that it has the same composition as the Sun, why does Regulus have a weaker Hα line? [5 pts] The weaker Hα line is due to the high temperature of Regulus s atmosphere, resulting in a higher fraction of ionized H atoms which cannot produce Hα emission. Page 4 of 11

5 (2) The temperature of the Earth. (a) Calculate the solar flux (in W/m 2 ) incident at the top of the Earth s atmosphere. This is known as the solar constant. [10 pts] Flux = L /4πd 2 = (3.839x10 26 W)/(4π x 1 AU 2 ) x (1 AU/1.496x10 11 m) 2 = 1365 W/m 2 (b) What is the total solar power incident on the Earth in W? Be sure to take into account the curvature of the Earth. [10 pts] From the parallel rays of the Sun reaching Earth, the latter appears as a circle with area πrearth 2. Therefore Power = Flux x Area = (1365 W/m 2 ) x π x (REarth) 2 x (6.378x10 6 m/rearth) 2 = 1.74x10 17 W Page 5 of 11

6 (c) Assume that all solar radiation energy incident on the Earth is absorbed and then re- radiated in all directions as thermal blackbody emission. Derive an expression for the blackbody temperature of Earth in terms of the Sun s photospheric temperature, radius and distance from Earth. [15 pts] Remembering that L = 4πR 2 σt 4 : Incident power from sun = (4πR 2 σt 4 /4πd 2 ) x πre 2 = πσt 4 R 2 RE 2 /d 2 Emitted power = 4πRE 2 σte 4 For energy balance, power in = power out so: πσt 4 R 2 RE 2 /d 2 = 4πRE 2 σte 4 => TE 4 = 1/4 T 4 R 2 /d 2 => TE = T (R /2d) 1/2 Note that this is generally applicable to other planets orbiting other stars (d) Evaluate the expression from part (c). Is this temperature greater than or less than the mean temperature on the surface of the Earth (287 K)? Why do you think this is the case? [10 pts] TE = (5800 K)(0.5 x 6.955x10 8 m /1.496x10 11 m) 1/2 = 280 K This is somewhat cooler than the Earth s surface, as it doesn t take into account heat retained by Earth s atmosphere; i.e., the greenhouse effect Page 6 of 11

7 (3) The following is the actual spectrum of the star 2MASS J in the red optical region, with some particular spectral features noted: H Spectral Flux Density Continuum TiO Li I TiO Wavelength (nm) (a) For each of the following features, indicate whether it arose from relatively thin or dense gas and relatively hot or cool gas: [5 pts] Feature Thin or Dense Gas? Hot or Cool Gas? Continuum Dense Hot Hα line Thin Hot Li I line Thin Cool TiO molecular bands Thin Cool Page 7 of 11

8 (b) Based on the presence of TiO molecular bands, what spectral class do you think this star is? [5 pts] Molecular bands are only seen in M, L and T dwarfs, but TiO is only seen in M- type dwarfs. A clue is also given in the next part where the temperature is given. (c) The photospheric temperature of this star is 3000 K. What fraction of all neutral Hydrogen atoms in the photosphere are in the n=3 state, and thus capable of producing an Hα emission line? To first order, you can assume nearly all H atoms are in ground state. Hint: this is going to be a very small number. [15 pts] We want to use the Boltzmann Equation. First compare n = 3 to n = 1: N3/N1 = g3/g1e - (E 3 - E 1 )/kt g1 = 2, g3 = 18 E3- E1/kT = ( ev) x (1/3 2-1)/(8.617x10-5 ev/k x 3000 K) = 52.6 x (1 1/9) = 46.8 => N3/N1 = 9e = 4.4x10-20 for completeness, we should also calculate N2/N1: g1 = 2, g2 = 8 E2- E1/kT = 52.6 x (1 1/4) = 39.5 N2/N1 = 4e = 2.95x10-17 <- note similar form as above <- also small So Ntotal N1 and hence N3/Ntotal N3/N1 4.4x10-20 Page 8 of 11

9 (d) The photospheric pressure of this star is about 10 6 N/m 2. Assuming a pure, neutral hydrogen ideal gas, what is the number density of H atoms at the photosphere? [5 pts] Assume an ideal gas, then P = nkt => n = P/kT = (10 6 N/m 2 )/[(1.381x10-23 J/K) x 3000 K] = 2.4 x m - 3 (e) What is the number density of H atoms in the n=3 state in this photosphere? [5 pts] Multiplying part d and part 3, we get: N3 = (4.4x10-20 ) x (2.4 x m - 3 ) = 1.1x10 6 m - 3 for comparison, interstellar space has a density of m - 3 (f) Based on your answers above, is it reasonable to believe that the emission from Hα comes from the photosphere? Use Kirchhoff s laws to explain where this emission is probably coming from. [10 pts] No, there are simply too few H atoms in the n=3 state to produce a measureable line. This feature mostly likely comes from hot thin gas above the photosphere, which is not dense enough to obscure light from the continuum below it but still produces emission in this line. [NOTE: this hot gas comes from magnetic interactions above the M dwarf photosphere] Page 9 of 11

10 (4) Assume a star of mass M and radius R has a constant density. (a) Write down an expression for the density ρ0 of the star. [7 pts] ρ0 = <ρ> = Mass/Volume = 3M/4πR 3 (b) Using the equation of mass conservation, write down an expression for M(r), the mass of the star interior to radius r. Assume that M(0) = 0. [10 pts] dm = 4πr 2 ρ(r)dr = 3 (M/R 3 ) r 2 dr => M(r) = M(r/R) 3 Page 10 of 11

11 (c) Using the equation of hydrostatic equilibrium, write down an expression for P(r), the interior pressure of the star as a function of radius. Assume that P(R) = 0. [10 pts] dp = - G (M(r)ρ(r)/r 2 ) dr = - GM (3M/4πR 3 ) (r/r) 3 /r 2 dr Evaluate from P(r) to P(R) and r to R: => P(R) P(r) = - (3GM 2 /8πR 6 ) (R 2 - r 2 ) => P(r) = (3GM 2 /8πR 6 ) (R 2 - r 2 ) = - (3GM 2 /4πR 6 ) r dr (d) Write down an expression for P(0), the core pressure of the star. [8 pts] P(r=0) = (3/8π)(GM 2 /R 4 ) Notice that this expression is proportional to GM 2 /R 4, which we found dimensionally in class (e) Using the Sun s mass and radius, evaluate P(0) and compare it to the known central pressure, 2x10 11 N/m 2. Explain the difference. [10 pts] P(0) = (3/8π) x (6.673x10-11 N m 2 kg - 2 ) x (1.989x10 30 kg) 2 x (6.955x10 8 m) - 4 = 1.35x10 14 N/m 2 This is about 3 orders of magnitude too large. The reason is that the Sun is more centrally concentrated, so more of the mass lies closer to the center and therefore exerts less pressure on the core. Page 11 of 11