Ay13 Set 1 solutions Mia de los Reyes October 18 1. The scale of the Sun a Using the angular radius of the Sun and the radiant flux received at the top of the Earth s atmosphere, calculate the effective temperature of the Sun. First, relate the angular radius of the Sun θ to the physical radius R: R a sin θ, 1 where a is the distance from the Earth to the Sun. Note that you can also use the small-angle approximation sin θ θ. You ll get basically the same answer. Then, from conservation of energy i.e., inverse square law: L πa f And from the Stefan-Boltzmann equation: L πr σ B T eff 3 Set equations and 3 equal, plug in equation 1 for R and solve for T eff : Plugging in given values yields T eff 5775K. πa f πr σ B T eff a f a sin θ σ B Teff 5 1/ f T eff σ B sin 6 θ b Using Venus orbital period of 5 days and Kepler s laws, what is the semi-major axis of Venus in AU? Start with Kepler s third law: Ω GM + M a 3 7 First, consider the Earth-Sun system, assuming that M E M : Ω E GM a 3 E 8 Now, Ω E can be calculated from known values Earth s period is 1 year and a E is known it s 1 AU, so can solve for M : M Ω E a3 E G Then consider the Venus-Sun system, again assuming that M V M : 9 Ω V GM a 3 V 1 1
Then solve for a V, substituting equation 9 for M and converting angular velocities to periods P π Ω : a V 1/3 GM 11 Ω V Ω E a 3 1/3 E 1 PV Ω V P E /3 a E 13 Plug in known values P E 365.5 days, a E 1 AU and given values P V 5 days, and find that a V.7 AU. c At conjunction with the Sun, it takes astronomers on Earth 76s to detect the radio waves that reflect off Venus. Assuming circular orbits for the Earth and Venus, compute the distance between in 1 AU. Note: conjunction Venus is directly between Earth and the Sun. Call the distance between Earth and Venus d V. The light takes time t to travel distance d V : From the previous problem, we can compute d V d V ct.1 11 cm 1 in terms of AU: d V a E a V 1.7 AU.76 AU 15 Then combine equations 1 and 15 to convert AU to cm: 1 AU 1.5 1 13 cm. d Use your results above to compute the absolute mass, radius, and luminosity of the Sun in cgs. Inverse square law: L πa E f Plug in a E 1 AU 1.5 1 13 cm and f from problem 1a. Find that L 3.93 1 33 erg/s. Kepler s third law: M Ω E a3 E G Plug in known values in cgs; find that M. 1 33 g. Definition of angular size: R a E sin θ. Stars are gases Plug in known values in cgs; find that R 6.98 1 1 cm. a Provide a quantitative relation between temperature and density of a star that indicates when we can treat it as a gas, and show that it holds at the center of the Sun. To check if the center of the Sun can be treated as a gas, we can compare Coulomb energy to thermal energy. The ideal gas law is reasonable when the thermal energy T is larger than the Coulomb energy Ze /r. This occurs when kt Ze /r 16 r Ze /kt 17 where r is the interparticle distance and Z is the charge of the ion Z 1 for a gas composed only of ionized hydrogen. By thinking of the number density n as r 3, we can write r in terms of mass density: r n 1/3 ρ/ 1/3. Then we can rewrite equation 17: ρ k/e 3 T 3 18
For atomic hydrogen, µ.5. Plugging this in, our condition for treating a star as an ideal gas is in cgs units: ρ 1.8 1 16 T 3 19 The sun s central temperature is T c 1 7 K. By equation 19, we require ρ c 6 1 5 g cm 3. Since the sun s central density ρ c is only 15 g cm 3, we may treat the sun as an ideal gas throughout, and need not consider Coulomb interactions between particles. b Use a scaling argument to determine the stellar mass at which the ideal gas assumption breaks down. Now we want to know how ρ scales with stellar mass M. In the following derivation we only care about approximate scaling arguments, so don t worry about prefactors. The internal energy can be approximated as U T M at the central temperature T. Now recall that by the virial theorem, the internal energy is approximately the gravitational energy Ω GM /R. Solve for T and find that T M/R. We can then assume that all stars have roughly the same central temperature which is actually a good approximation for main-sequence stars, so the central density M R. Then ρ M/R 3 M, so M ρ 1/. 1/ Plugging in values for the Sun, we find M lim ρlim M. This yields a limiting mass of M.16M, which is about the mass of brown dwarfs or giant planets not stars!. Therefore, we will never have to consider Coulomb interactions for main sequence stars. 3. A toy star Assume that a star obeys the density model ρr ρ c 1 r. R a Find an expression for the central density in terms of R and M. Solve for total mass M by integrating over the density profile: M π π dφ π πρ c R 3 ρ sin θdθ ρ c r r3 R 3 R3 r ρrdr 1 dr 3 π 3 ρ cr 3 Then solve for central density: ρ c 3M πr 3. b Find the pressure as a function of radius. Doing the same integral as in the previous problem, we know that mr π 3 ρ cr 3 1 3r R 5 Now use the equation of hydrostatic equilibrium: dp dr Gm ρr 6 r G π 3 ρ cr 1 3r ρ c 1 r 7 R R π 3 Gρ cr 1 7r R + 3r R 8 3
Integrate equation 8 to get the total pressure: P r π 3 Gρ c π [ r 3 Gρ c r 7r R + 3r3 R dr 9 7r3 1R + 3r 16R + C Use the zero boundary condition P R to solve for the integration constant C: P R π [ 1 3 Gρ cr 7 1 + 3 16 + C ] R 31 C R 1 + 7 1 3 3 16 5 8 R 33 Now let s solve for the central pressure P c P r : ] 3 P c π 3 Gρ cc 3 5π 36 Gρ cr 35 Okay, finally we can substitute stuff into equation 3 to write the full equation for pressure: P r π [ r 3 Gρ c 7r3 1R + 3r 16R 5 ] 8 R 36 [ P c 1 r 8 r 3 9 r ] + 37 5 R 5 R 5 R So we find that P r P c f r R as expected. Now plug in the answer for part a to get P c as a function of M and R: We can simplify this to P c 5 GM π R. P c 5π 3M 36 G πr 3 R 38 c What is the central temperature T c, assuming an ideal gas equation of state? How does it scale with mean particle mass? Ideal gas: P ρt Solve this for central temperature, plugging in answer from b for P c and answer from a for ρ c : T c P c ρ c 39 5π 36 Gρ cr 5π 36 G 3M πr R µm 3 p 1 This simplifies to T c 5GM 1R which scales as T c. is the mean particle mass.
d Find the ratio of radiation pressure to gas pressure at the center of the star as a function of total stellar mass in M. At what mass does radiation pressure become comparable to the ideal gas pressure? Radiation pressure is given by 1 3 a ot. The ratio at the center of the star is therefore Then plug in T c from part c: 1 P gas 3 a Tc o 5 GM π R P gas 1 3 a o 5GM 1R 5 GM π 15π 1555 a og 3 M 3 R µmp Assuming solar composition µ.6, we can rewrite this in terms of solar masses as: P gas 7. 1 M M When P gas 1, the mass of the star is M 37M. Note that this is not an exact result, since our formula for T c assumes that radiation pressure is negligible. e Write the total gravitational potential energy of this toy star and verify the virial theorem. The total gravitational potential is Ω Gmr r dm. For simplicity s sake, let s convert this to an integral over radius so we can plug in the definition of mr from part b and the definition of ρr: Ω G 16π 3 ρ c G 16π 3 ρ c Gmrπrρdr 5 G π 3 ρ cr 3 1 3r R [ R 5 r 7r5 R + 3r6 R ] 5 7R6 R + 3R7 8R πrρ c 1 r dr 6 R dr 7 The gravitational potential is G 6π 315 ρ cr 5 Now let s do math with the other side of the virial theorem, plugging in P r from equation 3: 3 P dv 3 1π π dφ 8π 3 Gρ c 8π 3 Gρ c π r π 3 Gρ c r [ R 5 sin θdθ r 8 r P rdr 9 7r3 1R + 3r 16R 5 8 R 7r5 1R + 3r6 16R 5r 1 7R5 7 + 3R5 11 5R5 1 ] 8 R dr 5 dr 51 5 6π 315 Gρ cr 5 53 So 3 P dv Ω, and the virial theorem is exactly satisfied. 5
. Stellar coronae a Solve hydrostatic equilibrium for the density profile as a function of radius given a density ρ b at the base radius r b R. Start with hydrostatic equilibrium and assume an ideal gas equation of state with constant temperature T. Also assume that the mass of the corona is negligible compared to the mass of the star M, so mr M: dp dr T dρ dr dρ dr Gmr r ρr 5 GM ρr 55 r GM r ρr 56 T Equation 56 is a separable differential equation: dρ GM dr ρ T r 57 GMµmp ρr Aexp 58 r T Now use the given boundary condition to solve for integration constant A: GMµmp ρ b Aexp r b T A ρ b exp GMµm p r b T 59 6 So the final equation is ρr ρ b exp GMµm p GMµmp exp r b T r T 61 b What is the pressure in the corona as r? Comment on the implications of a low-density, low-pressure ISM surrounding the star. As r, we find the limit ρr ρ b exp GMµmp r b T. Since we ve assumed an ideal gas P ρt, this yields a pressure P ρ b T exp GMµmp r b T. This is finite! Since the stellar corona is surrounded by an ISM with much lower pressure, we expect the corona to be able to produce a stellar wind. 5. Using the MESA stellar evolution code a Run the default stellar model located in mesa/star/work/. At time step 1, what is the core temperature and surface temperature of the model? The core temperature is log T c 5.8 [K], or T c 3.1 1 5 K. The surface temperature is T eff 35 K. b Evolve a 1M model up the red giant branch. Make an HR diagram and plot log L as a function of time. Recall that an H-R diagram should have temperature increasing from right to left. Also note that in astronomy, log means log 1 and not ln. Finally, make sure to include units! 6