AST111 PROBLEM SET 6 SOLUTIONS

Transcription

1 AST111 PROBLEM SET 6 SOLUTIONS Homework problems 1. Ideal gases in pressure balance Consider a parcel of molecular hydrogen at a temperature of 100 K in proximity to a parcel of ionized hydrogen at a temperature of 10 4 K. Assume the two parcels are in pressure balance. That means both parcels have the same pressure. How much denser is the molecular hydrogen? Answer: Molecular hydrogen has a molecular weight of about 2 atomic mass units per particle. Ionized hydrogen has a molecular weight of about 1/2 atomic mass units per particle where we count the free electron. Using P = ρkt µ P H2 = ρ H2 2m H k100 P H = ρ H 0.5m H k10 4 Set the two equal and solve for the density ratio 2. Jeans escape ρ H2 = ρ H = 400 The timescale for Jean s escape depends on a unitless factor e Y /Y and a timescale h/v 0 which is of order a minute for most planets in the Solar system. Here h is the atmospheric scale height and v 0 = 2kT/m depends on particle mass and temperature is a characteristic thermal velocity. The parameter Y v 2 e/v 2 0 where v e = 2GM/R is the escape velocity. 1

2 2 AST111 PROBLEM SET 6 SOLUTIONS (a) Consider a molecular species with mass m that gives Y = 20. What is the timescale for thermal escape of this molecule? Should it be present in the planet s atmosphere? (b) How much more massive a molecule would allow it to remain in the atmosphere for 4 billion years? (c) What is the ratio of the escape velocity of Jupiter compared to that of the Earth? (d) Taking into account only the ratio of semi-major axis, what is the ratio of the equilibrium temperature of Jupiter compared to that of the Earth? (Ignore differences in albedo, emissivity and atmospheric opacity). (e) A molecule on the Earth that has Y = 20 has approximately what value on Jupiter? A light molecule that escapes from Earth can last forever in Jupiter s atmosphere. a) e 20 /20 = and when we multiply by 60 seconds and divide by the number of seconds in a year ( ) we find 50 years. This is so short that the molecule should not be present in the atmosphere. b) The thermal velocity v 0 = 2kT m m 1/2 whereas the escape velocity is independent of particle mass m. The parameter Y v 2 e/v 2 0 m. By trial and error I found Y = 39 gives an escape time of 4 billion years. This is about twice as large as Y = 20 which means the molecule mass is only twice as large! The lifetime is extremely sensitive to the mass! c) The escape velocity v e M 1 2 R 1 2 Jupiter is 318 times more massive than the Earth and its equatorial radius is 11.2 times larger. 318/11.2 = 5.3. The ratio of the escape velocity of Jupiter compared to that of the Earth is 5.3. d) Equilibrium temperature depends on T eq d 1 2 and Jupiter has a semimajor axis of 5.2 AU so its equilibrium temperature is about 1/2 that of the Earth.

3 e) AST111 PROBLEM SET 6 SOLUTIONS 3 Y = v2 e v 2 0 = GM kt R M(T R) 1 Temperature is 1/2 times smaller, mass is 318 times larger, radius is 11.2 times larger. Y is larger by a factor 318/( ) = /0.5 = 57. On Jupiter Y = = The molecule escapes from the Earth but not from Jupiter. 3. Thermal tides on Mars On Mars, the fractional change in atmosphere temperature between noon and midnight is about 40%. This temperature difference (sometimes called a thermal tide) can be estimated from the total heat absorbed by the entire atmosphere. The solar heat input into a planetary atmosphere per day is Q abs πr 2 (1 A b ) L 4πd 2 t day where A b is the bond albedo, R is the planet radius, d is the distance from planet to star, t day is the time length of a day (a full day, like 24 hours). (a) Show that the mass of the entire atmosphere per unit area is approximately P 0 /g where g is the gravitational acceleration and P 0 is the pressure at the base of the atmosphere. (b) Show that the total heat Q dt required to raise the temperature of the entire atmosphere by a temperature difference T is Q dt = c P P o g 4πR2 T. (c) Show that the fractional increase in temperature between day and night is approximately (Hint: set Q abs = Q dt ). T T L (1 A b )gt day 16πd 2 P 0 c P T. The bond albedo for Mars is 0.16, the equilibrium temperature is 222K, the mean heliocentric distance is AU, the mass of the planet is kg, the mean radius is 3390 km, the sidereal rotation period 24.6hours. The specific heat for CO2 is cal/g/c. The surface gravity is 370 cm s 2. The surface pressure is bar. The gravitational acceleration g = GM/R 2 = 370 cm s 2.

4 4 AST111 PROBLEM SET 6 SOLUTIONS The specific heat c P = 0.25cal/g/C erg/cal = erg/c. t day = s. Using the above equation, we get T/T = Force on surface is due to weight of gravity on the atmosphere. The total mass of atmosphere in a column with area A is m. The total weight is mg. The pressure on the surface is force divided by area or P 0 = mg/a. The mass per unit area is m/a = P 0 /g. The mass per unit area is P 0 /g so the total mass of the atmosphere is this times the surface area or 4πR 2 P 0 /g. The heat required to raise this by T depends on the specific heat or Heat absorbed is Q dt = c P P o g 4πR2 T Q abs πr 2 (1 A b ) L 4πd 2 t day We set this to be equal to that required to raise the temperature Q abs = Q a. and solve for T πr 2 (1 A b ) L 4πd 2 t day = c P P o g 4πR2 T T = gl (1 A b )t day c P P 0 16πd 2 Divide this by T to get the expression T T L (1 A b )gt day 16πd 2 P 0 c P T This expression assumes that all the energy during the day is going into heating up the atmosphere and none is going out as radiation during the daytime. A better calculation would take into account the energy emitted during the daytime. Workshop problems 1. Isothermal gas Explain why γ = 1 gives a gas that is isothermal under compression. In other you can compress it without changing its temperature.

5 AST111 PROBLEM SET 6 SOLUTIONS 5 Answer: Inverting the pressure/temperature relation dt dp = γ 1 T γ P = 0 That means that there is no change in temperature for a change in pressure. Likewise P V γ = P V = constant and P V = NkT together imply that T is a constant. Interstellar medium is often considered to be isothermal because its temperature is set by absorbing and emitting radiation. 2. The Dry Lapse Rate Consider a convecting atmosphere. (a) Show that the factor that the temperature changes per atmospheric scale height is equal to (1 γ)/γ. In other words what is T/T across a scale height h? (b) Show that the factor that the temperature changes per atmospheric scale height is usually less than 1. (c) Compare the temperature as a function of opacity for an optically thick atmosphere undergoing radiative energy transport to the temperature change across an atmospheric scale height for a conductive and adiabatic atmosphere computed in the previous problem. A dense atmosphere can have extremely high opacity. Explain why when there is a strong green house effect (and the planet s surface temperature is much higher than the effective temperature) that convective heat transport is unlikely. (d) Compute T/T across a scale height h for monatomic, diatomic and polyatomic gases. a,b ) As the atmosphere is convective the temperature drops with radius with the dry lapse rate. h = kt/(gµ) and dt dz = γ 1 γ µg k

6 6 AST111 PROBLEM SET 6 SOLUTIONS divide dt by h giving dz dt dz h = γ 1 γ T Across a scale height T T = γ 1 γ The adiabatic index is positive and usually greater than 1 and this means that the factor that the temperature changes in a scale height is less than 1. We used here the scale height computed with a fixed temperature. This is a reasonable approximation since T < 1 T c) The dry lapse rate (giving the temperature gradient during convection) is just too shallow to give a large temperature gradient. When a large temperature gradient is present likely energy transport is radiative. d) For monotomic T/T = 0.4 across a scale height, diatomic, 0.28, and for polyatomic, 0.25, using γ = 5/3, 7/5, and 4/3 respectively. For atmospheres with lots of molecules, the temperature gradient is very shallow. 3. Optically thick radiative transfer The absorption coefficient of gas in a circumstellar disk might be of order α 1g 1 cm 2. The optical depth through the disk τ = sαρ where s is the path length and ρ the density. Note that the product sρ is a mass per unit area. We can integrate ρdz through the disk to estimate a disk mass surface density. Here z is in the direction normal to the disk midplane. Integrating from the midplane upwards ρdz = Σ/2 where the factor of two arises because we only integrate 0 for positive z and Σ is the the mass per unit area in the disk. Taking into account only the integrated column through half of the disk (from the midplane upwards) the optical depth τ = 0 ραdz = ασ/2. (a) If the disk has a mass surface density of Σ = 20g cm 2 what is the optical depth through half of it? (using the above given absorption coefficient). (b) If turbulence in the disk and associated accretion heats the gas in the midplane it is hotter than the disk surface. If the disk is optically thick then energy through the disk may be transported by radiation, through absorption and reemission of photons. Both the top and bottom of the disk can radiate. What is the ratio of midplane to effective disk temperature?

7 AST111 PROBLEM SET 6 SOLUTIONS 7 The optical depth is τ = ασ/2 = 1 20/2 = 10. Which equation relating temperature to opacity should we use? We want effective temperature of radiation near the top related to the midplane temperature. Probably we should use something like the ground temperature equation, and as a quick approximation ignore that radiation from the ground only goes in one direction. Using that equation 3/4τ + 1 = 8.5. The temperature ratio goes as the 1/4th power of this or 1.7. The midplane is about twice as hot as the disk surface. There are many ways that reality is more complex that this. For example the absorption coefficient would be dependent on wavelength and temperature (and metalicity and ionization state etc etc...)