Question

. Phobos is the larger and closer of Mars’s two moons. It has no atmosphere, a mean radius of 11 km, an albedo of 0.07, and an emissivity of 1.0.

(i) Assuming that the radiative temperature of Phobos is 222 K and that it is spherical, what is its mean distance from the sun?

(ii) Mars is much cooler than the Sun (210 K) but much closer to Phobos (9.38*10^6 m). Its radius is 3.4*10^6 m. Does the radiation from Mars have a significant effect on the temperature of Phobos?

Answer 1

(i) Perform an energy balance using the irradiance of the sun at the Mars location (Hsun) and the thermal emission of Mars:
0.93* pi* R^2 * Hsun =
1.0 * sigma* T^4* 4*pi R^2
+ 0.07* pi* R^2 * Hsun

The R^2 terms cancel out. You know that T = 222 K. Solve for R, the mean distance from the sun.

Note that the intercepted solar flux area is pi R^2 but the areas that is doing the infrared emitting is 4 pi R^2.

"Sigma" is the Stefan-Boltzmann constant. Look it up if you don't know it.

ii) Compare the absorbed energy from the sun,
alphap* pi* Rp^2 * Hsun
with the absorbed infrared energy from Mars,
pi*Rm^2*sigma*Tm^4*(1/pi)*(pi*Rp^2/X^2)
Rm = Mars radius
Rp = Phobos radius
X = Mars-Phobos separation
Tm = Mars temperature
alphap = Phobos albedo (assume 0.5) They should have given you a value for alphap

The (1/pi) term in the last equation is necessary to convert total Mars emission per projected area to radiation per area per steradian.