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?
(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?
Answers
drwls
(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.
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.