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?
PHYSICS- SCIENCE - drwls, Saturday, January 23, 2010 at 5:31pm
(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.
sorry but i don't understand wat is R^2 wat does that represent? and wat is Hsun?
(i) To find the mean distance of Phobos from the Sun, we need to perform an energy balance using the irradiance of the Sun at the location of Mars (Hsun) and the thermal emission of Phobos.
The energy balance equation is given by:
0.93 * pi * R^2 * Hsun = 1.0 * sigma * T^4 * 4 * pi * R^2 + 0.07 * pi * R^2 * Hsun
In this equation, R represents the mean distance of Phobos from the Sun, T represents the radiative temperature of Phobos (222 K), Hsun represents the irradiance of the Sun at the location of Mars, and sigma represents the Stefan-Boltzmann constant.
Solving for R will give us the mean distance of Phobos from the Sun.
(ii) To determine if the radiation from Mars has a significant effect on the temperature of Phobos, we need to compare the absorbed energy from the Sun to the absorbed infrared energy from Mars.
The absorbed energy from the Sun is given by:
alphap * pi * Rp^2 * Hsun
The absorbed infrared energy from Mars is given by:
pi * Rm^2 * sigma * Tm^4 * (1/pi) * (pi * Rp^2 / X^2)
In these equations, alphap is the albedo of Phobos (0.5), Rp is the radius of Phobos, Rm is the radius of Mars, Tm is the temperature of Mars (210 K), and X is the separation between Mars and Phobos.
Comparing these two energies will help us determine the effect of radiation from Mars on the temperature of Phobos.