. 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) 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.
To answer these questions, we need to use the Stefan-Boltzmann law and the inverse-square law of radiation.
(i) To determine the mean distance of Phobos from the Sun, we can use the Stefan-Boltzmann law. The law states that the power radiated by an object is proportional to its temperature raised to the fourth power and its surface area. It can be expressed as:
P = σ * (T^4) * A
Where:
P is the power radiated
σ is the Stefan-Boltzmann constant (approximately 5.67 * 10^-8 W/(m^2*K^4))
T is the temperature in Kelvin (222 K in this case)
A is the surface area of Phobos (4 * π * R^2, where R is the radius of Phobos)
Since we are assuming Phobos is spherical, we can substitute the mean radius (11 km) into the equation to calculate the surface area. Remember to convert the radius to meters before the calculation.
A = 4 * π * (11000 m)^2
Once we have the surface area, we can rearrange the equation to solve for the power radiated:
P = (σ * T^4 * A)
Now, the power from the Sun that reaches Phobos is given by the inverse-square law. It states that the power received by an object is inversely proportional to the square of the distance from the source of radiation. The equation is:
P_sun = (L / (4π * d^2))
Where:
P_sun is the power received from the Sun
L is the luminosity of the Sun (approximately 3.8 * 10^26 W)
d is the mean distance of Phobos from the Sun (which we want to find)
Equating the power radiated by Phobos (P) to the power received from the Sun (P_sun) allows us to solve for the distance (d):
P = P_sun
(σ * T^4 * A) = (L / (4π * d^2))
You can now solve this equation for d to determine 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 can compare the power received from Mars to the power radiated by Phobos.
The power received from Mars is given by the inverse-square law, similar to the equation for the Sun:
P_mars = (L_mars / (4π * d_mars^2))
Where:
P_mars is the power received from Mars
L_mars is the luminosity of Mars (which is negligible compared to the Sun)
d_mars is the distance from Phobos to Mars (9.38 * 10^6 m)
We can compare this power to the power radiated by Phobos (P) using:
ΔP = P - P_mars
If ΔP is negligible compared to P, then the radiation from Mars has a negligible effect on the temperature of Phobos.
Plug in the appropriate values to calculate P_mars, and then calculate ΔP. Compare ΔP to P to determine if the radiation from Mars has a significant effect.
Remember to convert the temperatures to Kelvin and distances to meters in all calculations.