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?

To answer these questions, we need to use the Stefan-Boltzmann Law and the inverse square law for radiation.

(i) To find the mean distance of Phobos from the Sun, we can start by using the Stefan-Boltzmann Law, which relates the radiative flux (F) to the temperature (T) and the surface area (A) of an object:

F = εσT^4

Here, ε is the emissivity of the object, σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W m^-2 K^-4), and T is the temperature in Kelvin.

We know the radiative temperature (T) of Phobos is 222 K, and we can assume its emissivity (ε) is 1. The flux (F) from the Sun that reaches Phobos is given by:

F = L / (4πR^2)

Here, L is the solar luminosity (3.828 x 10^26 W) and R is the mean distance from the Sun to Phobos.

Setting the two expressions for flux equal to each other, we get:

εσT^4 = L / (4πR^2)

Simplifying, we can solve for R:

R = sqrt(L / (4πεσT^4))

Plugging in the values:

R = sqrt((3.828 x 10^26) / (4π(1)(5.67 x 10^-8)(222^4)))

Calculating this expression 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 can compare the flux received from Mars to the flux received from the Sun.

The flux from Mars to Phobos is given by the same equation:

F = L / (4πR^2)

Substituting the values:

F_mars = (3.828 x 10^26) / (4π(9.38 x 10^6)^2)

Comparing this flux to the flux received from the Sun, we can determine if it has a significant effect on the temperature of Phobos.