Question

The temperature of the Earth's surface is maintained by radiation from the Sun. By making the approximation that the Sun is a black body, but now assuming that the Earth is a grey body with albedo A (this means that it reflects a fraction A of the incident energy), show that the ratio of the Earth's temperature to that of the Sun is given by T_Earth = T_Sun (1 - A)^1/4 Squareroot R_Sun/2d, where R_Sun is the radius of the Sun and the Earth-Sun separation is D.

Answer 1

Answer:

T_t = Ts  (1-A$)^{1/4}$ √ (Rs/D)

Explanation:

The black body radiation power is given by Stefan's law

       P = σ A e T⁴

This power is distributed over a spherical surface, so the intensity of the radiation is

     I = P / A

Let's apply these formulas to our case. Let's start by calculating the power emitted by the Sun, which has an emissivity of one (e = 1) black body

    P_s = σ A_s 1 T_s⁴

This power is distributed in a given area, the intensity that reaches the earth is

     I = P_s / A

    A = 4π R²

The distance from the Sun Earth is R = D

     I₁ = Ps / 4π D²

     I₁ = σ (π R_s²) T_s⁴ / 4π D²

     I₁ = σ T_s⁴ R_s² / 4D²

Now let's calculate the power emitted by the earth

     P_t = σ A_t (e) T_t⁴

     I₂ = P_t / A_t

     I₂ = P_t / 4π R_t²2

     I₂ = σ (π R_t²) T_t⁴ / 4π R_t²2

     I₂ = σ T_t⁴ / 4

The thermal equilibrium occurs when the emission of the earth is equal to the absorbed energy, the radiation affects less the reflected one is equal to the emitted radiation

     I₁ - A I₁ = I₂

     I₁ (1 - A) = I₂

Let's replace

    σ T_s⁴  R_s²/4D²   (1-A)  = σ T_t⁴ / 4

    T_s⁴ R_s² /D²   (1-A) = T_t⁴

    T_t⁴ = T_s⁴  (1-A)  (Rs / D) 2

    T_t = Ts  (1-A$)^{1/4}$ √ (Rs/D)