Question

Distance between sun and earth is 2 × 10⁸ km, temperature of sun 6000 K, radius of sun 7 × 10⁵ km, if emmisivity of earth is 0.6, then find out temperature of earth in thermal equilibrium.

• A. 400 K
• B. 300 K
• C. 500 K
• D. 600 K

Answer 1

Answer: (B)

For thermal equilibrium
energy received by earth = Energy emmited by earth
$\frac{T_s^4.4\pi R_s^2}{4\pi d^2}\times \pi R{e}^2=\sigma .\rho .T{e}^4.4\pi R{e}^2$
$\frac{T_s^4\times R_s^2}{4\times d^2\times e}=T_e^4$
$\Rightarrow \frac{{\left(6000\right)}^4\times {\left(7\times 10^8\right)}^2}{4\times {\left(2\times 10^{11}\right)}^2\times 0.6}=T_e^4$
$\Rightarrow \frac{36\times 36\times 7\times 7}{4\times 4\times 0.6}\times 10^6=T_e^4$
$\Rightarrow 66.15\times 108=T_e^4$
$Te\approx 300K$