Question

Distance between sun and earth is $2\times {10}^8$km, temperature of sun $6000$K, radius of sun $7\times {10}^5$ km, if emissivity 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

The correct option is B $300$K

For thermal equilibrium,

Energy received by earth $=$ Energy emitted by earth

$\frac{T_s^4\cdot 4\pi R_s^2}{4\pi d^2}\times \pi R_e^2=\sigma \cdot \rho \cdot T_e^4\cdot 4\pi R_e^2$

$\frac{T_s^4\times R_s^2}{4\times d^2\times e}=T_e^4$ $(\text{Here, Emissivity of earth}e=\alpha \times \rho )$

$\Rightarrow \frac{(6000{)}^4\times (7\times {10}^8{)}^2}{4\times (2\times {10}^{11}{)}^2\times 0.6}=T_e^4$

$\Rightarrow \frac{36\times 36\times 7\times 7}{4\times 4\times 0.6}\times {10}^6=T_e^4$

$\Rightarrow 66.15\times 108=T_e^4$

$T_e=300K.$