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
$\frac{T^{ 4}_{s}.4\pi R^{ 2}_{s}}{4\pi d^{2}}\times\pi Re^{2}=\sigma.\rho.T e^{4}.4\pi Re^{2}$
$\frac{T^{ 4}_{s}\times R^{ 2}_{s}}{4\times d^{2}\times e}=T^{ 4}_{e}$
$\Rightarrow \frac{\left(6000\right)^{4}\times\left(7\times10^{8}\right)^{2}}{4\times\left(2\times10^{11}\right)^{2}\times0.6}=T^{ 4}_{e}$
$\Rightarrow \frac{36\times36\times7\times7}{4\times4\times0.6}\times10^{6}=T^{ 4}_{e}$
$\Rightarrow 66.15\times108=T^{ 4}_{e}$
$Te \approx300 K$