91Ó°ÊÓ

The below figure shows two blankets of the earth's atmosphere for receiving the energy from the sun.

In the above figure the upper blanket must send one unit of the infrared energy for every unit of energy absorbed from the sun.

Let the$30\%$of the solar light is reflected. So, the temperature of the upper blanket is

$T_{\text{upper}}={\left(\frac{(0.7)\text{solar constant}}{4\mathrm{σ}}\right)}^{1/4}$

where, $\mathrm{σ}$is the Stefan Boltzmann constant.

$\text{Putting}1370\mathrm{W}/{\mathrm{m}}^2\text{for the solar constant and}5.67×{10}^{-8}\mathrm{W}/{\mathrm{m}}^2{\mathrm{K}}^4\text{for}\mathrm{σ}\text{.}$

$T_{\text{upper}}={\left(\frac{(0.70)\left(1370\mathrm{W}/{\mathrm{m}}^2\right)}{4\left(5.67×{10}^{-8}\mathrm{W}/{\mathrm{m}}^2{\mathrm{K}}^4\right)}\right)}^{\frac{1}{4}}$

$=255\mathrm{K}$

So, the temperature of the upper blanket$255K$.

The temperature of the lower blanket is

$T_{\text{lower}}={\left(\frac{2(0.7)\text{solar constant}}{4\mathrm{σ}}\right)}^{1/4}$

$=(2{)}^{1/4}{\left(\frac{(0.7)\text{solar constant}}{4\mathrm{σ}}\right)}^{1/4}$

$\text{Putting}{T}_{\text{upper}}\text{for}{\left(\frac{(0.7)\text{solar constant}}{4\mathrm{σ}}\right)}^{1/4}$we get

$T_{\text{lower}}=2^{1/4}{T}_{\text{upper}}$

So, the temperature of the lower atmosphere blanket is$2^{1/4}$times of the temperature of the upper atmosphere.

$T_{\text{lower}}=2^{1/4}(255\mathrm{K})$

$=303\mathrm{K}$

$\text{Hence, the temperature of the lower atmosphere blanket is}303\mathrm{K}$.

The temperature of the earth is

$T_{\text{ground}}={\left(\frac{3(0.7)\text{solar constant}}{4\mathrm{σ}}\right)}^{1/4}$

$=(3{)}^{1/4}{\left(\frac{(0.7)\text{solar constant}}{4\mathrm{σ}}\right)}^{1/4}$

Putting $T_{\text{upper}}\text{for}{\left(\frac{(0.7)\text{solar constant}}{4\mathrm{σ}}\right)}^{1/4}$

$T_{\text{ground}}=3^{1/4}{T}_{\text{upper}}$

$T_{\text{ground}}=3^{\frac{1}{4}}(255\mathrm{K})$

$=335.6\mathrm{K}$

$=(335.6-273.15)°\mathrm{C}$

$=62°\mathrm{C}$

$\text{Hence, the equilibrium surface temperature of the earth is}62°\mathrm{C}\text{.}$