JEE MAIN - Physics (2006 - No. 26)
Where r0 is the radius of the Earth and $$\sigma $$ is Stefan's constant.
$$4\pi r_0^2{R^2}\sigma {{{T^4}} \over {{r^2}}}$$
$$\pi r_0^2{R^2}\sigma {{{T^4}} \over {{r^2}}}$$
$$r_0^2{R^2}\sigma {{{T^4}} \over {4\pi {r^2}}}$$
$${R^2}\sigma {{{T^4}} \over {{r^2}}}$$
Explicação
Total power radiated by Sun $$ = \sigma {T^4} \times 4\pi {R^2}$$
The intensity of power at earth's surface $$ = {{\sigma {T^4} \times 4\pi {R^2}} \over {4\pi {r^2}}}$$
Total power received by Earth $$ = {{\sigma {T^4}{R^2}} \over {{r^2}}}\left( {\pi r_0^2} \right)$$
The intensity of power at earth's surface $$ = {{\sigma {T^4} \times 4\pi {R^2}} \over {4\pi {r^2}}}$$
Total power received by Earth $$ = {{\sigma {T^4}{R^2}} \over {{r^2}}}\left( {\pi r_0^2} \right)$$