radiant power?

Question

the amount of radiant power produced by the sun is approximately 3.9 X 10^26 W. assuming the sun to be a perfect blackbody spear with a radius of 6.96 X 10^8m, find its surface temperature (in kelvins)

Answer 1

Power emitted by sun =
Ps= 3.9*10^26 = sigma* [Ts^4 – Te^4](4pi Rs^2) -----(1)

Where Rs is radius of sun and Ts its surface temp. here Te being earth’s surface temp This power is isotropic (same in all directions). But earth only receives fraction of it at a distance ‘d’

So power absorbed by earth Pae = 3.9*10^26* [pi*Re^2] / 4pi*d^2 ---(2)

Earth will absorb with pi Re^2 where sun throws in surface area 4pi d^2.

On the other hand earth will re-radiate it as a sphere. So its emitted will be

Pee= sigma*Te^4 * 4pi Re^2
In thermal equilibrium,
Pae = Pee
3.9*10^26* [pi*Re^2] / 4pi*d^2 = sigma*Te^4 * 4pi Re^2

= sigma*Te^4 * 4pi

sigma*Te^4 = 3.9*10^26 / 16 pi d^2 -----(3)
put in (1)

3.9*10^26 / (4pi Rs^2) = sigma* [Ts^4] – [3.9*10^26 / 16 pi d^2]

sigma* [Ts^4] = (1/4pi) [{3.9*10^26) /Rs^2) + (3.9*10^26) /d^2)]

Rs = 7*10^8 m, d=1.5 *10^11 m

sigma* [Ts^4] = (1/4pi) [{79590*10^4) + (1.73333*10^4)]
sigma* [Ts^4] = 6336.92 *10^4

[Ts^4] = 6336.92 *10^4 / [5.67*10^-8]
[Ts^4] = 1117.62 *10^12
[Ts] = 5.782 *10^3 = 5782.32 K
sun’s temp as Black body

this is some more than standard 5780 K

Answer 2

you need to use the Wien's law for the blackbody to solve the problem, the one which goes like T^4