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(C) The total power radiated by the sun is given by the Stefan-Boltzmann law: $P = \sigma T^4 (4 \pi R^2)$.This energy spreads out over a spherical su

Vedclass · Accepted Answer

$\frac{\pi r_0^2 R^2 \sigma T^4}{r^2}$

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(C) The total power radiated by the sun is given by the Stefan-Boltzmann law: $P = \sigma T^4 (4 \pi R^2)$.This energy spreads out over a spherical surface of radius $r$ as it travels from the sun to the earth. The intensity $I$ (power per unit area) at a distance $r$ from the sun is given by: $I = \frac{P}{4 \pi r^2} = \frac{\sigma T^4 (4 \pi R^2)}{4 \pi r^2} = \frac{\sigma R^2 T^4}{r^2}$.The earth acts as a disc of radius $r_0$ intercepting this radiation. The area of this disc is $A = \pi r_0^2$.Therefore,the total radiant power incident on the earth is $Q = I 	imes A = \left( \frac{\sigma R^2 T^4}{r^2} ight) 	imes (\pi r_0^2) = \frac{\pi r_0^2 R^2 \sigma T^4}{r^2}$.

Assuming the sun to be a spherical body of radius $R$ at a temperature of $T \ K$,evaluate the total radiant power incident on the earth at a distance $r$ from the sun. (Where $r_0$ is the radius of the earth and $\sigma$ is Stefan's constant.)

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