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(A) Let $S$ be the total energy emitted by the Sun per second and $r$ be the average distance between the Earth and the Sun. The intensity of solar ra

Vedclass · Accepted Answer

$10^{-7} \%$

Vedclass · Answer

(A) Let $S$ be the total energy emitted by the Sun per second and $r$ be the average distance between the Earth and the Sun. The intensity of solar radiation at the Earth's distance is $I = \frac{S}{4\pi r^2}$.The Earth intercepts this radiation over its cross-sectional area,which is $\pi R^2$,where $R$ is the radius of the Earth.Energy intercepted by Earth per second $= I 	imes \pi R^2 = \frac{S}{4\pi r^2} 	imes \pi R^2 = \frac{S R^2}{4 r^2}$.The percentage of the Sun's total energy reaching the Earth is given by:$	ext{Percentage} = \left( \frac{	ext{Energy intercepted}}{	ext{Total energy emitted}} ight) 	imes 100 = \left( \frac{S R^2 / 4 r^2}{S} ight) 	imes 100 = \frac{R^2}{4 r^2} 	imes 100$.Substituting the values $R \approx 6.4 	imes 10^6 \ m$ and $r \approx 1.5 	imes 10^{11} \ m$:$	ext{Percentage} = \frac{(6.4 	imes 10^6)^2}{4 	imes (1.5 	imes 10^{11})^2} 	imes 100 \approx \frac{40.96 	imes 10^{12}}{9 	imes 10^{22}} 	imes 100 \approx 4.5 	imes 10^{-10} 	imes 100 \approx 4.5 	imes 10^{-8} \%$.Given the standard approximation in physics problems of this type,the order of magnitude is $10^{-7} \%$.

The percentage of the Sun's total energy that reaches the Earth's surface is:

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