A certain stellar body has radius 50 ,R_{s} a | vedclass

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Question

(D) The intensity of radiation from the sun received on earth (solar constant) is given by:$S_{1} = \frac{P}{4 \pi R_{0}^{2}} = \frac{4 \pi R_{s}^{2}

Vedclass · Accepted Answer

$10^{-16}$

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(D) The intensity of radiation from the sun received on earth (solar constant) is given by:$S_{1} = \frac{P}{4 \pi R_{0}^{2}} = \frac{4 \pi R_{s}^{2} \cdot \sigma \cdot T_{s}^{4}}{4 \pi R_{0}^{2}} = \sigma \left( \frac{R_{s}}{R_{0}} ight)^{2} T_{s}^{4}$ ....................$(i)$where $R_{s}$ is the radius of the sun,$R_{0}$ is the earth-sun distance $(1 \,AU)$,$\sigma$ is the Stefan-Boltzmann constant,and $T_{s}$ is the temperature of the sun.Now,the intensity of radiation received from the stellar body on earth's surface is:$S_{2} = \frac{\sigma (50 R_{s})^{2}}{(2 	imes 10^{10} R_{0})^{2}} \cdot (2 T_{s})^{4}$$S_{2} = \frac{2500 \cdot R_{s}^{2}}{4 	imes 10^{20} \cdot R_{0}^{2}} \cdot 16 \cdot T_{s}^{4}$$S_{2} = \frac{2500 	imes 16}{4 	imes 10^{20}} \cdot \sigma \left( \frac{R_{s}}{R_{0}} ight)^{2} T_{s}^{4}$$S_{2} = \frac{40000}{4 	imes 10^{20}} \cdot S_{1} = 10^{4} 	imes 10^{-20} \cdot S_{1} = 10^{-16} S_{1}$Therefore,the ratio $\frac{S_{2}}{S_{1}} = 10^{-16}$.

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