A) \[4\pi r_{0}^{2}\,{{R}^{2}}\,\sigma {{T}^{4}}/\,{{r}^{2}}\]
B) \[\pi r_{0}^{2}\,{{R}^{2}}\,\sigma {{T}^{4}}/\,{{r}^{2}}\]
C) \[r_{0}^{2}\,{{R}^{2}}\,\sigma {{T}^{4}}/4\pi \,{{r}^{2}}\]
D) \[{{R}^{2}}\,\sigma {{T}^{4}}/\,{{r}^{2}}\]
Correct Answer: B
Solution :
[b] From Stefans law, the rate at which energy is radiated by sun at its surface is |
(Sun is a perfectly black body as it emits radiations of all wavelengths and so for it e = 1. |
The intensity of this power at earths surface [under the assumption \[r>>~{{r}_{0}}\]] is |
\[I=\frac{p}{4\pi {{r}^{2}}}=\,\frac{\sigma \times \,4\pi \,{{R}^{2}}{{T}^{4}}}{4\pi {{r}^{2}}}=\frac{\sigma {{R}^{2}}{{T}^{4}}}{{{r}^{2}}}\] |
The area of earth which receives this energy is only one-half of total surface area of earth, whose projection would be \[\pi r_{0}^{2}\]. |
\ Total radiant power as received by earth\[=\,\pi r_{0}^{2}\times \,1\] |
\[=\frac{\pi r_{0}^{2}\times \,\sigma {{R}^{2}}{{T}^{4}}}{{{r}^{2}}}\,=\,\frac{\pi r_{0}^{2}{{R}^{2}}\,\sigma {{T}^{4}}}{{{r}^{2}}}\] |
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