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Earth receives one part in 2.2 billion of the radiation emitted by the sun.
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Assuming the radiation from the sun is the same in all directions, at the distance of the Earth, it is spread over an area of
4πr2 ≈ 4π(9.3*107 miles)2 ≈ 3.5π*1016 square miles
The cross section of the Earth (as seen from the sun) is about
πr2 ≈ π(4000 miles)2 = 1.6π107 square miles
The proportion of the sun's energy seen by the Earth is approximately the ratio of these areas.
1.6π*107/(3.5π*1016) ≈ 4.6*10-10, about .00000005%
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There are a few tens of miles of atmosphere that also capture solar radiation, and the magnetosphere captures or deflects radiation, depending. Some percentage is simply reflected back into space, so is not absorbed by Earth. About all you can say about this calculation is that it may get you within an order of magnitude (or two) of the right number. (That's why I kept only one significant digit.) The "right number" will vary with time of year and cloud cover. Figuring it out more precisely is one of the things climate modelers do for a living.
4πr2 ≈ 4π(9.3*107 miles)2 ≈ 3.5π*1016 square miles
The cross section of the Earth (as seen from the sun) is about
πr2 ≈ π(4000 miles)2 = 1.6π107 square miles
The proportion of the sun's energy seen by the Earth is approximately the ratio of these areas.
1.6π*107/(3.5π*1016) ≈ 4.6*10-10, about .00000005%
_____
There are a few tens of miles of atmosphere that also capture solar radiation, and the magnetosphere captures or deflects radiation, depending. Some percentage is simply reflected back into space, so is not absorbed by Earth. About all you can say about this calculation is that it may get you within an order of magnitude (or two) of the right number. (That's why I kept only one significant digit.) The "right number" will vary with time of year and cloud cover. Figuring it out more precisely is one of the things climate modelers do for a living.
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