Howmuch energy fromcosmic radiation (not solar radiation) is absorbed by Earth? An answer in watts wouldbenice

Question

Any sources you have,please cite.

Thanks,

-David

Answer 1

"Of the total sunlight falling on Earth (100%), 31% is reflected back to Space and 69% is absorbed. The absorbed radiation is then emitted as infrared radiation. Scientists reach these conclusions by studying data gathered by satellites and computer simulations of the working of Earth's environment."

"The solar constant includes all types of solar radiation, not just the visible light. It is measured by satellite to be roughly 1366 watts per square meter."

So 69% of 3,166 W/m^2, or about 2,185 W/m^2 is absorbed by the Earth. Note that you must use Earth's cross-sectional area, and not the surface area to arrive at the total wattage absorbed. For the purist an additional multiplier of 0.999914776 may be applied. (the ratio of the Earth's apparent area at 93,000,000 miles to its actual area), but measurement errors will swallow up this error.

Edit:
Sorry, I misread your question. There is a background radiation that appears uniform in all directions equivalent to a 2.725°K black body. The watt density of this radiation is
j = σT^4 watts/m^2 when σ is the Stephan-Boltzman constant ant T is the temperature in °K
σ is 5.670 400(40)×10−8 W·m-2·K-4.

The total amount of radiation of this type impinging on Earth can be represented by 4πR^2σT^4

taking the Earth's radius as 6378.137 km, and remembering the 31% reflectivity from above, the amount of energy absorbed by the Earth is

P = (0.69)(4)π(6378.137km)^2(5.670 400(40)×10−8 W·m-2·K-4)(2.725°K)^4(1,000,000 m^2/km^2

P = 1.102,875,260 gigawatts