Insolation as Percent of Total Solar Output
Date: Summer 2012
What percentage of the sun's total output do we actually receive on Earth?
This looks like the best answer I can find on the Internet:
“So, the Earth intercepts about 5 x 10^7 / 3 x 10^16 = .0000002 % of the Sun's output of about 4 x 10^20 megaWatts, or about 7 x 10^11 megaWatts.”
I will walk you through the calculations, you can do the actual math yourself.
First imagine that there is a sphere with a radius (rS) that is equal to the distance from the Sun to the Earth. This sphere would be like a ball that around the Sun and the surface of this ball cuts through the center of the Earth. The total surface of this sphere (AS) can be calculated using the formula AS = 4(pi)(rS)^2.
Next we need to know the area of the cross-section of the Earth (AE). To do this we need to know the radius of the Earth (rE), and use the formula: AE = (pi)(rE)^2.
The percent of the Sun's energy that reaches the Earth is the ratio of the area of the cross-section of the Earth (AE) to the total area of the imagined sphere (AS) - since this is the amount of surface that the Earth takes up over the entire surface at the particular distance from the Sun. So percent = AE/AS * 100.
Now, this is the percent of energy that *reaches* the Earth. The percent that actually touches the *surface* of the Earth depends on clouds and intervening gases. This would also depend on whether the Sun is at its zenith or low on the horizon, etc. So you would need to find an average and then take that from the calculations in the previous paragraph.
Greg (Roberto Gregorius)
The answer will vary because of what you mean exactly by “Earth surface?” Calculated is the percent fromTotal Solar Radiance(TSR) and total output by Sun. The numbers are from NASA’s Solar Radiation and Climate Experiment (SORCE) satellite data.
6.94*1014 W TSR Earth ÷ 3.838*1026 W output by Sun = 1.8*10-10 percent.
The fraction that is used based on elevation, over water, etc may be substantially different.
Keep your Sunny side up!
Peter E. Hughes, Ph.D. Milford, NH
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Update: December 2011