# Lee's Back-of-the-Envelope Calculation

First, I'll define **irradiance** as the power of the radiation striking a surface per unit area. For this calculation, "surface" refers to either the Earth's surface or the "top" of the atmosphere.

Let's assume that the radiating temperature of a clear atmosphere (day and night) is 250 Kelvins. At this radiating temperature, I calculate that the irradiance at the Earth's surface (produced by radiation emitted by the atmosphere) is 220 watts per square meter. If I multiply this result by 24 hours, I get about 5280 watt-hours per square meter. Hold this result for a moment.

What about sunlight? First, let's assume that there are 12 hours of daylight (12 hours of sunlight). I'll also assume that the sky is clear. The irradiance at the top of the atmosphere (from solar energy) is 1360 watts per square meter. Accounting for the atmosphere back-scattering and absorbing incoming solar radiation, let's say that the irradiance at the Earth's surface is 1000 watts per square meter (a reasonable assumption).

Now the angle that incoming solar radiation strikes the carth's surface also governs irradiance. Let's assume that the average of the cosine of the angle of incidence is 1/2 during the 12 hours of daylight. As a result, the average irradiance produced by solar radiation at the Earth's surface is 500 watts per square meter. Now multiply this result by 12 hours and I get 6000 watt-hours per square meter.

Granted, this is an idealized analysis, but, at the very least, you can see that the radiation emitted to the Earth's surface by the atmosphere is in the same league as solar radiation. Now, if I add clouds to the picture, my calculation changes. Indeed, the amount of infrared radiation now increases (clouds act like space heaters). If I assume that the radiating temperature of the cloudy sky is 280 Kelvins, the Infrared contribution from the atmosphere increases by a factor of 1.6. More importantly, the amount of solar radiation reaching the Earth's surface decreases (a dramatic reduction if clouds are thick).

The bottom line is that the amount of radiation that the Earth's surface receives from the atmosphere in a 24-hour period is comparable to the amount it receives from the Sun, and, if clouds are present, the atmosphere's contribution likely exceeds that of the sun.