## Info

where Bv is the Planck function. The power absorbed by the Earth is then

^R0Rp d2

Equating these gives

We now look at the power radiated. We assume that the planet rotates fast enough that there is no great difference between day and night temperatures so we can treat the temperature of the planet as being the same everywhere. (This is a good approximation for the Earth but not for the Moon.) The power radiated per unit surface area is eaT^, where e is the emissivity. The emissivity can range from zero to one, and is one for a perfect blackbody. Multiplying this by the planet's surface area 4-n-Rp, we obtain the total power radiated:

If we equate the power absorbed with the power radiated, we solve for the equilibrium temperature of the planet, giving

This equation cannot be solved directly for TP even if we know av and ev. But since we know T0, we can evaluate the right-hand side of the equation. We then try different values of TP on the left-hand side, probably using a computer to evaluate the integral, until we find a value of TP which makes the left-hand side equal to the right-hand side. To be even more rigorous, we should also account for the fact that the temperature is not constant across the surface of a planet.

We can take advantage of the fact that most of the Sun's energy is in the visible and most of the energy given off by the Earth is in the infrared. We can assume that there is a constant albedo, aV, in the visible, and a constant emissivity, eIR, in the infrared. Since aV and eIR are constant over the region of each integral where Bv is significant, we can factor them out of the integral. The result is similar to equation (23.4):

This calculation doesn't account for the fact that the albedo and emissivity vary with wavelength. We must integrate the energy received over the spectral energy distribution of the Sun, and integrate the energy radiated over the spectral energy distribution of the planet. In each case, we incorporate a frequency dependent

On the Earth, the albedos are different for the oceans and for land. They are also different for cloud cover. When we take these into account, the equilibrium temperature is 246 K. However, this is still not the temperature we measure at the ground. We have not yet considered the u

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