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That is, $\gamma$ represents how likely a surface is to absorb some radiation that tries to pass through it; reflected energy never makes this attempt, and so does not matter here. This behavior is intuitive if we think, to begin, about the surface of the planet: while it has a non-negligible albedo (it reflects some radiation), it is effectively opaque. The planet's surface does reflect some energy outright, but virtually all of the energy it doesn't reflect is absorbed. Very little E/M radiation simply passes through the surface of the planet. We can thus set $\gamma _{s}=1$. We are interested in solving for $\gamma _{a}$—we're interested in figuring out just how opaque the atmosphere is. From all of this, we can deduce another equation: one for the energy emitted by the atmosphere ($F_{a}$).

$F_{a}=\gamma _{a}\sigma T_{a}^{4}$

(4e)

We have to include $\gamma$ in this equation, as (recall) the atmosphere is transparent (or nearly so) only with respect to incoming solar radiation. Radiation emitted both by the surface and by the atmosphere itself has a chance of being reabsorbed.

At last, then, we're in a position to put all of this together. We have an equation for the energy emitted by the atmosphere and an equation for the energy reaching the ground from the sun. For the purposes of this model, this exhausts all the sources of radiative forcing on the surface of the Earth. If we hold on to the supposition that things are at (or near) equilibrium, we know that the energy radiated by the surface (which we can calculate independently from the Stefan-Boltzmann law) must be in balance with these two sources. The full balance for the surface at equilibrium, then, is:

${\tfrac {S_{o}(1-\sigma )}{4}}+\gamma _{a}\sigma T_{a}^{4}=\gamma _{s}\sigma T_{s}^{4}$

(4f)

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