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where Trad is defined as before. Thus, the skin temperature is colder than the blackbody radiating temperature by a factor of 2-1. The skin temperature is the natural temperature the outer regions of an atmosphere would have in the absence of in situ heating by solar absorption or other means. Note that the skin layer does not need any interior heat transfer mechanism to keep it isothermal, since the argument we have applied to determine Tskin applies equally well to any sublayer of the skin layer.

A layer that has low emissivity, and hence low absorptivity, in some given wavelength band is referred to as being optically thin in this band. A layer could well be optically thick in the infrared, but optically thin in the visible, which is in fact the case for strong greenhouse gases.

Now let's suppose that the entire atmosphere is optically thin, right down to the ground, and compute the pure radiative equilibrium in this system in the absence of heat transfer by convection. We'll also assume that the atmosphere is completely transparent to the incident Solar radiation. Let S be the incident Solar flux per unit surface area, appropriate to the problem under consideration (e.g. 4Lq for the global mean or Lq for temperature at the subsolar point on a planet like modern Mars). Since the atmosphere has low emissivity, the heating of the ground by absorption of downwelling infrared emission coming from the atmosphere can be neglected to lowest order. Since the ground is heated only by absorbed Solar radiation, its temperature is determined by -T^ = (1 — a)S, just as if there were no atmosphere at all. In other words, prad = ps because the atmosphere is optically thin, so that the atmosphere does not affect the surface temperature no matter what its temperature structure turns out to be. Next we determine the atmospheric temperature. The whole atmosphere has small but nonzero emissivity so that the skin layer in this case extends right to the ground. The atmosphere is then isothermal, and its temperature Ta is just the skin temperature 2-1/4Ts.

The surface is thus considerably warmer than the air with which it is in immediate contact. There would be nothing unstable about this situation if radiative transfer were truly the only heat transfer mechanism coupling the atmosphere to the surface. In reality, the air molecules in contact with the surface will acquire the temperature of the surface by heat conduction, and turbulent air currents will carry the warmed air away from the surface, forming a heated, buoyant layer of air. This will trigger convection, mixing a deep layer of the atmosphere within which the temperature profile will follow the adiabat. The layer will grow in depth until the temperature at the top of the mixed layer matches the skin temperature, eliminating the instability. This situation is depicted in Figure 3.14. The isothermal, stably stratified region above the mixed region is the stratosphere in this atmosphere, and the lower, adiabatic region is the troposphere; the boundary between the two is the tropopause. We have just formulated a theory of tropopause height for optically thin atmospheres. To make it quantitative, we need only require that the adiabat starting at the surface temperature match to the skin temperature at the tropopause. Let ps be the surface

Figure 3.14: The unstable pure radiative equilibrium for an optically thin atmosphere (solid line) and the result of adjustment to the adiabat by convection (dashed line). The adjustment of the temperature profile leaves the surface temperature unchanged in this case, because the atmosphere is optically thin and has essentially no effect on the OLR.

Figure 3.14: The unstable pure radiative equilibrium for an optically thin atmosphere (solid line) and the result of adjustment to the adiabat by convection (dashed line). The adjustment of the temperature profile leaves the surface temperature unchanged in this case, because the atmosphere is optically thin and has essentially no effect on the OLR.

pressure and ptrop be the tropopause pressure. For the dry adiabat, the requirement is then

Ts(ptrop/Ps)R/Cp = Tskin. Since Ts = 21/4Tsfci„, the result is ptrop = 2-(3.25)

Note that the tropopause pressure is affected by R/cp, but is independent of the insolation S.

The stratosphere in the preceding calculation differs from the observed stratosphere of Earth in that it is isothermal rather than warming with altitude. The factor we have left out is that real stratospheres often contain constituents that absorb solar radiation. To rectify this shortcoming, let's consider the effect of solar absorption on the temperature of the skin layer. Let eir be the infrared emissivity, which is still assumed small, and asw be the shortwave (mostly visible) absorptivity, which will also be assumed small. Note that Kirchhoff's Law does not require eir = asw, as the emissivity and absorptivity are at different wavelengths. The solar absorption of incident radiation is aswS. We'll assume that the portion of the solar spectrum which is absorbed by the atmosphere is absorbed so strongly that it is completely absorbed before reaching the ground. This is in fact the typical situation for solar near-infrared and ultraviolet. In this case, one need not take into account absorption of the upwelling solar radiation reflected from the surface.

Exercise 3.6.1 Show that if the atmosphere absorbs uniformly throughout the solar spectrum, then the total absorption in the skin layer is (1 + (1 — asw)a.g)aswS, where ag is the solar albedo of the ground. Show that the planetary albedo - i.e. the albedo observed at the top of the atmosphere

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