Let's start with consider a zero layer model for the earth (e.g. no atmosphere).
Solar Constant (So)is 1370 Watts per sq meter. Stefan-Boltzmann constant is 5.67x10-8 Watts per sq meter per degree:
T4 = (1/4*1370/a)(1-A)
T = 278*(1-A)1/4
For A = 0.3 one gets T = 254K
One Layer Model (e.g. our atmosphere)
We assume the following:
Let TA be the temperature of the atmosphere and TS be the temperature of the surface of the planet. Since all the longwave radiation emitted by the planet is absorbed by the atmosphere, none of it leaves the top of the atmosphere. Thus, the energy balance at the top of the atmosphere is given by the same equations as the zero layer model and
The energy emitted upwards but the ground scale as (TS)4 and the energy emitted upwards and downwards by the atmosphere will scale as (TA )4
In particular, not all the thermal IR emitted by the planet is absorbed by the atmosphere. The oceans are important here and have been ignored.
Now finally let's consider a two layer model :
The top layer has temperature T1 and the bottom layer has temperature of T2.
Again, energy conservation requires that T1 = TTe
For the top layer:
Energy out goes as T14(emitted up) + T14(emitted down)
The Energy input comes from energy emitted upwards from the second layer so that
Energy in goes as T24
Energy out goes as T24(emitted up) + T24(emitted down)
Energy in = (T1)4 (emitted down by layer 1) + (TS)4 (emitted up by surface)
Using the above relationships get,
2(T2)4 = 4(TE)4 = (TE)4 + (TS)4,
which finally gives
TS = 31/4 TE.
So, a two layer model of the earth would give a temperature of about 336 K, which is much too high.For an N-layered atmosphere (relevant to Venus), one can derive the general expression:
TS = (N+1)1/4 TE
The key here, is that atmospheric gases are weak absorbers in the visible but strong absorbers in the infrared. So incoming radiation easily reaches the ground, but outgoing radiation has a more difficult time escaping back to space, and some fraction of it is re-radiated back to the ground.
Atmosphere is thin layer with absorptivity = 0.1 for visible radiation but 0.8 for infrared radiation.
Let x = radiation emitted by the earth's surface and y = radiation emitted (upwards and downwards) by the atmosphere. E = the net solar radiation absorbed by the earth-atmosphere system.
At the earth's surface, radiative equilibrium demands (since 10% of E is absorbed by the atmosphere)
The absorptivity of the atmosphere in the IR is 0.8 therefore 0.2x escapes back to space. This means
or substituting for x
And clearly x = 1.58E
Where E = So/4(1-A) or 241 watts per sq meter for A= 0.3.
So for the surface temperature one gets
aT4 = 1.58 x 241 277 K
for the atmosphere 0.8aT4 = 0.68 x 241 245 K
Let's consider the impact of radiation on surfaces. If a surface is opaque then the radiation is either absorbed or reflected. The relative amounts of absorption and reflection for any material are highly wavelength dependent. But at any time:
where w is wavelength and a and r are absorptivity and reflectivity coefficients for that wavelength.
In the case where the surface is non-opaque, then indicident radiation can be transmitted through the surface (e.g. glass). In that case we have the more general expression:
where t is the transmission coefficient.
Back to real physics in next lecture when we talk about scattering!