Lectures - Monday and Wednesday, 11:00 AM - 12:15 PM
Lab - Tuesday, 4:10 PM -7 PM
This summary is meant to explain the way the Earth mean temperature is maintained through a radiative balance with the incoming solar radiation. The explanation is done using a simple model, much less complex than the way the real climate system works, but the model contains the basic ingredients needed to explain the problem.
The model we are using is of a spherical planet that rotates on its axis and revolves around the Sun at the same time. We are also studying the average state of the planet, without addressing the details of the distribution of the climate with latitude. What we are seeking is to find under these conditions what is the average temperature near the surface of the planet. We will consider also how this temperature depends on the presence of an atmosphere. In our calculations will make use of the laws of radiative heat transfer as summarized in the notes for Lecture 1.
In the lecture notes we have already reviewed the subject of effective temperature. Here we repeat the derivation for the sake of completion. As stated, the effective temperature of a planet is the temperature that would exist near its surface under radiative balance with the incoming solar radiation if the planet had no atmosphere. The only transformation to the radiation from the Sun considered in this case is reflection. Reflection reduces the amount of heat the planet has to come to equilibrium with.
Consider the following case:
With the help of Figure 1 below, we can see that the planet's area that collects the sunlight is effectively the area of the planet's shadow on a plane perpendicular to the incoming sunlight. Considering that a fraction, A, is being reflected back into space, the overall energy received by the planet is:
(1) Es= (1-A) SπR2
where Es is the total solar (shortwave) energy collected by the planet per unit time (in units of W).
Figure 1: The area collecting sunlight is the cross sectional area of the spherical planet. The area emitting planetary radiation is the entire surface of the sphere.
Over a period of one planetary rotation around its axis, the entire surface of the planet is exposed to the incoming radiation and warms up to a absolute temperature, T. Averaged over the length of a full number of rotation period the entire plant's surface (equal to 4πR2) emits energy, which according to the Stefan-Boltzman law is proportional to the fourth power of its temperature:
(2) Ep= 4πR2 σT4
where EP is the planetary (longwave) radiation (in W) and σis the Stefan-Boltzman constant. The temperature increases until a radiative balance is reached and then it remains constant. This occurs when:
(4) 4πR2 σT4 = (1-A) S πR2
Eliminating identical terms from both sides of the equation we obtain:
(5) G = σTe4 = (1-A) S / 4
Where we used the letter G to express the radiative flux at the ground and identified the effective temperature by the subscript e.
The radiative flux (energy per unit area) emitted by the planet is thus 4 times smaller than the amount of solar flux it absorbs.
As described in the notes to lecture 1, when we apply this equation to Earth the calculated temperature we obtain is much lower than the observed temperature. As explained in the notes this is because the Earth's atmosphere contains gases that absorb the longwave radiation emitted from the Earth's surface. Let us now examine what this atmospheric absorption implies.
Figure 2: A plant with an atmosphere containing a single layer of IR absorbing matter.
Consider Figure 2 above. It shows the planet's surface still receiving a solar flux of (1-A) S (accept for reflection, expressed in the albedo A, the solar flux is hardly affected by the presence of the atmosphere because only small segments of the shortwave spectrum are absorbed by its constituents). The planet warms up and emits a flux of longwave (IR) radiation that we will denote by the letter G. In the atmosphere there are gases that absorb longwave radiation. We will consider for a moment that these gases are all arranged in a layer of some thickness above the surface and spread equally over the entire planet. We will also assume that all the longwave radiation is absorbed in that layer. As it absorbs the longwave radiation the layer also warms up and by radiative laws also emits longwave radiation. We will denote this flux by the letter H. Because the layer is elevated from the surface, it emits radiation through both its upper and lower surface in equal amounts. Thus overall the layer emits a total flux of 2H W m-2. Using these considerations we can find if the addition of longwave radiation absorbing gases to the atmosphere changes the equilibrium surface temperature.
In this new situation the outer surface of the absorbing layer becomes the outer surface of the planet, and as described in section 1 above, it must come to radiative balance with the radiation absorbed by the planet. Thus, in radiative balance and following equation (5) above, we have:
(6) H = (1-A) S / 4
where the division by 4 represents the ratio between the effective planetary area for absorption of solar radiation and the planetary area emitting longwave radiation (Figure 1 above).
At the surface, the energy balance is:
(7) G = (1-A) S / 4 + F
This is because all the incoming shortwave radiation, not reflected back to space, is absorbed by the Earth's surface. Substituting for H from equation (6) we get:
(8) G = 2 (1-A) S / 4
The addition of one IR absorbing layer has thus changed the balance of energy at the ground. It now must balance twice as much radiation than before! We can now use the Stefan-Boltzman law to find out the new equilibrium temperature (the one affected by the greenhouse effect) and get:
(9) sTg4 = 2 (1-A) S / 4
Which implies that in the presence of one effective layer of an atmospheric absorbent the temperature ( Tg) is larger than the equivalent temperature ( Te) by a factor of 21/4 (the fourth root of 2 or approximately 1.2).
When this calculation is repeated adding another layer, the temperature at the equilibrium temperature at the ground increases even more. With N layers, equation 9 becomes:
(10) σTg4 = (1+N) (1-A) S / 4
and TG becomes (1+N)1/4 times larger than Te.
Text by Yochanan Kushnir, 2000.