# 7.3: Atmospheric Radiation and Earth’s Climate

Let’s first look at the general energy balance—the radiative equilibrium—of the Earth system (see figure below). The solar irradiance is essentially composed of parallel radiation beams (or radiances) that strike half the globe. At the same time, outgoing infrared radiation is emitted to space in all directions from both the sunlit and dark sides of the globe. At the top of the atmosphere, the difference of the incoming solar radiation energy minus the amount of solar radiation energy that is scattered back to space (this difference being the amount of solar radiation energy absorbed by the Earth system) must balance the emitted infrared radiation energy for radiative equilibrium to hold. The total amount of solar radiation energy striking Earth per second is equal to the solar irradiance, $$F\left(\mathrm{W} \mathrm{m}^{-2}\right),$$ times the Earth's cross sectional area, $$\pi R_{E a r t h}^{2}\left(\mathrm{m}^{2}\right)$$. Some of the solar radiation energy is reflected by clouds, aerosols, snow, ice, and the land surface back to space and is not absorbed, hence does not contribute energy to raise Earth’s temperature. The fraction that is reflected is called the albedo, and we can account for it by subtracting the albedo from 1 and multiplying $$F \pi R_{E a r t h}^{2}$$ times the difference: $$F \pi R_{E a r t h}^{2}(1-a) .$$ The albedo has been estimated to be 0.294 (Stephens et al., 2012, Nature Geoscience 5, p. 691). On the other hand, Earth and its atmosphere radiate in all directions and the radiation can be described by the Stefan–Boltzmann Law, which, recall, is the integral of the Planck function over all wavelengths. Thus the emitted infrared energy per unit area (or emitted infrared irradiance) out the top of the atmosphere is $$\sigma T_{\text {top}}^{4}\left(W m^{-2}\right),$$ where we have assumed an emissivity of 1 for the atmosphere at all emitted infrared radiation wavelengths. To get the total energy we must multiply this irradiance by the Earth’s total surface area, 4$$\pi R_{\text {Farth }}^{2}$$ . The top of the atmosphere is at an altitude of $$\sim 50-100$$ km above the surface, compared to Earth’s radius of 6400 km, so we will ignore this small difference.

See the video (1:37) below for a more detailed explanation:

Earth Energy Balance

To calculate the average temperature at the top of Earth's atmosphere, we need to look at the balance between the solar radiation energy coming into the Earth's system against the infrared radiation going out of the Earth's system. The solar irradiance is essentially parallel by the time it gets to Earth, so it is intercepted by Earth's cross-section, which is just pi r Earth squared. Since a fraction of the solar radiation is immediately reflected and scattered back out into space-- this is what we call the albedo-- we have to correct the amount of radiation energy that Earth's system absorbed by subtracting off the albedo. On the other hand, Earth radiates in all directions. So assuming Earth's emissivity, than Earth's irradiance is in watts per meter squared. And if we multiply by Earth's surface area, really the surface area at the top of the atmosphere, then we get the energy leaving the Earth's system every second. That is in watts. We can use the laws of exponents to rearrange this equation to get an equation for temperature. When we put in typical values for the earth's system and solar irradiance, we calculate that the radiating temperature at the top of the atmosphere is 255 Kelvin or minus 18 degrees C or 0 Fahrenheit.

Equating the solar radiation energy absorbed by the Earth system to the infrared radiation energy emitted by the Earth system to space gives the equation:

\begin{align} \pi R_{E a r t h}^{2} F(1-a) &=4 \pi R_{E a r t h}^{2} \sigma T_{t o p}^{4} \\[4pt] T_{t o p} &=\left(\frac{(F / 4)(1-a)}{\sigma}\right)^{1 / 4} \label{eq3} \end{align}

But what is the temperature at the top of the atmosphere, Ttop? Put in the values F = 1361 W m–2, a = 0.294, and σ = 5.67 x 10–8 W m–2 K–4. Therefore,

\begin{align} T_{t o p} &=\left(\frac{\left(1361 \mathrm{Wm}^{-2} / 4\right)(1-0.294)}{5.67 \times 10^{-8} \mathrm{W} \mathrm{m}^{-2} \mathrm{K}^{-4}}\right)^{1 / 4} \\[4pt] &=255 \mathrm{K} \end{align}

The temperature at the top of the atmosphere is 255 K, which equals –18 oC or 0 oF. It is substantially less than Earth’s average surface temperature of 288 K, which equals 15 oC or 59 oF. This top-of-the-atmosphere temperature is the same as what the Earth’s surface temperature would be if Earth had no atmosphere but had the same albedo. It is clear from these calculations that the atmosphere, modeled with an emissivity, and hence absorptivity, of 1 over all emitted infrared radiation wavelengths, is creating a difference between the temperature at the top of the atmosphere and the temperature at Earth’s surface.

In particular, let’s look at only the vertical energy balance averaged over the entire globe. We will think of everything in terms of the SI units of irradiance (or energy per second per unit area), which is W m–2. Consider two idealized cases first before examining the actual atmosphere.

Let’s build a simple, flat atmosphere with all solar and infrared radiation energy moving only vertically (see figure below). We will make the following physical assumptions:

• solar irradiance is mostly in the visible (even though in reality half the solar irradiance is in the infrared with wavelengths greater than 0.7 µm);
• Earth's surface and atmosphere emitted radiation is in the infrared;
• the atmosphere is transparent to visible radiation;
• the atmosphere is opaque (that is, has an absorptivity of 1) to infrared radiation;
• radiation energy flowing up must equal radiation energy flowing down at every level in the atmosphere.

Please watch the following video (2:18)

Climate Model

In the no-atmosphere model, the only radiating bodies are the Sun and the Earth. (By the way, if Earth had a pure nitrogen atmosphere, the results would be very similar to the no atmosphere scenario.) The solar radiation passes through the altitude levels where a stratosphere and troposphere would be and the fraction 1 – a of it is absorbed by the Earth’s surface. We assume that Earth’s albedo is still 0.294 so that 0.706, or 70.6%, of the solar radiation is absorbed at the surface with the rest reflected back to space. The Earth’s surface radiates infrared radiation energy back out to space with no absorption at the levels where the stratosphere and troposphere would be. The surface temperature in this model is such that the infrared radiation energy leaving the surface balances the incoming solar radiation energy absorbed by the surface. In terms of the arrows in the figure, there is one down arrow and one up arrow at every level.

Comparison of Interface Fluxes in Two Radiation Energy Balance Models
model no atmosphere   atmosphere transparent in visible opaque in infrared
interface down arrows up arrows down arrows up arrows
space–stratosphere     1 1
stratosphere–troposphere     2 2
troposphere–surface     3 3
space–surface 1 1

So what would the temperature at Earth’s surface be if there was no atmosphere? Equation [7-2] applies to the no-atmosphere case and hence the Earth with no atmosphere has a surface temperature of 255 K. This temperature is the same as the radiating temperature at the top of our Earth with an atmosphere whose absorptivity, hence emissivity, is 1 at all emitted infrared radiation wavelengths. The surface would be so cold that any water on it would freeze and stay frozen.

Now consider Earth with an idealized atmosphere identical to that used to derive Equation [7-2] but now paying attention to the temperature of Earth’s surface under such an atmosphere. As before, this atmosphere is transparent to all solar radiation energy coming down to Earth’s surface and is opaque to all infrared radiation. “Opaque” means that the infrared radiation is completely absorbed over very short distances (i.e., the absorptivity, hence emissivity is 1, and the absorption optical depth is great, so by Beer’s Law, very little infrared radiation is transmitted). The atmosphere itself is strongly emitting in all directions, both up and down, and the only infrared radiation that does not get absorbed is that emitted out the top of the stratosphere to space.

We know that the infrared radiation energy leaving the Earth system must come close to balancing the solar radiation energy absorbed by the Earth system. Otherwise, the temperatures of Earth’s surface and atmosphere would adjust until this condition was true. So, we will assume radiative equilibrium. Our model is a two layer model—an upper layer and a lower layer—with a solid Earth beneath them. We are assuming that each layer is at a constant temperature and absorbs all infrared radiation energy impinging on it, and then emits infrared radiation out its top and its bottom in equal amounts (because the layer emits infrared radiation energy in both directions equally). The amount of infrared radiation energy emitted by the layer is determined by its temperature only because its emissivity is set to 1 at all wavelengths. Thus between the upper layer and space, we have one arrow going down and one arrow going up: the outgoing emitted infrared radiation energy exactly balances the incoming solar radiation energy that is absorbed.

The upper layer thus also emits one arrow of infrared radiation down. So, at the interface between the upper and lower layer, the solar radiation and the upper layer's infrared radiation are going down (two arrows), so to be in radiative equilibrium there must be enough upwelling infrared radiation from the lower layer to equal the incoming solar radiation energy that is absorbed and the downward infrared radiation emitted by the upper layer (two arrows). But that means that the lower layer must also be emitting the same amount of infrared radiation down to Earth’s surface (two arrows).

Mathematically, we can account for this extra energy near Earth's surface by simply multiplying the solar radiant energy by an IR multiplier, multiplierIR = 3, in Equation \ref{eq3}:

$T_{\text {Earth}}=\left(\frac{\text {multiplier }_{I R}(F / 4)(1-a)}{\sigma}\right)^{1 / 4}=\left(\frac{3\left(1361 \mathrm{Wm}^{-2} / 4\right)(1-0.294)}{5.67 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{K}^{-4}}\right)^{1 / 4}=336 \mathrm{K}$

This temperature (336 K = 63 oC = 145 oF) is deadly and much higher than Earth’s actual surface temperature, 288 K. So this model also fails to simulate the real Earth. The no-atmosphere model is too cold while the model with a two-layer, infrared-opaque atmosphere is too hot. So we can guess that something in between might be just right.

Indeed this is the case! If you look at the infrared absorption spectrum in Lesson 6, you will recall that there are some wavelengths at which all the infrared is absorbed and others, called windows, at which only a small fraction of the infrared radiation is absorbed. So, we find that a mix of total absorption, partial absorption, and no absorption at various wavelengths gives an atmosphere that allows Earth’s surface to radiate much radiation directly to space at some wavelengths but not at other wavelengths, where troposphere absorption is strong. But a large absorptivity implies a large emissivity so that at those wavelengths for which there is strong absorption there is also emission; however, given that the troposphere is cooler than the surface, the troposphere emits less upwelling infrared radiation energy than it absorbs from the warmer surface underneath. But irrespective of wavelength, emission by the troposphere is downwards as well as upwards, and provides another radiation energy source to heat Earth’s surface. This is called the greenhouse effect, which is poorly named because a greenhouse warms the Earth by suppressing heat loss by convection whereas the troposphere warms the Earth by emitting infrared radiation.

A study by Kiehl and Trenberth (1997, Bulletin of the American Meteorological Society 78, p. 197) determined the contributions to the greenhouse effect. It was shown that 81% of the greenhouse effect is due to greenhouse gases and 19% is due to clouds. Of the greenhouse effect resulting from gases, 60% is contributed by water vapor, 26% by carbon dioxide, and 14% by ozone, nitrous oxide, and methane.

In parts of the spectrum where water vapor, carbon dioxide and other gases absorb more weakly, the atmosphere is less opaque. However, if the amounts of these gases are increased, then they will absorb more strongly and thus start emitting more strongly, thus increasing the radiation emitted by the atmosphere to the surface and thus increasing the surface temperature in order for the surface to come into radiative equilibrium. Remember that the energy going out of the top of the atmosphere is still essentially the same as the solar radiation energy coming into the atmosphere that is absorbed. In a sense, by adding carbon dioxide and other greenhouse gases to the atmosphere, we are moving Earth’s surface temperature from being closer to the no-atmosphere model to being closer to the infrared-opaque model.