Simple Earth Climate Model: Single-Layer Imperfect Greenhouse Atmosphere

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Why does the emission of carbon dioxide influence our climate?

Big Ideas: 

Whether radiation is absorbed or transmitted depends on the wavelength.
The absorption of the IR radiation in the atmosphere depends mainly on the presence of water vapor, ozone, and carbon dioxide.
Increasing the concentration of carbon dioxide will increase the emissivity and the surface temperature.

The atmosphere is not a perfect absorber for all radiation. We know already that the atmosphere is transparent for sunlight but it is also transparent for some of the thermal infrared radiation emitted from the Earth's surface. Consequently, only a fraction of the thermal IR radiation is absorbed by the atmosphere, which means that the emissivity ε is not equal to one. Therefore, an observer in space would detect IR radiation emitted from the surface as well as from the atmosphere, rather than just Earth's atmosphere (Fig. 1). Our balanced equation for the conservation of energy on Earth's surface is

<br />
\begin{eqnarray}<br />
   I_{in} &=& I_{out} \nonumber \\<br />
   \dfrac{S}{4}(1-A) + \epsilon \sigma T_a^4 &=& \sigma T_e^4 \nonumber<br />
   \end{eqnarray}<br />

As before (see main article), S is the solar constant (S = 1367 W/m2), A is the albedo (A = 0.3), and σ is the Stefan-Boltzmann constant (σ = 5.67 x 10-8 W/m2 K). Notice that the emissivity is still equal to 1 for the surface (so we did not write it explicitly in the second equation) but is less than 1  for the atmosphere now. The balanced equation for the conservation of energy of Earth's atmosphere becomes

<br />
\begin{eqnarray}<br />
   I_{in} &=& I_{out} \nonumber \\<br />
   \epsilon \sigma T_e^4 &=& 2 \epsilon \sigma T_a^4 \nonumber\\<br />
  T_e &=& 1.19 T_a \nonumber<br />
   \end{eqnarray}<br />

We can solve for either the surface temperature or the atmosphere temperature by combining the equations for the surface and the atmosphere. We leave this up to the reader as an exercise. You can study influence of the emissivity on the surface temperature with our spreadsheet  Earths_Surface_Temperature_Spreadsheet.xls.

 

Figure 1. A diagram of the exchange of EM radiation between the Sun, Earth, and Earth's atmosphere. All three objects are assumed to be black bodies, and so energy is conserved. The green arrows represent the incident solar intensity, which is not absorbed by Earth's atmosphere as the solar EM radiation spectrum consists of 37% visible, 51% near IR, and 12% UV radiation. The red arrows represent IR radiation, which is emitted by both Earth and Earth's atmosphere. The difference in the wavelength of EM radiation is due to the temperature of the radiating object. The red equations represent the intensities that are either emitted (outgoing arrows) or absorbed (incoming arrows).

 

So what is a reasonable value for the emissivity of the atmosphere? Based on measured spectra 1, we know that the atmosphere is transparent for some wavelength, even in the thermal IR. An example is shown below.

 

Figure 2. The graph 1shows the total outgoing flux measured at the top of Earth's atmosphere (blue curve). This is compared to the radiation of a perfect blackbody corresponding to a temperature of 294 K (red curve). The difference between the red and the blue curve is due to absorption. Most of the absorption is due to the presence of water vapor, ozone, and carbon dioxide. 

 

An estimate of the difference between the measured flux and the flux of an ideal blackbody from figure 2 yields roughly 35%. (For this you compare the areas under the two curves.) This is the fraction of the Earth’s thermal radiation that is not absorbed by the Earth atmosphere: So the measured flux is 65% of the flux we expect from a perfect blackbody. Looking at the flux diagram, the measured flux should be 

<br />
     I_{measured} = (1 - \epsilon) \sigma T_e^4 + \epsilon \sigma T_a^4 \nonumber\\<br />

and also

<br />
I_{measured} = 0.65 \sigma T_e^4<br />

 

Combining these equations and using the relationship between surface and atmosphere temperature yields ε = 0.7. Entering the data into our spreadsheet yields a surface temperature of Te = 285 K, close to the current measured value of 288 K.

A more refined analysis 2 yields ε = 0.78 and a temperature of 288 K.

 

Influence of CO2 and other greenhouse gases

The concepts developed above allow us now to understand the influence of carbon dioxide and other greenhouse gases on our climate. The absorption of radiation is due to the molecules in our atmosphere. Most of the absorption is due to water, ozone, and carbon dioxide, as shown in the spectrum above. If we double the concentration of CO2 in the atmosphere, a simple model predicts that the emissivity increases from ε = 0.78 to ε = 0.80 2 3.

Using our spreadsheet again, we see that the surface temperature would increase by 1.2 K. Additional effects such as ‘positive feedback’ due to increased water vapor lead to an increase in emissivity by another increment of 0.02 3. In a static model this would raise the Earth's temperature to 292K. However, more sophisticated climate models indicate that further positive feedbacks would cause the temperature to go on rising for many centuries. [IPC ref.]

 

To take a more extensive look at how changing different variables effects Earth's surface temperature, check out Earths_Surface_Temperature_Spreadsheet.xls.

 

Comments

Me again - I think my last

Me again -

I think my last question about this model is with respect to the surface vs atmosphere temperature. It was always my thought that, as the surface temperature increases, the atmosphere temperature decreases. According to your result, Te = 1.19Ta, a linear relationship. Since the solar flux isn't changing, shouldn't we see this inverse relationship to keep the energy balance?

Thanks,
Brian

Hi Brian, Good that you bring

Hi Brian,
Good that you bring this up. It seems that the Earth energy increases due to the atmosphere although the solar flux remains the same. This seems to be in contradiction with the idea of an energy balance and with conservation of energy.
However, if we consider the atmosphere as part of the Earth system, then everything remains the same. What's radiated back into space all comes from the atmosphere that is at 255 K - the same temperature that Earth would have without an atmosphere. This choice of system makes sense since the incoming energy comes from "above" the atmosphere. We also have to consider the energy leaving the system above the atmosphere. You will get contradictions if you don't use the same definition of your system for incoming and outgoing flux.
The diagram explains why the temperatures of the atmosphere has to be proportional to the Earth's surface: It absorbs the radiation coming from the surface, and the warmer the surface is, the more radiation it emits. That in turn means that the atmosphere absorbs more radiation leading to a higher atmosphere temperature.

Your question makes sense.

Your question makes sense. The answer is more obvious if we go from flux back to power. So if we start with the power balance for the atmosphere P_in = P_out, we would write Stefan's law for both sides. P_in is due to the Earth's surface:
P_in = (epsilon)(sigma)(area_Earth)Ts^4.
For P_out, we can write a similar equation and take the surface area of the atmosphere.
P_out = epsilon)(sigma)(area_atmosphere)Ta^4
If we visualize the atmosphere as a thin layer similar to the glass in a greenhouse, it would have a top and a bottom surface. Since the atmosphere is not far above the surface of the Earth (~80 km, compared to the Earth's radius of 6400 km), the radius of the atmosphere is in good approximation the same as the Earth's radius and we get the same surface area, except that you have a top and a bottom surface that both radiate. So area_atmosphere = 2 area_Earth.
The thing to realize is that in our model the entire atmosphere layer is at the same temperature. So it is not that the bottom absorbs the radiation from Earth and immediately radiates it back. In our simple model, the energy gets absorbed, heats up the entire atmosphere, and the atmosphere simply radiates according to its temperature from its entire surface.

Brian, Thank you for your

Brian,
Thank you for your question. The factor of 2 is due to the fact that half of the radiation absorbed in the atmosphere is radiated back to Earth and half is radiated up into space.
So from the point of view of the Earth's surface, only half of the radiation emitted into the atmosphere comes back.
Another way to look at this problem is to consider the Earth's atmosphere as a thin slab (similar to a window) with a top and bottom surface. The radiation from the Earth's surface is absorbed at the bottom surface. Inside the slab absorption and emission processes re-distribute the absorbed heat, so the entire slab is at the same temperature. Finally the radiation is emitted from the top and bottom surfaces.

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