Simple Earth Climate Model: Single-Layer Imperfect Greenhouse Atmosphere

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 of the atmosphere $ε$ 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), as indeed Earth-sensing satellites do. Our balanced equation for the conservation of energy on Earth’s surface is

$$\begin{eqnarray}
I_{in} &=& I_{out} \nonumber \tag{1}\\
\dfrac{S}{4}(1-A) + ϵ σ T_a^4 &=& σ T_e^4 \nonumber\tag{2}
\end{eqnarray}$$

As before, $S$ is the solar constant ($S = 1367\text{ W}\text{ m}^{\text{-}2}$), $A$ is the albedo ($A = 0.3$), and $σ$ is the Stefan-Boltzmann constant ($σ = 5.67 \cdot 10^{\text{-}8}\text{ W}\text{ m}^{\text{-}2}\text{ K}^{\text{-}1}$). 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

$$\begin{eqnarray}
I_{in} &=& I_{out} \nonumber \tag{3}\\
ϵ σ T_e^4 &=& 2 ϵ σ T_a^4 \nonumber\tag{4}\\
T_e &=& 1.19 T_a \nonumber\tag{5}
\end{eqnarray}$$

We can solve for either the surface temperature or the atmosphere temperature by combining the equations for the surface and the atmosphere.

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[note] Science Briefs – Taking the Measure of the Greenhouse Effect, Gavin Schmidt, http://www.giss.nasa.gov/research/briefs/schmidt_05 [2019-10-11].[/note], we know that the atmosphere is transparent for some wavelength, even in the thermal IR. An example is shown below.

Figure 2. The graph shows the total outgoing flux measured at the top of Earth’s atmosphere (blue curve). A wavenumber of 1000 cm$^{-1}$ is a wavelength of 10 μm; 500 cm$^{-1}$ is a wavelength of 20 μm.

This is compared to the radiation of a perfect blackbody corresponding to a temperature of $294\text{ K}$ (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 vapour, 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

$$I_{measured} = (1 – \epsilon) \sigma T_e^4 + \epsilon \sigma T_a^4 \tag{6}$$

and also

$$I_{measured} = 0.65 \sigma T_e^4\tag{7}$$

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 $T_e = 285\text{ K}$, close to the current measured value of $288\text{ K}$.

A more refined analysis[note] http://en.wikipedia.org/wiki/Idealized_greenhouse_model [2019-10-11].[/note] yields $ε = 0.78$ and a temperature of $288\text{ K}$.

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 $ε = 0.80$[note] http://en.wikipedia.org/wiki/Radiative_forcing [2019-10-11].[/note].

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

Can Volcanoes Change the Climate?

Fig.1. Krakatau in 2011[note]Image of Krakatau courtesy of Michael Dalton-Smith http://digitalcrossing.ca/[/note]

Volcanoes do a number of things that can affect the climate:

1. They belch greenhouse gases that can potentially warm the earth.

2. They emit clouds of ash and sulphur dioxide, which reflect away incoming sunlight.

3. The ash can darken snow cover and glaciers, which absorbs more sunlight than clean snow and ice.

4. They stop jet flights, thus preventing a load of greenhouse gas from entering the atmosphere.

We can dispose of the first easily. Humans add almost 30 gigatonnes of carbon dioxide to the atmosphere annually. Mount Pinatubo’s eruption in 1991, the biggest in recent years, added only about 40 megatonnes. So humans are the equivalent of several hundred Mount Pinatubos, each year. If volcanoes did add significant quantities to the atmosphere, we would see big spikes in global carbon dioxide concentration measurements. We don’t[note]Earth System Research Laboratory.  Recent Global CO₂ (online).  http://www.esrl.noaa.gov/gmd/ccgg/trends/co2_data_mlo.html [2019-10-15].[/note].

But the second is more serious (and much bigger than (3) and (4), which we won’t consider here). It is also amenable to some straightforward physics, which is what this C21 website is all about. The earth is heated by incoming sunlight, and cooled by outgoing thermal radiation. The hotter the earth gets, the more it radiates. This is the primary negative feedback mechanism that keep the earth’s temperature (sort-of) stable.

Volcanoes emit enormous quantities of ash and sulphur dioxide, both of which reflect sunlight, increasing our albedo A, currently measured to be about 0.3. The intensity of sunlight measured at the top of our atmosphere is called the solar constant, S, and its value is about 1.367 kW/m². The mean incident solar intensity, I_in, is simply this divided over the entire surface area of Earth. In other words, it is the power per area that the Earth’s surface would receive if the Sun was shining equally on every surface of the Earth (Eqn.1).

$I_{in} = \dfrac{S}{4}(1-A)\tag{1}$

This incoming intensity is matched by the outgoing thermal radiation, which is simply a function of surface temperature, according to the Stefan-Boltzmann law (Eqn.2):

$I_{out} = \sigma T_{e}^{4}\tag{2}$

Here σ is a universal constant. For a rigorous analysis T_e should be the radiation temperature of the upper atmosphere, but for the simple proportional treatment we adopt here, we can use it to approximate the change in the Earth’s surface temperature. To reach a stable temperature after a change in A, the earth’s temperature changes until I_in = I_out. Because we assume nothing else is changing, we can say that T_e⁴ is proportional to (1-A), i.e.T_e ∝ (1-A)^1/4. Or, by rewriting Eqns. 1 and 2 in terms of a new albedo A’ and a new temperature T_e‘ and dividing the two to get rid of terms we don’t care about (Eqn.3):

$\dfrac{T’_{e}}{T_{e}} = \left( \dfrac{1-A’}{1-A}\right )^{1/4}\tag{3}$

Fig.2. The area affected by the Eyjafjallajökull eruption of 2010 as shown in the media[note]http://news.bbc.co.uk/2/hi/8623534.stm[2019-10-15].[/note]

So how much did A change with the Eyjafjallajökull eruption? Let’s take a worst-case scenario and assume the plume from the volcano reflects all sunlight. The area, judging from Fig.2, appears to be about twice that of Scandinavia, or 2 million km². Viewed from the Sun, this plume always appears at an angle. Its mean latitude is about 60 deg, so we divide this area by 2 (because the cosine of 60deg is 0.5), and the resulting 1 million km² is only 0.2% of the earth’s surface (510 million km²). If the albedo of the plume rises from 0.3 to a maximum possible of 1.0 that means the total albedo of earth becomes 0.3 + 1.0×0.7×0.2/100 = 0.3014. Putting this in equation 3 (don’t round off and throw out the baby with the bath water!) and assuming that Te is the present value of 287.5K, yields T_e‘ = 287.36K, or a drop of 0.14K (or Celsius). This is comparable to typical year-to-year variations in the mean surface temperature of the Earth. However, the change is unlikely to be anything like this big, because the time-constant for the earth to respond to such changes is about 5 years[note]Stephen E. Schwartz, Heat capacity, time constant, and sensitivity of Earth’s climate system, Brookhaven National Laboratory report 2007, https://agupubs.onlinelibrary.wiley.com/doi/10.1029/2007JD008746 [2019-10-15].[/note], so to achieve this change in temperature, the ash plume would have to stay in the atmosphere for about a decade.

The much larger Mount Pinatubo did cool the Earth by about 0.5C in the early 90s.

Updated (CEW) 2019-10-15