Layer Model with Greenhouse Effect

The model

The figure above depicts the previously described first basic climate model with bare-rock surface. The incoming energy budget is represented by $L(1-\alpha )/4$ in the figure, where L is the solar constant, α is the albedo of Earth, and number 4 represents the geometry of the system (we’re considering the entire planet, so it’s a sphere, keep this in mind because we’ll deal with different problems that consider different conditions). This energy would be balanced by the energy that leaves the system, as indicated by $ϵ\sigma T_{ground}^4$, where ε, σ and T are the emissivity of Earth, the Stefan-Boltzmann constant, and the Earth’s ground temperature, respectively.

However, we discovered that the given model is rather cold in comparison to real-world values. This is due mostly to the presence of greenhouse gases in the atmosphere, which absorb and emit radiation as well.

For the sake of simplicity, imagine all of the gases that absorb and emit as a pane of glass in the atmosphere. In this layer model, the atmosphere is assumed to be a blackbody in the infrared that absorbs and emits all frequencies of infrared light.

We will also make the following assumptions for this model:

• The pane of glass does not absorb sunlight.
• The glass absorbs infrared radiated by the earth.
• The glass pane itself emits infrared light in both directions, up and down, according to its own temperature T_a.

We can then obtain some formulae for the various types of energy in the model:

• $\frac{L(1-\alpha )}{4}$ is the incoming energy budget that the ground absorbs.
  
  
• $ϵ\sigma T_g^4$ represents energy emitted by the ground.
  
  
• $ϵ\sigma T_a^4$ represents energy absorbed and emitted by the pane of glass (assuming that the glass has good blackbody properties, ε ≈ 1)

Then we get the following table for the energy budgets that enter and exit the system:

Energy budget	IN	OUT
Ground	$\frac{L(1-\alpha )}{4}+ϵ\sigma T_a^4$	$ϵ\sigma T_g^4$
Atmosphere	$ϵ\sigma T_g^4$	$2ϵ\sigma T_a^4$

To get the energy budget for the Earth system overall, we could add them up and obtain:

$$\frac{L(1-\alpha )}{4}=ϵ\sigma T_a^4$$

However, there is a simpler, much less complicated solution: create a general budget that depicts the whole system. We may draw a line above the atmosphere representing the border to space (as depicted in the figu re). And the results would be the same as what we obtained previously.

Energy budget	IN	OUT
Overall System	$\frac{L(1-\alpha )}{4}$	$ϵ\sigma T_a^4$

This is still not close to reality, of course. It turns out that the atmosphere cannot be assumed as a blackbody in the infrared that absorbs and emits all frequencies of infrared light. In fact, gases absorbs infrared light selectively, and most gases in the atmosphere doesn’t interact with IR light at all.

Problems:

How hot is The Moon?

The layer model assumes that the temperature of the body in space is all the same. This isn’t really very accurate, as you know that it’s colder at the poles than it is at the equator. For a bare rock with no atmosphere or ocean, like the moon, the situation is even worse, because fluids like air and water are how heat is carried around on the planet. So let’s make the other extreme assumption, that there is no heat transport on a bare rock like the moon. Assume for comparability that the albedo of this world is 0.33, and the solar constant is 1350 Watts/m2, same as for Earth.

a) What would be the equilibrium temperature of the surface of the moon, where influx equals outflux, on the equator, at local noon, when the sun is directly overhead, in Kelvins?

Since the assumption is that the moon’s surface is bare rock with no atmosphere or ocean, we can use the simple model for bare-rock layer on the equator at local noon (where the location in consideration is perpendicular to the sunlight direction). As we only want to find the temperature at the equator, not on the whole spherical surface, we can ignore the geometric quantity.

We have: L = 1350 W/m², α = 0.33.

The equilibirum equation for energy:

$$E_{in}=E_{out}$$ $$L(1-\alpha )=ϵ\sigma T_g^4$$ $$T_g\approx 355.4$$

b) What would be the equilibrium temperature, where energy outflow equals energy inflow, on the moon at night, in Kelvins?

At night there is no sunlight directed towards the surface, so basically no energy flow. The temperature would be 0K.

A Stronger Greenhouse Effect

Insert another atmospheric layer into the model, just like the first one. Arrow a represents the energy inflow and arrow f represents the energy outflow of the overall system. We have:

$$\frac{L(1-\alpha )}{4}=ϵ\sigma T_{layer2}^4$$ $$T_{layer2}=254.1K$$

Similarly, the energy budget of layer 2 at equilibrium:

$$ϵ\sigma T_{layer1}^4=2ϵ\sigma T_{layer2}^4$$ $$(\frac{T_{layer1}}{T_{layer2}}{)}^4=2$$ $$\frac{T_{layer1}}{T_{layer2}}=2^{1/4}$$

Energy budget for layer 1:

$$ϵ\sigma (T_g^4+T_{layer2}^4)=2ϵ\sigma T_{layer1}^4$$ $$\frac{T_g^4+T_{layer2}^4}{T_{layer1}^4}=2$$ $$(\frac{T_g}{T_{layer1}}{)}^4=2-(\frac{T_{layer2}}{T_{layer1}}{)}^4=2-\frac{1}{2}=\frac{3}{2}$$

Nuclear Winter

Go back to the 1-layer model, but change it so that the atmospheric layer absorbs visible light rather than allowing to pass through. This could happen if the upper atmosphere were filled with dust. For simplicity, assume that the albedo of the earth remains at 30%, even though in the real world it might change with a dusty atmosphere. What is the ratio of T_g / T_a in this case?

The answer is 1. We can see that the energy budget for the Earth at equilibrium would be:

$$ϵ\sigma T_g^4=ϵ\sigma T_a^4$$

So apparently, $T_g=T_a$.