Earth's Energy Balance

Now that we have discussed how electromagnetic radiation can be a vector for heat transfer, we now have the information necessary to determine how heat transfer radiation affects the temperature of the Earth.

The way we can estimate the Earth's expected temperature from solar radiation is by using an energy balance equation. Recall that the emissivity of a body is equal to the absorptivity of a body. Also, if we assume that the Earth in in thermal equilibrium, an energy balance equation will state that the electromagnetic radiation energy that goes into the Earth will equal the energy that leaves the Earth. So by these two statements, the energy that is absorbed by the Earth will equal the energy that is emitted by the Earth.

Eabsorbed=Eemitted{E}_{absorbed}={E}_{emitted}

First, lets find the energy that is absorbed by the Earth. Based on observations, the amount of energy from the sun that intercepts the Earth is about equal to 1361 Wm−21361 W {m}^{-2} (watts per meter^2), also called the solar insolation (Ks{K}_{s}). We also need the surface area of the earth that the intercepts the sun's radiation. This can be estimated by showing that the amount of the Earth that intercepts the sun's energy is exactly the same amount that would be intercepted by a flat disk that faces the sun.

Incident radiation of a sphere vs. a disk

Incident radiation from the sun to the Earth can be modeled with a flat disk. (Interaction of EMR with Earth's Surfaces)

Using these assumptions, we can write the energy intercepted by the Earth to be:

Eintercepted=πRearth2×Ks{E}_{intercepted}=π{R}_{earth}^{2}×{K}_{s}

However, since Earth is not a perfect blackbody, not all radiation that is intercepted by the Earth is absorbed. Some of the incident energy is reflected, so we will need to take this into account. We quantify an object's reflectance with albedo, with a value between 0 and 1, where 0 would be the albedo of a perfect blackbody. Because the Earth very thick and no radiation will be transmitted through it, we can assume that any energy not reflected by the Earth is abosorbed. Given that the Earth's average albedo is about 0.31, the absorbtivity of Earth is would be (1-0.31), or 0.69. Given this absorbtivity, we can then calculate the energy absorbed:

Eabsorbed=πRearth2×Ks×(1−albedo){E}_{absorbed}=π{R}_{earth}^{2}×{K}_{s}×(1-albedo)

Next to find the energy emitted by the Earth, we will use the Stefan-Boltzmann equation:

Eemitted=Q•=σATs4{E}_{emitted}=\overset{•}{Q}=σA{T}_{s}^{4}

We cannot assume that the Earth will act like a disk for energy emitted, so the surface area of the Earth will be equal to a sphere, which is equal to 4πr24π{r}^{2}

Eemitted=σ4πRearth2Ts4{E}_{emitted}=σ4π{R}_{earth}^{2}{T}_{s}^{4}

For this demonstration we are calculating the effective temperature of a planetary body, which assumes that the body emits energy as if it were a perfect blackbody. This is obviously not true for the Earth, or any other body, so adding the emissivity of the Earth into the equation would yield a more accurate answer.

By substituting the equations above into the energy balance equation, we will get:

(1−albedo)KsπRearth2=4σπRearth2Ts4(1-albedo){K}_{s}π{R}_{earth}^{2}=4σπ{R}_{earth}^{2}{T}_s^4

The πRearth2π{R}_{earth}^2 will cancel out, and solving for Ts{T}_s gives:
Ts=Ks(1−albedo)4σ4T_s=\sqrt[4]{\frac{{K}_{s}(1-albedo)}{4σ}}
Plugging in our numbers:

Ts=1361×(1−0.31)4×5.6704×10−84=254  Kelvin
T_s=\sqrt[4]{\frac{1361×(1-0.31)}{4×5.6704×10^{-8}}}=254 K

Which is equal to -19 C. This is quite a bit less than the Earth's actual average surface temperature, which is about 14 C. In addition to being influenced by radiation, Earth's climate is also influenced by the greenhouse effect that traps much of the incident and emitted radiation from the Earth and lowers the emissivity value that we assumed was 1 in the above demonstration.

This process for estimating the temperature at the surface of a planet will also work for other planetary bodies. All that needs to be done is changing the variables to their respective values for whatever planetary body is being inspected.

There are many other variables that can affect the climate of a planet that were unaccounted for in this analysis, but it still gives a good estimate for what a body's expected temperature will be.


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