# Climate Physics Chapter 3: Radiation Balance

Planets absorb energy as light from stars and radiate energy out in to space at a rate that increases with temperature, allowing it to keep in balance, at some equilibrium temperature, when the energy gain equals the energy loss.

A sufficiently hot body emits light in the form of blackbody radiation:

$$B(v,T)=\frac{2hv^{3}}{c^{2}}\frac{1}{e^{\frac{hv}{kT}}-1}\\$$

This gives the irradiance, or the energy flux per area per frequency, of the radiation emitted from the body. In this equation, v is the frequency of the emitted radiation, h is Planck’s constant, c is the speed of light, k is Boltzmann’s constant, and T is the temperature of the body.
Integrating this equation over every frequency gives us the total power per unit area:

$$F=\sigma T^{4}\\$$

where $$\sigma\approx 5.67*10^{-8}$$ W/m$${}^{2}$$K$${}^{4}$$ is the Stefan-Boltzmann constant.

We can calculate the temperature of a planet based on a simple accounting of the energy entering and exiting the system.
A planet receives energy from its star. This energy takes the form of black body radiation leaving the photosphere of the star. This means we can write the sun’s emitted energy as $$4\pi r_{s}^{2}\sigma T_{s}^{4}$$, where $$r_{s}$$ and $$T_{s}$$ are the sun’s radius and photospheric temperature. At a distance r from the star, that radiation has spread out over a sphere of surface area $$4\pi r^{2}$$, so a planet at that distance sees a total flux of $$L=\sigma T_{s}^{4}\frac{r_{s}}{r}$$. The planet receives this energy over its cross sectional area. If the planet’s radius is a, this area is $$\pi a^{2}$$, so it receives $$\pi a^{2}L$$ energy per unit time. However, due to the planet’s reflectivity, not all of that energy will be absorbed. The proportion of sunlight reflected by the planet is the planetary albedo, denoted $$\alpha$$. A planet with albedo of 1 would reflect all light, while a planet with $$\alpha=0$$ would absorb all light. Therefore, the rate of energy absorption of a planet with albedo $$\alpha$$ and radius a is $$(1-\alpha)\pi a^{2}L$$.
The planet also emits energy by radiating as a black body from its entire surface. The rate of energy loss is then $$4\pi a^{2}\sigma T^{4}$$, where T is the planet’s surface temperature. In equilibrium, the rate of energy gain must equal the energy loss, so in equilibrium our energy balance is:

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

or, solving for the planet’s surface temperature,

$$T=\frac{1}{\sqrt{2}}(1-\alpha)^{1/4}\sqrt{\frac{r_{s}}{r}T_{s}}\\$$

The factor of $$\frac{1}{4}$$ comes from the ratio of the planet’s cross sectional area, which receives energy, and its total surface area, which radiates energy. This is because we have averaged the energy budget over the planet’s surface. For a planet that doesn’t receive energy evenly over its surface, for example because it has no atmosphere to transport heat, it doesn’t make sense to average over the full surface. Instead, we might want to consider a point on the surface where the sun is directly overhead. Looking at the energy per unit area for that spot, we would get an energy budget of $$\sigma T^{4}=(1-\alpha)L_{s}$$, without the factor of 4.

## Example Problem: Energy Balance

A spherical planet at the orbit of Mercury has a nitrogen atmosphere that has no effect on IR, but is so good at mixing heat that it keeps the planet’s surface isothermal. If the surface temperature is 300K, what is the albedo?
From energy balance:

$$\sigma T^{4}=\frac{1}{4}L(1-\alpha)=\frac{1}{4}\sigma T_{star}^{4}\frac{r_{star}^{2}}{r^{2}}(1-\alpha)\\ \textrm{ Plugging in values: }T_{star}=5777\textrm{ K},r_{star}=695700\textrm{ km},r=57.91e6\textrm{ km},T=300\textrm{ K}\\ \alpha=1-\frac{4\sigma T^{4}}{\sigma T_{star}^{4}\frac{r_{star}^{2}}{r^{2}}}=1-\frac{4*(300\textrm{K})^{4}}{(5777\textrm{K})^{4}\frac{(695700\textrm{km})^{2}}{(57.91e6\textrm{km})^{2}}}=0.7984\nonumber \\$$

## Example Problem: Energy Balance

Consider a planet covered in water ice with a uniform albedo of 0.7. The same face of the planet always faces the sun, and the atmosphere has negligible greenhouse effect. Compute the solar constant needed to begin melting ice if:
a) the entire surface of the planet has the same temperature
This means we need to average the temperature over the whole planet, giving a factor of 4 between the cross-sectional area which receives radiation and the surface area that radiates OLR:

$$L=\frac{4\pi a^{2}\sigma T^{4}}{\pi a^{2}(1-\alpha)}=\frac{4*5.67*10^{-8}\textrm{ W}/\textrm{m}^{2}\textrm{K}^{4}*(273\textrm{ K})^{4}}{(0.3)}=4200\textrm{ W/m}^{2}\nonumber \\$$

b) the dayside temperature is uniform but no heat is carried to the nightside
In this case we only need to average the temperature over half the surface, so we have a factor of 2 in our energy balance equation.

$$L=\frac{2\pi a^{2}\sigma T^{4}}{\pi a^{2}(1-\alpha)}=\frac{2*5.67*10^{-8}\textrm{ W}/\textrm{m}^{2}\textrm{K}^{4}*(273\textrm{ K})^{4}}{(0.3)}=2100\textrm{ W/m}^{2}\nonumber \\$$

c) there is no atmosphere, so each bit of the surface is in equilibrium with the solar radiation it absorbs
We can find the maximum temperature by looking at a patch where the sun is directly overhead, and we don’t have to average the temperature over the planet at all, so the area of the patch receiving the radiation is the same patch that’s emitting radiation, leaving no extra factor in our energy equation.

$$L=\frac{\sigma T^{4}}{(1-\alpha)}=\frac{5.67*10^{-8}\textrm{ W}/\textrm{m}^{2}\textrm{K}^{4}*(273\textrm{ K})^{4}}{(0.3)}=1050\textrm{ W/m}^{2}\nonumber \\$$