How Do You Calculate a Planet's Equilibrium Temperature From Energy Balance?

SL and HL: A planet's equilibrium temperature is calculated by equating the incoming solar power it absorbs to the outgoing thermal power it radiates. This gives

$$T=\left(\frac{(1-\alpha)S}{4\varepsilon\sigma}\right)^{1/4}.$$

Here, $S$ is the solar intensity at the planet's orbit, $\alpha$ is its albedo, $\varepsilon$ is its emissivity, and $\sigma$ is the Stefan and Boltzmann constant.

The model makes three simplifying assumptions: the planet is in steady thermal equilibrium, absorbed energy is redistributed across the entire surface, and the planet emits as a grey body. Consequently, the result is an effective radiating temperature, not necessarily the temperature at every location or altitude.

Before calculating, convert all quantities to SI units and use kelvin for temperature. Increasing albedo or emissivity lowers the predicted temperature, while increasing solar intensity raises it. Because temperature depends on the fourth root, even a large percentage change in absorbed intensity produces a smaller percentage change in temperature.

The Energy-Balance Mechanism

A planet of radius $R$ intercepts sunlight over its projected circular area, $\pi R^2$. If its albedo is $\alpha$, the absorbed power is

$$P_{\text{absorbed}}=(1-\alpha)S\pi R^2.$$

The planet radiates thermal energy over its whole spherical surface area, $4\pi R^2$. Using the Stefan and Boltzmann law, its emitted power is

$$P_{\text{emitted}}=4\pi R^2\varepsilon\sigma T^4.$$

At thermal equilibrium, $P_{\text{absorbed}}=P_{\text{emitted}}$. Cancelling $\pi R^2$ and rearranging produces the equilibrium-temperature equation above. The factor of 4 comes from the difference between the projected area receiving sunlight and the total surface area emitting radiation.

For Earth, take $S=1.36\times10^3\ \text{W m}^{-2}$, $\alpha=0.30$, , and :

$$T=\left(\frac{0.70(1.36\times10^3)}{4(5.67\times10^{-8})}\right)^{1/4}\approx255\ \text{K}.$$

This is about $-18\,^{\circ}\text{C}$.

A common misconception is that this must equal Earth's average surface temperature. It does not: $255\ \text{K}$ is the effective radiating temperature without accounting fully for atmospheric absorption and re-emission. The greenhouse effect raises the average surface temperature.

Exam Technique

For an IB Calculate question, show absorbed power, emitted power, the equilibrium condition, substitution, and the final temperature in kelvin. Do not omit the factor of 4 or use albedo as the absorbed fraction, the absorbed fraction is $1-\alpha$.