Planetary equilibrium temperature

March 24, 2023

Temperature is one of the most important variables in assessing an object's habitability. If an object is too hot, it may not have liquid water, and if it's too cold, then the chemical reactions needed to sustain life may not happen quickly enough on it.

In this post, I explore how astronomers can estimate the temperature of an object by calculating its equilibrium temperature, which is one of the tools astronomers can use to estimate an object's surface conditions.

Prerequisites

In order to understand the content in this post, you should have a basic understanding of the following topics:

• Energy, power, and flux/intensity
• Inverse-square law
• Stefan-Boltzmann Law

If you want to brush up on these topics, I highly recommend reading Section 3.2 in A Student's Guide to the Mathematics of Astronomy by Daniel Fleisch and Julia Kregenow.

Setup

Consider a star of radius $R_{s}$ and temperature $T_{s}$. An object (say, a planet) of radius $R_p$ is in a circular orbit around the star at a distance $D$. The planet has a Bond albedo $\alpha$.¹ We are trying to determine the temperature of the planet when it is in thermal equilibrium.

The Energy Balance

In order for our planet to be in thermal equilibrium, the power emitted by the planet has to be equal to the power the planet absorbs,² as such:

$P_{\rm emitted} = P_{\rm absorbed}$

This is a consequence of conservation of energy, which states that energy cannot be created or destroyed. If we can derive expressions for $P_{\rm emitted}$ and $P_{\rm absorbed}$, we can set them equal to each other and solve for the equilibrium temperature of the planet. But, what are $P_{\rm emitted}$ and $P_{\rm absorbed}$?

Finding $P_{\rm emitted}$

To find $P_{\rm emitted}$, we treat the planet as a black body of uniform temperature. The power emitted by the planet is, by definition, the luminosity of the planet, since we are neglecting any internal heat. Using the Stefan-Boltzmann Law, we find that it is

$P_{\rm emitted} = L_p = \boxed{4\pi R_p^2 \sigma T_p^4}$

where $\sigma$ is the Stefan-Boltzmann constant (approximately $5.67 \times 10^{-8} \text{ W}\cdot\text{m}^{-2}\cdot\text{K}^{-4}$) and $R_p$ and $T_p$ are the radius and temperature of the planet, respectively.

Finding $P_{\rm absorbed}$

Using the same procedure as above, we know that the luminosity of the star is

$L_s = 4\pi R_s^2 \sigma T_s^4$

The inverse-square law tells us that the light from the star will be spread out over a spherical surface as it travels through space. As a result, the energy flux from the star at a distance $D$ will be

$F = \dfrac{L_s}{4\pi D^2} = \dfrac{R_s^2 \sigma T_s^4}{D^2}$

The light from the star will hit the half of the planet facing it, which has an area of $2\pi R_p^2$. However, the intensity of the light falling on the hemisphere is not spread out evenly, as it is falling on a spherical surface, not a flat surface.

At the location where the light is directly overhead, the planet is receiving the maximum intensity of light possible. However, at any other location, the light strikes the planet at an angle. As a result, the light is spread out over a larger area, decreasing the intensity of the light. As we move farther away from the "center", the angle becomes more and more pronounced. In the most extreme case, near the poles of the planet, the light hits the planet at such an oblique angle that almost none of it is absorbed by the planet. Overall, it turns out that the average flux over the entire hemisphere is $F/2$.³ Therefore, the power absorbed by the planet (before accounting for albedo) is

$\overbrace{2\pi R_p^2}^{\substack{\text{area of} \\ \text{hemisphere}}} \times \overbrace{\dfrac{R_s^2 \sigma T_s^4}{2 D^2}}^{\substack{\text{average} \\ \text{flux}}} = \dfrac{\pi R_p^2R_s^2 \sigma T_s^4}{D^2}$

There's an easier way to do this. The light that hits the surface of the planet has to be equal to the light that is blocked by the planet. The amount of power blocked by the planet is the product of its cross-sectional area, $\pi R_p^2$, and the flux at the planet, $F$. Multiplying these values doesn't tell us how the energy is distributed across the surface, but it gives us the same answer as before, which is all we need for this derivation.

$\overbrace{\pi R_p^2}^{\substack{\text{cross-} \\ \text{sectional}\\ \text{area}}} \times \overbrace{\dfrac{R_s^2 \sigma T_s^4}{D^2}}^{\substack{\text{full flux}}} = \dfrac{\pi R_p^2R_s^2 \sigma T_s^4}{D^2} \leftarrow \text{same result!}$

Finally, we consider the Bond albedo of the planet. By definition, the fraction of light the planet will absorb will be $(1-\alpha)$. Putting it all together, we find that the power absorbed by the planet is

$P_{\rm absorbed} = \dfrac{\pi R_p^2R_s^2 \sigma T_s^4}{D^2} \times (1-\alpha)= \boxed{\dfrac{\pi R_p^2 R_s^2 \sigma T_s^4}{D^2}(1-\alpha)}$

Putting it all together

Now, we set $P_{\rm emitted}$ and $P_{\rm absorbed}$ equal to each other, since the planet is in thermal equilibrium, and solve for $T_p$:

$\boxed{T_p = T_s \sqrt{\dfrac{R_s}{2D}} (1-\alpha)^{1/4}}$

And there we have it! If, instead, you are given the flux received at the planet, $F$, instead of any of the properties about the star, you can use the same procedure to find that

$\boxed{T_p = \left( \dfrac{F (1-\alpha)}{4 \sigma} \right)^{1/4}}$

Limitations

Heat redistribution

One of the assumptions our derivation makes is that the planet is able to redistribute heat so that it is all the same temperature. The planet can do this through rotating quickly (relative to the length of its revolution) or through convection/advection in its atmosphere, among other things. Many of the planets we've found in the habitable zone are tidally locked to their parent stars, which could result in one side of the planet becoming much hotter than the other.

Atmospheres

Although our derivation takes into account the albedo of the object, it does not fully capture the effect of an atmosphere on the planet, espcially at the surface, which may vary significantly from the atmospheric conditions. For example, the calculated equilibrium temperature of Venus is about 230 Kelvin, while the surface temperature is over 700 Kelvin. It is possible to include a simple atmosphere in the derivation for equilibrium temperature, but modeling anything complex can become intractable quickly.

Closing thoughts

Overall, the equilibrium temperature of an object is a theoretical construct that is not based in any actual measurements about the object itself, like its size or flux measurements. As a result, it is difficult to make any definitive statements about an object's habitability solely based on its equilibrium temperature, and when astronomers use equilibrium temperature calculations to estimate the habitable zones of stars, the boundaries are intentionally flexible. Nevertheless, it is still an extremely useful metric to make reasonable guesses about the surface/atmospheric conditions of an object.