How to calculate the average temperature of the hemispheres of a planet tidally locked to its star?

You asked "Is there any way to calculate it?". The answer is yes, but it's not going to be as simple as plugging a few numbers into a simple formula.

You'll need a general circulation model.

The reason is that the heat transport around the planet involves the atmosphere and the oceans, and these require dealing with fluid dynamics. This is going to bring in dependencies like the topography of the planet (the wind is going to be affected by obstacles such as mountain ranges, likewise the oceans are going to respond to the shape of the ocean basins). And just to make things worse, the oceans and atmosphere are coupled. You'll also need to deal with annoying things that aren't very well constrained, like cloud formation, which affects the albedo of the planet.

Needless to say, this is rather computationally intensive (do you have a supercomputer to hand?), and even if you find an available GCM you'll likely have to do a lot of modifications so it can be applied to a tidally-locked exoplanet, especially if the atmosphere is non-Earthlike as well.

One model I've seen being used for a bunch of exoplanet studies is LMDZ4, as used e.g. for Proxima b. Not sure if the source code is freely available though and even if it were, I'm not sure whether it would be executable on standard desktop hardware.

Otherwise you could try to fudge it by throwing in a simple redistribution factor and emissivity into the usual effective temperature formula. With stellar luminosity $L_\ast$, planet-star distance $d$, emissivity $\epsilon$, albedo $A$ and the fraction of energy distributed to the nightside $f \in [0, 0.5]$ where 0.5 means an equal fraction of energy distributed to both hemispheres, equating received and emitted power and you end up with:

$$\begin{align} T_\mathrm{d} & = \left[\frac{L_\ast (1-A) (1-f)}{8 \pi d^2 \cdot \sigma \epsilon_\mathrm{d}}\right]^{1/4} \\ T_\mathrm{n} & = \left[\frac{L_\ast (1-A) f}{8 \pi d^2 \cdot \sigma \epsilon_\mathrm{n}}\right]^{1/4} \end{align}$$

Where $\sigma$ is the Stefan-Boltzmann constant. The d and n suffixes represent the day and nightside, and I've allowed for different emissivities of both hemispheres (e.g. due to cloud buildup on the dayside versus clearer skies at night).

But figuring out what the appropriate values for $A$, $\epsilon_\mathrm{d,n}$ and $f$ are basically requires doing things properly.

Derivation of the formulae:

For a planet orbiting at distance $d \gg R_\ast$ where $R_\ast$ is the radius of the star (i.e. negligible illumination of the far hemisphere, light rays can be treated as parallel), the fraction of the power output of the star intercepted is the ratio of the area of the planetary disc, $\pi R_\mathrm{p}^2$, where $R_\mathrm{p}$ is the planetary radius, to the area over which the star's radiation is distributed across, i.e. a sphere of radius $d$, which has area $4\pi d^2$. The albedo $A$ represents the fraction of this reflected back into space, so the absorbed power is:

$$P_\mathrm{abs} = L_\ast (1-A) \left(\frac{R_\mathrm{p}^2}{4d^2}\right)$$

For the planet to be in equilibrium, the power radiated must equal the power absorbed. Assume the planet has two hemispheres, with uniform properties across each hemisphere. Energy balance gives

$$P_\mathrm{rad,d} + P_\mathrm{rad,n} = P_\mathrm{abs}$$

So representing the fraction of the power transferred to the nightside by $f$, we can write:

$$\begin{align} P_\mathrm{rad, d} & = (1-f)P_\mathrm{abs} \\ P_\mathrm{rad, n} & = fP_\mathrm{abs} \end{align}$$

The next stage is to write the greybody emission law for each hemisphere. The total area of each hemisphere is $2\pi R_\mathrm{p}^2$, the power per unit area at a given temperature $T$ is $\sigma T^4$, and we scale by the emissivity $\epsilon$:

$$\begin{align} P_\mathrm{rad,d} & = 2\pi R_\mathrm{p}^2 \cdot \epsilon_\mathrm{d} \sigma T_\mathrm{d}^4 \\ P_\mathrm{rad,n} & = 2\pi R_\mathrm{p}^2 \cdot \epsilon_\mathrm{n} \sigma T_\mathrm{n}^4 \end{align}$$

Substituting these expressions into the previous ones gives the formulae in the text above.

posted about 5 years ago

antispinwards‭

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