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

I have a planet orbiting a red dwarf and, as expected, it is tidally locked to its star. I know that these planets will have a very significant temperature difference between the diurnal and nocturnal hemispheres, but I want to know how different it will be, that is to say, what will be the average temperature of both hemispheres if you take into account the warming of the diurnal side and the cooling of the nocturnal side. Is there any way to calculate it?

Keep in mind that:

1. The host star has a bolometric luminosity of 0.01 (Sun=1).

2. The planet's insolation is 4.85 (Earth=1).

3. The planet's semi-major axis is 0.05 AU.

4. The planet has an argon atmosphere, with small amounts of other gases (CO2, CH4, NH3, etc.).

5. The surface gravity is 0.75 (Earth=1)

6. The surface atmospheric pressure is 0.6 atm.

7. The albedo of the diurnal hemisphere is 0.4 and the albedo of the nocturnal hemisphere is 0.9 (Bond albedo).

8. The equilibrium temperature of the planet is 356 K for the hemisphere with the albedo of 0.4 and 228 K for the hemisphere with the albedo of 0.9.

• @011358smell Of course, I will provide more details about the planet and its star. – URIZEN Oct 12 '19 at 19:33
• To explain @011358smell's comment. Earth, as seen from space, has an average temperature of about −18 °C (0 °F); if it didn't reflect any of the sunlight received, it would have, as seen from space, an average temperature of about 5 °C (41 °F). However, because of the way Earth atmosphere and water cycle work, the actual average temperature at the surface of Earth is about 14 °C (57 °F). And this average is very misleading, masking differences between over 40 °C (105 °F) in Arabia in summer and less than −20 °C (-4 °F) in Novosibirsk in winter. And those are neither Antarctica nor Sahara. – AlexP Oct 12 '19 at 19:40
• @AlexP I am aware that the temperature on Earth varies greatly depending on location and season. I am also aware that greenhouse gases (and other factors) modify the average temperature of the planet. But I am trying to give a simple model, without considering so many factors. Basically, I am referring to the warming and cooling of the surface in each hemisphere as a function of time. – URIZEN Oct 12 '19 at 20:16
• @011358smell The equilibrium temperature? No. The temperature on the diurnal side will be much higher than this and on the nocturnal side it will be much lower. I'm talking about how constant sunshine (and lack of it) affects the average temperature in each hemisphere. – URIZEN Oct 12 '19 at 20:22
• This has the potential to be a really good question, I'm concerned that it might get closed as opinion based or too broad . Please be patient, there may be some here who can give you an answer. – Rottweiler on market-day. Oct 12 '19 at 20:43

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.

• You got my attention with those formulas, where did you get them exactly? – URIZEN Oct 14 '19 at 0:22
• @URIZEN - added the derivation, hope that helps. – user66717 Oct 14 '19 at 18:45