What would the temperature variation be on this planet? (High eccentricity & slow rotation)

Question

I've been trying to figure out a model (or find a relevant paper) for calculating the approximate day/night and periapsis/apoapsis temperature variation for an earth-like planet that has a high eccentricity and a slow rotation rate. The planet orbits an M-class star, and is slightly smaller than earth with a thicker atmosphere of primarily nitrogen and oxygen.

Mass of the atmosphere and the size of the oceans play a key role in the retention and redistribution of heat. However I've been unable to locate any research papers or explicit models that would help me figure out specifics of how much of the temperature variation would be mitigated by retention and redistribution.

Here are the key characteristics of the planet:

• 27-day orbit
• 13.5-day (324 hour) rotation (2:1 orbital resonance)
• 27-day solar day (sunrise to sunrise)
• 0 obliquity
• 0.24 eccentricity
• 3 bar surface atmospheric pressure
• 1476 W/m^2 solar flux (based on semimajor axis)
• apoapsis: 0.16 AU
• periapsis: 0.099 AU
• semimajor axis: 0.13 AU
• 11.5C mean temperature (based on semimajor axis, flux, etc.).

Anything...even simply links to applicable research papers would be immensely helpful. Thanks!

Answer 1

First off, I'm not a climate scientist, so take this with a grain of salt. It's also a very rough estimate and doesn't take climate or emissivity changes, etc into account.

Anyways, here goes: According to the Stefan-Boltzmann law, a black body radiates at a rate proportional to the fourth power of its temperature. Now, to be in equilibrium, the planet has to radiate as much as it absorbs by the star. So, say that on the periapsis, the total insolation of your planet is $I_p$, while at apoapsis it is $I_a$. The insolation is inversely proportional to the square of the distance to the star, so $$I_a/I_p=(0.099/0.16)^2$$ which gives $$I_a = 0.3828I_p$$

Now, to be in equilibrium at both periapsis and apoapsis, we must have $$I_p = A \sigma \epsilon T_p^4$$ $$I_a =A \sigma \epsilon T_a^4$$ where $T_p,T_a$ are global temperatures at periapsis and apoapsis, $A$ is surface area, $\sigma$ is the stefan-boltzmann constant, and $\epsilon$ is emissivity, which we are assuming is constant. Dividing the third and fourth equations and combining with the second equation, we obtain $$(T_a/T_p)^4=0.3828 \implies T_a = 0.7866T_p$$ Finally, you state the mean temperature is $284.5K$. I'm going to assume this is the equally weighted average of the apoapsis and periapsis temperatures, ie $$(T_p+T_a)/2=284.5$$ Solving the previous two equations simultaneously, we obtain $$T_p = 318.5K$$ $$T_a = 250.5K$$

Considering that these should be treated as the global average temperature, we can see that this is very extreme from an Earthly perpective, where the extremes in global temperature have been within $12K$ for the last million years.