Where on a tidally locked planet with a 25 °C maximum is the 0 °C isotherm?

This is just a basic answer, but I used Samuel et al. (2014), who in turn cited the climate model of Léger et al. (2011) for a tidally-locked planet with no atmosphere. They give the formula for surface temperature as $$\begin{array}{}\text{(1)}& T_s={\left(\frac{{ϵ}_5}{{ϵ}_2}\right)}^{\frac{1}{4}}{\left(\frac{R_{\ast }}{a}\right)}^{\frac{1}{2}}{T}_{\ast }{\cos }^{\frac{1}{4}}\theta \end{array}$$ Substituting in $T_s=298$ (in Kelvin) and $\theta =0$ means that $${\left(\frac{{ϵ}_5}{{ϵ}_2}\right)}^{\frac{1}{4}}{\left(\frac{R_{\ast }}{a}\right)}^{\frac{1}{2}}{T}_{\ast }=298\text{ K}$$ Therefore, we get an easy formula for $\theta$: $$\begin{array}{}\text{(2)}& \theta =\arccos \left[{\left(\frac{T_s}{298\text{ K}}\right)}^4\right]\end{array}$$ If we set $T_s=273$, we then get $\theta =45.25°$, which happens to be almost exactly what you predicted.

A very similar derivation of $(1)$ equation can be found here, although that $\theta$ is latitude, not the same as our $\theta$, and so $\sin \theta$ should be replaced by $\cos \theta$, leading to the above formula. A related equation is given such that the change in temperature from the greenhouse effects can be added in. I think we can do the same thing here: $$\begin{array}{}\text{(3)}& T_s={\left(\frac{{ϵ}_5}{{ϵ}_2}\right)}^{\frac{1}{4}}{\left(\frac{R_{\ast }}{a}\right)}^{\frac{1}{2}}{T}_{\ast }{\cos }^{\frac{1}{4}}\theta +\mathrm{\Delta }{T}_s\end{array}$$ where we can compute $\mathrm{\Delta }{T}_s$ assuming that it comes from radiative forcing: $$\mathrm{\Delta }{T}_s=\lambda \mathrm{\Delta }F$$ given the climate sensitivity $\lambda$ and radiative forcing $\mathrm{\Delta }F$. I wrote about this in a lot more depth in this answer. To accurately calculate both $\lambda$ and $\mathrm{\Delta }F$, we need to know a lot more about the planet, its atmosphere, and its crustal composition. Drawing from my previous answer, if we knew the concentrations of various greenhouse gases, we could use a table from the IPCC Second Assessment Report to calculate the relevant $\mathrm{\Delta }F$s. However, that would require a lot of guesswork on my part.

Now, perhaps we could neglect the greenhouse effect entirely. Yang et al. (2013) write

The greenhouse effect for tidally locked planets is much smaller than those for non-tidally locked planets (Fig.2d). This results from a low-level temperature inversion on the nightside of tidally-locked planets (see also Joshi et al. 1997; Leconte et al. 2013). The inversion is due to efficient radiative cooling by the surface on the nightside and strong atmospheric energy transport from the dayside to the nightside (Merlis & Schneider 2010). The outgoing infrared radiation to space is therefore similar to the near-surface upward infrared radiation, resulting in a small $G$.

Here, $G$ is a parameter - a temperature difference - between the upward infrared fluxes at the planet's surface and at the top of its atmosphere; it's essentially $\mathrm{\Delta }{T}_s$. The difference in how important $G$ is is shown in the author's Fig. 2:

The 1:1 data points are for a tidally locked planet with clouds; the 2:1 and 6:1 points are for slightly faster rotators and the other two tracks are variations on the tidally locked model. See how the track without clouds has an extremely large greenhouse effect.

Perhaps, then, at most stellar fluxes, we can neglect the greenhouse effect on our tidally locked planet.

Something assumptions I'm still making:

• There is little to no atmospheric circulation.
• The surface is approximately homogeneous.

Take this answer (v1) with a grain of salt. It's only a really, really basic approximation.

posted over 8 years ago

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