The stellar flux at the orbit is
$$
F_*=\sigma T_s^4\left(\frac{R_s}{a}\right)^2.
$$
Let $\epsilon$ be the fraction of the planet's total absorbed power transported to the night side, and let $A_B$ be its <Bond albedo>. Equating transported power to nightside blackbody emission gives
$$
2\pi R_p^2\sigma T_n^4
=\epsilon\pi R_p^2(1-A_B)F_*.
$$
Hence the <nightside equilibrium temperature> is
$$
\boxed{
T_n=T_s\sqrt{\frac{R_s}{a}}
\left[\frac{\epsilon(1-A_B)}2\right]^{1/4}}.
$$
Perfect global <day-night heat redistribution> has $\epsilon=1/2$ and gives $T_n=T_s\sqrt{R_s/a}[(1-A_B)/4]^{1/4}$; without redistribution, stellar heating alone gives $T_n=0$ in this idealization. Efficient atmospheric or oceanic transport prevents volatile cold trapping and broadens the habitable region of a tidally locked planet, whereas a thin atmosphere can leave a frozen night side even when the day side is temperate.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 315 / 3 / d / Solution

{title2=$T_n$}

If a fraction $\epsilon$ of total absorbed stellar power reaches the night side, blackbody balance gives
$$
T_n=T_*\sqrt{\frac{R_*}{a}}
\left[\frac{\epsilon(1-A_B)}2\right]^{1/4}.
$$


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 Incoming links: Nightside equilibrium temperature