Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 315 / 1 / a / Solution

The internal effective temperature of a planet $T_{\mathrm{int}}$ parametrizes intrinsic cooling, while the irradiation temperature $T_{\mathrm{irr}}$ parametrizes incident stellar flux before the redistribution factor $f$. The quantityis the visible-to-thermal mean-opacity ratio, and $\tau$ is downward thermal optical depth. Under the Eddington two-stream boundary condition,while the common semi-grey choice is $c=\sqrt{3}$.

For a young giant at $40\mathrm{au}$, stellar heating is weak and a still-large $T_{\mathrm{int}}$ dominates. Its pressure-temperature profile rises steadily inward, approximately as $T^4\propto b+\tau$, and joins a deep convective adiabat.

For a hot Jupiter close to a Sun-like star, $T_{\mathrm{irr}}\gg T_{\mathrm{int}}$. It has a broad, nearly isothermal irradiated radiative layer, followed by a deep rise where intrinsic flux and increasing opacity matter. If $\gamma >1$, absorption of starlight above the thermal photosphere can create an atmospheric thermal inversion.

For a temperate sub-Neptune around an M dwarf, irradiation and internal cooling can be more comparable. Its profile generally has a moderate radiative layer above a convective interior; near-infrared stellar radiation, molecular opacity, clouds, and hazes determine whether the upper profile is weakly inverted or decreases outward.

Solved by gpt-5.6-sol high.