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

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.

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At depths satisfying $\gamma c\tau \gg 1$, the exponential term is negligible. The atmosphere remains approximately isothermal while its intrinsic contribution is also small:Thus the plateau occupies approximatelyand has temperature

For example, take $T_{\mathrm{irr}}=2000\mathrm{K}$, $T_{\mathrm{int}}=200\mathrm{K}$, $\gamma =1$, $f=1/2$, and the Eddington constants. Then $T_{\mathrm{iso}}\simeq 1650\mathrm{K}$ and the formal plateau extends from $\tau$ of order unity to several thousand. At still greater depth,so the radiative solution rises as $T\propto {\tau }^{1/4}$ until convection replaces it.

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Differentiating the semi-grey irradiated atmosphere profile givesHydrostatic balance gives $d\tau /dP={\kappa }_{\mathrm{IR}}/g$ when the thermal mean opacity is locally constant. HenceFor variable opacity, replace ${\kappa }_{\mathrm{IR}}$ by its local value and integrate $d\tau ={\kappa }_{\mathrm{IR}}(P,T)dP/g$.

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A useful first estimate places the radiative-convective boundary where the intrinsic term becomes comparable to the deep irradiation plateau:ThusWith the assumptions of part (b), $g=10\mathrm{m}{\mathrm{s}}^{-2}$ and ${\kappa }_{\mathrm{IR}}=10^{-3}{\mathrm{m}}^2\mathrm{k}{\mathrm{g}}^{-1}$, one finds ${\tau }_{\mathrm{RCB}}\sim 6.2\times 10^3$ and $P_{\mathrm{RCB}}\sim 6.2\times 10^7\mathrm{Pa}\simeq 620\mathrm{bar}$.

More precisely one equates the local radiative logarithmic gradient to the adiabatic gradient. A constant-opacity grey profile approaches ${\nabla }_{\mathrm{rad}}=1/4$, so a diatomic adiabat with ${\nabla }_{\mathrm{ad}}=2/7$ requires the realistic increase of opacity with depth to become convective. The numerical pressure is therefore an order-of-magnitude estimate, sensitive mainly to $T_{\mathrm{int}}$ and deep opacity.

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A thermal inversion means temperature rises outward, so $dT/d\tau <0$. From part (c), near the observable atmosphere this requiresAt the top, the condition isIt states that shortwave absorption high in the atmosphere must overwhelm intrinsic heating.

Jupiter has substantial intrinsic flux and generally lacks enough persistent high-altitude visible opacity for a strong global inversion, although localized stratospheric heating occurs. An ultra-hot Jupiter around a $6500\mathrm{K}$ star receives intense optical and ultraviolet radiation; metals, TiO/VO where present, and continuum absorption can make $\gamma >1$, so inversions are common. A temperate sub-Neptune around a $2500\mathrm{K}$ star receives much of its stellar power in the near infrared, where the same molecules also emit thermally. This reduces the separation between shortwave and longwave opacity; clouds or photochemical hazes can still create upper heating, but an inversion is less automatic.

Solved by gpt-5.6-sol high.