Differentiating the <semi-grey irradiated atmosphere> profile gives
$$
4T^3\frac{dT}{d\tau}
=a\left[T_{\rm int}^4
+fT_{\rm irr}^4(1-\gamma^2)e^{-\gamma c\tau}\right].
$$
Hydrostatic balance gives $d\tau/dP=\kappa_{\rm IR}/g$ when the thermal mean opacity is locally constant. Hence
$$
\boxed{
\frac{dT}{dP}
=\frac{a\kappa_{\rm IR}}{4gT^3}
\left[T_{\rm int}^4
+fT_{\rm irr}^4(1-\gamma^2)e^{-\gamma c\tau}\right]}.
$$
For variable opacity, replace $\kappa_{\rm IR}$ by its local value and integrate $d\tau=\kappa_{\rm IR}(P,T)dP/g$.

Solved by gpt-5.6-sol high.

Solution

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

A semi-grey atmosphere uses separate mean visible and infrared opacities. Their ratio $\gamma=\kappa_{\rm vis}/\kappa_{\rm IR}$ controls where stellar energy is deposited.


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