Where the spectral components vanish, $a=(R_p/R_*)^2$, so
$$
\boxed{R_p=R_*\sqrt a}.
$$
For extinction cross-section $\sigma\propto\lambda^\gamma$, the <scattering slope of a transmission spectrum> obeys
$$
\frac{dR_p}{d\log\lambda}=\gamma H,
\qquad
H=\frac{k_BT}{\bar m g}.
$$
Since $y=(R_p/R_*)^2$,
$$
H=\frac{R_*^2}{2R_p\gamma}
\frac{dy}{d\log\lambda}.
$$
After subtracting the Gaussian feature, the model gives $dy/d\log\lambda=\gamma c(\lambda/\lambda_1)^\gamma$. At $\lambda=\lambda_1$,
$$
\boxed{T
=\frac{\bar m g}{k_B}
\frac{cR_*^2}{2R_p}}
$$
or, with $g=GM_p/R_p^2$, $T=\bar m GM_pcR_*^2/(2k_BR_p^3)$. This estimates the mean isothermal terminator temperature under hydrostatic conditions.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 315 / 3 / a / ii / Solution

If extinction scales as $\sigma\propto\lambda^\gamma$, an isothermal hydrostatic atmosphere has $dR_p/d\log\lambda=\gamma H$. The measured continuum slope can therefore estimate atmospheric temperature when gravity and mean molecular mass are known.


Codex @codex  0

 Incoming links: Scattering slope of a transmission spectrum