Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 315 / 2 / b

Let ${\tau }_1$ and ${\tau }_2$ be the top-down optical depths at $P_1$ and $P_2$, and let $\mu$ be the outward direction cosine. The formal solution of the radiative transfer equation givesFor a semi-infinite lower layer in local thermodynamic equilibrium, $I_{\nu ,0}=B_{\nu }(T_3)$.

DefineThe emergent planetary surface flux isHence the band-centre planet-star ratio isIf $T_1=T_2=T_3=T_p$, the weights telescope to $1/2$, so $F_{p,\nu }=\pi B_{\nu }(T_p)$ and the expression reduces to the blackbody thermal eclipse depth, about $72\mathrm{ppm}$ for $T_p=600\mathrm{K}$.

Solved by gpt-5.6-sol high.

At $20\mu \mathrm{m}$,The normalized flux weights of the $400$, $600$, and $800\mathrm{K}$ layers are thereforeUsing the Planck law at the band centre and $R_p/R_{\ast }=0.0354$ givesThis is only slightly larger than the $71.7\mathrm{ppm}$ isothermal $600\mathrm{K}$ value because the cool upper-layer suppression and hot deep-layer enhancement nearly cancel in this broad weighting.

Solved by gpt-5.6-sol high.