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

At full phase immediately before an exoplanet secondary eclipse, reflected and thermal light giveThe phase function equals one at secondary eclipse. For the stated model, the first term is constant below $1\mu \mathrm{m}$ and negligible above it. The $600\mathrm{K}$ thermal component is tiny in the visible, rises through the infrared, and peaks near $4.8\mu \mathrm{m}$ in $B_{\lambda }$, while the contrast continues to improve toward the mid-infrared because the stellar spectrum falls more rapidly.

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Assume full heat redistribution, soFor $T_{\ast }=5778\mathrm{K}$, $T_{\mathrm{eq}}=600\mathrm{K}$, and Bond albedo $A_B=0.5$,Taking $R_p=R_{\mathrm{Nep}}=0.0354R_{\ast }$ givesThe reported $50\mathrm{ppm}$ visible eclipse would therefore implywhich is impossible. At least one assumption, the measurement, or the stated system parameters must fail; even $A_g=1$ gives only about $1.2\mathrm{ppm}$.

At $20\mu \mathrm{m}$, reflected light is negligible. The thermal eclipse depth isWith $T_p=600\mathrm{K}$ this is

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A perfectly reflecting Lambertian surface receiving normal flux $F$ has radiancebecause integrating $I\cos \theta$ over the outward hemisphere returns $F$. A face-on disc of radius $R$ subtends solid angle $\pi R^2/d^2$. The observed flux is consequently

Solved by gpt-5.6-sol high.

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.