Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 315 / 1

Interpret every logarithm in the empirical profile as base ten with the dimensionless argument $P/(1\mathrm{mbar})$. PutThe upper atmosphere has $T=800\mathrm{K}$ for $x\le 0$, while integration of $dT/d{\log }_{10}P=Ax$ below it givesAssume that the stated $1250\mathrm{K}$ planetary equilibrium temperature is a reasonable brightness temperature at the $1\mathrm{bar}=10^3\mathrm{mbar}$ photosphere. ThenAt $10\mathrm{bar}$, $x=4$, and thereforeThis estimate neglects day-night variation, wavelength-dependent photospheric pressure, and a possible radiative-convective boundary; it treats the retrieved profile as representative of the dayside disk.

In a thin atmosphere in hydrostatic equilibrium, take gravity $g$ and mean molecular mass $\mu m_H$ as constant. The ideal gas equation of state and hydrostatic balance giveFor $P>1\mathrm{mbar}$,so the atmospheric lapse rate follows from the chain rule:It vanishes in the assumed isothermal region above $1\mathrm{mbar}$. If natural logarithms are used instead, the factor $\ln 10$ is absent and the fitted numerical value of $A$ changes accordingly.

Assume that both objects emit as blackbodies, take $T_{\ast }=5778\mathrm{K}$ and $R_p/R_{\ast }\simeq R_J/R_{\odot }=0.100$, and use $T_p=1250\mathrm{K}$ because the line-free window sees the $1$-bar continuum photosphere. Across a narrow bin, the thermal eclipse depth isAt $\lambda =17\mu \mathrm{m}$ this givesIntegrating the Planck law over the full $1\mu \mathrm{m}$ bin changes this narrow-bin estimate only slightly.

For $R_p=R_{\oplus }$, $R_{\ast }=0.1R_{\odot }$, $T_p=600\mathrm{K}$, and $T_{\ast }=3500\mathrm{K}$, the same thermal eclipse depth at $17\mu \mathrm{m}$ isThus a $100\mathrm{ppm}$ uncertainty givesThis assumes one independent eclipse measurement with that precision, negligible reflected light in the infrared, no atmosphere, and blackbody emission from the bare surface.