Assume that both objects emit as <blackbodies>, take $T_*=5778\ {\rm K}$ and $R_p/R_*\simeq R_J/R_\odot=0.100$, and use $T_p=1250\ {\rm K}$ because the line-free window sees the $1$-bar continuum photosphere. Across a narrow bin, the <thermal eclipse depth> is
$$
\frac{F_p}{F_*}\simeq
\left(\frac{R_p}{R_*}\right)^2
\frac{B_\lambda(T_p)}{B_\lambda(T_*)}
=\left(\frac{R_p}{R_*}\right)^2
\frac{e^{hc/(\lambda k_BT_*)}-1}
{e^{hc/(\lambda k_BT_p)}-1}.
$$
At $\lambda=17\,\mu{\rm m}$ this gives
$$
\boxed{\frac{F_p}{F_*}\simeq1.64\times10^{-3}
\simeq1640\ {\rm ppm}}.
$$
Integrating the <Planck law> over the full $1\,\mu{\rm m}$ bin changes this narrow-bin estimate only slightly.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-315/1/c/solution