For $R_p=R_\oplus$, $R_*=0.1R_\odot$, $T_p=600\ {\rm K}$, and $T_*=3500\ {\rm K}$, the same <thermal eclipse depth> at $17\,\mu{\rm m}$ is
$$
\frac{F_p}{F_*}
=\left(\frac{R_\oplus}{0.1R_\odot}\right)^2
\frac{e^{hc/(17\mu{\rm m}\,k_B3500{\rm K})}-1}
{e^{hc/(17\mu{\rm m}\,k_B600{\rm K})}-1}
\simeq7.40\times10^{-4}.
$$
Thus a $100\ {\rm ppm}$ uncertainty gives
$$
\boxed{{\rm SNR}\simeq\frac{740\ {\rm ppm}}{100\ {\rm ppm}}\simeq7.4}.
$$
This assumes one independent eclipse measurement with that precision, negligible reflected light in the infrared, no atmosphere, and blackbody emission from the bare surface.

\Image[../../../paper-315-earth-m-dwarf-eclipse.svg]
{title=Blackbody secondary-eclipse spectrum of a 600-kelvin Earth-size planet around the stated M dwarf}
{description=The planet-star ratio is extremely small on the Wien tail at short wavelength and rises toward about 740 parts per million at 17 micrometres.}

Solution

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

 Solution

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