Assume full heat redistribution, so
$$
T_{\rm eq}=T_*\sqrt{\frac{R_*}{2a_{\rm orb}}}
(1-A_B)^{1/4}.
$$
For $T_*=5778\,\mathrm K$, $T_{\rm eq}=600\,\mathrm K$, and <Bond albedo> $A_B=0.5$,
$$
\frac{a_{\rm orb}}{R_*}
=\frac12\left(\frac{T_*}{T_{\rm eq}}\right)^2
\sqrt{1-A_B}\simeq32.8.
$$
Taking $R_p=R_{\rm Nep}=0.0354R_*$ gives
$$
\left(\frac{R_p}{a_{\rm orb}}\right)^2
\simeq1.17\times10^{-6}.
$$
The reported $50\,\mathrm{ppm}$ visible eclipse would therefore imply
$$
\boxed{A_g\simeq\frac{50\times10^{-6}}{1.17\times10^{-6}}simeq43},
$$
which 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> is
$$
\left(\frac{R_p}{R_*}\right)^2
\frac{e^{hc/(\lambda k_BT_*)}-1}
{e^{hc/(\lambda k_BT_p)}-1}.
$$
With $T_p=600\,\mathrm K$ this is
$$
\boxed{F_p/F_*\simeq7.2\times10^{-5}\simeq72\,\mathrm{ppm}}.
$$

Solved by gpt-5.6-sol high.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-315/2/a/ii/solution