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

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> gives
$$
\boxed{
I_\nu(0,\mu)
=B_\nu(T_1)(1-e^{-\tau_1/\mu})
+B_\nu(T_2)(e^{-\tau_1/\mu}-e^{-\tau_2/\mu})
+I_{\nu,0}e^{-\tau_2/\mu}}.
$$
For a semi-infinite lower layer in <local thermodynamic equilibrium>, $I_{\nu,0}=B_\nu(T_3)$.

Define
$$
E_3(x)=\int_0^1\mu e^{-x/\mu}\,d\mu.
$$
The emergent planetary surface flux is
$$
F_{p,\nu}=2\pi\left[
B_1\left(\frac12-E_3(\tau_1)\right)
+B_2(E_3(\tau_1)-E_3(\tau_2))
+B_3E_3(\tau_2)\right].
$$
Hence the band-centre planet-star ratio is
$$
\boxed{
\frac{F_p}{F_*}
=\left(\frac{R_p}{R_*}\right)^2
\frac{2[B_1(1/2-E_3(\tau_1))
+B_2(E_3(\tau_1)-E_3(\tau_2))+B_3E_3(\tau_2)]}
{B_\nu(T_*)}}.
$$
If $T_1=T_2=T_3=T_p$, 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.

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

For blackbody star and planet, the monochromatic thermal eclipse depth is
$$
\frac{F_p}{F_*}=\left(\frac{R_p}{R_*}\right)^2
\frac{B_\lambda(T_p)}{B_\lambda(T_*)}.
$$


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 Incoming links: Thermal eclipse depth