= 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.
Back to article page
