A homogeneous interpretation assigns the planet a <brightness temperature> $T_b$ satisfying
$$
\boxed{
B_\nu(T_b)=fB_\nu(T_1)+(1-f)B_\nu(T_2),
\qquad f=\sin^2\theta_T}.
$$
At a fixed wavelength let $x=hc/(\lambda k_B)$ and
$$
S=\frac{f}{e^{x/T_1}-1}
+\frac{1-f}{e^{x/T_2}-1}.
$$
Inverting the <Planck law> gives
$$
\boxed{T_b=\frac{x}{\log(1+S^{-1})}}.
$$
At $\lambda=20\,\mu{\rm m}$, $x=719.4\,{\rm K}$. With $\theta_T=\pi/4$, $T_1=1500\,{\rm K}$, and $T_2=1000\,{\rm K}$, one obtains
$$
\boxed{T_b\simeq1251\,{\rm K}}.
$$
The <Rayleigh-Jeans law> would give the nearly identical area-weighted estimate $1250\,{\rm K}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 315 / 1 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-315/1/b/ii/solution