USAAAO 2022, I кръг — Задача 23

• D

This is the equilibrium temperature assuming zero albedo and full redistribution.

$L_{\star}/(4\pi d^2)\pi R_p^2 = 4\pi R_p^2\sigma T_p^4$.

First, solve for d using Kepler’s third law, assuming the planet’s mass is much smaller than the Sun:

$d(au)^3 = T(year)^2$

$d = 0.036AU$.

Solve for $T_p$ in the above expression:

$T_p = \left(\frac{L}{16\sigma\pi d^2}\right)^{1/4}$.

Plug in for $d$ and $L = 3.827 \times 10^{26}$ Watts to find the temperature:

$T_p = 1476K$

Answer: D