USAAAO 2026, I кръг — Задача 29

• A

Solution: Notice that if $L_{\odot}$ is the power radiated by the Sun, then the total solar power incident on the Earth is $P'_{\odot} = \frac{\pi R_{\oplus}^2}{4\pi D^2}L_{\odot}$, where $D$ is the Sun-Earth distance. On the other hand, the power incident on the Earth by CMB photons is the same power as if Earth was radiating at a temperature $T_c$ (because then the Earth would be in thermal equilibrium with the CMB and there would be no net thermal exchange), so the total CMB power incident on the Earth is $P'_c = (4\pi R_{\oplus}^2)\sigma T_c^4$. Lastly, notice that by the Wien Displacement Law, the mean/typical wavelength emitted are proportional to $T^{-1}$, meaning that the average photon energy is proportional to $T$, and so if $N$ is the numerical photon emission rate, then $P\propto NT \longrightarrow N\propto P/T$. The desired ratio is thus,

which is 51400 to three significant figures. Alternatively, using the formula $L_{\odot} = 4\pi\sigma R_{\odot}^2T_{\odot}^4$ gives the desired ratio as

which gives the same numerical result to within 3 significant figures. The factor of $R_{\odot}^2/4D^2$ in the latter solution can be interpreted as comparing the “fractions” of the sky taken up by each source. Additionally, it helps to see that the Earth poses an effective cross section of $\pi R_{\oplus}^2$ to solar radiation but all $4\pi R_{\oplus}^2$ of the Earth’s surface area uniformly receives CMB radiation, with the resulting factor of $\pi R_{\oplus}^2/4\pi R_{\oplus}^2 = 1/4$ being another way of intuiting the 4 in the denominator of our second final expression.

The numerical answer (from either method) is closest to $5\times10^{4}$,

Answer: A