# 天体物理学|PHYS1116 Astrophysics代写

The solution to the planetary tempearture comes from assuming a state of “balanced power”. The energy received per second by the planet from its host star, $\dot{E}{\text {in }}$ is balanced by the energy it radiates per second as a blackbody, $\dot{E}{\text {out }}$. If this were not so, then the temperature of the planet would rise or fall until the balance was achieved. This would take something like the heat capacity of the atmosphere divided by $E_{\text {in }}$ or in the case of the earth a few weeks.

The energy in is the cross sectional area of the planet as viewed from the star, a disk with radius $R_{p}$ intersects the radiation that would have passed througha disk of area $\pi R_{p}^{2}$ (not $2 \pi R_{p}^{2}$, then one would have to integrate $\cos \theta$ for the incident radiation, which adds work). The flux passing through each square $\mathrm{cm}$ of the disk is $L_{} /\left(4 \pi d^{2}\right)$ where $d$ is the distance from the star to the planet. Additionally it was specified that only $80 \%$ of the flux gets through to the planets surface and contributes to its warming, so $$\dot{E}{i n}=0.8 \frac{L{}}{4 \pi d^{2}}\left(\pi R_{p}^{2}\right)$$
The energy radiated depends on the temperature of the surface of the planet, $T_{p}$ – to be solved for – and the emitting area. If a rapid rotator with an atmosphere the temperature is pretty much the same wordwide so the emitting area is $4 \pi R_{p}^{2}$ and
$$\dot{E}{\text {out }}=\left(4 \pi R{p}^{2}\right)\left(\sigma T_{p}^{4}\right)$$
Setting $\dot{E}{\text {in }}=\dot{E}{\text {out }}$, we see that the planet’s radius drops out. It doesn’t matter if its a rapidly spinning baseball or an earth-sized planet, and
\begin{aligned} 0.8 \frac{L_{*}}{4 \pi d^{2}} &=(4)\left(\sigma T_{p}^{4}\right) \ T_{p} &=\left(\frac{0.8 L_{v}}{16 \pi \sigma d^{2}}\right)^{1 / 4} \ &=262 \mathrm{~K}=-11 \mathrm{C} \end{aligned}

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