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A planet of radius $\mathrm{R}_{\mathrm{p}}$ is revolving around a star of radius $\mathrm{R}^{*}$, which is at temperature $\mathrm{T}^{*}$. The distance between the star and the planet is $\mathrm{d}$. If the planet's temperature is $f \mathrm{~T}^{*}$, then $f$ is proportional to
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Verified Answer
The correct answer is:
$\sqrt{\mathrm{R}^{*} / \mathrm{d}}$
The planet should be in Thermal equilibrium with star.
Amount of heat energy emitted by start per second $\mathrm{E}_{1}=\left(\sigma \mathrm{T}^{* 4}\right)\left(4 \pi \mathrm{R}^{*^{2}}\right)$
$\because \mathrm{d}=$ distance between star and planet
Hence, amount of energy reaching planet per
$\begin{aligned}
\text { unit area per second } &=\frac{E_{1}}{4 \pi \mathrm{d}^{2}} \\
&=\frac{\left(\sigma \mathrm{T}^{* 4}\right)\left(4 \pi \mathrm{R}^{* 2}\right)}{4 \pi \mathrm{d}^{2}} \\
&=\frac{\sigma \mathrm{T}^{*^{4}} \mathrm{R}^{* 2}}{\mathrm{~d}^{2}}
\end{aligned}$
Hence energy received by planet per second
$\mathrm{E}_{2}=\left(\frac{\sigma \mathrm{T}^{* 4} \mathrm{R}^{* 2}}{\mathrm{~d}^{2}}\right) \pi \mathrm{R}_{\mathrm{P}}^{2}$ $\mathrm{~T}=$ Temperature of planet then amount of energy emitted by planet per second
$$
E_{3}=\left(\sigma T^{4}\right)\left(4 \pi R_{P}^{2}\right)
$$
For thermal equilibrium $E_{3}=E_{2}$
$$
\begin{array}{l}
\Rightarrow\left(\sigma \mathrm{T}^{4}\right)\left(4 \pi \mathrm{R}_{\mathrm{P}}^{2}\right)=\frac{\left(\sigma \mathrm{T}^{* 4} \mathrm{R}^{* 2}\right)}{\mathrm{d}^{2}}\left(\pi \mathrm{R}_{\mathrm{p}}^{2}\right) \\
\Rightarrow \mathrm{T}^{4}=\frac{\mathrm{T}^{* 4} \mathrm{R}^{* 2}}{4 \mathrm{~d}^{2}} \\
\Rightarrow \mathrm{T}=\mathrm{T}^{*} \sqrt{\frac{\mathrm{R}^{*}}{2 \mathrm{~d}}}=\mathrm{fT}^{*} \\
\Rightarrow \mathrm{f} \propto \sqrt{\frac{\mathrm{R}^{*}}{\mathrm{~d}}}
\end{array}
$$
Hence correct answer is (A)
Amount of heat energy emitted by start per second $\mathrm{E}_{1}=\left(\sigma \mathrm{T}^{* 4}\right)\left(4 \pi \mathrm{R}^{*^{2}}\right)$
$\because \mathrm{d}=$ distance between star and planet
Hence, amount of energy reaching planet per
$\begin{aligned}
\text { unit area per second } &=\frac{E_{1}}{4 \pi \mathrm{d}^{2}} \\
&=\frac{\left(\sigma \mathrm{T}^{* 4}\right)\left(4 \pi \mathrm{R}^{* 2}\right)}{4 \pi \mathrm{d}^{2}} \\
&=\frac{\sigma \mathrm{T}^{*^{4}} \mathrm{R}^{* 2}}{\mathrm{~d}^{2}}
\end{aligned}$
Hence energy received by planet per second
$\mathrm{E}_{2}=\left(\frac{\sigma \mathrm{T}^{* 4} \mathrm{R}^{* 2}}{\mathrm{~d}^{2}}\right) \pi \mathrm{R}_{\mathrm{P}}^{2}$ $\mathrm{~T}=$ Temperature of planet then amount of energy emitted by planet per second
$$
E_{3}=\left(\sigma T^{4}\right)\left(4 \pi R_{P}^{2}\right)
$$
For thermal equilibrium $E_{3}=E_{2}$
$$
\begin{array}{l}
\Rightarrow\left(\sigma \mathrm{T}^{4}\right)\left(4 \pi \mathrm{R}_{\mathrm{P}}^{2}\right)=\frac{\left(\sigma \mathrm{T}^{* 4} \mathrm{R}^{* 2}\right)}{\mathrm{d}^{2}}\left(\pi \mathrm{R}_{\mathrm{p}}^{2}\right) \\
\Rightarrow \mathrm{T}^{4}=\frac{\mathrm{T}^{* 4} \mathrm{R}^{* 2}}{4 \mathrm{~d}^{2}} \\
\Rightarrow \mathrm{T}=\mathrm{T}^{*} \sqrt{\frac{\mathrm{R}^{*}}{2 \mathrm{~d}}}=\mathrm{fT}^{*} \\
\Rightarrow \mathrm{f} \propto \sqrt{\frac{\mathrm{R}^{*}}{\mathrm{~d}}}
\end{array}
$$
Hence correct answer is (A)
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