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If $\rho$ is the density of the planet, the time period of nearby satellite is given by
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Verified Answer
The correct answer is:
$\sqrt{\frac{3 \pi}{G \rho}}$
Time period of nearby satellite
$$
\begin{aligned}
T &=2 \pi \sqrt{\frac{r^{3}}{G M}} \\
&=2 \pi \sqrt{\frac{R^{3}}{G M}} \\
&=\frac{2 \pi\left(R^{3}\right)^{1 / 2}}{\left[G \cdot \frac{4}{3} \pi R^{3} \rho\right]^{1 / 2}} \\
&=\sqrt{\frac{3 \pi}{G \rho}}
\end{aligned}
$$
$$
\begin{aligned}
T &=2 \pi \sqrt{\frac{r^{3}}{G M}} \\
&=2 \pi \sqrt{\frac{R^{3}}{G M}} \\
&=\frac{2 \pi\left(R^{3}\right)^{1 / 2}}{\left[G \cdot \frac{4}{3} \pi R^{3} \rho\right]^{1 / 2}} \\
&=\sqrt{\frac{3 \pi}{G \rho}}
\end{aligned}
$$
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