A satellite is revolving around a planet in a circular orbit close to its surface. Let ' $\rho$ ' be the…
A satellite is revolving around a planet in a circular orbit close to its surface. Let ' $\rho$ ' be the mean density and ' $R$ ' be the radius of the planet. Then the period of the satellite is ( $\mathrm{G}=$ universal constant of gravitation)
$\sqrt{\frac{4 \pi}{\rho G}}$
$\sqrt{\frac{\pi}{\rho G}}$
$\sqrt{\frac{3 \pi}{\rho G}}$
$\sqrt{\frac{2 \pi}{\rho G}}$
Solution
From Kepler's third law,
$\begin{aligned}
& \mathrm{T}^2=\frac{4 \pi^2 \mathrm{r}^3}{\mathrm{GM}} \\
& \therefore \quad \mathrm{~T} =2 \pi \sqrt{\frac{\mathrm{r}^3}{\mathrm{GM}}}
\end{aligned}$
As the satellite is very close to the planet, $r=R$.
$\therefore \quad \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}^3}{\mathrm{GM}}}...(i)$
$\begin{aligned}
& \text {We know, } \text {Mass } =\text { Volume } \times \text { Density }(\rho) \\
& =\frac{4}{3} \pi \mathrm{R}^3 \times \rho...(ii)
\end{aligned}$
Putting (ii) into (i)
$T=2 \pi \sqrt{\frac{R^3}{G \times \frac{4}{3} \pi R^3 \rho}}=\sqrt{\frac{3 \pi}{\rho G}}$