A satellite is orbiting just above the surface of the planet of density ' $\rho$ ' with periodic time ' $T$…
A satellite is orbiting just above the surface of the planet of density ' $\rho$ ' with periodic time ' $T$ '. The quantity $\mathrm{T}^2 \rho$ is equal to ( $\mathrm{G}=$ universal gravitational constant)
$\frac{4 \pi^2}{\mathrm{G}}$
$\frac{3 \pi^2}{\mathrm{G}}$
$\frac{3 \pi}{\mathrm{G}}$
$\frac{\pi}{G}$
Solution
Time period of a nearby satellite is given by,
$\begin{aligned}
& T=2 \pi \sqrt{\frac{R}{g}} \\
\therefore & T^2=4 \pi^2 \times \frac{R}{g} \\
& \text { But, } g=\frac{4}{3} \pi \rho G R . \\
\therefore & T^2=\frac{4 \pi^2 R}{\frac{4}{3} \pi \rho G R} \\
\therefore & T^2 \rho=\frac{3 \pi}{G}
\end{aligned}$