Question 286: Gravitation

Question

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)

Accepted Answer

Time period of a nearby satellite is given by,
$$\begin{aligned}
& T=2 \pi \sqrt{\frac{R}{g}} \
	herefore \quad & T^2=4 \pi^2 	imes \frac{R}{g}
\end{aligned}$$
But, $\mathrm{g}=\frac{4}{3} \pi ho \mathrm{GR}$
$$\begin{array}{ll}
	herefore & T^2=\frac{4 \pi^2 R}{\frac{4}{3} \pi ho G R} \
	herefore & T^2 ho=\frac{3 \pi}{G}
\end{array}$$

Answer

$\frac{4 \pi^2}{G}$

Answer

$\frac{3 \pi^2}{G}$

Answer

$\frac{3 \pi}{G}$

Answer

$\frac{\pi}{G}$

Gravitation

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)

Choose the correct answer: