Q 11-07-009NEETNEET 2023Top questionMedium
A satellite is orbiting just above the surface of the earth with period $T$. If $d$ is the density of the earth and $G$ is the universal constant of gravitation, the quantity $\dfrac{3\pi}{Gd}$ represents :
Answer: (B) $T^2$
For an orbit just above the surface (radius $R$):
$$T = 2\pi\sqrt{\frac{R^3}{GM}}, \qquad M = \frac{4}{3}\pi R^3 d$$
$$T^2 = \frac{4\pi^2R^3}{G \cdot \frac{4}{3}\pi R^3 d} = \frac{3\pi}{Gd}$$
So $\dfrac{3\pi}{Gd}$ is $T^2$.
Solution by Sreeraj P, M.Sc Physics