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) $T^2$
Concept. Near-surface orbit: $T = 2\pi\sqrt{\dfrac{R^3}{GM}}$ with $M = \dfrac43\pi R^3d$.
Step 1. Square.
$$T^2 = \frac{4\pi^2R^3}{G \cdot \frac43\pi R^3d} = \frac{3\pi}{Gd}$$
Answer. $T^2$ (Option B).
Solution by Sreeraj P, M.Sc Physics