2023 GravitationOrbital Motion of Satellites Medium Question 11 of 66 in Gravitation

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 $\frac{3\pi}{Gd}$ represents:

A $\sqrt{T}$
B $T$
C $T^2$
D $T^3$

The answer is C.

Solution

$T=2\pi\sqrt{\frac{R^3}{GM}}$ with $M=d\cdot\frac{4}{3}\pi R^3$ gives $T^2=\frac{3\pi}{Gd}$.

Time period of satellite: $T=2\pi\sqrt{\frac{R^3}{GM}}$ $=2\pi\sqrt{\frac{R^3}{Gd\,\frac{4}{3}\pi R^3}}$ $\Rightarrow T=\sqrt{\frac{3\pi}{Gd}}$ $\Rightarrow\frac{3\pi}{Gd}=T^2$

Concept time period of near-earth satellite

Gravitation in past papers

66 questions from this chapter have appeared across 26 exam years.