A satellite is orbiting just above the surface of the earth with period $T$. If $d$ is the density of the eart

Question

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:

Accepted Answer

1. **Time Period Formula:** The time period $T$ of a satellite orbiting very close to the surface of the Earth (radius $R$) is given by: 
   $$ T = 2\pi \sqrt{\frac{R^3}{GM}} $$
   Squaring both sides: 
   $$ T^2 = \frac{4\pi^2 R^3}{GM} $$
2. **Substitute Mass with Density:** Assuming the Earth is a sphere of uniform density $d$, mass $M = 	ext{Volume} 	imes 	ext{Density} = \frac{4}{3}\pi R^3 d$.
3. **Derivation:** Substitute the expression for $M$ into the $T^2$ equation:
   $$ T^2 = \frac{4\pi^2 R^3}{G \left( \frac{4}{3}\pi R^3 d ight)} $$
   $$ T^2 = \frac{4\pi^2 R^3}{\frac{4}{3}\pi G R^3 d} $$
   Canceling $4\pi$, $R^3$ terms:
   $$ T^2 = \frac{\pi}{\frac{1}{3}Gd} = \frac{3\pi}{Gd} $$
   Thus, the quantity $\frac{3\pi}{Gd}$ represents $T^2$.

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