Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

$\sqrt{T}$

$T$

$T^2$

$T^3$

Step-by-Step Solution

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: $T^2 = \frac{4\pi^2 R^3}{GM}$
Substitute Mass with Density: Assuming the Earth is a sphere of uniform density $d$ , mass $M = \text{Volume} \times \text{Density} = \frac{4}{3}\pi R^3 d$ .
Derivation: Substitute the expression for $M$ into the $T^2$ equation: $T^2 = \frac{4\pi^2 R^3}{G \left( \frac{4}{3}\pi R^3 d \right)}$ $T^2 = \frac{4\pi^2 R^3}{\frac{4}{3}\pi G R^3 d}$ 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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