NEET 2023 — Physics solution

NEET2023Physics

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 $\frac{3 π}{G d}$ represents:

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

Solution

For a satellite in a low-earth orbit, $T = 2 π \frac{R ^{3}}{GM}$ with $M = \frac{4}{3} π R^{3} d$ .

$T^{2} = \frac{4 π ^{2} R ^{3}}{G \cdot \frac{4}{3} π R ^{3} d} = \frac{3 π}{G d}$ .

So $\frac{3 π}{G d}$ has the dimensions of $T^{2}$ .

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