A satellite is orbiting just above the surface of the earth with period . If is the density of the earth and is the universal constant of gravitation, the quantity represents: [2023]

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Solution

Q.
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: [2023]

1 $T^{3}$

2 $\sqrt{T}$

3 $T$

4 $T^{2}$

Ans.
(4)

Let the planet is moving above the earth surface at $R$ distance.

$T = \frac{2 π R}{v}$ ...(i)

As, the centripetal force of satellite is due to gravitational force of earth.

$\frac{G M m}{R^{2}} = \frac{m v^{2}}{R} \Rightarrow \frac{G M}{R} = v^{2} or v = \sqrt{\frac{G M}{R}}$

Using (i), $T = \frac{2 π R \sqrt{R}}{\sqrt{G M}}$

$T^{2} = \frac{4 π^{2} R^{3}}{G (d \times \frac{4}{3} π R^{3})} = \frac{3 π}{G d} . Hence, \frac{3 π}{G d} = T^{2}$

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