Kelvi

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

hard

Gravitation

2023

physics

$T^2$

$T^3$

$\sqrt{T}$

Explanation

To solve this problem, we need to find the relationship between the orbital period $T$ of a satellite orbiting just above the Earth's surface and the given expression $\frac{3Gd}{\pi}. \\$ First, recall the formula for the orbital period $T$ of a satellite: $\\ T = 2\pi \sqrt{\frac{R^3}{GM}} \\$

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A satellite is orbiting just above the surface of the earth with period T. T. T. If d d d is the density of the earth and G G G is the universal constant of gravitation, the quantity 3Gdπ \frac{3Gd}{\pi} π3Gd​ represents:

Explanation

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