NEET · Physics · STD 11 - 7. gravitation
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 \(\frac{3 \pi}{G d}\) represents :
- A \(\sqrt{T}\)
- B \(T\)
- C \(T^2\)
- D \(T^3\)
Answer & Solution
Correct Answer
(C) \(T^2\)
Step-by-step Solution
Detailed explanation
\(T=\frac{2 \pi}{\sqrt{G M}} r^{3 / 2} \Rightarrow T^2=\frac{4 \pi^2 R^3}{G\left(\frac{4}{3} \pi R^3 d\right)} \quad( r = R )\) \(T ^2=\frac{3 \pi}{ Gd }\)
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