Gravitation

JEE Main level, single correct, medium

38 of 223

Question 38 of 223

A test particle is moving in a circular orbit in the gravitational field produced by a mass density ρ(r)=Kr2.\rho \left(r\right)=\dfrac{K}{{r}^{2}}. Identify the current relation between the radius RR of the particle’s orbit and its period T:T:

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Answer: (C) TR\dfrac{T}{R} is a constant

M=∫ρ4πr2drM=\int \rho 4\pi {r}^{2}dr
=∫0RKr24πr2dr={\int }_{0}^{R}\dfrac{K}{{r}^{2}}4\pi {r}^{2}dr
=4πKR=4\pi KR
Satellite velocity =GMR=4πK=\sqrt{\dfrac{GM}{R}}=\sqrt{4\pi K}
Time period, T=2πRv=2πR4πKT=\dfrac{2\pi R}{v}=\dfrac{2\pi R}{\sqrt{4\pi K}}
Now,
TR=2π4πK=\dfrac{T}{R}=\dfrac{2\pi }{\sqrt{4\pi K}}= constant

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