A test particle is moving in a circular orbit in the gravitational field produced by a mass density \rho (r) = K \over r^2 . Identify the correct relation between the radius R of the particle's orbit and its period T

Gravitation PYQ — JEE Main 2019 · 9 Apr · Shift 2 · Q53 with Solution

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Correct answer: D

The spherically symmetric mass density is

\rho(r)=\frac{K}{r^2}.

We need the relation between orbital radius $R$ and time period $T$ for a test particle in a circular orbit.

For a spherically symmetric distribution, only the mass inside radius $R$ contributes to the gravitational field at $R$ .

So,

M(R)=\int_0^R 4\pi r^2 \rho(r)\,dr.

Substitute $\rho(r)=K/r^2$ :

M(R)=\int_0^R 4\pi r^2 \left(\frac{K}{r^2}\right)dr =\int_0^R 4\pi K\,dr =4\pi K R.

Thus,

M(R)\propto R.

The gravitational force on a particle of mass $m$ at radius $R$ is

F=\frac{G M(R)m}{R^2}.

Using $M(R)=4\pi K R$ ,

F=\frac{G(4\pi K R)m}{R^2} =\frac{4\pi G K m}{R}.

For circular orbit,

\frac{mv^2}{R}=\frac{4\pi G K m}{R}.

Cancel $m$ and $R$ :

v^2=4\pi G K.

So $v$ is a constant, independent of $R$ .

For circular motion,

T=\frac{2\pi R}{v}.

Since $v$ is constant,

T\propto R.

Therefore,

\frac{T}{R}=\text{constant}.

A: $\dfrac{T^2}{R^3}=\text{constant}$
False, because here $T\propto R$ , so $T^2/R^3 \propto 1/R$ .
B: $TR=\text{constant}$
False, because $T\propto R$ , so $TR\propto R^2$ .
C: $\dfrac{T}{R^2}=\text{constant}$
False, because $T/R^2\propto 1/R$ .
D: $\dfrac{T}{R}=\text{constant}$
True.

The correct relation is

\boxed{\frac{T}{R}=\text{constant}}

so the correct option is D.