A particle of mass m is under the influence of the gravitational field of a body of mass M(gg m) . The particle is moving in a circular orbit of radius ro with time period T 0 around the mass M . Then, the particle is subjected to an additional central force, corresponding to the potential energy V c (r)=m a / r 3 , where α is a positive constant of suitable dimensions and r is the distance from the center of the orbit. If the particle moves in the same circular orbit of radius ro in the combined gravitational potential due to M and V c (r) , but with a new time period T 1 , then (T 1 2 -T 0 2 / T 1 2 . is given by [0pt] [ G is the gravitational constant.]

Physics · Gravitation

JEE Advanced 2024 — Paper 2 — Question 14

A particle of mass $m$ is under the influence of the gravitational field of a body of mass $M(\gg m)$ . The particle is moving in a circular orbit of radius ro with time period $\mathrm{T}_{0}$ around the mass M . Then, the particle is subjected to an additional central force, corresponding to the potential energy $V_{c}(r)=m a / r^{3}$ , where $\alpha$ is a positive constant of suitable dimensions and $r$ is the distance from the center of the orbit. If the particle moves in the same circular orbit of radius ro in the combined gravitational potential due to $M$ and $V_{c}(r)$ , but with a new time period $T_{1}$ , then $\left(T_{1}^{2}-T_{0}^{2} / T_{1}^{2}\right.$ is given by [0pt] [ G is the gravitational constant.]

Option A:
$\frac{3 \alpha}{\mathrm{GMr}_{0}^{2}}$
Correct
Option B:
$\frac{\alpha}{2 G M r_{0}^{2}}$
Option C:
$\frac{\alpha}{\mathrm{GMr}_{0}^{2}}$
Option D:
$\frac{2 \alpha}{\mathrm{GMr}_{0}^{2}}$

Answer: A

Step-by-step solution

$\quad \frac{G M m}{r_{0}^{2}}=m \omega^{2} r_{0}$

$\omega=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}_{0}^{3}}}$

\begin{gathered} \mathrm{T}_{0}^{2}=4 \pi^{2}\left(\frac{\mathrm{r}_{0}^{3}}{\mathrm{GM}}\right) \end{gathered}

$F=-\frac{d v_{c}(r)}{d r}$ $F=\frac{3 m \alpha}{r^{4}}$

Net force $\frac{G M m}{r_{0}^{2}}-\frac{3 m \alpha}{r_{0}^{4}}=m \omega^{2} r_{0}$

$\omega^{2}=\frac{G M}{r_{0}^{3}}-\frac{3 \alpha}{r_{0}^{5}}$

We know $\omega=\frac{2 \pi}{T_{1}}$

$\frac{4 \pi^{2}}{T_{1} 2}=\frac{G M}{r_{0}^{3}}-\frac{3 \alpha}{r_{0}^{5}}$

\begin{gathered} T_{1}^{2}=\frac{4 \pi^{2}}{\frac{G M}{r_{0}^{3}}-\frac{3 \alpha}{r_{0}^{5}}} \end{gathered}

From equation (1) and (2) $\frac{\mathrm{T}_{1}^{2}-\mathrm{T}_{0}^{2}}{\mathrm{~T}_{1}^{2}}=\frac{3 \alpha}{\mathrm{GMr}_{0}^{2}}$ .

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2024
Paper: Paper 2
Subject: Physics
Chapter: Gravitation
Topic: Planetary Motion & Binary Star System (Kepler's Law)