Consider a star of mass m 2 kg revolving in a circular orbit around another star of mass m 1 kg with m 1 gg m 2 . The heavier star slowly acquires mass from the lighter star at a constant rate of γ kg / s . In this transfer process, there is no other loss of mass. If the separation between the centers of the stars is r , then its relative rate of change 1/r frac d r d t (in s -1 ) is given by

Physics · Gravitation

JEE Advanced 2025 — Paper 2 — Question 4

Consider a star of mass $m_{2} \mathrm{~kg}$ revolving in a circular orbit around another star of mass $m_{1} \mathrm{~kg}$ with $m_{1} \gg m_{2}$ . The heavier star slowly acquires mass from the lighter star

at a constant rate of $\gamma \mathrm{kg} / \mathrm{s}$ . In this transfer process, there is no other loss of mass. If the separation between the centers of the stars is $r$ , then its relative rate of change $\frac{1}{r} \frac{\mathrm{~d} r}{\mathrm{~d} t}$ (in $\mathrm{s}^{-1}$ ) is given by

Option A:
$-\frac{3 \gamma}{2 m_{2}}$
Option B:
$-\frac{2 \gamma}{m_{2}}$
Correct
Option C:
$-\frac{2 \gamma}{m_{1}}$
Option D:
$-\frac{3 \gamma}{2 m_{1}}$

Answer: B

Step-by-step solution

$m_{2} \omega^{2} r=\frac{G m_{1} m_{2}}{r^{2}}$

$\omega=\sqrt{\frac{G m_{1}}{r^{3}}}$

$\mathrm{L}=\mathrm{m}_{2} \omega \mathrm{r}^{2}$

$=m_{2} \sqrt{\frac{G m_{1}}{r^{3}}} r^{2}$

$\mathrm{L}=\mathrm{m}_{2} \sqrt{\mathrm{Gm}_{1} \mathrm{r}}=$ const . $\ell \mathrm{nL}=\ell \mathrm{nm}_{2}+\ell \mathrm{nG}+\frac{1}{2} \ell \mathrm{~nm}_{1}+\frac{1}{2} \ell \mathrm{nr}$

$0=\frac{\mathrm{dm}_{2}}{\mathrm{~m}_{2}}+\frac{1}{2} \frac{\mathrm{dm}_{1}}{\mathrm{~m}_{1}}+\frac{1}{2} \frac{\mathrm{dr}}{\mathrm{r}}$

$\frac{\mathrm{dr}}{\mathrm{rdt}}=-\frac{2 \mathrm{dm}_{2}}{\mathrm{~m}_{2} \mathrm{dt}}-\frac{\mathrm{dm}_{1}}{\mathrm{~m}_{1} \mathrm{dt}} \approx-\frac{2 \gamma}{\mathrm{~m}_{2}} \quad$ Here $\left(\frac{\mathrm{dm}}{\mathrm{dt}}=\gamma\right)$

Answer key and solution verified before publishing.

Practise Gravitation

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Advanced 2025
Paper: Paper 2
Subject: Physics
Chapter: Gravitation
Topic: Planetary Motion & Binary Star System (Kepler's Law)

A geostationary satellite above the equator is orbiting around the earth at a fixed distance r 1 from the center of the earth. A second…

Answer: B

Step-by-step solution

Practise Gravitation

More Gravitation questions from this paper