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 $\gamma$ 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 $\frac{1}{r} \frac{dr}{dt}$ (in $s^{-1}$) is given by:

(B) $-\frac{2 \gamma}{m_{2}}$

(A) $-\frac{3 \gamma}{2 m_{2}}$

(C) $-\frac{2 \gamma}{m_{1}}$

(D) $-\frac{3 \gamma}{2 m_{1}}$

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Gravitation

Consider a star of mass

m_{2}

kg revolving in a circular orbit around another star of mass

m_{1}

kg with

m_{1} ≫ 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

\frac{1}{r} \frac{d r}{d t}

(in

s^{- 1}

) is given by:

Answer: (B)

The angular momentum of the system is conserved:

L = m_{2} ω r^{2} = m_{2} G m_{1} r

Taking the logarithm and differentiating:

ln L = ln m_{2} + \frac{1}{2} ln m_{1} + \frac{1}{2} ln r

0 = \frac{d m _{2}}{m _{2}} + \frac{1}{2} \frac{d m _{1}}{m _{1}} + \frac{1}{2} \frac{d r}{r}

Since

\frac{d m _{2}}{d t} = - γ

and

\frac{d m _{1}}{d t} = γ

, and

m_{1} ≫ m_{2}

, the second term is negligible:

\frac{1}{r} \frac{d r}{d t} \approx - \frac{2 γ}{m _{2}}

The answer is (B).

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Single correct option