Consider a star of mass m2 kg revolving in a circular orbit around another star of mass m1 kg with m1 \gg m2. 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 \frac1r\fracdrdt (in s−1) is given by:

Gravitation PYQ — JEE Advanced 2025 · Shift 2 · Q36 with Solution

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Correct answer: NONE OF THE GIVEN OPTIONS, IF THE INTENDED APPROXIMATION IS USED, THE ANSWER SHOULD BE $\DISPLAYSTYLE +\FRAC{2\GAMMA}{M_2}$

Given system

A lighter star of mass $m_2$ revolves around a much heavier star of mass $m_1$ , with $m_1 \gg m_2$ .

Mass is transferred from the lighter star to the heavier star at a constant rate $\gamma$ :

\frac{dm_1}{dt}=+\gamma, \qquad \frac{dm_2}{dt}=-\gamma

Since there is no mass loss from the system,

m_1+m_2=\text{constant}

We must find the relative rate of change of orbital radius $r$ .

Orbital angular momentum of the lighter star

Because $m_1 \gg m_2$ , the heavier star may be treated approximately as fixed, and the lighter star moves in a circular orbit of radius $r$ .

For circular motion,

\frac{G m_1 m_2}{r^2} = \frac{m_2 v^2}{r}

So,

v=\sqrt{\frac{Gm_1}{r}}

Hence the orbital angular momentum of the lighter star is

L = m_2 v r = m_2 \sqrt{Gm_1 r}

Conservation of angular momentum

There is no external torque on the two-star system, so orbital angular momentum is conserved:

L = m_2\sqrt{Gm_1 r} = \text{constant}

Squaring,

L^2 = G m_1 m_2^2 r = \text{constant}

Therefore,

m_1 m_2^2 r = \text{constant}

Taking logarithmic differentiation:

\frac{1}{m_1}\frac{dm_1}{dt} + 2\frac{1}{m_2}\frac{dm_2}{dt} + \frac{1}{r}\frac{dr}{dt} = 0

Thus,

\frac{1}{r}\frac{dr}{dt} = -\frac{1}{m_1}\frac{dm_1}{dt} - 2\frac{1}{m_2}\frac{dm_2}{dt}

Substitute

\frac{dm_1}{dt}=\gamma, \qquad \frac{dm_2}{dt}=-\gamma

we get

\frac{1}{r}\frac{dr}{dt} = -\frac{\gamma}{m_1} + \frac{2\gamma}{m_2}

Use $m_1 \gg m_2$

Since $m_1 \gg m_2$ , the term $\gamma/m_1$ is negligible compared to $2\gamma/m_2$ . Therefore,

\frac{1}{r}\frac{dr}{dt} \approx \frac{2\gamma}{m_2}

This is positive, meaning the orbital radius increases.

Compare with options

The derived result is

\boxed{\frac{1}{r}\frac{dr}{dt} \approx \frac{2\gamma}{m_2}}

But all given options are negative:

A: $-\dfrac{3\gamma}{2m_2}$
B: $-\dfrac{2\gamma}{m_2}$
C: $-\dfrac{2\gamma}{m_1}$
D: $-\dfrac{3\gamma}{2m_1}$

None matches the physically and mathematically derived answer.

So the stored answer appears missing/incorrect, and the option list likely has a sign error. The correct expression should be

\boxed{\frac{2\gamma}{m_2}}

under the approximation $m_1 \gg m_2$ .

(Exact expression: $\displaystyle \frac{1}{r}\frac{dr}{dt} = \frac{2\gamma}{m_2}-\frac{\gamma}{m_1}$ .)

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View written solutionFree

Correct answer: NONE OF THE GIVEN OPTIONS, IF THE INTENDED APPROXIMATION IS USED, THE ANSWER SHOULD BE $\DISPLAYSTYLE +\FRAC{2\GAMMA}{M_2}$

Given system

A lighter star of mass $m_2$ revolves around a much heavier star of mass $m_1$ , with $m_1 \gg m_2$ .

Mass is transferred from the lighter star to the heavier star at a constant rate $\gamma$ :

\frac{dm_1}{dt}=+\gamma, \qquad \frac{dm_2}{dt}=-\gamma

Since there is no mass loss from the system,

m_1+m_2=\text{constant}

We must find the relative rate of change of orbital radius $r$ .

Orbital angular momentum of the lighter star

Because $m_1 \gg m_2$ , the heavier star may be treated approximately as fixed, and the lighter star moves in a circular orbit of radius $r$ .

For circular motion,

\frac{G m_1 m_2}{r^2} = \frac{m_2 v^2}{r}

So,

v=\sqrt{\frac{Gm_1}{r}}

Hence the orbital angular momentum of the lighter star is

L = m_2 v r = m_2 \sqrt{Gm_1 r}

Conservation of angular momentum

There is no external torque on the two-star system, so orbital angular momentum is conserved:

L = m_2\sqrt{Gm_1 r} = \text{constant}

Squaring,

L^2 = G m_1 m_2^2 r = \text{constant}

Therefore,

m_1 m_2^2 r = \text{constant}

Taking logarithmic differentiation:

\frac{1}{m_1}\frac{dm_1}{dt} + 2\frac{1}{m_2}\frac{dm_2}{dt} + \frac{1}{r}\frac{dr}{dt} = 0

Thus,

\frac{1}{r}\frac{dr}{dt} = -\frac{1}{m_1}\frac{dm_1}{dt} - 2\frac{1}{m_2}\frac{dm_2}{dt}

Substitute

\frac{dm_1}{dt}=\gamma, \qquad \frac{dm_2}{dt}=-\gamma

we get

\frac{1}{r}\frac{dr}{dt} = -\frac{\gamma}{m_1} + \frac{2\gamma}{m_2}

Use $m_1 \gg m_2$

Since $m_1 \gg m_2$ , the term $\gamma/m_1$ is negligible compared to $2\gamma/m_2$ . Therefore,

\frac{1}{r}\frac{dr}{dt} \approx \frac{2\gamma}{m_2}

This is positive, meaning the orbital radius increases.

Compare with options

The derived result is

\boxed{\frac{1}{r}\frac{dr}{dt} \approx \frac{2\gamma}{m_2}}

But all given options are negative:

A: $-\dfrac{3\gamma}{2m_2}$
B: $-\dfrac{2\gamma}{m_2}$
C: $-\dfrac{2\gamma}{m_1}$
D: $-\dfrac{3\gamma}{2m_1}$

None matches the physically and mathematically derived answer.

So the stored answer appears missing/incorrect, and the option list likely has a sign error. The correct expression should be

\boxed{\frac{2\gamma}{m_2}}

under the approximation $m_1 \gg m_2$ .

(Exact expression: $\displaystyle \frac{1}{r}\frac{dr}{dt} = \frac{2\gamma}{m_2}-\frac{\gamma}{m_1}$ .)

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