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PhyHardMarks to AllAdv 2025 • Paper 2

Relative Rate of Separation Change in Mass Transfer Binary Star System

Consider a star of mass m2 kgm_2\text{ kg} revolving in a circular orbit around another star of mass m1 kgm_1\text{ kg} with m1≫m2m_1 \gg m_2. The heavier star slowly acquires mass from the lighter star at a constant rate of γ kg/s\gamma\text{ kg/s}. In this transfer process, there is no other loss of mass. If the separation between the centers of the stars is rr, then its relative rate of change 1rdrdt\frac{1}{r}\frac{dr}{dt} (in s−1\text{s}^{-1}) is given by:

Official Notice: Marks Awarded to All

This question was dropped / full marks were awarded to all candidates in the official answer key by the exam conducting body due to an ambiguity or error in the question or options.

Options

A

−3γ2m2-\frac{3\gamma}{2m_2}

B

−2γm2-\frac{2\gamma}{m_2}

C

−2γm1-\frac{2\gamma}{m_1}

D

−3γ2m1-\frac{3\gamma}{2m_1}

Topics & Concepts

GravitationGravitation

Step-by-Step Solution

To determine the relative rate of change of the separation rr between the two stars, we analyze the dynamics of mass transfer in a binary star system.

1. Binary System Framework

Consider two stars of masses m1m_1 and m2m_2 separated by a distance rr, where m1≫m2m_1 \gg m_2. The total mass of the system is: M=m1+m2M = m_1 + m_2

Mass is transferred slowly from the lighter star m2m_2 to the heavier star m1m_1 at a constant rate γ kg/s\gamma \text{ kg/s}, with no total mass loss from the system: dm1dt=γ,dm2dt=−γ,dMdt=0\frac{dm_1}{dt} = \gamma, \quad \frac{dm_2}{dt} = -\gamma, \quad \frac{dM}{dt} = 0

For a circular orbit, the orbital angular frequency ω\omega is governed by Kepler's Third Law: ω=GMr3\omega = \sqrt{\frac{GM}{r^3}}

The reduced mass of the system is: μ=m1m2m1+m2=m1m2M\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{m_1 m_2}{M}

The total orbital angular momentum JJ of the binary system is: J=μr2ω=m1m2Mr2GMr3=m1m2MGrJ = \mu r^2 \omega = \frac{m_1 m_2}{M} r^2 \sqrt{\frac{GM}{r^3}} = \frac{m_1 m_2}{\sqrt{M}} \sqrt{Gr}

2. Case A: Conservative Mass Transfer (Total Angular Momentum Conserved)

Assuming that total orbital angular momentum JJ is conserved during the mass transfer process (dJdt=0\frac{dJ}{dt} = 0), we take the natural logarithm of JJ: ln⁡J=ln⁡m1+ln⁡m2−12ln⁡M+12ln⁡G+12ln⁡r\ln J = \ln m_1 + \ln m_2 - \frac{1}{2}\ln M + \frac{1}{2}\ln G + \frac{1}{2}\ln r

Differentiating with respect to time tt: 1JdJdt=1m1dm1dt+1m2dm2dt−12MdMdt+12rdrdt\frac{1}{J}\frac{dJ}{dt} = \frac{1}{m_1}\frac{dm_1}{dt} + \frac{1}{m_2}\frac{dm_2}{dt} - \frac{1}{2M}\frac{dM}{dt} + \frac{1}{2r}\frac{dr}{dt}

Substituting dJdt=0\frac{dJ}{dt} = 0, dMdt=0\frac{dM}{dt} = 0, dm1dt=γ\frac{dm_1}{dt} = \gamma, and dm2dt=−γ\frac{dm_2}{dt} = -\gamma: 0=γm1−γm2+12rdrdt0 = \frac{\gamma}{m_1} - \frac{\gamma}{m_2} + \frac{1}{2r}\frac{dr}{dt}

Rearranging to solve for the relative rate of change of separation 1rdrdt\frac{1}{r}\frac{dr}{dt}: 1rdrdt=2γ(1m2−1m1)\frac{1}{r}\frac{dr}{dt} = 2\gamma \left( \frac{1}{m_2} - \frac{1}{m_1} \right)

Since m1≫m2m_1 \gg m_2, we have 1m1≪1m2\frac{1}{m_1} \ll \frac{1}{m_2}, yielding: 1rdrdt≈+2γm2\frac{1}{r}\frac{dr}{dt} \approx +\frac{2\gamma}{m_2}

3. Case B: Specific Angular Momentum Loss Model

If the transferred mass leaves star m2m_2 carrying the specific orbital angular momentum of star m2m_2, the rate of change of orbital angular momentum is: dJdt=dm2dt(r2v2)=−γGm1r\frac{dJ}{dt} = \frac{dm_2}{dt} (r_2 v_2) = -\gamma \sqrt{G m_1 r}

Since J≈m2Gm1rJ \approx m_2 \sqrt{G m_1 r}, differentiating JJ gives: dJdt=dm2dtGm1r+m2ddt(Gm1r)\frac{dJ}{dt} = \frac{dm_2}{dt}\sqrt{G m_1 r} + m_2 \frac{d}{dt}\left(\sqrt{G m_1 r}\right) −γGm1r=−γGm1r+m2Gm1r(12m1dm1dt+12rdrdt)-\gamma \sqrt{G m_1 r} = -\gamma \sqrt{G m_1 r} + m_2 \sqrt{G m_1 r} \left( \frac{1}{2m_1}\frac{dm_1}{dt} + \frac{1}{2r}\frac{dr}{dt} \right)

Simplifying: 0=γ2m1+12rdrdt  ⟹  1rdrdt=−γm10 = \frac{\gamma}{2m_1} + \frac{1}{2r}\frac{dr}{dt} \implies \frac{1}{r}\frac{dr}{dt} = -\frac{\gamma}{m_1}

Conclusion

Because the mechanism for angular momentum loss/transfer is not uniquely specified in the problem statement, different standard models yield different expressions (e.g., +2γm2+\frac{2\gamma}{m_2} vs. −γm1-\frac{\gamma}{m_1}). None of the options unambiguously capture all assumptions without ambiguity.

Therefore, the official decision for this question is MARKS TO ALL.

Relative Rate of Separation Change in Mass Transfer Binary Star System | Physics PYQ Solution - JEE Challenger