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…

• System: Binary star system with masses \({m}_{1}\)and\({m}_{2}\) (\({m}_{1}≫{m}_{2}\)).
• Process: Mass transfer from lighter star (\({m}_{2}\)) to heavier star (\({m}_{1}\)) at a rate \(\gamma\).
  • \({\overset{˙}{m}}_{1}=\gamma\)
  • \({\overset{˙}{m}}_{2}=−\gamma\)
  • Total mass \(M={m}_{1}+{m}_{2}\)is constant.
• Principle: Conservation of orbital angular momentum \(L\).

Step 1: Write the expression for Angular Momentum. The orbital angular momentum \(L\) for a binary system is given by: \(L=\mu \sqrt{GMr}\) where \(\mu =\frac{{m}_{1}{m}_{2}}{{m}_{1}+{m}_{2}}\) is the reduced mass, \(M\) is the total mass, and \(r\) is the separation.

Step 2: Differentiate with respect to time. Taking the natural logarithm of both sides: \(\text{ln}L=\text{ln}\mu +\frac{1}{2}\text{ln}\left(GM\right)+\frac{1}{2}\text{ln}r\) Since \(L\),\(G\), and \(M\) are constant, their derivatives are zero. Differentiating with respect to time \(t\) : \(0=\frac{1}{\mu }\frac{d\mu }{dt}+\frac{1}{2r}\frac{dr}{dt}\)Rearranging to solve for the relative rate of change of r : \(\frac{1}{r}\frac{dr}{dt}=−2\frac{1}{\mu }\frac{d\mu }{dt}\)

Step 3: Calculate \(\frac{1}{\mu }\frac{d\mu }{dt}\). Using the definition of reduced mass \(\mu =\frac{{m}_{1}{m}_{2}}{M}\): \(\text{ln}\mu =\text{ln}{m}_{1}+\text{ln}{m}_{2}−\text{ln}M\) \(\frac{1}{\mu }\frac{d\mu }{dt}=\frac{{\overset{˙}{m}}_{1}}{{m}_{1}}+\frac{{\overset{˙}{m}}_{2}}{{m}_{2}}\)Substitute \({\overset{˙}{m}}_{1}=\gamma\) and \({\overset{˙}{m}}_{2}=−\gamma\): \(\frac{1}{\mu }\frac{d\mu }{dt}=\frac{\gamma }{{m}_{1}}−\frac{\gamma }{{m}_{2}}=\gamma \left(\frac{1}{{m}_{1}}−\frac{1}{{m}_{2}}\right)\)

Step 4: Apply the approximation \({m}_{1}≫{m}_{2}\). Since \({m}_{1}\) is much larger than \({m}_{2}\), the term \(\frac{1}{{m}_{1}}\)is negligible compared to \(\frac{1}{{m}_{2}}.\frac{1}{\mu }\frac{d\mu }{dt}\approx \gamma \left(0−\frac{1}{{m}_{2}}\right)=−\frac{\gamma }{{m}_{2}}\)

Step 5: Final Result. Substitute this back into the equation for \(\frac{1}{r}\frac{dr}{dt}:\frac{1}{r}\frac{dr}{dt}=−2\left(−\frac{\gamma }{{m}_{2}}\right)=\frac{2\gamma }{{m}_{2}}\)

Note: While the derived physical result is positive (indicating orbit expansion), the options provided in the question all have a negative sign. This suggests a sign convention difference or a typo in the question statement regarding the direction of mass flow (usually Heavy \(\to\) Light causes shrinking/negative rate). However, the magnitude and dependency on \({m}_{2}\) clearly point to Option (B).

Answer: (B) \(−\frac{2\gamma }{{m}_{2}}\)