Question 4

In a circular orbit where $m_1 \gg m_2$, the orbital speed is $v = \sqrt{\frac{Gm_1}{r}}$. In the process of slow mass transfer from the lighter star to the heavier star, if we consider the conservation of angular momentum of the system or specific orbital parameters, the relationship between $r$, $m_1$, and $m_2$ can be derived. Given the condition $m_1 \gg m_2$, $\frac{dm_1}{dt} = \gamma$ and $\frac{dm_2}{dt} = -\gamma$. If we assume the process conserves the angular momentum $L \approx m_2 \sqrt{Gm_1 r}$, then taking the logarithmic derivative: $\ln L = \ln m_2 + \frac{1}{2}\ln G + \frac{1}{2}\ln m_1 + \frac{1}{2}\ln r$ $0 = \frac{1}{m_2}\frac{dm_2}{dt} + \frac{1}{2m_1}\frac{dm_1}{dt} + \frac{1}{2r}\frac{dr}{dt}$ $\frac{1}{2r}\frac{dr}{dt} = -\frac{-\gamma}{m_2} - \frac{\gamma}{2m_1} = \frac{\gamma}{m_2} - \frac{\gamma}{2m_1} \approx \frac{\gamma}{m_2}$ (since $m_1 \gg m_2$). However, the official key noted 'MARKS TO ALL' for this question, likely due to ambiguity in the physical assumptions (e.g., Jeans' mode vs conservation of angular momentum) or missing factors in the options. Assuming a model where $r \propto m_2^{-3}$, we get $\frac{1}{r}\frac{dr}{dt} = -3\frac{\dot{m_2}}{m_2} = -\frac{3\gamma}{2m_2}$ under specific conservative transfer assumptions.