To determine the relative rate of change of the separation between the two stars, let's first understand the relationship governing their motion.

Equilibrium of Forces:

$ m_2 \omega^2 r = \frac{G m_1 m_2}{r^2} $

Here, $ \omega $ is the angular velocity. From the balance of gravitational force and the centripetal force, we can solve for $ \omega $:

$ \omega = \sqrt{\frac{G m_1}{r^3}} $

Angular Momentum (L):

$ L = m_2 \omega r^2 = m_2 \sqrt{\frac{G m_1}{r^3}} r^2 $

Simplifying, we find:

$ L = m_2 \sqrt{G m_1 r} = \text{constant} $

Differential Relation:

Taking the natural logarithm of the angular momentum expression:

$ \ln L = \ln m_2 + \ln G + \frac{1}{2} \ln m_1 + \frac{1}{2} \ln r $

Since $ L $ is constant, differentiate both sides:

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

Rate of Change of Separation (r):

Solving for $\frac{dr}{r}$, we arrive at:

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

Since mass is transferred at a constant rate $ \gamma $, we have $ \frac{dm_1}{dt} = \gamma $ and $ \frac{dm_2}{dt} = - \gamma $.

Substitute the Variable Rates:

Substituting these values:

$ \frac{1}{r} \frac{dr}{dt} = -\frac{2(-\gamma)}{m_2} - \frac{\gamma}{m_1} \approx -\frac{2\gamma}{m_2} $

Thus, the relative rate of change of the separation between the stars is approximately $-\frac{2\gamma}{m_2}$ s$^{-1}$.