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(D) The gravitational force provides the centripetal force: $m_2 \omega^2 r = \frac{G m_1 m_2}{r^2}$.This simplifies to $\omega = \sqrt{\frac{G m_1}{r

Vedclass · Accepted Answer

$-\frac{2\gamma}{m_2}$

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(D) The gravitational force provides the centripetal force: $m_2 \omega^2 r = \frac{G m_1 m_2}{r^2}$.This simplifies to $\omega = \sqrt{\frac{G m_1}{r^3}}$.The angular momentum $L$ of the lighter star is $L = m_2 \omega r^2 = m_2 \sqrt{G m_1 r}$.Since there are no external torques,$L$ is constant: $L^2 = m_2^2 G m_1 r = 	ext{constant}$.Taking the natural logarithm on both sides: $2 \ln m_2 + \ln G + \ln m_1 + \ln r = 	ext{constant}$.Differentiating with respect to time $t$: $2 \frac{1}{m_2} \frac{dm_2}{dt} + \frac{1}{m_1} \frac{dm_1}{dt} + \frac{1}{r} \frac{dr}{dt} = 0$.Given $\frac{dm_2}{dt} = -\gamma$ and $\frac{dm_1}{dt} = \gamma$,we have: $2 \frac{(-\gamma)}{m_2} + \frac{\gamma}{m_1} + \frac{1}{r} \frac{dr}{dt} = 0$.Since $m_1 \gg m_2$,the term $\frac{\gamma}{m_1}$ is negligible compared to $\frac{2\gamma}{m_2}$.Thus,$\frac{1}{r} \frac{dr}{dt} = \frac{2\gamma}{m_2} - \frac{\gamma}{m_1} \approx \frac{2\gamma}{m_2}$.Wait,checking the sign: $\frac{1}{r} \frac{dr}{dt} = \frac{2\gamma}{m_2} - \frac{\gamma}{m_1}$. As $m_2$ decreases,$r$ must increase to conserve angular momentum,so $\frac{dr}{dt} > 0$. The correct relative rate is $\frac{2\gamma}{m_2}$.

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