Question

Consider a star of mass $m_2$ kg revolving in a circular orbit around another star of mass $m_1$ kg with $m_1 >> m_2$. The heavier star slowly acquires mass from the lighter star at a constant rate of $\gamma$ kg/s. In this transfer process, there is no other loss of mass. If the separation between the centers of the stars is $r$, then its relative rate of change $\frac{1}{r}\frac{dr}{dt}$ (in $s^{-1}$) is given by:

• A. $-\frac{3\gamma}{2m_2}$
• B. $-\frac{2\gamma}{m_2}$
• C. $-\frac{2\gamma}{m_1}$
• D. $-\frac{3\gamma}{2m_1}$

Answer 1

Answer: (B)

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

Answer 2

Assume the question intends mass transfer from the heavier star ($m_1$) to the lighter star ($m_2$) at rate $\gamma$, so $\frac{dm_1}{dt} = -\gamma$ and $\frac{dm_2}{dt} = \gamma$. Assuming the angular momentum of the lighter star orbiting the heavier star $L = m_2 \sqrt{G m_1 r}$ is conserved, differentiate $L$ with respect to time and set to zero. This leads to $\frac{1}{r}\frac{dr}{dt} = \frac{\gamma}{m_1} - \frac{2\gamma}{m_2}$. Given $m_1 >> m_2$, the term $\frac{\gamma}{m_1}$ is negligible compared to $-\frac{2\gamma}{m_2}$, so $\frac{1}{r}\frac{dr}{dt} \approx -\frac{2\gamma}{m_2}$. This matches option (B). Alternatively, assuming conservation of total angular momentum $L_{tot} = \frac{m_1 m_2}{m_1 + m_2} \sqrt{G(m_1+m_2)r}$ for mass transfer from $m_1$ to $m_2$, we get $\frac{1}{r} \frac{dr}{dt} = 2\gamma \left(\frac{1}{m_1} - \frac{1}{m_2}\right)$, which simplifies to $-\frac{2\gamma}{m_2}$ for $m_1 >> m_2$.