Question 436: Gravitation

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
\gg 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:

Accepted Answer

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} = 	ext{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$^{-1}$.

Answer

$-\frac{3\gamma}{2m_{2}}$

Answer

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

Answer

$-\frac{2\gamma}{m_{1}}$

Answer

$-\frac{3\gamma}{2m_{1}}$

Gravitation

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