Question 10.8: Escape from a free gravitating body Two particles P1 and P2,...
Escape from a free gravitating body
Two particles P_{1} \text { and } P_{2}, with masses m_{1} \text { and } m_{2} , can move freely under their mutual gravitation. Initially both particles are at rest and separated by a distance c. With what speed must P_{1} be projected so as to escape from P_{2}?
Since this is a mutual gravitation problem, we take our rule in the form: The motion of P_{1} \text { relative to } P_{2} \text { is the same as if } P_{2} were held fixed and the constant of gravitation G replaced by G’, where
G^{\prime}=\left(\frac{m_{1}+m_{2}}{m_{2}}\right) G.
From the one-body theory in Chapter 7, we know that P_{1} will escape from a fixed P_{2} if it has positive energy, that is if
\frac{1}{2} m_{1} V^{2}-\frac{m_{1} m_{2} G}{c} \geq 0.
Hence, when P_{2} is not fixed, P_{1} will escape if
\frac{1}{2} m_{1} V^{2}-\frac{m_{1} m_{2} G^{\prime}}{c} \geq 0,
that is, if
V^{2} \geq \frac{2\left(m_{1}+m_{2}\right) G}{c}.
This is the required escape condition.
