Gravitation Potential Energy — Question 8

Question 8

Escape of two free bodies depends on relative energy

Two isolated point masses $M$ and $m$ separate radially at distance $r$ with relative speed $v_{rel}$. Use reduced mass $\mu=Mm/(M+m)$ and internal energy $E_{int}=\mu v_{rel}^2/2-GMm/r$.

1. Find the threshold relative speed for unbounded separation.

2. Explain why giving both masses the same additional velocity cannot turn a bound relative motion into an escaping one.

Question 8 - Solution

Escape of two free bodies depends on relative energy

(a) Derivation and calculation

For radial separation, unbounded escape requires $E_{\mathrm{int}}\ge0$. At threshold, $$\frac12\mu v_{\mathrm{rel}}^2=\frac{GMm}r,\qquad \mu=\frac{Mm}{M+m}.$$ Therefore $$\boxed{v_{\mathrm{rel,esc}}=\sqrt{\frac{2G(M+m)}r}}.$$

(b) Result and interpretation

Adding the same velocity $\vec V$ to both bodies gives $$\vec v'_{\mathrm{rel}}=(\vec v_m+\vec V)-(\vec v_M+\vec V)=\vec v_{\mathrm{rel}}.$$ The total kinetic energy separates as $$K=\frac12(M+m)V_{\mathrm{CM}}^2+\frac12\mu v_{\mathrm{rel}}^2.$$ Only the first term changes under a common boost; internal energy and the bound/escaping classification stay unchanged.

Physical check

When $M\gg m$, the relative escape speed approaches the usual fixed-source value $\sqrt{2GM/r}$.