Solution 8: Two Particle Bound State
1. Energy in Reduced Mass Frame:
Reduced mass $\mu = m/2$. Relative velocity $v_{rel} = v_0$. \[ K_{rel} = \frac{1}{2} \mu v_0^2 = \frac{1}{2} \left(\frac{m}{2}\right) v_0^2 = \frac{mv_0^2}{4} \]
Reduced mass $\mu = m/2$. Relative velocity $v_{rel} = v_0$. \[ K_{rel} = \frac{1}{2} \mu v_0^2 = \frac{1}{2} \left(\frac{m}{2}\right) v_0^2 = \frac{mv_0^2}{4} \]
2. Bound Condition ($E < 0$):
\[ E = K_{rel} + U < 0 \] \[ \frac{mv_0^2}{4} - \frac{Gm^2}{r_0} < 0 \] \[ \frac{v_0^2}{4} < \frac{Gm}{r_0} \implies v_0 < 2\sqrt{\frac{Gm}{r_0}} \]
\[ E = K_{rel} + U < 0 \] \[ \frac{mv_0^2}{4} - \frac{Gm^2}{r_0} < 0 \] \[ \frac{v_0^2}{4} < \frac{Gm}{r_0} \implies v_0 < 2\sqrt{\frac{Gm}{r_0}} \]
Answer: (d)