Solution 9: Three Particle Bound State
1. System Analysis:
Rigid body AB ($2m$) + Free particle C ($m$). $\mu = \frac{2m \cdot m}{3m} = \frac{2}{3}m$.
Rigid body AB ($2m$) + Free particle C ($m$). $\mu = \frac{2m \cdot m}{3m} = \frac{2}{3}m$.
2. Kinetic Energy:
\[ K = \frac{1}{2} \mu v_0^2 = \frac{1}{2} \left(\frac{2}{3}m\right) v_0^2 = \frac{1}{3}mv_0^2 \]
\[ K = \frac{1}{2} \mu v_0^2 = \frac{1}{2} \left(\frac{2}{3}m\right) v_0^2 = \frac{1}{3}mv_0^2 \]
3. Potential Energy:
\[ U = -\frac{Gm^2}{r_{AC}} – \frac{Gm^2}{r_{BC}} = -\frac{2Gm^2}{r_0} \]
\[ U = -\frac{Gm^2}{r_{AC}} – \frac{Gm^2}{r_{BC}} = -\frac{2Gm^2}{r_0} \]
4. Condition:
\[ \frac{1}{3}mv_0^2 – \frac{2Gm^2}{r_0} < 0 \implies v_0 < \sqrt{\frac{6Gm}{r_0}} \]
\[ \frac{1}{3}mv_0^2 – \frac{2Gm^2}{r_0} < 0 \implies v_0 < \sqrt{\frac{6Gm}{r_0}} \]
Answer: (d)