Solution:
Using energy conservation in the center-of-mass frame: \(\frac{1}{2} \mu v_{\text{rel}}^2 = \frac{GMm}{d}\), where \(\mu = \frac{Mm}{M+m}\) is the reduced mass. Substituting \(\mu\) yields \(v_{\text{rel}} = \sqrt{\frac{2G(M+m)}{d}}\).
The gravitational force between two particles with masses \(m\) and \(M\), initially at rest at great separation, pulls them together. When their separation becomes \(d\), then speed of either particle relative to the other will be :
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