Relative Velocity of Approaching Particles – Rankers Physics

Gravitational Potential Energy: Practice Problem & Solution

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 :
\(\sqrt{G(M+m)/2d}\)
\(\sqrt{G(M+m)/d}\)
\(\sqrt{4G(M+m)/d}\)
\(\sqrt{2G(M+m)/d}\)

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Gravitational Potential Energy. Understanding the underlying formula is key to arriving at the correct answer below:

By conservation of mechanical energy, the relative speed is found using the reduced mass \(\mu = \frac{mM}{m+M}\). Thus, \(\frac{1}{2} mu v_{\text{rel}}^2 = \frac{GMm}{d}\), which simplifies to \(v_{\text{rel}} = \sqrt{\frac{2G(M+m)}{d}}\).

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