First slide

Gravitaional potential energy

Question

Two particles of mass m and M are initially at rest and infinitely separated from each other. Due to mutual interaction they approach each other. Their relative velocity of approach at a separation distance d between them is

Moderate

A

$2 G (\frac{d}{M + m})$
B

$\sqrt{[\frac{2 G (M + m)}{d}]}$
C

$\sqrt{[\frac{2 G (M - m)}{d}]}$
D

$\sqrt{(\frac{2 Gd}{M + m})}$

Solution

Here we assume that one mass is at rest and the other mass equal to reduced mass $μ [μ = Mm / (M + m)]$ is in motion. Using conservation of energy, the final energy is
$\frac{1}{2} {μv}^{2} - \frac{GMm}{d} 0$
or $V^{2} = \frac{2 GMm}{μd} = \frac{2 GMm (M + m)}{mMd}$
$= \frac{2 G (M + m)}{d}$
$V = \sqrt{[\frac{2 G (M + m)}{d}]}$

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