First slide

Gravitational potential and potential energy

Question

Two bodies of masses M₁ and M₂ are placed at a distance R apart. Then at the position where the gravitational field due to them is zero, the gravitational potential is

Difficult

A

$- G \frac{\sqrt{M_{1}}}{R}$
B

$- G \frac{\sqrt{M_{2}}}{R}$
C

$- {(\sqrt{M_{1}} + \sqrt{M_{2}})}^{2} \frac{G}{R}$
D

$− {(\sqrt{M_{1}} - \sqrt{M_{2}})}^{2} \frac{G}{R}$

Solution

$\begin{array}{l} \frac{{GM}_{1}}{x^{2}} = \frac{{GM}_{2}}{(R - x)^{2}} or \frac{M_{2}}{M_{1}} x^{2} = R^{2} + x^{2} - 2 Rx \\ Let \frac{M_{2}}{M} = k \\ x^{2} (k - 1) + 2 Rx - R^{2} = 0 \\ x = - \frac{2 R + \sqrt{4 R^{2} + 4 (k - 1) R^{2}}}{2 (k - 1)} = \frac{R \sqrt{M_{1}}}{\sqrt{M_{1}} + \sqrt{M_{2}}} \\ R - x = \frac{R \sqrt{M_{2}}}{\sqrt{M_{1}} + \sqrt{M_{2}}} \end{array}$
Gravitational potential at point P is
$\begin{array}{l} - (\frac{{GM}_{1}}{x} + \frac{{GM}_{2}}{R - x}) \\ = - (\frac{{GM}_{1} (\sqrt{M_{1}} + \sqrt{M_{2}})}{R \sqrt{M_{1}}} + \frac{{GM}_{2} (\sqrt{M_{1}} + \sqrt{M_{2}})}{R \sqrt{M_{2}}}] \\ = - [\frac{G (\sqrt{M_{2}} + \sqrt{M_{1}})}{R} (\sqrt{M_{1}} + \sqrt{M_{2}})] \\ = - \frac{G {(\sqrt{M_{1}} + \sqrt{M_{2}})}^{2}}{R} \end{array}$

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