SATHEE: Gravitation Question 230

Gravitation Question 230

Question: Two bodies of masses $[M_{1}]$ and $[M_{2}]$ are placed at a distance $[R]$ apart. Then at the position where the gravitational field due to them is zero, the gravitational potential is

Options:

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

Show Answer

Answer:

Correct Answer: C

Solution:

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

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