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}] $
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Answer:
Correct Answer: C
Solution:
- $ [\frac{GM_{1}}{x^{2}}=\frac{GM_{2}}{{{(R-x)}^{2}}}] $ or $ [\frac{M_{2}}{M_{1}}x^{2}=R^{2}+x^{2}-2Rx] $ Let $ [\frac{M_{2}}{M}=k] $ $ [x^{2}(k-1)+2Rx-R^{2}=0] $ $ [x=-\frac{2R+\sqrt{4R^{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 $ [-\left( \frac{GM_{1}}{x}+\frac{GM_{2}}{R-x} \right)] $ $ [=-\left[ \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}}} \right]] $ $ [=-\left[ \frac{G(\sqrt{M_{2}}+\sqrt{M_{1}})}{R}(\sqrt{M_{1}}+\sqrt{M_{2}}) \right]] $ $ [=-\frac{G{{(\sqrt{M_{1}}+\sqrt{M_{2}})}^{2}}}{R}] $
BETA
