Gravitation Question 361
Question: Two bodies of masses $ [M_{1}andM_{2}] $ are placed at a distanced/apart. What is the potential at the position where the gravitational field due to them is zero?
Options:
A) $ [-\frac{G}{d}(M_{1}+M_{2}+2\sqrt{M_{1}}\sqrt{M_{2}})] $
B) $ [-\frac{G}{d}(M_{1}+M_{2}-2\sqrt{M_{1}}\sqrt{M_{2}})] $
C) $ [-\frac{G}{d}(2M_{1}+M_{2}+2\sqrt{M_{1}}\sqrt{M_{2}})] $
D) $ [-\frac{G}{2d}(M_{1}+M_{2}+2\sqrt{M_{1}}\sqrt{M_{2}})] $
Show Answer
Answer:
Correct Answer: A
Solution:
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Let the gravitational field be zero at a point distant x from$ [M_{1}] $ .
$ [\frac{GM_{1}}{x^{2}}=\frac{GM_{2}}{{{\left( d-x \right)}^{2}}};\frac{x}{d-x}=\sqrt{\frac{M_{1}}{M_{2}}}] $ $ [x\sqrt{M_{2}}=\sqrt{M_{1}}d-x\sqrt{M_{1}}] $
$ [x\left[ \sqrt{M_{1}}+\sqrt{M_{2}} \right]=\sqrt{M_{1}}] $
$ [x=\frac{d\sqrt{M_{1}}}{\sqrt{M_{1}}+\sqrt{M_{2}}},d-x=\frac{d\sqrt{M_{2}}}{\sqrt{M_{1}}+\sqrt{M_{2}}}] $
Potential at this point due to both the masses will be $ [-\frac{GM_{1}}{x}-\frac{GM_{2}}{\left( d-x \right)}] $
$ [=-G\left[ \frac{M_{1}\left( \sqrt{M_{1}}+\sqrt{M_{2}} \right)}{d\sqrt{M_{1}}}+\frac{M_{2}\left( \sqrt{M_{1}}+\sqrt{M_{2}} \right)}{d\sqrt{M_{2}}} \right]] $
$ [=-\frac{G}{d}{{\left( \sqrt{M_{1}}+\sqrt{M_{2}} \right)}^{2}}] $
$ [=-\frac{G}{d}\left( M_{1}+M_{2}+2\sqrt{M_{1}}+\sqrt{M_{2}} \right)] $
