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) \[-\frac{G}{d}(2{{M}_{1}}+{{M}_{2}}+2\sqrt{{{M}_{1}}}\sqrt{{{M}_{2}}})\]
D) \[-\frac{G}{2d}({{M}_{1}}+{{M}_{2}}+2\sqrt{{{M}_{1}}}\sqrt{{{M}_{2}}})\]
Correct Answer: A
Solution :
[a] Let the gravitational field be zero at a point distant x from\[{{M}_{1}}\]. \[\frac{G{{M}_{1}}}{{{x}^{2}}}=\frac{G{{M}_{2}}}{{{\left( d-x \right)}^{2}}};\frac{x}{d-x}=\sqrt{\frac{{{M}_{1}}}{{{M}_{2}}}}\] \[x\sqrt{{{M}_{2}}}=\sqrt{{{M}_{1}}}d-x\sqrt{{{M}_{1}}}\] \[x\left[ \sqrt{{{M}_{1}}}+\sqrt{{{M}_{2}}} \right]=\sqrt{{{M}_{1}}}\] \[x=\frac{d\sqrt{{{M}_{1}}}}{\sqrt{{{M}_{1}}}+\sqrt{{{M}_{2}}}},d-x=\frac{d\sqrt{{{M}_{2}}}}{\sqrt{{{M}_{1}}}+\sqrt{{{M}_{2}}}}\] Potential at this point due to both the masses will be \[-\frac{G{{M}_{1}}}{x}-\frac{G{{M}_{2}}}{\left( d-x \right)}\] \[=-G\left[ \frac{{{M}_{1}}\left( \sqrt{{{M}_{1}}}+\sqrt{{{M}_{2}}} \right)}{d\sqrt{{{M}_{1}}}}+\frac{{{M}_{2}}\left( \sqrt{{{M}_{1}}}+\sqrt{{{M}_{2}}} \right)}{d\sqrt{{{M}_{2}}}} \right]\]\[=-\frac{G}{d}{{\left( \sqrt{{{M}_{1}}}+\sqrt{{{M}_{2}}} \right)}^{2}}\] \[=-\frac{G}{d}\left( {{M}_{1}}+{{M}_{2}}+2\sqrt{{{M}_{1}}}+\sqrt{{{M}_{2}}} \right)\]
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