question_answer

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

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) \[-{{(\sqrt{{{M}_{1}}}-\sqrt{{{M}_{2}}})}^{2}}\frac{G}{R}\]

Correct Answer: C

Solution :

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