question_answer

Four identical particles of mass M are located at the corners of a square of side 'a'. What should be their speed if each of them revolves under the influence of other's gravitational field in a circular orbit circumscribing the square? [JEE Main 8-4-2019 Morning]

A) \[1.21\sqrt{\frac{GM}{a}}\]              

B) \[1.41\sqrt{\frac{GM}{a}}\]

C) \[1.16\sqrt{\frac{GM}{a}}\]

D) \[1.35\sqrt{\frac{GM}{a}}\]

Correct Answer: C

Solution :

Net force on particle towards centre of circle is\[{{F}_{C}}=\frac{G{{M}^{2}}}{2{{a}^{2}}}+\frac{G{{M}^{2}}}{{{a}^{2}}}\sqrt{2}\]\[=\frac{G{{M}^{2}}}{{{a}^{2}}}\left( \frac{1}{2}+\sqrt{2} \right)\] This force will act as centripetal force. Distance of particle from centre of circle is\[\frac{a}{\sqrt{2}}.\] \[r=\frac{a}{\sqrt{2}},{{F}_{C}}=\frac{m{{\text{v}}^{2}}}{r}\] \[\frac{m{{\text{v}}^{2}}}{\frac{a}{\sqrt{2}}}=\frac{G{{M}^{2}}}{{{a}^{2}}}\left( \frac{1}{2}+\sqrt{2} \right)\] \[{{\text{v}}^{2}}=\frac{GM}{a}\left( \frac{1}{2\sqrt{2}}+1 \right)\] \[{{\text{v}}^{2}}=\frac{GM}{a}(1.35)\] \[\text{v}=1.16\sqrt{\frac{GM}{a}}\]