Question 42 of 223

Four particles, each of mass $M$ and equidistant from each other, move along a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is

A$\sqrt{\dfrac{\text{GM}}{\text{R}}}$B$\sqrt{2\sqrt{2}\dfrac{\text{GM}}{\text{R}}}$C$\sqrt{\dfrac{\text{GM}}{\text{R}}\left(1+2\sqrt{2}\right)}$D$\dfrac{1}{2}\sqrt{\dfrac{\text{GM}}{\text{R}}\left(1+2\sqrt{2}\right)}$

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Answer: (D) $\dfrac{1}{2}\sqrt{\dfrac{\text{GM}}{\text{R}}\left(1+2\sqrt{2}\right)}$

$\text{F}={\text{F}}_{1}={\text{F}}_{2}\text{=}\dfrac{{\text{GM}}^{2}}{2{\text{R}}^{2}}$

${\text{F}}_{3}=\dfrac{{\text{GM}}^{2}}{4{\text{R}}^{2}}$

$∴$ FTotal towards the centre.

$=\sqrt{2}\text{F}+{\text{F}}_{3}$

$=\sqrt{2}\dfrac{{\text{GM}}^{2}}{{\text{2R}}^{2}}+\dfrac{{\text{GM}}^{2}}{4{\text{R}}^{2}}$

$=\dfrac{{\text{GM}}^{2}}{4{\text{R}}^{2}}\left[2\sqrt{2}+1\right]$

${\text{F}}_{\text{T}}{=\text{F}}_{\text{CP}}$

$\dfrac{{\text{GM}}^{2}}{4{\text{R}}^{2}}\left[2\sqrt{2}+1\right]=\dfrac{{\text{Mv}}^{2}}{\text{R}}$

$∴\text{v}=\sqrt{\dfrac{\text{GM}}{4\text{R}}\left[2\sqrt{2}+1\right]}$.