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Solve the following :

Question:

Four particles of equal masses $M$ move along a circle of radius $R$ under the action of their mutual gravitational attraction. Find the speed of each particle.

Solution:

Here, $\mathrm{r}_{13}=2 \mathrm{R}$

$r_{14}=2 R \cos 45=\sqrt{2} R$

$r_{12}=2 R \sin 45=\sqrt{2} R$

Now, force on particle 1

Due to particle 4: $\mathrm{F}_{14}=\frac{\mathrm{GMM}}{(\sqrt{2} \mathrm{R})^{2}}=\frac{\mathrm{GM}^{2}}{2 \mathrm{R}^{2}}$

Due to particle 2: $\mathrm{F}_{12}=\frac{\mathrm{GMM}}{(\sqrt{2} \mathrm{R})^{\mathrm{2}}}=\frac{\mathrm{GM}^{2}}{2 \mathrm{R}^{2}}$

Due to particle 1: $F_{13}=\frac{\text { GMM }}{(2 R)^{2}}=\frac{G^{2}}{4 R^{2}}$

Resultant of $\mathrm{F}_{14} \& \mathrm{~F}_{12}=\sqrt{\mathrm{F}_{12}^{2}+\mathrm{F}_{14}^{2}+2\left(\mathrm{~F}_{12}\right)\left(\mathrm{F}_{14}\right) \cos 90^{\circ}}$

$=\sqrt{\left(\frac{\mathrm{GM}^{2}}{2 \mathrm{R}^{2}}\right)^{2}+\left(\frac{\mathrm{GM}^{2}}{2 \mathrm{R}^{2}}\right)^{2}}$

$\mathrm{F}_{\mathrm{R}}=\frac{\mathrm{GM}^{2}}{\sqrt{2} \mathrm{R}^{2}}$ (along $\mathrm{r}_{13}$ direction)

Net resultant force, $\mathrm{F}_{\mathrm{N}}=\mathrm{F}_{\mathrm{R}}+\mathrm{F}_{\mathrm{B}}$

Since particle is revolving in a circle. So, this is centripetal force.

$\mathrm{F}_{\mathrm{N}}=\frac{\mathrm{Mv}^{2}}{\mathrm{R}}$

$\frac{\mathrm{GM}^{2}}{4 \mathrm{R}^{2}}=\frac{\mathrm{Mv}^{2}}{\mathrm{R}}$

$\mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}\left(\frac{2 \sqrt{2}+1}{4}\right)}$

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