Four identical particles of equal masses

Question:

Four identical particles of equal masses $1 \mathrm{~kg}$ made to move along the circumference of a circle of radius $1 \mathrm{~m}$ under the action of their own mutual gravitational attraction. The speed of each particle will be

  1. (1) $\frac{\sqrt{(1+2 \sqrt{2}) G}}{2}$

  2. (2) $\sqrt{G(1+2 \sqrt{2})}$

  3. (3) $\sqrt{\frac{G}{2}(2 \sqrt{2}-1)}$

  4. (4) $\sqrt{\frac{G}{2}(1+2 \sqrt{2})}$


Correct Option: 1,

Solution:

$\Rightarrow$ By resolving force $\mathrm{F}_{2}$, we get $\Rightarrow \mathrm{F}_{1}+\mathrm{F}_{2} \cos 45^{\circ}+\mathrm{F}_{2} \cos 45^{\circ}$

$\Rightarrow F_{1}+2 F_{2} \cos 45^{\circ}=F_{c}$

$F_{\mathrm{c}}=$ centripital force $=\frac{\mathrm{MV}^{2}}{\mathrm{R}}$

$\Rightarrow \frac{\mathrm{GM}^{2}}{(2 \mathrm{R})^{2}}+\left[\frac{2 \mathrm{GM}^{2}}{(\sqrt{2} \mathrm{R})^{2}} \cos 45^{\circ}\right]=\frac{\mathrm{MV}^{2}}{\mathrm{R}}$

$\Rightarrow \frac{\mathrm{GM}^{2}}{4 \mathrm{R}^{2}}+\frac{2 \mathrm{GM}^{2}}{2 \sqrt{2} \mathrm{R}^{2}}=\frac{\mathrm{MV}^{2}}{\mathrm{R}}$

$\Rightarrow \frac{\mathrm{GM}}{4 \mathrm{R}}+\frac{\mathrm{GM}}{\sqrt{2} \cdot \mathrm{R}}=\mathrm{V}^{2}$

$\Rightarrow V=\sqrt{\frac{\frac{G M}{4 R}+\frac{G M}{\sqrt{2} \cdot R}}{\frac{G M}{R}\left[\frac{1+2 \sqrt{2}}{4}\right]}}$

$\Rightarrow V=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}(1+2 \sqrt{2})}$

$\Rightarrow \mathrm{V}=\frac{1}{2} \sqrt{\frac{\left(\frac{G}{R}\right)}{R}}$

(given : mass $=1 \mathrm{~kg}$, radius $=1 \mathrm{~m}$ )

$\Rightarrow \mathrm{v}=\frac{1}{2} \sqrt{\mathrm{G}(1+2 \sqrt{2})}$

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