Question 21 of 223

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

A$\sqrt{G\left(1+2\sqrt{2}\right)}$B$\sqrt{\dfrac{G}{2}\left(1+2\sqrt{2}\right)}$C$\sqrt{\dfrac{G}{2}\left(2\sqrt{2}-1\right)}$D$\dfrac{\sqrt{\left(1+2\sqrt{2}\right)G}}{2}$

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

${F}_{1}=\dfrac{Gmm}{{\left(2R\right)}^{2}}=\dfrac{G{m}^{2}}{4{R}^{2}}$
${F}_{2}=\dfrac{Gmm}{{\left(\sqrt{2}R\right)}^{2}}=\dfrac{G{m}^{2}}{2{R}^{2}}$
${F}_{3}=\dfrac{Gmm}{{\left(\sqrt{2}R\right)}^{2}}=\dfrac{G{m}^{2}}{2{R}^{2}}$
$\Rightarrow {F}_{net}={F}_{1}+{F}_{2}\cos 45^{\circ}+{F}_{3}\cos 45^{\circ}$
$=\dfrac{G{m}^{2}}{4{R}^{2}}+\dfrac{G{m}^{2}}{2{R}^{2}}\dfrac{1}{\sqrt{2}}+\dfrac{G{m}^{2}}{2{R}^{2}}\dfrac{1}{\sqrt{2}}$
$=\dfrac{G{m}^{2}}{{R}^{2}}\left(\dfrac{1}{4}+\dfrac{1}{\sqrt{2}}\right)=\dfrac{G{m}^{2}}{4{R}^{2}}\left(1+2\sqrt{2}\right)$
${F}_{net}=\dfrac{G{m}^{2}}{4{R}^{2}}\left(1+2\sqrt{2}\right)=\dfrac{m{v}^{2}}{R}$
$\Rightarrow v=\dfrac{\sqrt{G\left(1+2\sqrt{2}\right)}}{2}$