# 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 others' gravitational field in circular orbit circumscribing the square ? Option 1) $1.41\sqrt{\frac{GM}{a}}$ Option 2) $1.16\sqrt{\frac{GM}{a}}$ Option 3) $1.21\sqrt{\frac{GM}{a}}$   Option 4) $1.35\sqrt{\frac{GM}{a}}$

$\\2r=\sqrt{2}a\\\; \; \; \; \\r=\frac{a}{\sqrt{2}}\; \; \; \; \; \; \; (1)$

$\\F_{1}=\frac{Gmm}{a^{2}}(due \ \ to \ two \ adjacent \ \ masses)\\\; \; \; \\F_{2}=\frac{Gmm}{(\sqrt{2}a)^{2}} (Due \ \ to \ two \ masses \ which \ are \ dimetrically \ opposite)$

$\\F_{net}=\sqrt{2}\; F_{1}+F_{2}=\frac{mV^{2}}{r}$

$\Rightarrow \frac{\sqrt{2}\times Gm^{2}}{a^{2}}+\frac{Gm^{2}}{(\sqrt{2}\; a)^{2}}=\frac{mV^{2}}{(\frac{a}{\sqrt{2}})}$

$\Rightarrow \frac{Gm}{\sqrt{2}\; a}(\sqrt{2}+\frac{1}{2})=V^{2}$

$V^{2}=\frac{Gm(2\sqrt{2}+1)}{2\sqrt{2}\; a}$

$V=1.16\sqrt{\frac{Gm}{a}}$

Option 1)

$1.41\sqrt{\frac{GM}{a}}$

Option 2)

$1.16\sqrt{\frac{GM}{a}}$

Option 3)

$1.21\sqrt{\frac{GM}{a}}$

Option 4)

$1.35\sqrt{\frac{GM}{a}}$

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