Four masses ‘m’ each are orbiting in a circle of radius’r’ in the same direction under gravitational force. Velocity of each particle is

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$\sqrt{\frac{G m}{r}}$

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

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

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

answer is D.

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Detailed Solution

${F^{'}}_{n e t} o n A d u e t o C a n d B = 2 F \cos \frac{θ}{2} = 2 F \cos \frac{90}{2} = 2 F \cos 45 = 2 F \frac{1}{\sqrt{2}} = \sqrt{2} F = \sqrt{2} \frac{G m^{2}}{{(\sqrt{2} r)}^{2}} = \sqrt{2} \frac{G m^{2}}{2 r^{2}} = \frac{G m^{2}}{\sqrt{2} r^{2}} \Rightarrow F_{n e t} = F_{A D} + {F^{'}}_{n e t} \Rightarrow F_{n e t} p r o v i d e s c e n t r i p e t a l f o r c e \Rightarrow \frac{G m^{2}}{{(2 r)}^{2}} + \frac{G m^{2}}{\sqrt{2} r^{2}} = \frac{m v^{2}}{r} \Rightarrow \frac{G m}{4 r} + \frac{G m}{\sqrt{2} r} = v^{2} \Rightarrow v = \sqrt{\frac{G m}{r} (\frac{1}{4} + \frac{1}{\sqrt{2}})} \Rightarrow v = \sqrt{\frac{G m}{r} (\frac{1}{4} + \frac{2 \sqrt{2}}{2 \sqrt{2} \sqrt{2}})} = \sqrt{\frac{G m}{r} (\frac{1}{4} + \frac{2 \sqrt{2}}{4})} = = \sqrt{\frac{G m}{4 r} (1 + 2 \sqrt{2})}$