Four similar particles of mass m are orbiting | vedclass

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Question

(A) The centripetal force required for circular motion is provided by the net gravitational force acting on one particle due to the other three partic

Vedclass · Accepted Answer

${\left[ {\frac{{Gm}}{r}\,\left( {\frac{{1 + 2\sqrt 2 }}{4}} \right)} \right]^{1/2}}$

Vedclass · Answer

(A) The centripetal force required for circular motion is provided by the net gravitational force acting on one particle due to the other three particles.Let the particles be at the corners of a square of side length $a = \sqrt{2}r$.The net gravitational force $F_{net}$ on one particle towards the center is:$F_{net} = 2F \cos(45^{\circ}) + F_1$Where $F = \frac{Gm^2}{(\sqrt{2}r)^2} = \frac{Gm^2}{2r^2}$ is the force from the two adjacent particles,and $F_1 = \frac{Gm^2}{(2r)^2} = \frac{Gm^2}{4r^2}$ is the force from the diagonally opposite particle.$F_{net} = 2 \left( \frac{Gm^2}{2r^2} \right) \frac{1}{\sqrt{2}} + \frac{Gm^2}{4r^2} = \frac{Gm^2}{r^2} \left( \frac{1}{\sqrt{2}} + \frac{1}{4} \right) = \frac{Gm^2}{r^2} \left( \frac{2\sqrt{2} + 1}{4} \right)$.Equating this to the centripetal force $\frac{mv^2}{r}$:$\frac{mv^2}{r} = \frac{Gm^2}{r^2} \left( \frac{2\sqrt{2} + 1}{4} \right)$$v^2 = \frac{Gm}{r} \left( \frac{2\sqrt{2} + 1}{4} \right)$$v = \sqrt{\frac{Gm}{r} \left( \frac{1 + 2\sqrt{2}}{4} \right)}$

$\text{LIST-I}$	$\text{LIST-II}$
$A$. Gravitational constant	$I$. $[LT^{-2}]$
$B$. Gravitational potential energy	$II$. $[L^2 T^{-2}]$
$C$. Gravitational potential	$III$. $[ML^2 T^{-2}]$
$D$. Acceleration due to gravity	$IV$. $[M^{-1} L^3 T^{-2}]$

Four similar particles of mass $m$ are orbiting in a circle of radius $r$ in the same angular direction because of their mutual gravitational attractive force. The velocity of a particle is given by

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