Four similar particles of mass m are orbiting | vedclass

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(A) Consider the forces acting on one of the masses $m$. Let this mass be at the bottom position.The distance to the two adjacent masses is $d = \sqrt

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Consider the forces acting on one of the masses $m$. Let this mass be at the bottom position.The distance to the two adjacent masses is $d = \sqrt{r^2 + r^2} = r\sqrt{2}$.The gravitational force from each of these two masses is $F_1 = \frac{Gm^2}{(r\sqrt{2})^2} = \frac{Gm^2}{2r^2}$.The horizontal components of these two forces cancel out,while the vertical components add up towards the center.The vertical component of each force is $F_1 \cos(45^\circ) = \frac{Gm^2}{2r^2} 	imes \frac{1}{\sqrt{2}} = \frac{Gm^2}{2\sqrt{2}r^2}$.Total vertical force from these two masses is $2 	imes \frac{Gm^2}{2\sqrt{2}r^2} = \frac{Gm^2}{\sqrt{2}r^2}$.The distance to the mass at the top is $2r$.The gravitational force from the mass at the top is $F_2 = \frac{Gm^2}{(2r)^2} = \frac{Gm^2}{4r^2}$.The total net force towards the center is $F_{net} = \frac{Gm^2}{\sqrt{2}r^2} + \frac{Gm^2}{4r^2} = \frac{Gm^2}{r^2} \left( \frac{1}{\sqrt{2}} + \frac{1}{4} ight) = \frac{Gm^2}{r^2} \left( \frac{2\sqrt{2} + 1}{4} ight)$.This net force provides the necessary centripetal force: $\frac{mv^2}{r} = F_{net}$.$\frac{mv^2}{r} = \frac{Gm^2}{r^2} \left( \frac{1 + 2\sqrt{2}}{4} ight)$.$v^2 = \frac{Gm}{r} \left( \frac{1 + 2\sqrt{2}}{4} ight)$.$v = \sqrt{\frac{Gm}{r} \left( \frac{1 + 2\sqrt{2}}{4} ight)}$.

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

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