Question

four particles of equal mass are moving around a circle of radius r due to their mutual gravitational attraction.find the angular velocity of each particle

Answer 1

if particles have mutual gravitational interaction then they will remain diametrically opp.and G.F
is centripetal
F=Gm^2/4r^2
if speed is v then C.F
F=MV^2/r
v=√gm/4r
angular velocity=v/r= √gm/4r^3

Answer 2

Answer:

The angular velocity of each particle is $\sqrt{\dfrac{GM}{R}(\dfrac{2\sqrt{2}+1}{4})}$

Explanation:

Given that,

Four particles of equal mass are moving around a circle of radius r due to their mutual gravitational attraction.

So, each particle will experience the gravitational attraction force due to the other three particles.

We need to calculate the angular velocity of each particle

According to figure,

The gravitational force between A and B

$\vec{F_{AB}}=\dfrac{GM^2}{(\sqrt{2}R)^2}\hat{j}$

The gravitational force between A and C

$\vec{F_{AC}}=\dfrac{GM^2}{4R^2}\hat{i}$

The gravitational force between A and D

$\vec{F_{AD}}=-\dfrac{GM^2}{(\sqrt{2}R)^2}$

The net force acting along horizontal direction is zero.

Now, The resultant force is

$F=F_{AB}\cos45^{\circ}+F_{AC}\cos0^{\circ}+F_{AD}\cos45^{\circ}$

$F=-\dfrac{GM^2}{2R^2}\dfrac{1}{\sqrt{2}}+\dfrac{GM^2}{4R^2}\times1+\dfrac{GM^2}{2R^2}\times\dfrac{1}{\sqrt{2}}$

The magnitude of the resultant force is

$F=\dfrac{GM^2}{4R^2}(2\sqrt{2}+1)$

For moving along  the circle,

The centripetal force is defined as,

$F = \dfrac{Mv^2}{R}$

Put the value of F into the formula

$\dfrac{GM^2}{4R^2}(2\sqrt{2}+1)=\dfrac{Mv^2}{R}$

$v=\sqrt{\dfrac{GM}{R}(\dfrac{2\sqrt{2}+1}{4})}$

Hence, The angular velocity of each particle is $\sqrt{\dfrac{GM}{R}(\dfrac{2\sqrt{2}+1}{4})}$