IIT JEE , AIEEE Forum - Physics , Mechanics

Ank999

Four particles of equal masses M move along a circle of of radius R under the action of their mutual gravitational attraction. Find the speed of each particle

waterdemon

I have given the diagtram along with the solution :

Gravitational force on A due to B = GM²/(R

2)².
The direction being along "AB".

Gravitational Force on A due to D = GM²/(R

2)².
The Direction being along "AD".

Gravitational Force on A due to C = GM²/4R².
The Direction being along "AC".

Now we will take the Resultant of all the Forces along AC:
We get:

F_Res = 2*GM²/(R

2)²*Cos(45) + GM²/4R²
F_Res = GM²/R²{(2

2+1)/4}

Now to get the speed of each particle moving in the circle
of Radius "R".

MV²/R = GM²/R²{(2

2+1)/4}

From above we get,
V =

[GM/R {(2

2+1)/4}]

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