Question

(100 Points) There is square and M masses are placed at the Vertexes. What will be the velocity with which they will revolve around in a Circle(under the influence of gravitational force) ?

Answer 1

Hey user

Since all the four particles are of same mass, I think their mutual interaction will bring them to the corners of a square.

Diagonal AD = 2r,Side AB = AE=(2r)2−r2=3–√r

Further, AB2+BD2=AD2=>2AB2=4r2=>AB=2–√r

Thus AB=BD=AE=2–√r

Consider the forces on any particle, say A due to other three particles.

FB=Gm22r2 and is directed towards B.

FE=Gm22r2 and is directed towards E.

FD=Gm24r2 and is directed towards D.

The net force F towards center C is then given by

F=FBcos45+FEcos45+FD. Or,

F=Gm22r2cos45+Gm22r2cos45+Gm24r2=Gm24r2(22–√+1)

This resultant force F provides the necessary centripetal force FC given by

FC=mrω2 . All the particles will revolve around their common centre of mass.

Setting the two forces equal, we have

mrω2=Gm24r2(22–√+1)

This gives

ω=Gm4r3(22–√+1)

Answer 2

Answer:

Explanation:

The net gravitational force = centripetal force

|F1||F1| force between A and B

=GMM(2–√r)2=GMM(2r)2

|F2||F2| force between A and D

=GMM(2–√r)2=GMM(2r)2

|F3||F3| force between A and C

=GMM(2r)2=GMM(2r)2

The components of F1F1 and F2F2 along the radius

|F1|cos45|F1|cos⁡45 and |F2|cos45[|F1|=|F2|=F]|F2|cos⁡45[|F1|=|F2|=F]

Net force =2fcos45+F3=2fcos⁡45+F3

=2GM2(2–√r)2×12+GM24r2=2GM2(2r)2×12+GM24r2

Mv20r=GM24r2Mv02r=GM24r2[22–√+1][22+1]

v0=[GM4rv0=[GM4r(22–√+1)]1/2(22+1)]1/2

Hence a is the orrect answer.