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Question

Three bodies of same masses are kept at the top of an equilateral triangle with side a. What speed would the three bodies be moved in a circle so that the triangle moves in the circular orbit and there should be no change in the side of the triangle?

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Solution

The gravitational force acting between each couples of masses
F=GMMa2=GM2a2
Each internal angle of equilateral triangle is of 60.
Therefore the resultant of forces acting on each mass will be directed towards the center O of the circle and the same will the role of centripetal force for circular motion of the whole triangle.

Centripetal force = Resultant force
or FC=FR=F2+F2+2F2cos60
or Mv2r=F3
or Mv2r=3GM2a2

or v2r=3GMa2 ..............(1)

Now from right angled triangle ABD,
AD2+BD2=AB2

or AD2+a24=a2
or AD2=a2a24=34a2

AD=a32

According to geometrical property of median of a equilateral triangle.
r=AO=23AD=23×a23=a3

r=AO=23AD=23×a23=a3

Therefore , from equation (1), we have
v2r=3GMa2v2=r3GMa2

or v2=a3×3GMa2=GMa

v=GMa

1752784_1027858_ans_d9544bda2b834161a315fe44d84d57a7.png

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