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3 particles of equal mass M eacha re moving on a circular path with a radius R under their mutual gravitational attraction. what is the speed of each particle. ans given : root[(GM)/1.732R]
dear arjit
place these masses at the corners of equilateral triangle of side a... then draw circumcircle of this triangle....
this is the circle in which these particles are rotating...now calculate length of side in terms of radius by using simple geometry...
a = sqrt3.R (after solving)
now ,let force bw any of the pair is F then
F=GM/a2
F=GM/3R2 ..........1
now calculate net force due to two particles on other particle ,let it be F1 ..
F1 = (F2 +F2 +2F2 cos60)1/2
F1=sqrt3.F or
F1 =sqrt3.GM/3R2 ...................2
now this resultant force will provide necessary centripital force for each particle......
so F1 = Mv2 /R (Fc)
sqrt3.GM/3R2 = Mv2 /R
v = (GM/sqrt3R)1/2 sqrt3 =1.732
so v= (GM/1.732R)1/2 ans
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