Questions
Four particles, each of mass M, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is
detailed solution
Correct option is D
F12=GM22R2F14=GM22R2The resultant of these two forces is2GM2/2R2. Now, F13=GM2/4R2The combined resultant of all the forces is2GM22R2+GM24R2 or GM2R222+14Equating this with centripetal force, we getMv2R=GM2R222+14or v2=GMR22+14 or v=GMR22+14Talk to our academic expert!
Similar Questions
A large spherical mass M is fixed at one position and two identical point masses m are kept on a line passing through the centre of M (see figure). The point masses are connected by a rigid massless rod of length l and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to M is at a distance from M , the tension in the rod is zero for . The value of k is
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