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Four identical particles of mass M are located at the corners of a square of side ‘a’. What should be their speed if each of them revolves under the influence of other’s gravitational field in a circular orbit circumscribing the square?(A) 1.35√(GM/a)(B) 1.16√(GM/a)(C) 1.41√(GM/a)(D) 1.21 √(GM/a) |
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Answer» C. is the CORRECT answers gravitational ENERGY = kinetic energy |
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