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NTA Abhyas JEE Main2020PhysicsGravitationPractice

Three particles each of mass m = 1 kg , are located at the vertices of an equilateral triangle of side a = 4 m . The speed at which they must move to revolve in a circular orbit, under the influence of mutual gravitational force, circumscribing and preserving the equilateral triangle is v = n G , then the value of n is

Correct answer

0.50

Step-by-step solution

The net gravitational force on any particle due to the other two is F g = 2 G m 2 a 2 cos 30 ° = 3 G m 2 a 2 Now, r = a 3 For circular motion, F g = 3 G m 2 a 2 = m v 2 r Solving it we obtain, v = 0.50 G

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