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NEET2016PhysicsGravitationActual

Time period of pendulum, on a satellite orbiting the earth, is

Options

  1. A1 /
  2. Bzero
  3. Cinfinity

Correct answer

3

Step-by-step solution

On an artificial satellite orbiting the earth the acceleration is given by G M R^2 towards the centre of the earth. Now for a body of mass m on the satellite the graviational force due to earth is G M m R^2 towards the centre of the earth. Let the reaction force on the surface of the satellite be N , then G M m R^2 -N=m ( G M R^2 ) N=0 That is on the satellite there is a state of weightlessness or g=0 The time period of the simple pendulum, T=2 l g =

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