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JEE MainPhysicsGravitation

A satellite revolves in a circular orbit very close to the surface of the Earth with an orbital speed v_o . A hypothetical planet X has a mass 8 times the mass of the Earth and a radius 2 times the radius of the Earth. The escape velocity from the surface of planet X will be

Options

  1. A2 v_o
  2. B2v_o
  3. Cv_o 2
  4. D2 2 v_o

Correct answer

D. 2 2 v_o

Step-by-step solution

The orbital speed of a satellite revolving very close to the surface of the Earth is given by v_o = GM_e R_e where M_e is the mass of the Earth and R_e is its radius. The escape velocity from the surface of planet X is given by v_ ex = 2GM_x R_x We are given that the mass of planet X is M_x = 8M_e and its radius is R_x = 2R_e . Substituting these into the escape velocity formula for planet X yields: v_ ex = 2G(8M_e) 2R_e v_ ex = 8GM_e R_e v_ ex = 8 GM_e R_e Since GM_e R_e = v_o , we get: v_ ex = 2 2 v_o Answer: 2 2

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