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NEETPhysicsGravitation

A hypothetical planet has an acceleration due to gravity at its surface that is 4 times that of the Earth, and its radius is half that of the Earth. If the escape velocity on the surface of the Earth is v_e , what is the escape velocity on the surface of this hypothetical planet?

Options

  1. A2 v_e
  2. B8 v_e
  3. Cv_e 2 2
  4. D2 v_e

Correct answer

D. 2 v_e

Step-by-step solution

The escape velocity v on the surface of a planet is related to its acceleration due to gravity g and radius R by the formula v = 2gR . For the Earth, v_e = 2g_e R_e . For the hypothetical planet, the acceleration due to gravity is g_p = 4g_e and the radius is R_p = R_e 2 . The escape velocity on the planet is: v_p = 2g_p R_p Substituting the given values: v_p = 2(4g_e) ( R_e 2 ) = 4(g_e R_e) = 2 2g_e R_e = 2 v_e . Thus, the escape velocity on the hypothetical planet is 2 v_e . Answer: 2 v_e

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