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NTA Abhyas JEE Main2020PhysicsGravitationPractice

A body is projected upwards with a velocity of 4 × 11.2 k m s - 1 from the surface of the earth. The velocity of the body when it escapes from the gravitational pull of earth will be

Options

  1. A11.2 k m s - 1
  2. B2 × 11.2 k m s - 1
  3. C3 × 11.2 k m s - 1
  4. D15 × 11.2 k m s - 1

Correct answer

D. 15 × 11.2 k m s - 1

Step-by-step solution

The minimum velocity of projection to achieve escape velocity can be calculated as, Initial KE = 1 2 mv 2 = 1 2 × m 4 × 11 .2 2 = 16 × 1 2 mv e 2 As 1 2 mv e 2 energy is used up in coming out from the gravitational pull of the earth, so Final KE should be 15 × 1 2 mv e 2 Hence, 1 2 m v ' 2 = 15 × 1 2 mv e 2 ∴ v ′ 2 = 15 v e 2 or v ′ = 15 v e = 15 × 11 .2 km s - 1

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