JEE Main 2020 · 4 September · evening · Single correct · +4 / −1

A body is moving in a low circular orbit about a planet of mass M and radius R. The radius of the orbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is:

  1. (a) 2
  2. (b) 1
  3. (c) √2
  4. (d) 1/√2
Answer and solution

Answer (d) 1/√2

Solution

Physics sir · 3 steps

Write both speeds from GM/R: orbit needs √(GM/R), escape needs √(2GM/R).

  1. Orbital speed. In a circular orbit of radius R, gravity supplies the centripetal force: GMm/R² = mvₒ²/R. This gives vₒ = √(GM/R).
  2. Escape speed. To escape from the surface, the kinetic energy must cancel the potential energy: 1/2mvₑ² = GMm/R. This gives vₑ = √(2GM/R).

    vₒ/vₑ = √(GM/R)/√(2GM/R) = 1/√2

  3. So. The orbital speed is 1/√2 times the escape speed, which is option (d).

The idea to keep. At the same radius, escape speed is always √2 times the circular orbital speed. Remember vₑ = √2 vₒ and you will not need to derive it each time.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 160 times in JEE Main since 2020. This question is from 2020.