JEE Main 2023 · 13 April · morning · Single correct · +4 / −1

A planet having mass 9 Me and radius 4 Rₑ, where Me and Re are mass and radius of earth respectively, has escape velocity in km / s given by:
(Given escape velocity on earth Vₑ=11.2 × 10³ m / s )

  1. (a) 33.6
  2. (b) 11.2
  3. (c) 16.8
  4. (d) 67.2
Answer and solution

Answer (c) 16.8

Solution

Physics sir · 3 steps

Escape velocity goes as √(M/R), so take the ratio with Earth and every constant drops out.

  1. Formula. A body escapes when its kinetic energy equals the gravitational binding energy, 1/2mv² = GMm/R. That gives vₑ = √(2GM/R).

    vₚ/vₑ = √(Mₚ/Mₑ×Rₑ/Rₚ) = √(9/1×1/4) = 3/2

  2. Put in numbers. vₚ = 1.5 × 11.2 km s⁻¹ = 16.8 km s⁻¹. Note that 11.2 × 10³ m s⁻¹ is the same as 11.2 km s⁻¹.
  3. So. The escape velocity on the planet is 16.8 km/s, which is option (c).

The idea to keep. For planet comparison questions, write vₑ ∝ √(M/R) and take the ratio with Earth. Mass goes on top, radius goes below, and the square root covers both.

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 2023.