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⚛️ Physics  ·  Gravitation  ·  NEET & JEE

The escape velocity from a planet is 11.2 km/s. If its radius is doubled while its mean density stays the same, the new escape velocity is:

Answer: 22.4 km/s.

  • A 22.4 km/s
  • B 11.2 km/s
  • C 5.6 km/s
  • D 44.8 km/s

Correct answer: A. 22.4 km/s

Explanation: At constant density, escape velocity is proportional to the radius (v = R√(8πGρ/3)). Doubling R doubles it to 22.4 km/s.

Variation of g with Height and Depthgg_surfaceAbove surface: g ∝ 1/r² (falls off curving down)Below surface: g ∝ r (falls off LINEARLY)Earth's surface (r = R)centre (r=0): g=0g is MAXIMUM exactly at the surface - it decreases in both directions, but by different laws

g is maximum at Earth's surface; going up, it falls off as 1/r² (inverse-square); going down, it falls off linearly with depth (since only the mass enclosed within radius r contributes), reaching zero at the centre.

Concept context

Universal gravitation, orbital mechanics, gravitational potential, and escape velocity.

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