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⚛️ Physics  ·  Gravitation  ·  NEET & JEE

Escape velocity on the Moon (g<sub>moon</sub> = g/6, R<sub>moon</sub> = R/4):

Answer: ve*sqrt(6)/4.

  • A ve/sqrt(24)
  • B ve*sqrt(6)/4
  • C ve/sqrt(6)
  • D ve/4

Correct answer: B. ve*sqrt(6)/4

Explanation: ve_moon = sqrt(2 x g<sub>moon</sub> x R<sub>moon</sub>) = sqrt(2 x g/6 x R/4) = sqrt(gR/12) = ve x sqrt(1/24) = ve/sqrt(24). Hmm. ve = sqrt(2gR). ve_moon = sqrt(2(g/6)(R/4)) = sqrt(gR/12) = sqrt(gR)/sqrt(12). ve_moon/ve = sqrt(1/12)/sqrt(2) = 1/sqrt(24). So ve_moon = ve/sqrt(24).

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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