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⚛️ Physics  ·  Gravitation  ·  NEET & JEE

At what height h above Earth's surface is g reduced to one-fourth its surface value?

Answer: R.

  • A R/2
  • B R
  • C 2R
  • D 3R

Correct answer: B. R

Explanation: g at height h: g' = g/(1+h/R)<sup>2.</sup> For g' = g/4: (1+h/R)<sup>2</sup> = 4. 1+h/R = 2. h = R.

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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