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⚛️ Physics  ·  Gravitation  ·  NEET & JEE

Gravitational field intensity at surface of Earth (radius R, mass M) is:

Answer: GM/R 2.

  • A G/R<sup>2</sup>
  • B GM/R<sup>2</sup>
  • C GMm/R<sup>2</sup>
  • D 2GM/R<sup>2</sup>

Correct answer: B. GM/R<sup>2</sup>

Explanation: Gravitational field intensity (force per unit mass) = GM/R<sup>2</sup> = g (acceleration due to gravity at surface).

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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