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⚛️ Physics  ·  Gravitation  ·  NEET & JEE

If Earth's mass is M, radius R, orbital velocity at height h = R (i.e., r = 2R) in terms of g and R:

Answer: sqrt(gR/2).

  • A sqrt(gR/2)
  • B sqrt(gR)
  • C sqrt(2gR)
  • D sqrt(gR/4)

Correct answer: A. sqrt(gR/2)

Explanation: vo = sqrt(GM/(2R)). GM = gR<sup>2.</sup> vo = sqrt(gR<sup>2</sup>/(2R)) = sqrt(gR/2).

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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