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⚛️ Physics  ·  Gravitation  ·  NEET & JEE

A satellite orbits at height equal to Earth's radius R (i.e., at distance 2R from center). Orbital velocity compared to surface orbital velocity:

Answer: 1/sqrt(2) times.

  • A Same
  • B 1/sqrt(2) times
  • C 1/2 times
  • D sqrt(2) times

Correct answer: B. 1/sqrt(2) times

Explanation: vo = sqrt(GM/r). At 2R: vo_new = sqrt(GM/2R) = vo_surface/sqrt(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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