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⚛️ Physics  ·  Gravitation  ·  NEET & JEE

Orbital velocity of satellite at height h from Earth (radius R, mass M):

Answer: sqrt(GM/(R+h)).

  • A sqrt(GM/R)
  • B sqrt(GM/(R+h))
  • C sqrt(2GM/(R+h))
  • D sqrt(gR<sup>2</sup>/(R+h))

Correct answer: B. sqrt(GM/(R+h))

Explanation: Orbital velocity: v<sub>o</sub> = sqrt(GM/(R+h)) for a satellite at height h.

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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