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⚛️ Physics  ·  Gravitation  ·  NEET & JEE

The total mechanical energy of a satellite of mass m in a circular orbit of radius r around a planet of mass M is:

Answer: −GMm/2r.

  • A −GMm/r
  • B −GMm/2r
  • C +GMm/2r
  • D −GMm/4r

Correct answer: B. −GMm/2r

Explanation: E = KE + PE = GMm/2r − GMm/r = −GMm/2r.

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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