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⚛️ Physics  ·  Gravitation  ·  NEET & JEE

Minimum energy needed to launch a satellite from Earth's surface to circular orbit at radius 2R:

Answer: 3GMm/(4R).

  • A GMm/(2R)
  • B 3GMm/(4R)
  • C 5GMm/(4R)
  • D GMm/R

Correct answer: B. 3GMm/(4R)

Explanation: E<sub>surface</sub> = -GMm/R. E<sub>orbit</sub>(2R) = -GMm/4R. Delta E = -GMm/4R - (-GMm/R) = GMm(1/R - 1/4R) = 3GMm/(4R).

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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