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⚛️ Physics  ·  Gravitation  ·  NEET & JEE

A body is dropped from rest at a height equal to the Earth radius R above the surface. Its speed on reaching the surface is about (g = 10 m/s<sup>2</sup>, R = 6.4×10<sup>6</sup> m):

Answer: 8 km/s.

  • A 8 km/s
  • B 2 km/s
  • C 4 km/s
  • D 11 km/s

Correct answer: A. 8 km/s

Explanation: Energy conservation: ½v<sup>2</sup> = GM(1/R - 1/2R) = gR/2, so v = √(gR) = √(6.4×10<sup>7</sup>) = 8000 m/s = 8 km/s.

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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