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⚛️ Physics  ·  Gravitation  ·  NEET & JEE

Time period of satellite at height h from Earth's surface:

Answer: 2*pi*sqrt((R+h) 3 /(GM)).

  • A 2*pi*sqrt(R/g)
  • B 2*pi*sqrt((R+h)<sup>3</sup>/(GM))
  • C 2*pi*sqrt((R+h)/g)
  • D 2*pi*sqrt(GM/R)

Correct answer: B. 2*pi*sqrt((R+h)<sup>3</sup>/(GM))

Explanation: T = 2*pi*sqrt(r<sup>3</sup>/(GM)) where r = R+h. Or T = 2*pi*sqrt((R+h)<sup>3</sup>/GM).

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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