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⚛️ Physics  ·  Gravitation  ·  NEET & JEE

Binary stars of mass M each orbit their center of mass at distance d apart. Orbital period T in terms of G, M, d:

Answer: 2*pi*sqrt(d 3 /(2GM)).

  • A 2*pi*sqrt(d<sup>3</sup>/(2GM))
  • B 2*pi*sqrt(d<sup>3</sup>/(GM))
  • C pi*sqrt(d<sup>3</sup>/(2GM))
  • D 2*pi*sqrt(d<sup>3</sup>/(4GM))

Correct answer: A. 2*pi*sqrt(d<sup>3</sup>/(2GM))

Explanation: Each star orbits at distance d/2 from center. Centripetal force = GM<sup>2</sup>/d<sup>2.</sup> (d/2)x omega<sup>2</sup> x M = GM<sup>2</sup>/d<sup>2.</sup> omega<sup>2</sup> = 2GM/d<sup>3.</sup> T = 2*pi/omega = 2*pi*sqrt(d<sup>3</sup>/(2GM)).

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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