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Two stars of equal mass m orbit their common centre of mass, with their centres separated by a distance d (d much larger than either star's size). (a) Derive the period of revolution T in terms of d, m and G. (b) Find the ratio L_M/Lₘ of their angular momenta about the centre of mass. (c) Find the ratio Kₘ/K_M of their kinetic energies.

1. T = 2*pi*sqrt(d³/(2Gm)); L_M/Lₘ = 1; Kₘ/K_M = 1
2. T = 2*pi*sqrt(d³/(Gm)); L_M/Lₘ = 1; Kₘ/K_M = 1
3. T = 2*pi*sqrt(d³/(4Gm)); L_M/Lₘ = 2; Kₘ/K_M = 2
4. T = pi*sqrt(d³/(2Gm)); L_M/Lₘ = 1; Kₘ/K_M = 1

Correct answer: T = 2*pi*sqrt(d³/(2Gm)); L_M/Lₘ = 1; Kₘ/K_M = 1

Solution

With equal masses m, each star moves on a circle of radius d/2. Centripetal equation: Gm²/d² = m*omega²*(d/2), so omega² = 2Gm/d³ and T = 2*pi/omega = 2*pi*sqrt(d³/(2Gm)). By symmetry, both stars have equal speed, radius, and mass, so L_M/Lₘ = 1 and Kₘ/K_M = 1.

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