Answer to Question #283010 in Mechanics | Relativity for mew

Question

A pair of stars revolves about their common center of mass

as in Fig. 1. One of the stars has a mass M that is twice the

mass m of the other star. Their centers are a distance d apart

(d being large compared to the size of either star). All units

are in SI.

(a) (4 marks) Derive an expression for the period of

revolution of the stars about their common center of mass in

terms of d, m, and G.

(b) (3 marks) Compare the angular momenta of the two stars

about their common center of mass by calculating the ratio

Lm /LM .

(c) (3 marks) Compare the kinetic energies of the two stars by

calculating the ratio Km /KM .

Answer 1

a)

stars will rotate about their c.m. position of centre of mass:

$0=\frac{2m(-x)+m(d-x)}{2m+m}\implies x=d/3$

force of atraction will provide required centripetal force:

$\frac{2mv^2_M}{d/3}=\frac{GMm}{d^2}\implies v_M=\sqrt{\frac{Gm}{3d}}$

period of revolution of the stars:

$T=\frac{2\pi (d/3)}{v_M}=\frac{2\pi d^{3/2}}{\sqrt{3Gm}}$

b)

for angular momentum:

$\frac{L_M}{L_m}=\frac{Mv_1d_1}{mv_2d_2}=\frac{M}{m}(\frac{m}{M})^2=m/M=1/2$

c)

for kinetic energies:

$\frac{K_M}{K_m}=\frac{Mv_1^2}{mv_2^2}=\frac{M}{m}(\frac{m}{M})^2=m/M=1/2$