Problem 11.45

11.45.

To see why the hint is true, let the center of mass be at the origin. Then
$0 = \frac{(Mr_{2} - mr_{1})}{M+m}$
and it follows directly that Mr₂ = mr₁.
Since each revolves around the other with the same angular velocity $\omega$, then if we solve for the gravitational force on each one we find
$\frac{MGm}{d^{2}} = \frac{mv_{1}^{2}}{d} ~=~ mr_{1}\omega^{2}$
and
$\frac{MGm}{d^{2}} = \frac{Mv_{2}^{2}}{d} ~=~ Mr_{2}\omega^{2}$
dividing out an m from the first equation and an M from the second and adding together the two equations, we find
$r_{1}\omega^{2} + r_{2}\omega^{2} ~=~ \frac{G(m+M)}{d^{2}}$
$(r_{1}+r_{2})\omega^{2} = \frac{G(m+M)}{d^{2}}$
and since d = r₁+r₂,
$\omega^{2} = \frac{G(m+M)}{d^{3}}$
and since
$T = \frac{2\pi}{\omega}$
$T^{2} = \frac{4\pi^{2}}{\omega^{2}}$
substitution yields the desired result,
$T^{2} = \frac{4\pi^{2}d^{3}}{G(M+m)}$