A pair of stars rotates about a common centre of mass. One of the stars has a mass M and the other m. Their centres are distance d apart, d being large compared to the size of either star. Then

The centre of mass of system divides the distance between the stars in the inverse ratio of their masses. If d₁ and d₂ are the distance force for their circular motion.
${\mathrm{d}}_{\mathrm{1}}\mathrm{=}\frac{\mathrm{d}}{\mathrm{M}+\mathrm{m}}\times \mathrm{m}$ and ${\mathrm{d}}_{\mathrm{1}}\mathrm{=}\frac{\mathrm{d}}{\mathrm{M}+\mathrm{m}}\times \mathrm{m}\times \mathrm{M}\Rightarrow \frac{{\mathrm{d}}_{\mathrm{1}}}{{\mathrm{d}}_2}=\frac{\mathrm{m}}{\mathrm{M}}$
stars will rotate in circles of radii d₁ and d₂ about their centre of mass. 

The same force of attraction provides the necessary centripetal force for their circular motion.
$\mathrm{\therefore }\frac{\mathrm{GmM}}{\mathrm{D}}={\mathrm{M\omega }}_1^2{\mathrm{d}}_1={\mathrm{m\omega }}_2^2{\mathrm{d}}_2$
$\Rightarrow$ ${\mathrm{\omega }}_1^2=\frac{\mathrm{Gm}}{{\mathrm{d}}^2{\mathrm{d}}_1}=\frac{\mathrm{Gm}}{{\mathrm{d}}^2}\times \frac{\mathrm{M}+\mathrm{m}}{\mathrm{d}\times \mathrm{m}}=\frac{\mathrm{G}\left(\mathrm{M}+\mathrm{m}\right)}{{\mathrm{d}}^3}$ and ${\mathrm{\omega }}_1={\mathrm{\omega }}_2=\sqrt{\frac{\mathrm{G}\left(\mathrm{M}+\mathrm{m}\right)}{{\mathrm{d}}^3}}$
From the fact that the moment of momentum is also the angular momentum
$\frac{{\mathrm{L}}_{\mathrm{M}}}{{\mathrm{L}}_{\mathrm{m}}}=\frac{\left({\mathrm{Mv}}_1\right){\mathrm{d}}_1}{\left({\mathrm{mv}}_2\right){\mathrm{d}}_2}=\frac{\mathrm{M}}{\mathrm{m}}\times \frac{{\mathrm{d}}_1^2}{{\mathrm{d}}_2^2}\Rightarrow \frac{{\mathrm{L}}_{\mathrm{M}}}{{\mathrm{L}}_{\mathrm{m}}}=\frac{\mathrm{m}}{\mathrm{M}}$
$\Rightarrow \frac{{\mathrm{K}}_{\mathrm{M}}}{{\mathrm{K}}_{\mathrm{m}}}=\frac{\frac{1}{2}{\mathrm{Mv}}_1^2}{\frac{1}{2}{\mathrm{mv}}_2^2}=\frac{\mathrm{M}}{\mathrm{w}}\frac{{\mathrm{\omega }}_1^2{\mathrm{D}}_1^2}{{\mathrm{\omega }}_2^2{\mathrm{d}}_2^2}=\frac{\mathrm{M}}{\mathrm{m}}\frac{{\mathrm{d}}_1^2}{{\mathrm{d}}_2^2}\hspace{0.17em}\hspace{0.17em}\left(\because \hspace{0.17em}\hspace{0.17em}{\mathrm{\omega }}_1={\mathrm{\omega }}_2\right)$
Therefore, $\frac{{\mathrm{K}}_{\mathrm{M}}}{{\mathrm{K}}_{\mathrm{m}}}=\frac{\mathrm{M}}{\mathrm{m}}{\left(\frac{\mathrm{m}}{\mathrm{M}}\right)}^2=\frac{\mathrm{m}}{\mathrm{M}}$