Two stars of masses m and 2m are co-rotating about their centre of mass. Their centres are at a distance r apart. If r is much larger than the sizes of the stars( r being large compared to size of either star) than the kinetic energies of stars of masses m and 2m are in the ratio 

 The distance x of the star of mass m from the centre of mass is given by

                $\frac{m}{x}=\frac{2m}{\left(r-x\right)}$

which gives x = $\frac{r}{3}$. The orbital speed v₁ of the star of mass m₁ = m is given by (here m₂ = 2m) 

              $\frac{G{m}_1{m}_2}{r^2}=\frac{m_1{v_1}^2}{x}=\frac{m_1{v_1}^2}{r/3}$

which gives       $v_1=\sqrt{\frac{G{m}_2}{3r}}=\sqrt{\frac{2Gm}{3r}} \left(i\right)$

        The orbital speed v₂ of the star of mass m₂ = 2m is given by 

      $$\begin{gathered} \frac{G{m}_1{m}_2}{r^2}=\frac{m_2{v_2}^2}{\left(r-x\right)} or \frac{Gm\left(2m\right)}{r^2}=\frac{2m_2{v_2}^2}{2r/3} \\ or v_2=\sqrt{\frac{2Gm}{3r}}=v_1 \left[see Eq.\left(i\right)\right] \\ Now K_1=\frac{1}{2}{m}_1{v_1}^2=\frac{1}{2}m{v_1}^2 and K_2=\frac{1}{2}{m}_2{v_2}^2=\frac{1}{2}\left(2m\right){v_2}^2 \end{gathered}$$

$\therefore \frac{K_1}{K_2}=\frac{{v_1}^2}{2{v_2}^2}=\frac{1}{2},$

since v₁ = v₂. Hence the correct choice is ( 3 ).