Mass of each star, M $= 2 \times 10^{30}$ kg Initial distance between two stars, $r = 10^{9}$ km $= 10^{12}$ m Initial potential energy of the system $=-\frac{G M M}{r}$ Total $\text{K.E.}$ of the stars $= \frac{1}{2} M v^{2} + \frac{1}{2} M v^{2} = M v^{2}$ where$v$ is the speed of stars with which they collide. When the stars are about to collide, the distance between their centres, $r' = 2R$ ∴ Final potential energy of two stars $=-\frac{G M M}{2 R}$ Since gain in K.E. is equal to loss in P.E. $\therefore M v^{2} = -\frac{G M M}{r} - \left( -\frac{G M M}{2 R} \right)$ $= \frac{-G M M}{r} + \frac{G M M}{2 R}$ $\Rightarrow v^{2} = \frac{-G M}{r} + \frac{G M}{2 R}$ or $v^{2} = 6.67 \times 10^{12}$$\Rightarrow v = 2.6 \times 10^{6} \,\mathrm{m} \,\mathrm{s}^{-1}$