Two stars of masses $3\times10^{31}\ \text{kg}$ each, and at distance $2\times10^{11}\ \text{m}$ rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the stars' rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is (Take gravitational constant $G = 6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}$)
Answer: (C) $2.8\times10^5\ \text{m s}^{-1}$
O is $10^{11}\ \text{m}$ from each star. To escape, $\tfrac12mv^2 + mV_O = 0$ with
$$V_O = -\frac{2GM}{r} = -\frac{2\times6.67\times10^{-11}\times3\times10^{31}}{10^{11}} = -4.0\times10^{10}\ \text{J/kg}$$
$$v = \sqrt{2\times4.0\times10^{10}} \approx 2.8\times10^5\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics