Two stars of masses $$3 \times 10^{31}$$ kg each, and at distance $$2 \times 10^{11}$$ m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star's 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 $$G = 6.67 \times 10^{-11}$$ N m$$^2$$ kg$$^{-2}$$)

Let the mass of each star be $$M = 3 \times 10^{31}\,$$kg and let the distance between their centres be $$d = 2 \times 10^{11}\,$$m.

The two stars are identical, so their common centre of mass O lies exactly midway between them. Therefore the distance of O from the centre of either star is

$$r = \dfrac{d}{2} = \dfrac{2 \times 10^{11}}{2} = 1 \times 10^{11}\,{\rm m}.$$

For a body of test mass $$m$$ placed at O, the gravitational potential due to one star is

$$V_1 = -\,\dfrac{G M}{r}.$$

The potential is negative because gravity is attractive. Since the contributions of the two stars simply add, the total gravitational potential at O is

$$V = V_1 + V_1 = -\,\dfrac{G M}{r} - \dfrac{G M}{r} = -\,\dfrac{2 G M}{r}.$$

To escape to infinity, the meteorite must have a total mechanical energy (kinetic + potential) that is zero or positive. The condition for minimum (escape) speed $$v_{\text{esc}}$$ at O is therefore

$$\dfrac{1}{2} m v_{\text{esc}}^{\,2} + m V = 0.$$

Substituting $$V = -\,\dfrac{2 G M}{r}$$ we obtain

$$\dfrac{1}{2} m v_{\text{esc}}^{\,2} - m \left(\dfrac{2 G M}{r}\right) = 0.$$

Dividing by $$m$$ and multiplying by 2,

$$v_{\text{esc}}^{\,2} = \dfrac{4 G M}{r}.$$

Taking the square root,

$$v_{\text{esc}} = 2 \sqrt{\dfrac{G M}{r}}.$$

Now we substitute the numerical values $$G = 6.67 \times 10^{-11}\,{\rm N\,m^2\,kg^{-2}},\; M = 3 \times 10^{31}\,{\rm kg},\; r = 1 \times 10^{11}\,{\rm m}:$$

$$\dfrac{G M}{r} = \dfrac{(6.67 \times 10^{-11})(3 \times 10^{31})}{1 \times 10^{11}} = 6.67 \times 3 \times 10^{-11 + 31 - 11} = 20.01 \times 10^{9} = 2.001 \times 10^{10}\,{\rm m^2\,s^{-2}}.$$

Hence

$$v_{\text{esc}} = 2 \sqrt{2.001 \times 10^{10}} = 2 \times \left(\sqrt{2.001}\right) \times 10^{5} \approx 2 \times 1.414 \times 10^{5} = 2.828 \times 10^{5}\,{\rm m\,s^{-1}}.$$

Written to two significant figures this is $$2.8 \times 10^{5}\,{\rm m\,s^{-1}}.$$

Hence, the correct answer is Option C.