Q. 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 $G = 6.67 \times 10^{-11} \; Nm^2 \; kg^{-2}$)

Solution:

By energy convervation between $0 \; \& \; \infty$
$- \frac{GMm}{r} + \frac{-GMm}{r} + \frac{1}{2} mV^{2} = 0+0 $
[M is mass of star m is mass of meteroite)
$ \Rightarrow v = \sqrt{\frac{4GM}{r}} =2.8 \times10^{5} m/s $

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