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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 (in $m / s$ ) that meteorite should have at $O$ is ......... $\times 10^{5}$:
(Take Gravitational constant $G =66 \times 10^{-11} Nm ^2 kg ^{-2}$ )

Gravitation

Solution:

Let $M$ is mass of star $m$ is mass of meteroite By energy convervation between 0 and $\infty$,
$-\frac{G M m}{r}+\frac{-G M m}{r}+\frac{1}{2} m V_{\text {ese }}^2=0+0$
$\therefore v =\sqrt{\frac{4 G M}{r}}=\sqrt{\frac{4 \times 6.67 \times 10^{-11} \times 3 \times 10^{31}}{10^{11}}} $
$\simeq 2.8 \times 10^5\, m / s$