Question

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 $G=66\times 10^{-11}N{m}^{2}k{g}^{-2}$ )

Answer 1

Answer: 2.8

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