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 that meteorite should have at $O$ is : (Take Gravitational constant $G=6.6\times 10^{-11}N{m}^{2}k{g}^{-2}$)

• A. $1.4\times 10^{5}m/s$
• B. $24\times 10^{4}m/s$
• C. $3.8\times 10^{4}m/s$
• D. $2.8\times 10^{5}m/s$

Answer 1

Answer: (D)

By energy convervation between $0\&\infty$
$-\frac{GMm}{r}+\frac{-GMm}{r}+\frac{1}{2}m{V}^2=0+0$
[M is mass of star m is mass of meteroite)
$\Rightarrow v=\sqrt{\frac{4GM}{r}}=2.8\times 10^{5}m/s$