Question

Two stars of masses $3\times {10}^{31}\text{kg}$ each, and at distance $2\times {10}^{11}\text{m}$ 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 $G=6.67\times {10}^{-11}{\text{Nm}}^2{\text{kg}}^{-2})$

• A. $2.4\times {10}^4\text{m/s}$
• B. $1.4\times {10}^5\text{m/s}$
• C. $3.8\times {10}^4\text{m/s}$
• D. $2.8\times {10}^5\text{m/s}$

Answer 1

The correct option is D $2.8\times {10}^5\text{m/s}$

Let $M$ is mass of star,
$m$ is mass of meteorite

By energy convervation between $O$ and $\infty$
$-\frac{GMm}{r}+\frac{-GMm}{r}+\frac{1}{2}m{V}^2=0+0$
$\therefore V=\sqrt{\frac{4GM}{r}}=\sqrt{\frac{4\times 6.67\times {10}^{-11}\times 3\times {10}^{31}}{{10}^{11}}}$
$V\approx 2.8\times {10}^5\text{m/s}$.