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 $G=6.67\times {10}^{(-11)}N{m}^{2}k{g}^{(-2)})$

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

Answer 1

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

Formula used: $U=\frac{-GMm}{r},KE=\frac{1}{2}m{v}^2$

Given,
$M$ is mass of star $=3\times {10}^{31}kg$

$m$ is mass of meteorite $=3\times {10}^{31}kg$

Distance$=2\times {10}^{11}m$
By energy conservation between $O$ and $\infty$.

$-\frac{GMm}{r}+-\frac{-GMm}{r}+\frac{1}{2}m{V}^2=0+0$

$v=\sqrt{\frac{4GM}{r}}=2.8\times {10}^{5}m/s$

Final Answer:(d)