Question

Two stars of masses $3\times {10}^{31}\text{kg}$ 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. $3.8\times {10}^{5}m/s$
• B. $2.8\times {10}^{5}m/s$
• C. $1.4\times {10}^{5}m/s$
• D. $2.4\times {10}^{5}m/s$

Answer 1

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


M is mass of star and m is mass of meteorite.
$-\frac{GMm}{r}+\frac{-GMm}{r}+\frac{1}{2}m{V}^2=0$
$\frac{2GMm}{r}=\frac{1}{2}m{v}^2$
$\sqrt{\frac{4GM}{r}}=v$
$\sqrt{\frac{6.67\times {10}^{-11}\times 4\times 3\times {10}^{31}}{1\times {10}^{11}}}$
$v=2.8\times {10}^5\text{m/sec}$