Question

two stars of masses 3*10[ power]31 kg each , and at distance 2*10[power]11m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the stars rotation plane. In order to escape from gravitation field of this double star,the minimum speed that meteorite should have at O is [take gravitational constant G=6.67*10[power]{-11}

Answer 1

Answer:

The minimum speed of meteorite at the center is $2.8\times10^{5}\ m/s$

Explanation:

Given that,

Mass of stars$M =3\times10^31\ kg$

Distance $d =2\times10^{11}\ m$

The radius will be

$R=\dfrac{d}{2}=\dfrac{2\times10^{11}}{2}=1\times10^{11}\ m$  

Potential energy of the meteorite while passing through the center

P.E=$\dfrac{-2GMm}{R}$

Let v is the minimum speed of meteorite.

The total energy is

Total energy = potential energy +kinetic energy

$T.E=\dfrac{-2GMm}{R}+\dfrac{1}{2}mv^{2}$

To escape the gravitational field of both starts , meteor must have total energy equal to zero.

$\dfrac{-2GMm}{R}+\dfrac{1}{2}mv^{2}=0$

$v=\sqrt{\dfrac{4GM}{R}}$

Put the value into the formula

$v=\sqrt{\dfrac{4\times6.67\times10^{-11}\times3\times10^{31}}{1\times10^{11}}}$

$v=2.8\times10^{5}\ m/s$

Hence, The minimum speed of meteorite at the center is $2.8\times10^{5}\ m/s$