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) $1.4 \times {10^5}m/s$
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

First, calculate the total energy of the entire system given in the question when the meteorite is at point O. Then calculate the total energy of the entire system at infinity (it is 0) and using the law of conservation of energy, equate these two energies and simplify to get answer.

Formula Used:
Gravitational potential Energy exerted by body 1 on body 2,
${U_{12}} = - \dfrac{{GMm}}{r}$.
Where, $G$ is the universal gravitational constant, $M$ is the mass of body 1, $m$ is the mass of body 2, $r$ is the distance between the two bodies.
Kinetic energy of a body in motion,
$KE = \dfrac{1}{2}m{v^2}$
Where, $m$ is the mass of the body, $v$ is the velocity of the body
Complete Step by Step Solution:
Since we are given that the mass of both the stars is same, let the mass of the stars be equal to $M$
These stars are separated by a distance $R = 2 \times {10^{11}}m$ . Since they have equal masses, their centre of mass will lie at the centre of the line joining the centres of the two spherical stars.

Hence, the centre of mass will lie at a distance of $1 \times {10^{11}}m$ from each of the stars on the line joining their centres. Let this point be O. The stars are rotating in a 2D plane around this point O.

Now the energy of this entire system when the meteorite is passing through O will be equal to the sum of gravitational potential energy due to each of the stars on the meteorite and the kinetic energy of the meteorite.
That is,
$\dfrac{{ - GMm}}{r} + \dfrac{{ - GMm}}{r} + \dfrac{1}{2}m{v^2}$ (considering the mass of meteorite is $m$ )
Simplifying,
$(2 \times \dfrac{{ - GMm}}{r}) + \dfrac{1}{2}m{v^2} = \dfrac{{ - 2GMm}}{r} + \dfrac{1}{2}m{v^2}$
And the energy of this system when the meteorite is at infinity will be 0. (energy of any system at infinity is 0)
According to the law of conservation of energy, the total energy of an isolated system remains constant.
Hence, to satisfy this law, the energy of this system when the meteorite is at O must be equal to the energy of the system when the meteorite is at infinity.
i.e. $\dfrac{{ - 2GMm}}{r} + \dfrac{1}{2}m{v^2} = 0$
or,
$\dfrac{1}{2}m{v^2} = \dfrac{{2GMm}}{r}$ $\Rightarrow {v^2} = \dfrac{{4GM}}{r}$ ( $M$ gets cancelled)
which gives,
$v = \sqrt {\dfrac{{4GM}}{r}} = \sqrt {\dfrac{{4 \times 6.67 \times {{10}^{ - 11}} \times 3 \times {{10}^{31}}}}{{1 \times {{10}^{11}}}}}$ (value of $G$ and $M$ is given) (meteorite is at the centre of mass of the stars so its distance from the stars will be the same as that of centre of mass)
on solving this, we get $v = 2.8 \times {10^5}m/s$

Hence, option D is the correct answer.

Note: For law of conservation of energy, we take any point into consideration. It is not necessary to solve the question considering the system at infinity. However, the energy of the system at infinity gets 0 and the calculations get easier. Hence, it is a good option to consider this.