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

Solution can be obtained by considering the potential conservation formula applied on all the parameters viz, stars, meteorite which is given as below,
$\dfrac{{ - GMm}}{r}$ for a star and meteorite where, G is gravitational constant, M is mass of star, m is mass of meteorite and r is radius of the rotation plane.

Complete step by step answer:
Given:
Stars of masses $3 \times {10^{31}}kg$, distance $2 \times {10^{11}}m$, Gravitational constant $G = 6.67 \times {10^{11}}N{m^2}k{g^{ - 2}}$
A meteorite passes through O moving perpendicular to the star's rotation plane.
The quantity of work done in shifting a unit take a look at mass from infinity into the
Gravitation has an impact on supply mass is called gravitational potential.
Basically, it's the gravitational potential energy possessed through a unit mass
$\Rightarrow {\rm{ }}V{\rm{ }} = {\rm{ }}U/m$
$\Rightarrow {\rm{ }}V{\rm{ }} = {\rm{ }} - GM/r$
For star and meteorite energy conservation we have to put potential energy conservation formula as, -$\dfrac{{ GMm}}{r}$
So, for two stars we can get,
-$\dfrac{{ GMm}}{r} + \dfrac{{ - GMm}}{r} + \dfrac{1}{2}m{V^2} = 0 + 0$
So we get, $V = 4GM/r$
Therefore put values of Gravitational constant, Masses of star and meteorite and radius of rotation path we get,
$2.8 \times {10^5}m/s = v$

So answer is option D. $2.8 \times {10^5}m/s$.

In order to escape from the gravitational field of this double star, the minimum speed that a meteorite should be $2.8 \times {10^5}m/s$.

Note: Gravitational Potential Energy at Height $\;h$, Derive $U{\rm{ }} = {\rm{ }}mgh$, and $v = - GM/r$ When, $h$ is less than $2$. The weight of a frame on the centre of the earth is zero because of the reality that the cost of $g$ on the centre of the earth is zero.