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

Hint: 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.