Gravitation 5 Question 3

3. Two stars of masses 3×1031kg each and at distance 2×1011 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=6.67×1011Nm2kg2 ) (Main 2019, 10 Jan II)

(a) 2.8×105m/s

(b) 3.8×104m/s

(c) 2.4×104m/s

(d) 1.4×105m/s

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Answer:

Correct Answer: 3. (a)

Solution:

  1. Let us assume that stars are moving in xy-plane with origin as their centre of mass as shown in the figure below

According to question, mass of each star, M=3×1031kg and diameter of circle,

2R=2×1011mR=1011m

Potential energy of meteorite at O, origin j^ is, Utotal =2GMmr

If v is the velocity of meteorite at O then Kinetic energy K of the meteorite is

K=12mv2

To escape from this dual star system, total mechanical energy of the meteorite at infinite distance from stars must be at least zero. By conservation of energy, we have

12mv22GMmR=0v2=4GMR=4×6.67×1011×3×10311011v=2.8×105m/s



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