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Two stars of masses 3 × 10³¹ kg 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 system the minimum speed that meteorite should have at O is - (Take gravitational constant: G = 6.67 × 10⁻¹¹ N m² kg⁻²)

1. 2.4 × 10⁴ m/s
2. 1.4 × 10⁵ m/s
3. 3.8 × 10⁴ m/s
4. 2.8 × 10⁵ m/s

Correct answer: 2.8 × 10⁵ m/s

Solution

Each star (M = 3x10^31 kg) is at distance r = d/2 = 1x10^11 m from the centre O. Potential energy per unit mass at O = -2GM/r. For escape, (1/2)v^2 = 2GM/r, so v = sqrt(4GM/r) = sqrt(4*6.67e-11*3e31/1e11) = approximately 2.8x10^5 m/s.

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