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Question:

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, the minimum speed that meteorite should have at O is : (take Gravitational constant G=6.67×10⁻¹¹ Nm²/kg²)

3.8×10⁴m/s

1.4×10⁵m/s

24×10⁴m/s

2.8×10⁵m/s

Solution:

By energy conservation between 0 → ∞
-GMm/r + ½mv² = 0 + 0
[M is mass of star, m is mass of meteorite]
→v = √(4GM/r) = 2.8×10⁵m/s