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Two stars, each of mass 3*10³¹ kg, are separated by 2*10¹¹ m and revolve about their common centre of mass O. A meteorite crosses O travelling perpendicular to the plane of rotation of the stars. Find the minimum speed the meteorite must have at O to escape the gravitational field of this binary system. (G = 6.67*10⁻¹¹ N*m²/kg²)

1. 2.8*10⁵ m/s
2. 1.4*10⁵ m/s
3. 24*10⁴ m/s
4. 3.8*10⁴ m/s

Correct answer: 2.8*10⁵ m/s

Solution

At O each star is at distance r = d/2 = 1*10¹¹ m. Potential energy of meteorite (mass m) = -2*G*M*m/r. For escape: (1/2)*m*v² = 2*G*M*m/r, so v = sqrt(4*G*M/r). Substituting M = 3*10³¹, r = 1*10¹¹: v = sqrt(4*6.67*10⁻¹¹*3*10³¹/10¹¹) = sqrt(8.004*10¹⁰) approx 2.83*10⁵ m/s.

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