Two stars of masses 3×1031kg each, and at distance 2×1011m 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=6.67×1011Nm2kg−2)
3.8×104m/s
24×104m/s
1.4×105m/s
2.8×105m/s
A
24×104m/s
B
3.8×104m/s
C
1.4×105m/s
D
2.8×105m/s
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Solution
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By energy convervation between 0 & ∞ −GMmr+−GMmr+12mV2=0+0 [M is mass of star m is mass of meteroite) ⇒v=√4GMr=2.8×105m/s
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