To escape, the total energy of small particle must be zero.
TE=KE+U
0=21mV2+(2d−GMm)×2
⇒V=d8GM
=2×10118×6.67×10−11×3×1031
=2.8×105ms−1
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 stars,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=6.67×10−11Nm2kg−2)
Held on 10 Jan 2019 · Verified 6 Jul 2026.
2.4×104ms−1
3.8×104ms−1
2.8×105ms−1
1.4×105ms−1
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Work through every JEE Main Mechanics PYQ, year by year.