Two stars of masses 3 × 1031 kg each, and at distance 2 × 1011 m… | JEE Main 2019 (10 Jan, Shift 2)

Q66JEE Main 2019 · 10 JAN · SHIFT 2Difficult

Escape Velocity

Two stars of masses 3 $\times$ 1031 kg each, and at distance 2 $\times$ 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 $\times$ 10–11 Nm2 kg–2)

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Answer: (d)

O is the midpoint, $d = 10^{11}\text{ m}$ from each star: $V_O = -\dfrac{2Gm}{d}$ .
To escape: $v^2 = 2|V_O| = \dfrac{4Gm}{d}$
$v^2 = \dfrac{4 \times 6.67\times10^{-11} \times 3\times10^{31}}{10^{11}} = 8.0 \times 10^{10}$
$v \approx 2.8 \times 10^{5}\text{ m/s}$ , so the answer is (d).

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Escape Velocity question from JEE Main 2019 (10 Jan, Shift 2)

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