Two stars of masses 3×10^{31}, kg each, and at distance 2×10^{11}, m rotate in a plane about their

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7.Gravitation

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Two stars of masses $3\times10^{31}\, kg$ each, and at distance $2\times10^{11}\, m$ 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 $G = 6.67\times10^{-11}\, Nm^2\, kg^{-2}$)

A

$2.4\times10^4\, m/s$

B

$1.4\times10^5\, m/s$

C

$3.8\times10^4\, m/s$

D

$2.8\times10^5\, m/s$

(JEE MAIN-2019)

Solution

$\frac{1}{2}m{v^2} + \frac{{2\left( { – GMm} \right)}}{r} = 0$

${V^2} = \frac{{4GM}}{r} = \frac{{4 \times 6.67 \times {{10}^{ – 11}} \times 3 \times {{10}^{31}}}}{{2 \times {{10}^{11}}}}$

$V = 20\sqrt 2 \times {10^4}\,m/s$

$ = 2.828 \times {10^5}\,m/s$

Standard 11

Physics

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