JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2019
- Medium
- MCQ
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 \(G = 6.67\times10^{-11}\, Nm^2\, kg^{-2}\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(2.8\times10^5\, m/s\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{2}m{v^2} + \frac{{2\left( { - GMm} \right)}}{r} = 0\)
\({V^2} = \frac{{4GM}}{r} = \frac{{4 \times 6.67 \times {{10}^{ - 11}} \times 3 \times {{10}^{31}}}}{{2 \times {{10}^{11}}}}\)
\(V = 20\sqrt 2 \times {10^4}\,m/s\)
\( = 2.828 \times {10^5}\,m/s\)
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