Two bodies each of mass M, are kept fixed with a separation 2L. A particle of mass m is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is G. The correct statement (s) is (are)

Engineering

Physics

Physical Quantities

Question

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 2 $\sqrt{\frac{GM}{L}}$

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 4 $\sqrt{\frac{GM}{L}}$

The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is $\sqrt{\frac{2 GM}{L}}$

The energy of the mass m remains constant

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$\frac{GMm}{L} \times 2 = \frac{1}{2} {mv}^{2}$

$v = 2 \sqrt{\frac{am}{L}}$

energy of system is constant, not of the mass m

Please subscribe our Youtube channel to unlock this solution.