Q44 / 60 · JEE Advanced 2013 (Paper 2) · Physics

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

A
The minimum initial velocity of the mass $m$ to escape the gravitational field of the two bodies is $4 \sqrt{\frac{G M}{L}}$ .
Correct answer
B
The minimum initial velocity of the mass $m$ to escape the gravitational field of the two bodies is $2 \sqrt{\frac{G M}{L}}$ .
C
The minimum initial velocity of the mass $m$ to escape the gravitational field of the two bodies is $\sqrt{\frac{2 G M}{L}}$ .
Correct answer
D
The energy of the mass $m$ remains constant.

Solution

Applying mechanical energy conservation,

$U_{i} + K E_{i} = U_{f} + K E_{f}$
$- \frac{G M m}{L} \times 2 + \frac{1}{2} m v^{2} = 0 + 0$
$\Rightarrow v = \sqrt{\frac{4 G M}{L}}$

$\Rightarrow v = 2 \sqrt{\frac{G M}{L}}$

Hence, the correct answer is options: 'B', 'D'.

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

Solution

Relevant Questions

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