Gravitation Ques 46
- Two bodies, each of mass $M$, are kept fixed with a separation $2 L$. A particle of mass $m$ is projected from the mid-point of the line joining their centres, perpendicular to the line. The gravitational constant is $G$. The correct statement(s) is (are)
(2013 Adv.)
(a) The minimum initial velocity of the mass $m$ to escape the gravitational field of the two bodies is $2 \sqrt{\frac{G M}{L}}$
(b) The minimum initial velocity of the mass $m$ to escape the gravitational field of the two bodies is $2 \sqrt{\frac{G M}{R}}$
(c) The minimum initial velocity of the mass $m$ to escape the gravitational field of the two bodies is $\sqrt{\frac{2 G M}{R}}$
(d) The energy of the mass $m$ remains constant
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Answer:
Correct Answer: 46.(b, d)
Solution:
Formula:
Let $v$ be the minimum velocity. From energy conservation,
$$ \begin{alignedat} U_{c}+K_{c} & =U_{\infty}+K_{\infty} \\ \therefore \quad m V_{c}+\frac{1}{2} m v^{2} & =0+0 \\ \therefore \quad v & =\sqrt{-2 V_{c}}=\sqrt{(-2) \frac{-2 G M}{L^2}} \\ & = 2 \sqrt{\frac{G M}{L}} \end{aligned} $$
BETA
