Question

A particle is projected from the mid-point of the line joining two fixed particles each of mass m. If the separation between the fixed particles is $l$ , the minimum velocity of projection of the particle so as to escape is equal to

• A. $\sqrt{\frac{Gm}{l}}$
• B. $\sqrt{\frac{Gm}{2l}}$
• C. $\sqrt{\frac{2Gm}{l}}$
• D. $2\sqrt{\frac{2Gm}{l}}$

Answer 1

Answer: (D)

The gravitational potential at the mid-point $P=V=V_1+V_2$
$=\frac{-Gm}{\left(\frac{l}{2}\right)}-\frac{Gm}{\left(\frac{l}{2}\right)}=-\frac{4Gm}{l}$
The gravitational potential energy $U=-4\frac{Gm{m}_0}{l},m_0$ = mass of particle
When it is projected with a speed v, it just escapes to infinity, and the potential and kinetic energy will become zero.
$\Rightarrow \Delta K.E.+\Delta P.E.=0$
$\Rightarrow \left(0-\frac{1}{2}{m}_0{v}^2\right)+\left\{0-\left(-\frac{4Gm{m}_0}{l}\right)\right\}=0$
$\Rightarrow v=2\sqrt{\frac{2Gm}{l}}$