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

$2 \sqrt{\frac{2 G M}{\ell}}$

The gravitational potential at the mid-point P = V = V₁ + V₂

$=\frac{-G m}{(\ell / 2)}-\frac{G m}{(\ell / 2)}=-\frac{4 G m}{\ell}$

⇒ The gravitational potential energy

$=U=-\frac{4 G m_0}{\ell}, m_0$ = mass of particle

When it is projected with a speed v, it just escapes to infinity, and the potential & kinetic energy will become zero.

$\Rightarrow \Delta KE+\Delta PE=0$

$\Rightarrow\left(0-\frac{1}{2} m_0 v^2\right)+\left\{0-\left(-\frac{4 Gmm_{0}}{\ell}\right)\right\}=0$

$\Rightarrow v=2 \sqrt{\frac{2 Gm}{\ell}}$