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
Answer & explanation
Correct answer: option 4
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.
$\Rightarrow \Delta KE+\Delta PE=0$
$\Rightarrow\left(0-\frac{1}{2} m_0 v^2\right)+\left\{0-\left(-\frac{4 Gmm_{0}}{\ell}\right)\right\}=0$
$\Rightarrow v=2 \sqrt{\frac{2 Gm}{\ell}}$