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  4. 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

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Q. 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

NTA AbhyasNTA Abhyas 2020Gravitation

Solution:

Solution
The gravitational potential at the mid-point $P=V=V_{1}+V_{2}$
$=\frac{- G m}{\left(\frac{l}{2}\right)}-\frac{G m}{\left(\frac{l}{2}\right)}=-\frac{4 G m}{l}$
The gravitational potential energy $U=-4\frac{G m m_{0}}{l}, \, m_{0}$ = 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.
$\Rightarrow \, \Delta K.E.+\Delta P.E.=0$
$\Rightarrow \, \left(0 - \frac{1}{2} m_{0} v^{2}\right)+\left\{0 - \left(- \frac{4 G m m_{0}}{l}\right)\right\}=0$
$\Rightarrow v=2\sqrt{\frac{2 G m}{l}}$