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 partic

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

TestHub · Accepted Answer

The gravitational potential at the mid-point $P = V = V_{1} + V_{2}$

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

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.

$\Rightarrow ∆ K . E . + ∆ P . E . = 0$

$\Rightarrow (0 - \frac{1}{2} m_{0} v^{2}) + \{0 - (- \frac{4 G m m_{0}}{l})\} = 0$

$\Rightarrow v = 2 \sqrt{\frac{2 G m}{l}}$

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