Physics - Gravitation Question with Solution | TestHub

The P.E. of the body at C is given by

$U=\frac{G{M}_{1}m}{\frac{d}{2}}-\frac{G{M}_{2}m}{\frac{d}{2}}=-\frac{2Gm}{d}\left(M_1+M_2\right)$

Initially it is at rest, hence its K.E. = 0

$\therefore$Total energy of the body$=-\frac{2Gm}{d}\left(M_1+M_2\right)$and Binding energy$=\frac{2Gm}{d}\left(M_1+M_2\right)$

Let$v_e$be the escape velocity that should be given to the body and its K.E. = Binding energy

$\therefore \frac{1}{2}m{v}_e^2=\frac{2Gm}{d}\left(M_1+M_2\right)$

$\therefore v_e=2\sqrt{\frac{G\left(M_1+M_2\right)}{d}}$