QJEEM14Ph8c85: JEE Main 2014 Physics - Class 11 JEEM14Ph8c85 Solution: Four particles, each of mass and equidistant from each other,…

All those particles are moving due to their mutual gravitational attraction.

The force between each masses are repulsive force.

On mass M at C, due to mass at D the repulsive force is F in the vertical direction.

On mass M at C, due to mass at B the repulsive force is F in the horizontal direction.

On mass M at C, due to mass at A the repulsive force is F'

Net force acting on particle at C,

= $2F\,\cos \,{45^ \circ } + F'$

Where $F = {{G{M^2}} \over {{{\left( {\sqrt 2 R} \right)}^2}}}$ and $F' = {{G{M^2}} \over {4{R^2}}}$

$\Rightarrow {{2 \times G{M^2}} \over {\sqrt 2 {{\left( {R\sqrt 2 } \right)}^2}}} + {{G{M^2}} \over {4{R^2}}}$

$\Rightarrow {{G{M^2}} \over R^2}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} \right]$

This net force will balance by the centripetal force F_cp = ${{M{v^2}} \over R}$

$\therefore$ ${{M{v^2}} \over R} =$ ${{G{M^2}} \over R^2}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} \right]$

$\Rightarrow$ $v = \sqrt {{{Gm} \over R}\left( {{{\sqrt 2 + 4} \over {4\sqrt 2 }}} \right)}$

$= {1 \over 2}\sqrt {{{Gm} \over R}\left( {1 + 2\sqrt 2 } \right)}$