Four identical masses m each are kept at points A, B, C & D shown in figure. Gravitational force on mass at point D (body centre) is -

Options

A. `(3Gm^(2))/(a^(2))`
B. `(12Gm^(2))/(a^(2))`
C. `(4Gm^(2))/(a^(2))`
D. `(4Gm^(2))/(3a^(2))`

Answer

4

Explanation

Let us put identical mass at E. Due to symmetry net force on mass at 'D' is equal to zero.

`therefore` Required force = Force due to mass placed at E ltbvrgt `=("Gm"^(2))/((sqrt3a//2)^(2))=("4Gm"^(2))/(3a^(2))`