If four different masses `m_1, m_2,m_3` and `m_4` are placed at the four corners of a square of side `a`, the resultant gravitational force on a mass `m` kept at the centre is

Explanation

The force on m due to `m_(1)` and `m_(3)` is `F_(1)=(2Gm)/(a^(2))(m_(1)-m_(3))` along the diagonal towards
`m_(1)` [if `m_(1)gtm_(3)`]

The force on m due to `m_(2)` and `m_(4)` is
`F_(2)=(2Gm)/(a^(2))(m_(2)-m_(4))` along the diagonal towards
`m_(2)` [ if `m_(2)gtm_(4)`]
The resultant force is `sqrt(F_(1)^(2)+F_(2)^(2))=F`
`F=(2Gm)/(a^(2))sqrt((m_(1)-m_(3))^(2)+(m_(2)-m_(4))^(2))`
and the resultant force makes an angle `theta` with `F_(1)` where `theta=tan^(-1)(F_(2))/F_(1))`