Four point mass bodies of masses as shown in Fig. are placed at the vertices of a square `ABCD`, gravitational force on the body at `A`. Given, `G = 6.6 xx 10^(-11) Nm^(2) kg^(-2)`

Explanation

Here, `AC = sqrt(AB^(2) + BC^(2)) = sqrt(1^(2) + 1^(2)) = sqrt(2) m`
Gravitational force on body `A` due to body at `B` is
`F_(1) (Gm_(1)m_(2))/(r^(2)) = (6.6 xx 10^(-11) xx 2xx 4)/(1^(2))`
`= 52.8 xx 10^(-11) N along AB`

Gravitational force on body at `A` due to body at `C`
is `F_(2) = (Gm_(1)m_(3))/(r_(1)^(2)) = (6.6 xx 10^(-11) xx 2xx 2)/((sqrt(2))^(2))`
`= 13.2 xx 10^(-11) N along AC`
Gravitational force on body at `A` due to body at `AC`
is `F_(3) = (Gm_(1)m_(4))/(r^(2)) = (6.6 xx 10^(-11) xx 2xx 4)/(1^(2))`
`= 52.8 xx 10^(-11) N along AD`
As `vec(F_(1))` and `vec(F_(3))` are perpendicular to each other and equal in magnitude, their effective gravitational force is
`F' = sqrt(F_(1)^(2) + F_(3)^(2)) = sqrt(F_(1)^(2) + F_(1)^(2))`
`= F_(1) sqrt(2) = (52.8 xx 10^(-11)) sqrt(2) N along AC`
Resultant gravitaitonal force on body at `A` is
`F = F' + F_(2) = (5.8 xx 10^(-11) xx sqrt(2)) + 13.2 xx 10^(-11)`
`= (74.66 + 13.2) xx 10^(-11)`
`= 87.86 xx 10^(-11) N along AC`