Four particles of masses m, 2m, 3m and 4m are placed at the four corners of a square of side a. Find the gravitational force on a particle of mass m placed at the centre of the square.

Explanation

Let O be the centre of the square ABCD of side a [Fig.1.13].

`therefore BD=sqrt(a^2+a^2)=sqrt(2)a`
Here , OA =OB=OC=OD
`=(BD)/2 =a/sqrt(2)=x` (say).
Suppose, the particle of mass m placed at O experiences gravitational forces `F_1,F_2,F_3,F_4` due to the particles placed at A ,B,C,D respectively.
Now, `F_1=G(mxxm)/(x^2)"along" vec(OA)`
`F_2=G(mxx2m)"along"vec(OB)`
`F_3=G(mxx3m)"along"vec(OC)`
`F_4=G(mxx4m)"along"vec(OD)`
`therefore` The resultant of `F_1 and F_3`,
`F^.=G(mxx3m)/(x^2)-G(mxxm)/(x^2)=G(2m^2)/(x^2) "along"vec(OC)`
and the resultant of `F_2 and F_4`,
`F^..=G(mxx4m)/(x^2)-G(mxx2m)/(x^2)=G(2m^2)/(x^2) "along"vec(OD)`.
Here `F^. and F^..` are equal in magnitude and at right angles to each other. The resultant of `F^. and F^..`,
`F=sqrt(F^(.2)+F^(..2)=F^.sqrt(2)`
`=(G.2m^2)/(x^2)xxsqrt(2)=(G.2m^2)/(a^2/2)xxsqrt(2)=4sqrt(2)(Gm^2)/(a^2)`