We consider four masses of mass m1,m2,m3,m4 located in space at coordinates (x1,y1,z1), (x2,y2,z2),(x3.y3,z3) and (x4,y4,z4) respectively. Also there is another particle of mass M located at (x,y,z). What is the magnitude and direction of the net force acting on M ?

Suppose there are several point masses at different coordinates. Then as per center of mass concept, the situation is equivalent to having the aggregate mass located at the coordinates Xcm,Ycm,Zcm. Suppose there is another mass M located at a specific coordinate. Then the aggregate Force (inverse square force such as gravitational or electrostatic or magnetostatic) F2 on M due to several point masses is the vector sum of the forces due to individual point masses. We can also conjure a force F1 between the M and m=(m1+m2+m3+m4+.......) located at Xcm,Ycm,Zcm. Are Both the forces equal ? No. They are not.

F2 =KM[(m1Δx1 / l1³ +(m2Δx2 / l2³  +(m3Δx3 / l3³  + (m4Δx4 / l4³)i +

           (m1Δy1 / l1³ +(m2Δy2 / l2³  +(m3Δy3 / l3³  + (m4Δy4 / l4³)j +

           (m1Δz1 / l1³ +(m2Δz2 / l2³  +(m3Δz3 / l3³  + (m4Δz4 / l4³)k] =

     KM[Xi +Yj+Zk]

|F| =KM√(X² +Y² + Z²)

Xcm = (m1x1+m2x2+m3x3+m4x4)/(m1+m2+m3+m4)

Ycm = (m1y1+m2y2+m3y3+m4y4)/(m1+m2+m3+m4)

Zcm = (m1z1+m2z2+m3z3+m4z4)/(m1+m2+m3+m4)

 F1 =KMm / (r*r)  where

r= (X-Xcm)i +(Y - Ycm)j +(Z-Zcm)k

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One Dimensional Case :

We take 2 masses m1,m2 at (x1,0) and (x2,0) respectively. Then, there is a mass M at (x,0)

Then F1=KM(m1+m2)³ / [m1(x-x1)+m2(x-x2)]² where F1 is the force on M due to the mass (m1+m2) located at center of mass

         F2=KM[m1(x-x2)² ± m2(x-x1)²] / (x-x1)²(x-x2)² ; F2 is the net gravitational force on M due to m1 and m2.

We take m2=Xm1, d1=x-x1 , d2=x-x2, d2=yd1

then F1/F2 = (1+X)³y² /(1+Xy)² (y²±X)

When F1=F2,

X=(2y³-3y²+1) /(-y⁴ +3y²-2y)

y=+1 not permitted, and for any real value of y, X=-Ve . Hence for all real mass and locations F1 and F2 cannot become equal.