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^{2}) by putting a mass of 1 kg of sphere on the surface of earth.

ACTIVE PART OF EQUATION - IMAGINE 1 kg SPHERE ON EARTH

**F1=Weight of sphere=W=GMm/R**,

^{2}=mg_{1}= 9.8 NewtonWhere M= mass of earth, m= mass of sphere, R = is c/c distance b/w sphere and earth, G= gravitational constant, g

_{1}=GM/R

^{2}= gravitational acceleration say 9.8 m/s/s

REACTIVE PART OF EQUATION - IMAGINE EARTH ON 1 kg SPHERE

F2=Weight of Earth on sphere=W=GmM/R

^{2}=Mg

_{2}Newton

Where g

_{2}=Gm/R

^{2}= sphere's gravitational pull at height R or accelaration of earth due to gravity of sphere = 1.6395685e -18 m/s/s.

**F2=W=g**

_{2}M= 1.6395e-18 x 5.9742e+24 = 9795110.168782 NewtonOR

It would be difficult to find gravitatinal pull of sphere if we put another same sphere on aforementioned sphere. But anyway if g

_{2 }= Gm/r

^{2}=6.67x10

^{-11}m/s/s

where m= 1 kg and radius of sphere = 1 m then

**F2= 398479140000000 Newton**

So in both cases F2>F1, Since both forces F1 and F2 should equal in magnitude (g

_{1}m=g

_{2}M)

**why F2 > F1 ?**