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## Main Question or Discussion Point

Here is the succinct story of apple which fell on Sir Newton. Since both apple and earth attract each other with a force equal in magnitude from both sides but apple move faster towards earth because of its greater acceleration than negligible acceleration of earth. So Newton derived the equation of univesal law of gravitation (F= GMm/R

ACTIVE PART OF EQUATION - IMAGINE 1 kg SPHERE ON EARTH

Where M= mass of earth, m= mass of sphere, R = is c/c distance b/w sphere and earth, G= gravitational constant, g

REACTIVE PART OF EQUATION - IMAGINE EARTH ON 1 kg SPHERE

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

Where g

OR

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

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

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

^{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/sREACTIVE PART OF EQUATION - IMAGINE EARTH ON 1 kg SPHERE

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

^{2}=Mg_{2}NewtonWhere 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/swhere 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 ?**