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Let us consider a body of mass $ m $ falling from a height $ h $ on earth’s surface.

Hence the weight of the body is $ mg $

Where, $ g $ is the acceleration due gravity of the earth.

Now we know that, if there are two objects of masses $ M $ and $ m $ respectively placed at a distance of $ r $ then the force acting between them will be-

$ \Rightarrow F = \dfrac{{GMm}}{{{r^2}}} $

Where, $ G $ is the Newton's gravitational constant, $ G{\text{ }} = 6.67{\text{ }}x{\text{ }}{10^{ - 11}}{\text{N}}{{\text{m}}^{\text{2}}}{\text{/k}}{{\text{g}}^{\text{2}}} $

So here in this case,

The force of attraction with which the earth attracts the falling object towards itself is,

$ \Rightarrow F = \dfrac{{GMm}}{{{{\left( {R + h} \right)}^2}}} $

Where, $ R $ is the radius of the earth and h is the distance above the earth’s surface from where the object is being thrown.

But, as we know $ R > > > > h $ , hence the above equation boils down to-

$ \Rightarrow F = \dfrac{{GMm}}{{{{\left( R \right)}^2}}} $

Now equating it with the weight of the object it becomes,

$ \Rightarrow \dfrac{{GMm}}{{{{\left( R \right)}^2}}} = mg $

Solving further we get,

$ \Rightarrow \dfrac{{GM}}{{{R^2}}} = g $

Which is the value of the acceleration due to gravity and from the above equation it is clear that $ g $ is independent from the mass of the falling object.

Please note that the distance between the object’s surface is considered negligible, since the radius of the Earth is very large. If the same concept is applied to a large considerable distance, then the value of g will be a different quantity. Also, we can derive from the above activity that whether a soft feather is being thrown down to earth or a heavy metal ball is thrown, the acceleration due to gravity will be the same in both cases.

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