Question

Derive the expression of the acceleration due to gravity by earth and define its
dependency on mass of falling object.​

Answer 1

What universal law of gravitation says is that the gravitational force of attraction between two bodies of masses m_1 and m_2 separated by a distance d from each other is given by,

$F=\dfrac {Gm_1m_2}{d^2}$

In the case of the gravitational force of attraction between a body of mass m and radius r placed on the surface of the earth and the earth of mass M, if the radius of the earth is R, then,

$F=\dfrac {GMm}{(R+r)^2}$

But r << R. Thus,

$F=\dfrac {GMm}{R^2}$

Well, the weight of the body completely provides the gravitational force of attraction, since the weight of the body is also acting towards the earth. Thus,

$mg=\dfrac {GMm}{R^2}$

where 'g' is the acceleration due to gravity, and 'mg', we know, is the weight of the body. Then, we get that,

$\boxed {\mathbf{g=\dfrac {GM}{R^2}}}$

Thus an expression for the acceleration due to gravity is derived!

From the expression it is clear that the acceleration due to gravity does not depend on the mass of the body, m. This implies every particle falling on the earth have the same acceleration, so that they reach the surface of earth at the same time, in the absence of air, when dropped from the same height, irrespective of their mass.

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