Consider a body of mass m on the surface of the Earth as shown in the Fig.. Its distance from the centre of the Earth is R (radius of the Earth).
The gravitational force experienced by the body is F = GMm / R2 where M is the mass of the Earth.

**Acceleration due to gravity at the surface of the Earth**

Consider a body of mass m on the
surface of the Earth as shown in the Fig.. Its distance from the centre of the
Earth is *R* (radius of the Earth).

The gravitational force
experienced by the body is *F = **GMm / R ^{2}
*where

From Newton's second law of
motion,

Force *F* = *mg*.

Equating the above two forces,
GMm/R^{2} = mg

G=GM/R^{2}

This equation
shows that *g* is independent of the
mass of the body *m*. But, it varies
with the distance from the centre of the Earth. If the* *Earth is assumed to be a sphere of radius *R*, the value of *g* on the
surface of the Earth is given by *g* =
GM/R^{2}.

*Mass of the Earth*

From the expression *g* = GM/R^{2} mass of the Earth
can be calculated as follows :

M =gR^{2}/G = 5.98 × 10^{24} kg

**Acceleration due to gravity**

Galileo was the
first to make a systematic study of the motion of a body under the gravity of
the Earth. He dropped various objects from the leaning tower of Pisa and made
analysis of their motion under gravity. He came to the conclusion that *'in the absence of air, all bodies* *will fall at the same rate'. *It is the
air resistance that slows down a piece* *of
paper or a parachute falling under gravity. If a heavy stone and a parachute
are dropped where there is no air, both will fall together at the same rate.

Experiments showed that the
velocity of a freely falling body under gravity increases at a constant rate.
(i.e) with a constant acceleration. The acceleration produced in a body on
account of the force of gravity is called *acceleration
due to gravity*. It is denoted by *g*.
At a given place, the value of *g* is
the same for all bodies irrespective of their masses. It differs from place to
place on the surface of the Earth. It also varies with altitude and depth.

The value of *g*
at sea−level and at a latitude of 45^{o} is taken as
the standard (i.e) *g* = 9.8 m s^{-2}

** **

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