When objects fall on the Earth, the acceleration of the object is towards the Earth. From Newton’s second law, an object is accelerated only under the action of a force.

**ACCELERATION
DUE TO GRAVITY OF THE EARTH**

When
objects fall on the Earth, the acceleration of the object is towards the Earth.
From Newton’s second law, an object is accelerated only under the action of a
force. In the case of Earth, this force is the gravitational pull of Earth.
This force produces a constant acceleration near the Earth’s surface in all bodies,
irrespective of their masses. The gravitational force exerted by Earth on the
mass m near the surface of the Earth is given by

Now
equating Gravitational force to Newton’s second law,

The
acceleration experienced by the object near the surface of the Earth due to its
gravity is called acceleration due to gravity. It is denoted by the symbol g.
The magnitude of acceleration due to gravity is

It
is to be noted that the acceleration experienced by any object is independent
of its mass. The value of g depends only on the mass and radius of the Earth.
Infact, Galileo arrived at the same conclusion 400 years ago that *all objects fall towards the* *Earth with the same acceleration *through* *various quantitative experiments. The
acceleration due to gravity g is found to be 9.8 m s^{−2} on the
surface of the Earth near the equator.

Consider
an object of mass m at a height h from the surface of the Earth. Acceleration
experienced by the object due to Earth is

If
*h* << *R _{e}*

We
can use Binomial expansion. Taking the terms upto first order

We
find that *g’<* *g *. This means that as* *altitude
h increases the acceleration due to gravity g decreases.

1.
Calculate the value of g in the following two cases:

a)
If a mango of mass ½ kg falls from a tree from a height of 15 meters, what is
the acceleration due to gravity when it begins to fall?

b)
Consider a satellite orbiting the Earth in a circular orbit of radius 1600 km
above the surface of the Earth. What is the acceleration experienced by the
satellite due to Earth’s gravitational force?

The
above two examples show that the acceleration due to gravity is a constant near
the surface of the Earth.

Consider
a particle of mass m which is in a deep mine on the Earth. (Example: coal mines
in Neyveli). Assume the depth of the mine as d. To calculate *g*′ at a depth d, consider the
following points.

The
part of the Earth which is above the radius (*R _{e}* −

Here
*M* ’
is the mass of the Earth of radius (*R _{e}-d)*

Assuming
the density of Earth ρ to be constant,

where
*M* is the mass of the Earth and V its
volume, Thus,

Thus

Here
also g ′ < g . As depth increases, g′
decreases. It is very interesting to know that acceleration due to gravity is
maximum on the surface of the Earth but decreases when we go either upward or
downward.

Whenever we analyze the motion of objects in rotating frames
[explained in chapter 3] we must take into account the centrifugal force. Even
though we treat the Earth as an inertial frame, it is not exactly correct
because the Earth spins about its own axis. So when an object is on the surface
of the Earth, it experiences a centrifugal force that depends on the latitude
of the object on Earth. If the Earth were not spinning, the force on the object
would have been *mg*. However, the
object experiences an additional centrifugal force due to spinning of the
Earth.

This
centrifugal force is given by *m*ω^{2}*R*’ .

where
λ is the latitude. The component of centrifugal acceleration experienced by the
object in the direction opposite to g is

Therefore,

From
the expression (6.52), we can infer that at equator, λ = 0; *g’= g *- ω^{2}*R*. The acceleration due to gravity is
minimum. At poles λ= 90; *g’=g*, it is maximum. At the equator, *g’* is minimum.

Find
out the value of *g*′ in your school laboratory?

Calculate
the latitude of the city or village where the school is located. The
information is available in Google search. For example, the latitude of Chennai
is approximately 13 degree.

*g
***′ =***
g ***−ω*** *^{2}* R ***cos ^{2} **

Here
*ω*^{2}*R* = (2x3.14/86400)^{2} x
(6400x10^{3}) = 3.4x10^{−2} m s^{−2}.

It
is to be noted that the value of λ should be in radian and not in degree. 13
degree is equivalent to 0.2268 rad.

*g*′ =* *9.8* *−* *(*
*3.4* *×*
*10^{−}^{2}* *) ×*
*(* *cos 0.2268)^{2}

*g* = 9.7677 m s^{−2}

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11th Physics : UNIT 6 : Gravitation : Acceleration Due to Gravity of the Earth |

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