# Acceleration due to Gravity Below and Above the Surface of Earth

Weight of a body is defined as the force with which a body is attracted towards the centre of earth. Its units and dimensions are the same as those of force.

The force of attraction between any two material bodies in the universe is known as force of gravitation. However, if one of the attracting bodies is earth or some other planet or natural satellite, then the force of attraction is called force of gravity. So, gravity is merely a special case of gravitation.

Variation of 'g' above and below the surface of earth

Variation of g with height (altitude)
Let P be a point on the surface of Earth and Q be a point at an altitude h. Let the mass of the Earth be M and radius of the Earth be R. Consider the Earth as spherical shaped body.
The acceleration due to gravity at point P on the surface of the earth is
g = ----- (i)

Let the body be placed at Q at a height h from the surface of the Earth. The acceleration due to gravity at Q is
---- (ii)

Dividing (ii) by (i),

By simplifying and expanding using binomial theorem,
The value of acceleration due to gravity decreases with increase in height above the surface of the Earth.

# Variation of g with depth

Assuming the earth to be a homogenous sphere (having uniform density) of radius R and mass M. Let ρ be the mean density of earth. Let a body be lying on the surface of earth where the value of acceleration due to gravity is "g".
Then g =
g = â€¦â€¦â€¦.(8.16)

Let the body be now taken to a depth d below the free surface of earth where the value of acceleration due to gravity gd.

Here, the force of gravity acting on the body is only due to the inner solid sphere of radius (R-d).
gd =

Where M' is the mass of the inner solid sphere of radius (R-d).
or gd = or
gd =
= =
gd =
g-gd=g
Here (g - gd) gives the decrease in the value of g.
Since g is constant at a given place of the earth and R is also a constant.
g - gd d
From the above equation, it is clear that if d increases, gd must decrease because g is constant. Thus the value of acceleration due to gravity decreases with the increase of depth.

Note 1. The fractional decrease in the value of g is given by =
Note 2. The percentage decrease in the value of g is 100.

Weight of the body at the centre of earth:
At a depth d below the free surface of earth.
gd = g
At the centre of earth, d = R
∴ gd =g = 0

If m is the mass of a body lying at the centre of the earth, then its weight = mgd =0.
Hence the weight of body at the centre of earth is equal to zero.

Comparison of height and depth for the same change in 'g':

As we have seen above, the value of g decreases as we go above the surface of earth or when we go below the surface of earth. This can be taken to mean that value of g is maximum on the surface of earth.
Now gh = g
and gd = g
When gh - gd, then
= 1 -
= -
d = 2h
Thus, the value of acceleration due to gravity at a height h is same as the value of acceleration due to gravity at a depth d (=2h). But this is true if h is very small.

# Variation of g with latitude

The value of acceleration due to gravity changes with latitude due to:
(a) shape of earth and
(b) rotation of the earth about its own axis.

1. Shape of Earth
The earth is not a perfect sphere. It is flattened at the poles (where latitude is 900) and bulges out at the equator (where latitude is O0). The equatorial radius Re is greater than the polar radius Rp by nearly 21 kilometer.

We know that g =
Since G and M are constants,
g
Thus the value of g at a place on the surface of earth varies inversely as the square of the radius of earth at that place. Since the value of R is greatest, at the equator, therefore; the value of g is least at the equator. Also, the value of R is least at the poles and hence g is maximum at the poles.
In other words, the value of g increases as we move from the equator to the pole. The value of g at the poles, at sea level, is greater than the value of g at the equator at sea level by nearly 1.80 cm s-2.
2. Rotation of Earth
Let us assume that the earth is a homogenous sphere of mass M and radius R. The earth rotates about its polar axis from west to east. Consider a particle of mass m lying at a point P on the surface of earth. Let φ be the latitude of the point. If the earth were at rest, the particle at P would have been attracted towards the centre O of earth. So, the true weight of the particle is directed towards O.

Due to rotation of earth, the particle at P will describe a circle of radius r. For moving in a circle, the particle requires a centripetal force mrω 2. Here, ω is not only the angular velocity of the particle but also the angular velocity of earth.
The true weight mg of the particle can be resolved into two rectangular components -mrω2 and mg . Here, mg is the apparent weight of the particle at P. Due to rotation of earth, the particle at P is attracted towards a point O , which is close to O.
Applying cosine law, (mg )2 = (mg)2 + (mrω 2)2 â€“ 2mg mrω 2 cos φ
on simplification, g = g
But r = R cos φ
g = g
Let us now calculate the value of .
We know that R = 6378 km = 6378 103 m
ω = rad s-1 g = 9.81 m s-2
=
Since is very small, therefore, squares and higher powers of can be neglected.
∴γ = g
Expanding the right hand side of the above equation by Binomial Theorem and neglecting squares and higher powers, we get
g = g
or g = g
or g = g - Rω 2 cos2φ
As φ increases, cos φ decreases and g will increase. So, the value of g increases as we move from equator to pole.
At equator, φ = 00
ge = g - Rω 2
At poles, φ = 900
gp = g

So, the value of acceleration due to gravity is maximum at poles. Moreover, the value of g at the poles will remain the same whether the earth is rotating or stationary.
The difference in the values of acceleration due to gravity at the pole and equator is given by
gp - ge = g-(g-Rω 2)
gp - ge = Rω 2When a body of mass m is moved from equator to either pole, the weight of the body increases by m(gp - gp) i.e. mRω 2.