Variation of acceleration due to gravity :
(i) Variation of g with altitude
(ii) Variation of g with depth
(iii) Variation of g with latitude
(iv) Variation of g with latitude (Rotation of the Earth)

**Variation of acceleration due to
gravity**

*(i) Variation of g with altitude*

Let *P* be a point on the surface of the 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 a spherical
shaped body.

The acceleration due to gravity
at P on the surface is

g =GM/R^{2 } ……………….. (1)

*Q* at a height *h* from the surface of the Earth. The acceleration due to gravity at
*Q* is

g_{h}=GM/(R+h)^{2} ……………….(2)

dividing (2) by (1) g_{h}/g
= R^{2}/(R+h)^{2}

By simplifying and expanding
using binomial theorem,

g_{h}= g(1-(2h/R))

The value of acceleration due to
gravity decreases with increase in height above the surface of the Earth.

*(ii) Variation of g with depth*

Consider the Earth to be a
homogeneous sphere with uniform density of radius *R* and mass *M*.

Let *P* be a point on the surface of the Earth and *Q* be a point at a depth *d*
from the surface.

The acceleration due to gravity
at *P* on the surface is *g* = g=GM/R^{2}

If ρ be the
density, then, the mass of the Earth is *M*
= *4/3 *π *R ^{3}*

*G= 4/3 G* π *R**ρ*

The acceleration due to gravity
at Q at a depth d from the surface of the Earth is

g_{d}=GMd^{2} /
(R-d)^{2}

where M_{d} is the mass
of the inner sphere of the Earth of radius (R− d).

Md = 4/3 π (R − d)^{3}ρ

g_{d} = 4/3 G π (R − d)ρ

dividing (2) by (1)

g_{d} = R-d/R

g_{d} = g(1-d/R)

The value of acceleration due to
gravity decreases with increase of depth.

**(iii) Variation of g with latitude **(Non−sphericity of the Earth) The
Earth is not a perfect sphere. It is an ellipsoid as shown in the
Fig.. It is flattened at the poles where the latitude is 90^{o} and
bulged at the equator where the latitude is 0^{o}.

The radius of the Earth at
equatorial plane Re is greater than the radius along the poles Rp by about 21
km.

We know that g = GM/R^{2}

g α 1/R^{2}

The value of g varies inversely
as the square of radius of the Earth. The radius at the equator is the
greatest. Hence the value of g is minimum at the equator. The radius at poles
is the least. Hence, the value of *g*
is maximum at the poles. The value of *g*
increases from the equator to the poles.

*(iv) Variation of g with latitude
***(Rotation of the Earth)**

Let us consider the Earth as a
homogeneous sphere of mass *M and radius
R. The Earth rotates about an axis passing through its north and south poles.
The Earth rotates from west to east in 24 hours. Its angular velocity is 7.3 ×
10 ^{−5} rad s^{−1}.*

Consider a body of mass m on the
surface of the Earth at P at a latitude θ. Let ω be the angular velocity. The
force (weight) F = mg acts along

PO. It could be resolved into two
rectangular components (i) mg cos θ along

PB and (ii) mg sin θ along PA
(Fig.).

From the ∆OPB, it is found that
BP = R cos θ. The particle describes a circle with B as centre and radius BP =
R cos θ.

The body at P experiences a
centrifugal force (outward force) F_{C} due to the rotation of the
Earth

(i.e) F_{C} = mRω^{2}
cos θ

The net force along PC = mg cos θ
− mRω^{2} cos θ

The body is acted upon by two
forces along PA and PC.

The resultant of these two forces
is

F= √(mg sinθ) 2+(mg cosθ−mRω^{2}
cosθ)^{2}

The force, F = mg g √[ 2Rω^{2}cos^{2}θ
/g] …………(1)

f g′ is the acceleration of the
body at P due to this force F, we have,

F = mg′ …………(2)

by equating (2) and (1)

g′ = g (1-(Rω^{2}cos^{2}θ
/g))

**Case (i)** At the poles, θ = 90^{o} ; cos θ = 0

g′ = g

**Case (ii)** At the equator, θ = 0 ; cos θ = 1

g′ = g(1- Rω^{2}/g)

So, the value of acceleration due
to gravity is maximum at the poles.

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