Earth rotates once about its axis of rotation in 1 day and moves around Sun in 365 days. Since Earth and a particle on Earth both move together with a constant speed around Sun, there is no effect in the measured acceleration due to gravity on the account of Earth’s translational motion. The curved path around Sun can be approximated to be linear for distances under consideration. Hence, Earth can serve as inertial frame of reference for the application of Newton’s law of motion, irrespective of its translational motion.
However, consideration of rotation of Earth about its axis changes the nature of Earth’s reference. It is no more an inertial frame. A particle at a point, “P”, is rotating about the axis of rotation. Clearly, a provision for the centripetal force should exist to meet the requirement of circular motion. We should emphasize here that centripetal force is not an additional force by itself, but is a requirement of circular motion, which should be met with by the forces operating on the particle. Here, gravitational force meets this requirement and, therefore, gets modified to that extent.
Here, we shall restrict our consideration specifically to the effect of rotation. We will ignore other factors that affect gravitational acceleration. This means that we consider Earth is a solid uniform sphere. If it is so then, measured value of acceleration is equal to reference gravitational acceleration (
g
0
g
0
) as modified by rotation.
As we have studied earlier, we can apply Newton’s law in a noninertial reference by providing for pseudo force. We should recall that pseudo force is applied in the direction opposite to the direction of acceleration of the frame of reference, which is centripetal acceleration in this case. The magnitude of pseudo force is equal to the product of mass of the particle and centripetal acceleration. Thus,
F
P
=
m
ω
2
r
F
P
=
m
ω
2
r
After considering pseudo force, we can enumerate forces on the particle at “P” at an latitude “φ” as shown in the figure :
 Pseudo force (
m
ω
2
r
m
ω
2
r
)
 Normal force (N)
 gravitational force (
m
g
0
m
g
0
)
The particle is subjected to normal force against the net force on the particle or the weight as measured. Two forces are equal in magnitude, but opposite in direction.
N
=
W
=
m
g
N
=
W
=
m
g
It is worthwhile to note here that gravitational force and normal force are different quantities. The measured weight of the particle is equal to the product of mass and the measured acceleration. It is given by the expression “mg”. On the other hand, Gravitational force is given by the Newton’s equation, which considers Earth at rest. It is equal to “
m
g
0
m
g
0
”.
⇒
F
G
=
G
M
R
2
=
m
g
0
⇒
F
G
=
G
M
R
2
=
m
g
0
Since particle is stationary on the surface of Earth, three forces as enumerated above constitute a balanced force system. Equivalently, we can say that the resultant of pseudo and gravitational forces is equal in magnitude, but opposite in direction to the normal force.
N
=
F
G
+
F
P
N
=
F
G
+
F
P
In other words, resultant of gravitational and pseudo forces is equal to the magnitude of measured weight of the particle. Applying parallelogram theorem for vector addition of gravitational and pseudo forces, the resultant of the two forces is :
N
2
=
m
g
0
2
+
m
ω
2
R
2
+
2
m
2
ω
2
g
0
R
cos
180
−
φ
N
2
=
m
g
0
2
+
m
ω
2
R
2
+
2
m
2
ω
2
g
0
R
cos
180
−
φ
Putting N = mg and rearranging, we have:
⇒
m
2
g
2
=
m
2
g
0
2
+
m
2
ω
4
R
2
−
2
m
2
ω
2
g
0
R
cos
φ
⇒
m
2
g
2
=
m
2
g
0
2
+
m
2
ω
4
R
2
−
2
m
2
ω
2
g
0
R
cos
φ
Angular velocity of Earth is quite a small value. It may be interesting to know the value of the term having higher power of angular velocity. Since Earth completes one revolution in a day i.e an angle of “2π” in 24 hrs, the angular speed of Earth is :
⇒
ω
=
2
π
24
X
60
X
60
=
7.28
X
10

5
rad
/
s
⇒
ω
=
2
π
24
X
60
X
60
=
7.28
X
10

5
rad
/
s
The fourth power of angular speed is almost a zero value :
⇒
ω
4
=
2.8
X
10

17
⇒
ω
4
=
2.8
X
10

17
We can, therefore, safely neglect the term “
m
2
ω
4
r
2
m
2
ω
4
r
2
”. The expression for the measured weight of the particle, therefore, reduces to :
⇒
g
2
=
g
0
2
−
2
ω
2
g
0
R
cos
φ
⇒
g
2
=
g
0
2
−
2
ω
2
g
0
R
cos
φ
⇒
g
=
g
0
1
−
2
ω
2
R
cos
φ
g
0
1
/
2
⇒
g
=
g
0
1
−
2
ω
2
R
cos
φ
g
0
1
/
2
Neglecting higher powers of angular velocity and considering only the first term of the binomial expansion,
⇒
g
=
g
0
1
−
1
2
X
2
ω
2
R
cos
φ
g
0
⇒
g
=
g
0
1
−
1
2
X
2
ω
2
R
cos
φ
g
0
⇒
g
=
g
0
−
ω
2
R
cos
φ
⇒
g
=
g
0
−
ω
2
R
cos
φ
This is the final expression that shows the effect of rotation on gravitational acceleration (
m
g
0
m
g
0
). The important point here is that it is not only the magnitude that is affected by rotation, but its direction is also affected as it is no more directed towards the center of Earth.
There is no effect of rotation at pole. Being a point, there is no circular motion involved and hence, there is no reduction in the value of gravitational acceleration. It is also substantiated from the expression as latitude angle is φ = 90° for the pole and corresponding cosine value is zero. Hence,
⇒
g
=
g
0
−
ω
2
R
cos
90
0
=
g
0
⇒
g
=
g
0
−
ω
2
R
cos
90
0
=
g
0
The reduction in gravitational acceleration is most (maximum) at the equator, where latitude angle is φ = 0° and corresponding cosine value is maximum (=1).
⇒
g
=
g
0
−
ω
2
R
cos
0
0
=
g
0
−
ω
2
R
⇒
g
=
g
0
−
ω
2
R
cos
0
0
=
g
0
−
ω
2
R
We can check approximate reduction at the equator, considering R = 6400 km = 6400000 m =
6.4
X
10
6
6.4
X
10
6
m.
⇒
ω
2
R
=
7.28
X
10

5
2
X
6.4
X
10
6
=
3.39
X
10

3
m
/
s
2
=
0.0339
m
/
s
2
⇒
ω
2
R
=
7.28
X
10

5
2
X
6.4
X
10
6
=
3.39
X
10

3
m
/
s
2
=
0.0339
m
/
s
2
This is the maximum reduction possible due to rotation. Indeed, we can neglect this variation for all practical purposes except where very high accuracy is required.