Earth is not uniform. Its density varies as we move from its center to the surface. In general, Earth can be approximated to be composed of concentric shells of different densities. For all practical purpose, we consider that the density gradation is radial and is approximated to have an equivalent uniform density within these concentric shells in all directions.

Figure 1: Density of Earth varies radially.

Constitution of the Earth

The main reason for this directional uniformity is that bulk of the material constituting Earth is fluid due to high temperature. The material, therefore, has a tendency to maintain uniform density in a given shell so conceived.

We have discussed that the density variation has no effect on a point on the surface or above it. The variation of density, however, impacts gravitational acceleration, when the point in the question is at a point below Earth’s surface.

In order to understand the effect, let us have a look at the expression of gravitational expression :

g 0 = F m = G M R 2 g 0 = F m = G M R 2

The impact of moving down below the surface of Earth, therefore, depends on two factors

A point inside a deep mine shaft, for example, will result in a change in the value of gravitational acceleration due to above two factors.

We shall know subsequently that gravitational force inside a spherical shell is zero. Therefore, mass of the spherical shell above the given point does not contribute to gravitational force and hence acceleration at that point. Thus, the value of “M” in the expression of gravitational acceleration decreases as we go down from the Earth’s surface. This, in turn, decreases gravitational acceleration at a point below Earth’s surface. At the same time, the distance to the center of Earth decreases. This factor, in turn, increases gravitational acceleration.

If we assume uniform density, then the impact of “decrease in mass” is greater than that of impact of “decrease in distance”. We shall prove this subsequently when we consider the effect of vertical position. As such, acceleration is expected to decrease as we go down from Earth’s surface.

In reality the density is not uniform. Crust being relatively light and thin, the impact of first factor i.e. “decrease in mass” is less significant initially and consequently gravitational acceleration actually increases initially for some distance as we go down till it reaches a maximum value at certain point below Earth’s surface. For most of depth beyond, however, gravitational acceleration decreases with depth.

Earth is not a sphere. It is an ellipsoid. Its equatorial radius is greater than polar radius by 21 km. A point at pole is closer to the center of Earth. Consequently, gravitational acceleration is greater there than at the equator.

Besides, some part of Earth is protruded and some part is depressed below average level. Once again, factors of mass and distance come into picture. Again, it is the relative impact of two factors that determine the net effect. Consider a point right at the top of Mt. Everest, which is about 8.8 km from the mean sea level. Imagine incrementing radius of Earth’s sphere by 8.8 km. Most of the volume so created is not filled. The proportionate increase in mass (mass of Everest mountain range) is less than that in the squared distance from the center of Earth. As such, gravitational acceleration is less than its average value on the surface. It is actually 9.80 m / s 2 m / s 2 as against the average of 9.81 m / s 2 m / s 2 , which is considered to be the accepted value for the Earth’s surface.

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 serves as inertial frame of reference for the application of Newton’s law of 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 non-inertial 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 :

Figure 2: The particle is at rest under action of three forces.

Forces on the particle on the surface of Earth

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 as sphere. 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 :

Figure 3: The particle is at rest under action of three forces.

Forces on the particle on the surface of Earth

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 precision is required.

In this section, we shall discuss the effect of the vertical position of the point of measurement. For this, we shall consider Earth as a perfect sphere of radius “R” and uniform density, “ρ”. Further, we shall first consider a point at a vertical height “h” from the surface and then a point at a vertical depth “d” from the surface.

Gravitational acceleration at a height

Gravitational acceleration due to Earth on its surface is equal to gravitational force per unit mass and is given by :

g 0 = F m = G M R 2 g 0 = F m = G M R 2

Figure 4: Distance between center of Earth and particle changes at an altitude.

Gravity at an altitude

where “M” and “R” are the mass and radius of Earth. It is clear that gravitational acceleration will decrease if measured at a height “h” from the Earth’s surface. The mass of Earth remains constant, but the linear distance between particle and the center of Earth increases. The net result is that gravitational acceleration decreases to a value “g’” as given by the equation,

⇒ g ′ = F ′ m = G M R + h 2 ⇒ g ′ = F ′ m = G M R + h 2

We can simplify this equation as,

⇒ g ′ = G M R 2 1 + h R 2 ⇒ g ′ = G M R 2 1 + h R 2

Substituting for the gravitational acceleration at the surface, we have :

⇒ g ′ = g 0 1 + h R 2 ⇒ g ′ = g 0 1 + h R 2

This relation represents the effect of height on gravitational acceleration. We can approximate the expression for situation where h<< R.

⇒ g ′ = g 0 1 + h R 2 = g 0 1 + h R - 2 ⇒ g ′ = g 0 1 + h R 2 = g 0 1 + h R - 2

As h<<R, we can neglect higher powers of “h/R” in the binomial expansion of the power term,

⇒ g ′ = g 0 1 − 2 h R ⇒ g ′ = g 0 1 − 2 h R

We should always keep in mind that this simplified expression holds for the condition, h << R. For small vertical altitude, gravitational acceleration decreases linearly with a slope of “-2/R”. If the altitude is large as in the case of a communication satellite, then we should resort to the original expression,

⇒ g ′ = G M R + h 2 ⇒ g ′ = G M R + h 2

If we plot gravitational acceleration .vs. altitude, the plot will be about linear for some distance and thereafter parabolic in nature.

Figure 5: The plot shows variations in gravitational acceleration as we move vertically upwards from center of Earth.

Acceleration .vs. linear distance

Gravitational acceleration at a depth

In order to calculate gravitational acceleration at a depth “d”, we consider a concentric sphere of radius “R-d” as shown in the figure. Here, we shall make use of the fact that gravitational force inside a spherical shell is zero. It means that gravitational force due to the spherical shell above the point is zero. On the other hand, gravitational force due to smaller sphere can be calculated by treating it as point mass. As such, net gravitational acceleration at point “P” is :

Figure 6: Distance between center of Earth and particle changes at an altitude.

Gravity at a depth

⇒ g ′ = F ′ m = G M ′ R - d 2 ⇒ g ′ = F ′ m = G M ′ R - d 2

where “M’” is the mass of the smaller sphere. If we consider Earth as a sphere of uniform density, then :

M = V ρ M = V ρ

⇒ ρ = M 4 3 π R 3 ⇒ ρ = M 4 3 π R 3

Hence, mass of smaller sphere is equal to the product :

M ′ = ρ V ′ M ′ = ρ V ′

⇒ M ′ = 4 3 π R − d 3 X M 4 3 π R 3 = R − d 3 X M R 3 ⇒ M ′ = 4 3 π R − d 3 X M 4 3 π R 3 = R − d 3 X M R 3

Substituting in the expression of gravitational acceleration, we have :

⇒ g ′ = G R − d 3 X M R − d 2 R 3 ⇒ g ′ = G R − d 3 X M R − d 2 R 3

Inserting gravitational acceleration at the surface ( g 0 = G M / R 2 g 0 = G M / R 2 ), we have :

⇒ g ′ = g 0 R − d 3 R − d 2 R = g 0 R − d R ⇒ g ′ = g 0 R − d 3 R − d 2 R = g 0 R − d R

g ′ = g 0 1 − d R g ′ = g 0 1 − d R

This is also a linear equation. We should note that this expression, unlike earlier case of a point above the surface, makes no approximation . The gravitational acceleration decreases linearly with distance as we do down towards the center of Earth. Conversely, the gravitational acceleration increases linearly with distance as we move from the center of Earth towards the surface.

Figure 7: The plot shows variations in gravitational acceleration as we move away from center of Earth.

Acceleration .vs. linear distance

The plot above combines the effect of altitude and the effect of depth along a straight line, starting from the center of Earth.