Brief Notes on Variation of G due to depth, the shape of the earth, and Rotation of Earth

The value of gravitational push is determined by the division of the mass within the earth or any celestial body. 

[g means gravity = universal force of attraction residing among every matter present as a part of the universe]

Gravity is considered the driving force to keep everything close and bound. Acceleration or movement on which objects fall freely is used to calculate gravity G. It is calculated at approximately 9.8 m/s² at the earth’s surface. This G may face variations which can be determined by depth, the shape of the earth, and the rotation of the earth.

The Formula of Acceleration due to Gravity

Force is denoted as-

F = mg

F denotes acting force, g is the acceleration because of G, and m denotes the mass of the body.

As per the universal law of gravitation, F = GMm/(r+h)²

[Denotations]

F = force between two bodies 

G = gravitational constant

m = mass of an item

M = mass of the earth 

r = radius of the earth 

h = the height above the surface of the earth

As the height has little difference from the radius of the earth, rewrite the above expression as F = GMm / r²

On equating the two expressions, the following is achieved:

mg = GMm / r²

g = GM / r²

Factors influencing the movement of G (Gravity)

• Variation of g with height: 

• It decreases with an increase in height and rises with a fall in height above the earth’s surface.

• Variation of g with depth:

• The value of g increases with the rise in depth and gets equal to zero at the earth’s centre because it is directly relevant to the depth below the earth’s surface.

• Variation of g due to the shape of the earth: 

• Its value is less than at the pole as compared to the equator.

• Variation of g due to Rotation of Earth: 

• It decreases as the rotation of the Earth increases.

Variation of g with Depth

Denotations: 

A body of mass = m, is at a point B

B is at a depth of h, which is the height from the earth’s surface

The distance from the earth’s centre = R – d

gd = g (R – d)/R

This movement decreases while moving to the earth’s centre, and that is experienced only in depth. In this way, the value of g is influenced by height and depth. It may also vary at the surface of the earth as g is lowest on the equator and the highest on the poles.

Variation of g due to Shape of the Earth

The earth’s shape is not exactly spherical but is a somewhat oblate spheroid. The radius near the poles, namely the polar radius, is 21 km smaller than its radius near the equator, namely the equatorial radius. According to the derived formula, radius (r) is less than the value of g at the pole than that at the equator.

On considering the earth’s shape to be a little oval, not circle. So, we get different distances from both the equator(RE) and the pole(RP) from the centre as:

RE > RP

The observed relation between RE and RP can be denoted as,

g ∝ 1/R²

Then, on considering G and M as constants, then,

gP ∝ 1/RP² and 

gE ∝ 1/RE²

Hence, the gravitational accelerations on both the equator and pole are given as,

gP > gE

So, from these distances, gravitational acceleration’s relationship is established as, 

gP / gE = RE² / RP²

Variation of g due to Rotation of Earth

While the earth is rotating, every object tends to face a centrifugal force that does not act in the gravitational direction. 

Contemplate a sample mass (m), forming an angle with the equator, located on a latitude. As observed earlier, every particle in the body gets engaged in a circular movement when it rotates around the axis. As the earth’s rotation takes place with a continual angular velocity, on the other hand, the test mass moves with the same velocity of radius ‘r’ in a round path.

Since G is a frame of reference that moves with the object, it holds a centrifugal force which gets implemented on the sample mass. G pulls the test mass towards the planet’s centre (mg). As these two mentioned forces, namely centrifugal and gravitational, which work from the same point, are considered co-initial forces and are considered coplanar forces on falling over the same reference point.

As per the vectors’ parallelogram law, if two sides of a parallelogram are formed by the two coplanar vectors, the resultant vector will always be the parallelogram’s diagonal.

The magnitude of the value of the latitudinal gravitational force can be calculated as:

(mg′)² = (mg)² + (mrω²)²+ 2(mg) (mrω²) cos(180 – θ)

Here, r denotes the radius of the circular path, that is, r = R cosθ.

So, the above expression gets converted as the following,

g’ = g – Rω² cos²θ

Here, 

g′ = apparent amount of acceleration for latitudinal gravity to the earth’s rotation.

g = actual value of the latitudinal gravity without considering the rotation of the earth.

Both the g and g’ belong to different positions on the earth.

Solved Questions

1. The Earth is supposed to be spherically similar. Calculate the acceleration if the gravity is at a depth of 2000 km under the earth’s surface. [Use: R = 6400 km and g = 9.8 m/s².]

As Given: d (the depth) = 2000 km

R (the Earth’s radius) = 6400 km

So gd (acceleration due to gravity ) = 6,738 m/s².

2. For an individual who weighs 60 kg on the earth’s surface, how much will be the individual’s mass on the surface of Neptune, with 1.2 times acceleration of gravity as compared to that of the earth.

The individual will weigh 60 kg on Neptune’s surface as the force of gravity brings a change in the weight of the object, not in its mass. The mass of an item remains the same without being influenced by the acting forces upon it.

3. It is observed that the value of g, at a place above the earth’s surface, is 0.2 m/s². Find out the height above the earth’s surface where the g is obtained. 

Considering the height as h’

g′ / g = (1− 2h / R)

or

g′ = g (1− 2h / R)

Switch the above given values as,

0.2 m/s² / 9.8 m/s² = (1− 2h / R)

2h / R = 48 / 49

h = 3134. 7 km 

Conclusion

So, it is justified that G may face variations which can be determined by depth, the shape of the earth, and the rotation of the earth. It decreases with an increase in height and rises with a fall in height above the earth’s surface. Similarly, It increases with the rise in depth and gets equal to zero at the earth’s centre because it is directly relevant to the depth below the earth’s surface. Its value is less than at the pole as compared to the equator. Besides these, It decreases as the rotation of the Earth increases.