Introduction
The acceleration due to gravity (g) is a fundamental concept in physics, affecting various physical phenomena. However, g is not constant everywhere on Earth. One of the key reasons for this variation is the rotation of the Earth. In this article, we will explore how Earth’s rotation influences g, supported by explanations, formulas, FAQs, and MCQs.
Variation of Acceleration Due to Gravity (g) due to Rotation of Earth
The variation in g is due to the centrifugal force acting on the rotation of the earth. When the earth is rotating, all the objects tend to experience a centrifugal force that won’t act in the direction of gravity.
Consider a sample mass (m) that is located on a latitude that forms an angle (θ) with the equator. As we have seen, when a body rotates, every particle in the body moves in a circular motion around the axis of revolution. In this instance, the earth rotates with a constant angular velocity while the test mass travels with an angular velocity in a circular path of radius ‘r’.
When a body is under rotation, every particle in the body makes circular motions about the axis of rotation. In the present case, the earth is under rotation with a constant angular velocity ω, then the test mass moves in a circular path of radius ‘r’ with an angular velocity ω.
This is the case of a non-inertial frame of reference, so there exists a centrifugal force on the test mass (mrω2). Gravity is acting on the test mass towards the centre of the earth (mg).
As both these forces are acting from the same point, these are known as co-initial forces, and as they lie along the same plane, they are termed co-planar forces.
According to the parallelogram rule of vectors, if two coplanar vectors form two sides of a parallelogram, the resultant of those two vectors will always be along the parallelogram’s diagonal. Using the parallelogram law of vectors, we can calculate the magnitude of the apparent value of gravitational force at the latitude:
(mg′)2 = (mg)2 + (mrω2)2 + 2(mg) (mrω2) cos(180 – θ)
Where r is the radius of the circular path followed, which is r = R cosθ. Therefore, the above expression becomes,
g’ = g – Rω2 cos2θ
Where g′ is the apparent amount of acceleration due to gravity at the latitude owing to earth rotation, and g is the real value of gravity at the latitude without considering earth rotation. Also, both g and g’ are related at a different position on earth, given as,
| Position at the Earth’s Surface | Value of θ (in °) | Relation between g and g’ |
| At the Equator | 0 | g′= g – Rω2 |
| At Pole | 90o | g’ = g |
Effect of Rotation on Weight of Objects
- Objects weigh more at the poles than at the equator due to the variation in g because g is more at poles than at equator.
- If Earth stops rotating, the value of g will be uniform everywhere because the centrifugal force would disappear.
Hence weight of body increases when earth rotates from equator to poles.
Why Centrifugal Force Affect g Due to Rotation of Earth ?
- The Earth rotates about its axis, creating a centrifugal force that acts outward.
- This force is maximum at the equator and zero at the poles.
- Due to this force, the effective value of g decreases as we move from the poles to the equator.
Value of g at the Equator and Poles
- At the poles, centrifugal force is zero, so g is maximum.
- At the equator, centrifugal force is maximum, reducing g slightly.
Frequently Asked Questions (FAQs)
Q1: Why does g decrease from the poles to the equator?
A: Due to Earth’s rotation, the centrifugal force reduces the effective gravity at the equator.
Q2: What is the maximum and minimum value of g on Earth?
A: Maximum at the poles (≈9.83 m/s²), minimum at the equator (≈9.78 m/s²).
Q3: How does latitude affect the value of g?
A: As latitude increases from 0° (equator) to 90° (poles), the value of g increases.
Q4: What happens to g if Earth stops rotating?
A: The centrifugal force would disappear, and g would be equal everywhere.
Multiple Choice Questions (MCQs)
Q1: Which of the following is a reason for the variation of g on Earth? A) Earth’s rotation
B) Earth’s revolution
C) Presence of the Moon
D) Magnetic field
Answer: A) Earth’s rotation
Explanation: The centrifugal force due to Earth’s rotation affects g.
Q2: Where is the acceleration due to gravity maximum on Earth? A) Equator
B) Poles
C) 45° latitude
D) 30° latitude
Answer: B) Poles
Explanation: The absence of centrifugal force at the poles results in the maximum value of g.
Q3: How does the value of g change with an increase in latitude? A) Increases
B) Decreases
C) Remains constant
D) None of the above
Answer: A) Increases
Explanation: The centrifugal force reduces g at lower latitudes, making it higher at higher latitudes.
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Written by: Neeraj Anand
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