Variation of g with Earth's rotation
Variation of g with Earth's Rotation
The acceleration due to gravity, denoted as ( g ), is not constant over the Earth's surface. It varies depending on several factors, including latitude, altitude, and the Earth's rotation. In this article, we will focus on how the Earth's rotation affects the value of ( g ).
Understanding Gravity and Centrifugal Force
Gravity is the force that attracts a body towards the center of the Earth. The acceleration due to gravity at the Earth's surface is approximately ( 9.81 \, \text{m/s}^2 ). However, the Earth's rotation generates a centrifugal force that acts outwardly and opposes gravity. This centrifugal force reduces the effective acceleration due to gravity, especially at the equator.
Theoretical Background
The centrifugal acceleration at a point on the Earth's surface due to its rotation is given by:
[ a_c = \omega^2 r ]
where:
- ( \omega ) is the angular velocity of the Earth's rotation (( \omega = \frac{2\pi}{T} ), with ( T ) being the period of rotation, approximately 24 hours)
- ( r ) is the distance from the axis of rotation to the point on the surface
The effective acceleration due to gravity, ( g' ), at a latitude ( \lambda ) can be expressed as:
[ g' = g - a_c \cos^2(\lambda) ]
where:
- ( g ) is the acceleration due to gravity without the Earth's rotation
- ( \lambda ) is the latitude (at the equator, ( \lambda = 0 ); at the poles, ( \lambda = 90^\circ ))
Variation of g with Latitude
The following table summarizes how the value of ( g ) varies with latitude due to the Earth's rotation:
| Latitude (λ) | Centrifugal Effect | Effective g (g') |
|---|---|---|
| 0° (Equator) | Maximum | Minimum |
| 90° (Poles) | None | Maximum |
At the equator, the centrifugal force is at its maximum because the distance ( r ) from the axis of rotation is greatest. As a result, the effective ( g ) is at its minimum. At the poles, there is no centrifugal effect because ( r ) is zero; hence, ( g ) is at its maximum.
Examples
At the Equator (λ = 0°):
- Centrifugal acceleration: ( a_c = \omega^2 r )
- Since ( r ) is maximum at the equator, ( a_c ) is maximum.
- Effective ( g ) is minimum.
At the Poles (λ = 90°):
- Centrifugal acceleration: ( a_c = 0 ) (since ( \cos(90^\circ) = 0 ))
- Effective ( g ) equals the theoretical ( g ) without rotation.
At Mid-Latitudes (e.g., λ = 45°):
- Centrifugal acceleration: ( a_c = \omega^2 r \cos^2(45^\circ) )
- Effective ( g ) is between the values at the equator and poles.
Mathematical Derivation
The Earth's rotation causes a centrifugal acceleration that reduces the effective gravitational acceleration. The centrifugal acceleration at the Earth's surface at a latitude ( \lambda ) is given by:
[ a_c = \omega^2 R \cos(\lambda) ]
where ( R ) is the Earth's radius. The effective acceleration due to gravity is then:
[ g' = g - \omega^2 R \cos^2(\lambda) ]
The value of ( g ) also changes slightly with altitude and the Earth's shape, but these factors are beyond the scope of this article.
Conclusion
The Earth's rotation causes a variation in the acceleration due to gravity across its surface. This variation is dependent on latitude, with ( g ) being least at the equator and greatest at the poles. Understanding this variation is crucial for accurate measurements and calculations in physics and engineering, as well as for explaining phenomena such as the Earth's oblateness and differences in weight at different latitudes.