What is the reason that the value of g (acceleration due to gravity) becomes greater at the poles than at the equator?
Radius of the earth increases from the poles to the equator
The acceleration due to gravity, denoted by 'g', is not constant across the surface of the Earth. It varies from place to place. One of the main reasons for this variation is the shape of the Earth itself.
The Earth is not a perfect sphere. It is actually an oblate spheroid, meaning it is flattened at the poles and bulges at the equator. This shape has a direct impact on the value of 'g'.
The formula for the acceleration due to gravity on the surface of a spherical body is given by:
\(g = \frac{GM}{R^2}\)
Where:
From this formula, we can see that the acceleration due to gravity \(g\) is inversely proportional to the square of the distance from the center of the Earth (\(R\)).
\(g \propto \frac{1}{R^2}\)
This means that if the distance \(R\) increases, the value of \(g\) decreases, and if \(R\) decreases, the value of \(g\) increases.
Due to the Earth's oblate spheroid shape:
Specifically, the equatorial radius is approximately 6378 km, while the polar radius is approximately 6357 km. This difference is about 21 km.
Since the radius \(R\) is smaller at the poles compared to the equator:
Therefore, the acceleration due to gravity 'g' is greater at the poles than at the equator primarily because the Earth's radius is smaller at the poles and larger at the equator.
While the difference in radius is a significant factor, the Earth's rotation also contributes to the variation in 'g'. The centrifugal force due to rotation acts outwards, opposing gravity, especially at the equator where its effect is maximum. This makes the *effective* gravity slightly less at the equator compared to the poles (where the effect is zero or minimal).
However, the question focuses on the options provided, and the option related to the radius variation is the key reason explained by the Earth's shape.
| Location on Earth | Distance from Center (R) | Value of g (\(g \propto 1/R^2\)) |
|---|---|---|
| Poles | Smaller (minimum) | Greater (maximum) |
| Equator | Larger (maximum) | Smaller (minimum) |
| Factor | Effect on 'g' | Explanation |
|---|---|---|
| Altitude (Height above surface) | Decreases | Distance from center increases. |
| Depth (Below surface) | Decreases (initially linear) | Mass pulling inwards decreases. |
| Shape of Earth (Radius) | Varies (Poles > Equator) | Equatorial radius > Polar radius. |
| Rotation of Earth | Decreases (max at Equator) | Centrifugal force effect. |
Gravity meters, called gravimeters, are used to measure the local acceleration due to gravity. These measurements are important for geological surveys, studying Earth's interior, and other scientific applications. The variations in 'g' also need to be accounted for in precise navigation and satellite trajectories.
Understanding the variation of acceleration due to gravity (g) is key in physics and geophysics. The shape of the Earth, its rotation, altitude, and local mass distribution all play a role in determining the precise value of 'g' at any given point on the surface.
Which of the following statement is correct?
I. Gravitation is a weak force unless large masses are involved.
II. The weight is equal to the product of mass and acceleration due to gravity.
The mass of the Earth is ________.
If the acceleration due to gravity on the surface of earth is g, then the acceleration due to gravity on the surface of a planet whose mass is same as that of earth and radius is twice as that of earth is ___________.
A man's mass is 72 kg on the earth. His mass on the moon will be:
The weight of 6 kg mass of a body on the surface of moon is about (for earth, \(g = 10 m/s^2 \) )