The value of acceleration due to gravity

Options

A. Is same on equator and poles
B. Is least on poles
C. Is least on equator
D. Increases from pole to equator

Explanation

To determine the value of acceleration due to gravity at different locations on Earth, we can analyze the effects of Earth's rotation and the shape of the Earth. Here’s a step-by-step solution:

### Step 1: Understand the Formula
The formula for the effective acceleration due to gravity (G') at a point on the Earth's surface is given by:
\[ G' = G - \omega^2 R \sin(\theta) \]
where:
- \( G \) is the standard acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( \omega \) is the angular velocity of the Earth,
- \( R \) is the radius of the Earth,
- \( \theta \) is the latitude.

### Step 2: Analyze at the Poles
At the poles, the latitude \( \theta = 0^\circ \):
- Therefore, \( \sin(0^\circ) = 0 \).
- Substituting into the formula:
\[ G' = G - \omega^2 R \cdot 0 = G \]
This means that the effective acceleration due to gravity at the poles is equal to \( G \).

### Step 3: Analyze at the Equator
At the equator, the latitude \( \theta = 90^\circ \):
- Thus, \( \sin(90^\circ) = 1 \).
- Substituting into the formula:
\[ G' = G - \omega^2 R \cdot 1 = G - \omega^2 R \]
This indicates that the effective acceleration due to gravity at the equator is less than \( G \).

### Step 4: Compare Values
From the analysis:
- At the poles, \( G' = G \) (maximum value).
- At the equator, \( G' < G \) (minimum value).

### Step 5: Conclusion
Thus, the acceleration due to gravity is maximum at the poles and minimum at the equator. Therefore, the correct option is:
**C: Least at the equator.**