The Value of g on the surface of earth is smallest at the equator because

Options

A. The centripetal force maximum at equator
B. The centripetal force is least at equator
C. The centrifugal force is maximum at equator
D. The angular speed of earth is least at equator

Answer

To solve the question regarding why the value of \( g \) (acceleration due to gravity) on the surface of the Earth is smallest at the equator, we can break down the explanation into several steps:

### Step-by-Step Solution:

1. **Understanding the Formula for \( g' \)**:
The effective acceleration due to gravity at a point on the Earth's surface can be expressed as:
\[ g' = g - r \omega^2 \cos^2 \lambda \]
where:
- \( g' \) is the effective gravity,
- \( g \) is the gravitational acceleration,
- \( r \) is the radius of the Earth,
- \( \omega \) is the angular velocity of the Earth’s rotation,
- \( \lambda \) is the latitude.

2. **Analyzing the Latitude**:
At the equator, the latitude \( \lambda \) is \( 0^\circ \). Therefore, \( \cos^2(0) = 1 \). At the poles, \( \lambda \) is \( 90^\circ \), and \( \cos^2(90) = 0 \).

3. **Substituting Values**:
Substituting \( \lambda = 0 \) into the equation gives:
\[ g' = g - r \omega^2 \cdot 1 = g - r \omega^2 \]
At the poles, substituting \( \lambda = 90^\circ \) gives:
\[ g' = g - r \omega^2 \cdot 0 = g \]

4. **Centrifugal Force Consideration**:
The term \( r \omega^2 \) represents the centrifugal force due to the Earth's rotation. This force acts outward and reduces the effective weight (or apparent gravity) experienced by an object.

5. **Comparing Equator and Poles**:
Since the centrifugal force is maximum at the equator (where \( \cos^2 \lambda \) is maximum), it results in a larger reduction of \( g \) at the equator compared to the poles. Thus:
\[ g'_{\text{equator}} < g'_{\text{poles}} \]

6. **Conclusion**:
Therefore, the value of \( g \) on the surface of the Earth is smallest at the equator due to the maximum centrifugal force acting against the gravitational force.

### Final Answer:
The value of \( g \) on the surface of the Earth is smallest at the equator because the centrifugal force due to Earth's rotation is maximum at the equator, which reduces the effective gravitational force experienced there.