The acceleration due to gravity at the pols and the equator is `g_(p) and g_(e)` respectively. If the earth is a sphere of radius `R_(E)` and rotating about its axis with angular speed `omega and g_(p)-g_(e)` given by

Options

A. `(omega^(2))/(R_(E))`
B. `(omega^(2))/(R_(E)^(2))`
C. `omega^(2)R_(E)^(2)`
D. `omega^(2)R_(E)`

Explanation

To solve the problem of finding the difference in acceleration due to gravity at the poles and the equator, we need to analyze the effects of Earth's rotation on gravity at these two locations.

### Step-by-Step Solution:

1. **Understanding the Acceleration Due to Gravity**:
The acceleration due to gravity at any point on Earth can be expressed as:
\[ g' = g - R \omega^2 \cos \lambda \]
where:
- \( g' \) is the effective acceleration due to gravity,
- \( g \) is the standard acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)),
- \( R \) is the radius of the Earth,
- \( \omega \) is the angular speed of Earth's rotation,
- \( \lambda \) is the latitude.

2. **Calculating Gravity at the Poles**:
At the poles, the latitude \( \lambda \) is \( 90^\circ \):
\[ g_p = g - R \omega^2 \cos(90^\circ) \]
Since \( \cos(90^\circ) = 0 \):
\[ g_p = g - R \omega^2 \cdot 0 = g \]

3. **Calculating Gravity at the Equator**:
At the equator, the latitude \( \lambda \) is \( 0^\circ \):
\[ g_e = g - R \omega^2 \cos(0^\circ) \]
Since \( \cos(0^\circ) = 1 \):
\[ g_e = g - R \omega^2 \]

4. **Finding the Difference \( g_p - g_e \)**:
Now, we can find the difference between the gravity at the poles and the equator:
\[ g_p - g_e = g - (g - R \omega^2) \]
Simplifying this:
\[ g_p - g_e = g - g + R \omega^2 = R \omega^2 \]

5. **Conclusion**:
Therefore, the difference in acceleration due to gravity at the poles and the equator is given by:
\[ g_p - g_e = R \omega^2 \]

### Final Answer:
The value of \( g_p - g_e \) is \( R \omega^2 \).