If the earth were a perfect sphere of radius `6.37 xx 10^(6) m`, rotating about its axis with a period of 1 day `( = 8.64 xx 10^(6) s)`, how much would the acceleration due to gravity differ from the poles to the equator.

Explanation

To find the difference in acceleration due to gravity from the poles to the equator, we can follow these steps:

### Step 1: Understand the formula for gravity at the poles and equator
The acceleration due to gravity at the poles (g') is given by:
\[ g'_{p} = g \]
where \( g \) is the acceleration due to gravity at the surface of the Earth.

At the equator, the formula for gravity (g') is modified due to the Earth's rotation:
\[ g'_{e} = g - R \omega^2 \]
where:
- \( R \) is the radius of the Earth,
- \( \omega \) is the angular velocity of the Earth.

### Step 2: Calculate the angular velocity (ω)
The angular velocity \( \omega \) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
where \( T \) is the period of rotation (1 day = 86400 seconds).

Substituting the values:
\[ \omega = \frac{2\pi}{86400 \, \text{s}} \]

### Step 3: Calculate the difference in gravity from poles to equator
The difference in gravity from the poles to the equator can be expressed as:
\[ \Delta g = g'_{p} - g'_{e} = R \omega^2 \]

### Step 4: Substitute the values into the equation
Using the radius of the Earth \( R = 6.37 \times 10^6 \, \text{m} \) and the calculated \( \omega \):
1. First, calculate \( \omega \):
\[ \omega = \frac{2\pi}{86400} \approx 7.272 \times 10^{-5} \, \text{rad/s} \]

2. Now calculate \( \omega^2 \):
\[ \omega^2 \approx (7.272 \times 10^{-5})^2 \approx 5.29 \times 10^{-9} \, \text{rad}^2/\text{s}^2 \]

3. Now substitute \( R \) and \( \omega^2 \) into the equation for \( \Delta g \):
\[ \Delta g = R \omega^2 = (6.37 \times 10^6) \times (5.29 \times 10^{-9}) \]

### Step 5: Calculate the final difference
Calculating the above expression:
\[ \Delta g \approx 6.37 \times 10^6 \times 5.29 \times 10^{-9} \approx 3.37 \times 10^{-2} \, \text{m/s}^2 \]

### Final Answer
The difference in acceleration due to gravity from the poles to the equator is approximately:
\[ \Delta g \approx 3.37 \times 10^{-2} \, \text{m/s}^2 \]