At what rate should the earth rotate so that the apparent g at the equator becomes zero? What will be the length of the day in this situation?

Explanation

To solve the problem of determining the rate at which the Earth should rotate so that the apparent gravitational acceleration (g') at the equator becomes zero, we can follow these steps:

### Step 1: Understand the relationship between apparent gravity and rotation
The apparent gravitational acceleration at the equator can be expressed as:
\[ g' = g - \omega^2 r \]
where:
- \( g \) is the actual gravitational acceleration (approximately \( 9.8 \, \text{m/s}^2 \)),
- \( \omega \) is the angular velocity of the Earth's rotation,
- \( r \) is the radius of the Earth (approximately \( 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \)).

### Step 2: Set the apparent gravity to zero
To find the rate of rotation where the apparent gravity becomes zero, we set:
\[ g' = 0 \]
Thus, we have:
\[ 0 = g - \omega^2 r \]
Rearranging this gives:
\[ \omega^2 r = g \]
From this, we can solve for \( \omega \):
\[ \omega = \sqrt{\frac{g}{r}} \]

### Step 3: Substitute the values
Substituting the values of \( g \) and \( r \):
- \( g \approx 9.8 \, \text{m/s}^2 \)
- \( r \approx 6400 \times 10^3 \, \text{m} \)

We get:
\[ \omega = \sqrt{\frac{9.8}{6400 \times 10^3}} \]

### Step 4: Calculate \( \omega \)
Calculating \( \omega \):
1. Calculate \( 6400 \times 10^3 \):
\[ 6400 \times 10^3 = 6.4 \times 10^6 \]
2. Now, calculate \( \frac{9.8}{6.4 \times 10^6} \):
\[ \frac{9.8}{6.4 \times 10^6} \approx 1.53125 \times 10^{-6} \]
3. Taking the square root:
\[ \omega \approx \sqrt{1.53125 \times 10^{-6}} \approx 0.001237 \, \text{rad/s} \]

### Step 5: Find the length of the day
The length of the day \( T \) is related to the angular velocity \( \omega \) by:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{0.001237} \]

### Step 6: Calculate \( T \)
Calculating \( T \):
1. Calculate \( 2\pi \):
\[ 2\pi \approx 6.2832 \]
2. Now, calculate \( T \):
\[ T \approx \frac{6.2832}{0.001237} \approx 5084.48 \, \text{s} \]

### Step 7: Convert seconds to hours
To convert seconds into hours:
\[ T \approx \frac{5084.48}{3600} \approx 1.41 \, \text{hours} \]

### Final Answer
The Earth should rotate at an angular velocity of approximately \( 0.001237 \, \text{rad/s} \) to make the apparent gravitational acceleration at the equator zero, resulting in a length of day of approximately \( 1.41 \, \text{hours} \).