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 at the equator becomes zero, we can follow these steps:

### Step 1: Understand the concept of apparent gravity
The apparent acceleration due to gravity (\(g'\)) at the equator, when considering the Earth's rotation, is given by the formula:
\[ g' = g - \omega^2 r \]
where:
- \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\)),
- \(\omega\) is the angular velocity of the Earth in radians per second,
- \(r\) is the radius of the Earth (approximately \(6400 \, \text{km} = 6.4 \times 10^6 \, \text{m}\)).

### Step 2: Set the apparent gravity to zero
To find the angular velocity at which the apparent gravity becomes zero, we set \(g' = 0\):
\[ 0 = g - \omega^2 r \]
This can be rearranged to find \(\omega\):
\[ \omega^2 r = g \implies \omega^2 = \frac{g}{r} \implies \omega = \sqrt{\frac{g}{r}} \]

### Step 3: Substitute the values
Now we substitute the known values of \(g\) and \(r\):
\[ g = 9.8 \, \text{m/s}^2, \quad r = 6.4 \times 10^6 \, \text{m} \]
Calculating \(\omega\):
\[ \omega = \sqrt{\frac{9.8}{6.4 \times 10^6}} \approx \sqrt{1.53125 \times 10^{-6}} \approx 1.237 \times 10^{-3} \, \text{rad/s} \]

### Step 4: Calculate the length of the day
The length of the day (\(T\)) can be found using the relationship between angular velocity and the period:
\[ \omega = \frac{2\pi}{T} \]
Rearranging gives:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \(\omega\):
\[ T = \frac{2\pi}{1.237 \times 10^{-3}} \approx 5086.4 \, \text{s} \]
To convert seconds into hours:
\[ T \approx \frac{5086.4}{3600} \approx 1.41 \, \text{hours} \]

### Final Answer
Thus, the Earth should rotate at an angular velocity of approximately \(1.237 \times 10^{-3} \, \text{rad/s}\) for the apparent gravitational acceleration at the equator to be zero, and the length of the day in this situation would be approximately \(1.41\) hours.

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