How many hours would make a day if the earth were rotating at such a high speed that the weight of a body on the equator were zero

Options

A. `6.24`
B. `1.4h`
C. `28 h`
D. `5.6h`

Explanation

To solve the problem of how many hours would make a day if the Earth were rotating at such a high speed that the weight of a body on the equator were zero, we can follow these steps:

### Step 1: Understand the condition for zero weight
When the weight of a body at the equator is zero, it means that the gravitational force acting on the body is balanced by the centrifugal force due to the Earth's rotation. Mathematically, this can be expressed as:
\[ mg = m \omega^2 r \]
where:
- \( m \) is the mass of the body,
- \( g \) is the acceleration due to gravity,
- \( \omega \) is the angular velocity of the Earth,
- \( r \) is the radius of the Earth.

### Step 2: Simplify the equation
Since the mass \( m \) appears on both sides of the equation, we can cancel it out:
\[ g = \omega^2 r \]

### Step 3: Solve for angular velocity \( \omega \)
Rearranging the equation gives:
\[ \omega = \sqrt{\frac{g}{r}} \]

### Step 4: Relate angular velocity to the time period
The angular velocity \( \omega \) is related to the time period \( T \) (the time for one complete rotation) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting this into our equation gives:
\[ \frac{2\pi}{T} = \sqrt{\frac{g}{r}} \]

### Step 5: Solve for the time period \( T \)
Rearranging the equation to solve for \( T \) results in:
\[ T = \frac{2\pi}{\sqrt{\frac{g}{r}}} \]
This can be rewritten as:
\[ T = 2\pi \sqrt{\frac{r}{g}} \]

### Step 6: Substitute values for \( r \) and \( g \)
The average radius of the Earth \( r \) is approximately \( 6400 \) km, which is \( 6400 \times 10^3 \) meters. The acceleration due to gravity \( g \) is approximately \( 9.8 \, \text{m/s}^2 \).

Substituting these values into the equation:
\[ T = 2\pi \sqrt{\frac{6400 \times 10^3}{9.8}} \]

### Step 7: Calculate \( T \)
Calculating the value:
1. Calculate \( \frac{6400 \times 10^3}{9.8} \):
\[ \frac{6400 \times 10^3}{9.8} \approx 653061.2245 \]
2. Take the square root:
\[ \sqrt{653061.2245} \approx 808.5 \]
3. Multiply by \( 2\pi \):
\[ T \approx 2\pi \times 808.5 \approx 5085.7 \, \text{seconds} \]

### Step 8: Convert seconds to hours
To convert seconds to hours:
\[ \text{Hours} = \frac{5085.7}{3600} \approx 1.41 \, \text{hours} \]

### Conclusion
Thus, if the Earth were rotating at such a speed that the weight of a body on the equator were zero, a day would be approximately **1.41 hours** long.