The rotation of the earth about its axis speeds up such that a man on the equator becomes weightness. In such a situation, what would be the duration of one day ?

Options

A. `2pi sqrt((R)/(g))`
B. `(1)/(2pi)sqrt((R)/(g))`
C. `2pi sqrt(Rg)`
D. `(1)/(2pi)sqrt(Rg)`

Explanation

To solve the problem, we need to determine the duration of one day (T) when a man at the equator experiences weightlessness due to the Earth's rotation. Here’s the step-by-step solution:

### Step 1: Understand the condition for weightlessness
When a person at the equator feels weightless, the effective gravitational force acting on them becomes zero. The effective gravitational force (G') can be expressed as:
\[ G' = G - R \omega^2 \]
where:
- \( G \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( R \) is the radius of the Earth (approximately \( 6.37 \times 10^6 \, \text{m} \)),
- \( \omega \) is the angular velocity of the Earth's rotation.

For weightlessness, we set \( G' = 0 \):
\[ G - R \omega^2 = 0 \]
This leads to:
\[ G = R \omega^2 \]

### Step 2: Express angular velocity in terms of the period
The angular velocity \( \omega \) can also be expressed in terms of the period \( T \) (the duration of one day) as:
\[ \omega = \frac{2\pi}{T} \]

### Step 3: Substitute \( \omega \) into the equation
Substituting \( \omega \) into the equation from Step 1 gives:
\[ G = R \left(\frac{2\pi}{T}\right)^2 \]
This can be rearranged to find \( T \):
\[ G = R \frac{4\pi^2}{T^2} \]
\[ T^2 = \frac{4\pi^2 R}{G} \]
\[ T = 2\pi \sqrt{\frac{R}{G}} \]

### Step 4: Substitute known values
Now, we substitute the known values for \( R \) and \( G \):
- \( R \approx 6.37 \times 10^6 \, \text{m} \)
- \( G \approx 9.81 \, \text{m/s}^2 \)

Calculating \( T \):
\[ T = 2\pi \sqrt{\frac{6.37 \times 10^6}{9.81}} \]

### Step 5: Calculate the numerical value
Calculating the square root:
\[ \sqrt{\frac{6.37 \times 10^6}{9.81}} \approx \sqrt{649,000} \approx 805.6 \]
Now, substituting back:
\[ T \approx 2\pi \times 805.6 \approx 5066.3 \, \text{s} \]

### Step 6: Convert seconds to hours
To convert seconds into hours:
\[ T \approx \frac{5066.3}{3600} \approx 1.41 \, \text{hours} \]

### Conclusion
Thus, the duration of one day when a man on the equator becomes weightless would be approximately **1.41 hours**.

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