How many times faster than its present speed the earth should rotate so that the apparent weight of an object at equator becomes zero ? Given radius of the earth = `6.37 xx 10^(6)` m . What would be the duration of the day in that case ?

Answer

To solve the problem of how many times faster the Earth should rotate so that the apparent weight of an object at the equator becomes zero, we can follow these steps:

### Step 1: Understand the Concept of Apparent Weight
The apparent weight of an object is the normal force acting on it. At the equator, the apparent weight becomes zero when the gravitational force is exactly balanced by the centrifugal force due to the Earth's rotation.

### Step 2: Write the Equation for Apparent Weight
At the equator, the apparent weight \( W' \) can be expressed as:
\[ W' = mg - m\omega^2 r \]
where:
- \( m \) = mass of the object
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( \omega \) = angular velocity of the Earth
- \( r \) = radius of the Earth (given as \( 6.37 \times 10^6 \, \text{m} \))

To make the apparent weight zero:
\[ mg - m\omega^2 r = 0 \]
This simplifies to:
\[ g = \omega^2 r \]

### Step 3: Substitute Known Values
We know:
- \( g \approx 9.81 \, \text{m/s}^2 \)
- \( r = 6.37 \times 10^6 \, \text{m} \)

Substituting these values into the equation gives:
\[ 9.81 = \omega^2 (6.37 \times 10^6) \]

### Step 4: Solve for \( \omega^2 \)
Rearranging the equation to find \( \omega^2 \):
\[ \omega^2 = \frac{9.81}{6.37 \times 10^6} \]

Calculating this:
\[ \omega^2 \approx 1.54 \times 10^{-6} \]

### Step 5: Calculate \( \omega \)
Taking the square root to find \( \omega \):
\[ \omega \approx \sqrt{1.54 \times 10^{-6}} \approx 0.00124 \, \text{rad/s} \]

### Step 6: Find the Current Angular Velocity
The current angular velocity of the Earth is:
\[ \omega_0 = \frac{2\pi}{T} \]
where \( T \) is the current duration of the day (approximately \( 86400 \, \text{s} \)):
\[ \omega_0 \approx \frac{2\pi}{86400} \approx 7.27 \times 10^{-5} \, \text{rad/s} \]

### Step 7: Calculate the Required Speed Ratio
To find how many times faster the Earth should rotate:
\[ \text{Ratio} = \frac{\omega}{\omega_0} \]
Substituting the values:
\[ \text{Ratio} = \frac{0.00124}{7.27 \times 10^{-5}} \approx 17.05 \]

### Step 8: Calculate the New Duration of the Day
The new duration of the day \( T' \) can be calculated using:
\[ T' = \frac{T}{\text{Ratio}} \]
Substituting the values:
\[ T' \approx \frac{86400}{17.05} \approx 5065.5 \, \text{s} \]

### Final Answer
Thus, the Earth should rotate approximately **17 times faster** than its current speed, and the new duration of the day would be approximately **5065.5 seconds** or about **1.4 hours**.

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