How mcuh faster than its present rate should the earth rotate about its axis so that the weight of a body at the equator becomes zero? Also calculate the new length of the day. (c ) What would happen if rotation becomes faster ?

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Understanding the Condition for Weight to Become Zero
The weight of an object at the equator becomes zero when the apparent gravity (g') is equal to zero. The formula for apparent gravity is given by:
\[ g' = g - \omega^2 R_e \]
where:
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( \omega \) is the angular velocity of the Earth,
- \( R_e \) is the radius of the Earth (approximately \( 6.4 \times 10^6 \, \text{m} \)).

### Step 2: Setting Up the Equation
Since we want the weight to become zero, we set \( g' = 0 \):
\[ 0 = g - \omega^2 R_e \]
This leads to:
\[ g = \omega^2 R_e \]

### Step 3: Solving for the Angular Velocity
Rearranging the equation gives us:
\[ \omega^2 = \frac{g}{R_e} \]
Taking the square root:
\[ \omega = \sqrt{\frac{g}{R_e}} \]

### Step 4: Substituting Values
Substituting the known values:
- \( g = 9.81 \, \text{m/s}^2 \)
- \( R_e = 6.4 \times 10^6 \, \text{m} \)

Calculating \( \omega \):
\[ \omega = \sqrt{\frac{9.81}{6.4 \times 10^6}} \]
\[ \omega \approx 0.001237 \, \text{rad/s} \]

### Step 5: Finding the Ratio of New Angular Velocity to Current Angular Velocity
The current angular velocity of the Earth (\( \omega_{current} \)) is approximately:
\[ \omega_{current} \approx 7.3 \times 10^{-5} \, \text{rad/s} \]

Now, we find the ratio:
\[ \frac{\omega}{\omega_{current}} = \frac{0.001237}{7.3 \times 10^{-5}} \approx 16.9 \]

This means the Earth should rotate approximately **17 times faster** than its current rate.

### Step 6: Calculating the New Length of the Day
The length of the day is related to angular velocity by the formula:
\[ T = \frac{2\pi}{\omega} \]

Substituting the new angular velocity:
\[ T = \frac{2\pi}{0.001237} \]
Calculating \( T \):
\[ T \approx 84.6 \, \text{minutes} \]
or approximately **1.4 hours**.

### Step 7: Consequences of Faster Rotation
If the Earth rotates faster than this new rate, the centrifugal force at the equator would exceed the gravitational force, causing objects to start lifting off the surface.

### Summary of the Solution
1. The Earth should rotate approximately **17 times faster** than its current rate.
2. The new length of the day would be approximately **1.4 hours**.
3. If the rotation becomes faster, objects at the equator would start to leave the surface.