What should be the velocity of earth due to rotation about its own axis so that the weight at equator become 3/5 of initial value. Radius of earth on equator is 6400 km

Options

A. `7.4xx10^(-4) rad//sec`
B. `6.7xx10^(-4) rad//sec`
C. `7.8xx10^(-4) rad//sec`
D. `8.7xx10^(-4) rad//sec`

Answer

To solve the problem of finding the velocity of Earth due to its rotation about its own axis so that the weight at the equator becomes \( \frac{3}{5} \) of its initial value, we can follow these steps:

### Step 1: Understand the Forces Involved
At the equator, an object experiences two forces: the gravitational force \( F_g \) acting downwards and the centripetal force \( F_c \) required for circular motion due to the Earth's rotation. The effective weight \( W' \) at the equator can be expressed as:
\[ W' = W - F_c \]
where \( W \) is the initial weight.

### Step 2: Express the Forces Mathematically
The gravitational force acting on an object of mass \( m \) at the surface of the Earth is given by:
\[ W = mg \]
where \( g \) is the acceleration due to gravity. The centripetal force required for circular motion is given by:
\[ F_c = m \omega^2 r \]
where \( \omega \) is the angular velocity and \( r \) is the radius of the Earth.

### Step 3: Set Up the Equation for Effective Weight
According to the problem, the effective weight at the equator becomes \( \frac{3}{5} \) of its initial value:
\[ W' = \frac{3}{5} W \]
Substituting the expressions for weight, we have:
\[ W - m \omega^2 r = \frac{3}{5} W \]

### Step 4: Rearrange the Equation
Rearranging the equation gives:
\[ mg - m \omega^2 r = \frac{3}{5} mg \]
This can be simplified to:
\[ mg - \frac{3}{5} mg = m \omega^2 r \]
\[ \frac{2}{5} mg = m \omega^2 r \]

### Step 5: Cancel Mass and Solve for Angular Velocity
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{2}{5} g = \omega^2 r \]
Now, solving for \( \omega^2 \):
\[ \omega^2 = \frac{2g}{5r} \]

### Step 6: Substitute Known Values
We know:
- \( g \approx 9.8 \, \text{m/s}^2 \)
- \( r = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \)

Substituting these values:
\[ \omega^2 = \frac{2 \times 9.8}{5 \times 6400 \times 10^3} \]

### Step 7: Calculate \( \omega \)
Calculating \( \omega \):
\[ \omega = \sqrt{\frac{2 \times 9.8}{5 \times 6400 \times 10^3}} \approx 7.8 \times 10^{-4} \, \text{rad/s} \]

### Step 8: Find the Linear Velocity
The linear velocity \( v \) at the equator is given by:
\[ v = \omega r \]
Substituting \( \omega \) and \( r \):
\[ v = (7.8 \times 10^{-4}) \times (6400 \times 10^3) \approx 4992 \, \text{m/s} \]

Thus, the required velocity of Earth due to its rotation so that the weight at the equator becomes \( \frac{3}{5} \) of its initial value is approximately **4992 m/s**.