The angular velocity of the earth about its polar axis so that the weight of the body at the equator will be zero is

Options

A. `1.25 xx10^(-3) " rad"s^(-1)`
B. `2 " rad s"^(-1)`
C. `1 " rad s"^(-1)`
D. 0

Answer

To find the angular velocity of the Earth about its polar axis such that the weight of a body at the equator becomes zero, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Forces at Play**:
At the equator, the effective weight of a body is influenced by two forces: the gravitational force acting downwards and the centripetal force due to the Earth's rotation acting upwards. For the weight to be zero, these two forces must balance each other.

2. **Write the Equation for Effective Weight**:
The effective gravitational acceleration at the equator can be expressed as:
\[ g' = g - R \omega^2 \]
where:
- \( g' \) is the effective gravitational acceleration at the equator,
- \( g \) is the acceleration due to gravity at the poles,
- \( R \) is the radius of the Earth,
- \( \omega \) is the angular velocity of the Earth.

3. **Set Effective Weight to Zero**:
For the weight to be zero at the equator:
\[ g' = 0 \implies g - R \omega^2 = 0 \]
Rearranging gives:
\[ g = R \omega^2 \]

4. **Solve for Angular Velocity (\(\omega\))**:
Rearranging the equation to solve for \(\omega\):
\[ \omega^2 = \frac{g}{R} \]
\[ \omega = \sqrt{\frac{g}{R}} \]

5. **Substitute Known Values**:
We know:
- \( g \) (acceleration due to gravity at the poles) is approximately \( 9.8 \, \text{m/s}^2 \),
- \( R \) (radius of the Earth) is approximately \( 6.4 \times 10^6 \, \text{m} \).

Substituting these values into the equation:
\[ \omega = \sqrt{\frac{9.8}{6.4 \times 10^6}} \]

6. **Calculate \(\omega\)**:
Performing the calculation:
\[ \omega = \sqrt{\frac{9.8}{6.4 \times 10^6}} \approx \sqrt{1.53125 \times 10^{-6}} \approx 1.24 \times 10^{-3} \, \text{rad/s} \]

7. **Final Result**:
Thus, the angular velocity of the Earth about its polar axis so that the weight of a body at the equator will be zero is approximately:
\[ \omega \approx 1.25 \times 10^{-3} \, \text{rad/s} \]