The angular speed of earth is `"rad s"^(-1)`, so that the object on equator may appear weightless, is (radius of earth = 6400 km)

Options

A. `1.23 xx 10^(-3)`
B. `6.20 xx 10^(-3)`
C. `1.56`
D. `1.23 xx 10^(-5)`

Explanation

To find the angular speed of the Earth at which an object on the equator may appear weightless, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Concept**:
An object appears weightless when the gravitational force acting on it is balanced by the centripetal force required to keep it in circular motion. This occurs when the effective gravitational acceleration \( g' \) at the equator becomes zero.

2. **Setting Up the Equation**:
The effective gravitational acceleration \( g' \) 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 (which is \( 6400 \, \text{km} = 6400000 \, \text{m} \)),
- \( \omega \) is the angular speed in radians per second.

3. **Condition for Weightlessness**:
For the object to appear weightless, we set \( g' = 0 \):
\[ 0 = g - r \omega^2 \]
Rearranging gives:
\[ g = r \omega^2 \]

4. **Solving for Angular Speed \( \omega \)**:
We can express \( \omega \) in terms of \( g \) and \( r \):
\[ \omega^2 = \frac{g}{r} \]
Taking the square root of both sides:
\[ \omega = \sqrt{\frac{g}{r}} \]

5. **Substituting Values**:
Now we substitute the known values:
\[ g = 9.81 \, \text{m/s}^2, \quad r = 6400000 \, \text{m} \]
\[ \omega = \sqrt{\frac{9.81}{6400000}} \]

6. **Calculating \( \omega \)**:
Performing the calculation:
\[ \omega = \sqrt{\frac{9.81}{6400000}} \approx \sqrt{1.53 \times 10^{-6}} \approx 0.00123 \, \text{rad/s} \]

### Final Answer:
The angular speed of the Earth at which an object on the equator may appear weightless is approximately:
\[ \omega \approx 1.23 \times 10^{-3} \, \text{rad/s} \]