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 determine 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. **Understand the Concept of Apparent Weightlessness**:
An object appears weightless when the net force acting on it is zero. At the equator, the apparent weight \( z' \) is given by the equation:
\[ z' = z - r \omega^2 \]
where:
- \( z \) is the gravitational acceleration (approximately \( 9.8 \, \text{m/s}^2 \)),
- \( r \) is the radius of the Earth,
- \( \omega \) is the angular speed of the Earth.

2. **Set Up the Equation for Weightlessness**:
For an object to appear weightless, we set \( z' = 0 \):
\[ 0 = z - r \omega^2 \]
Rearranging this gives:
\[ r \omega^2 = z \]

3. **Solve for Angular Speed \( \omega \)**:
We can express \( \omega \) in terms of \( z \) and \( r \):
\[ \omega^2 = \frac{z}{r} \]
Taking the square root of both sides, we find:
\[ \omega = \sqrt{\frac{z}{r}} \]

4. **Substitute the Values**:
- The radius of the Earth \( r = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \).
- The gravitational acceleration \( z = 9.8 \, \text{m/s}^2 \).

Now substituting these values into the equation:
\[ \omega = \sqrt{\frac{9.8}{6400 \times 10^3}} \]

5. **Calculate \( \omega \)**:
- First, calculate \( 6400 \times 10^3 = 6.4 \times 10^6 \, \text{m} \).
- Now, substitute:
\[ \omega = \sqrt{\frac{9.8}{6.4 \times 10^6}} \approx \sqrt{1.53125 \times 10^{-6}} \approx 1.236 \times 10^{-3} \, \text{rad/s} \]

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