What is the angular speed of rotation about its polar axis, so that the bodies on its equator would fell no weight?
(`g=9.8 m//s^(2),R=6.4xx10^(6)m)`

Options

A. `1.237xx10^(3) rad//s`
B. `1.237xx10^(-4) rad//s`
C. `7.27xx10^(-3) rad//s`
D. `7.27xx10^(-4) rad//s`

Explanation

To find the angular speed of rotation about the polar axis such that bodies on the equator feel no weight, we can follow these steps:

### Step 1: Understand the Forces Acting on the Bodies
When a body is at the equator of the rotating Earth, two main forces act on it:
1. The gravitational force (weight) acting downwards, which is given by \( F_g = mg \), where \( m \) is the mass of the body and \( g \) is the acceleration due to gravity.
2. The centrifugal force acting outward due to the rotation of the Earth, which is given by \( F_c = m \omega^2 R \), where \( \omega \) is the angular speed and \( R \) is the radius of the Earth.

### Step 2: Set Up the Equation for No Weight
For the bodies at the equator to feel no weight, the centrifugal force must balance the gravitational force. Therefore, we can set up the equation:
\[ mg = m \omega^2 R \]

### Step 3: Cancel the Mass
Since the mass \( m \) appears on both sides of the equation, we can cancel it out:
\[ g = \omega^2 R \]

### Step 4: Solve for Angular Speed \( \omega \)
Rearranging the equation gives us:
\[ \omega^2 = \frac{g}{R} \]
Taking the square root of both sides, we find:
\[ \omega = \sqrt{\frac{g}{R}} \]

### Step 5: Substitute the Values
Now, we can substitute the given values into the equation. We know:
- \( g = 9.8 \, \text{m/s}^2 \)
- \( R = 6.4 \times 10^6 \, \text{m} \)

Thus, we have:
\[ \omega = \sqrt{\frac{9.8}{6.4 \times 10^6}} \]

### Step 6: Calculate the Value
Now, we can calculate \( \omega \):
\[ \omega = \sqrt{\frac{9.8}{6.4 \times 10^6}} = \sqrt{1.53125 \times 10^{-6}} \approx 0.001237 \, \text{rad/s} \]

### Final Answer
The angular speed of rotation about its polar axis, so that the bodies on its equator would feel no weight, is approximately:
\[ \omega \approx 1.237 \times 10^{-3} \, \text{rad/s} \]

---