If the earth has no rotational motion, the weight of a person on the equation is W. Detrmine the speed with which the earth would have to rotate about its axis so that the person at the equator will weight `(3)/(4)` W. Radius of the earth is 6400 km and g = 10 `m//s^(2)`.

Options

A. `0.63 xx 10^(-3) "rad/s"`
B. `0.28 xx 10^(-3) "rad/s"`
C. `1.1 xx 10^(-3) "rad/s"`
D. `0.83 xx 10^(-3) "rad/s"`

Answer

To solve the problem, we need to determine the speed at which the Earth would have to rotate such that a person at the equator weighs \( \frac{3}{4} W \) instead of \( W \).

### Step-by-Step Solution:

1. **Understanding Weight and Effective Gravity**:
- The weight of a person is given by \( W = mg \), where \( m \) is the mass of the person and \( g \) is the acceleration due to gravity.
- When the Earth rotates, the effective gravity \( g_{\text{effective}} \) at the equator is given by:
\[ g_{\text{effective}} = g - \omega^2 R \]
where \( \omega \) is the angular velocity of the Earth and \( R \) is the radius of the Earth.

2. **Setting Up the Equation**:
- According to the problem, when the Earth rotates, the weight of the person becomes \( \frac{3}{4} W \). Therefore:
\[ mg_{\text{effective}} = \frac{3}{4} mg \]
- This simplifies to:
\[ g_{\text{effective}} = \frac{3}{4} g \]

3. **Substituting into the Effective Gravity Equation**:
- Substitute \( g_{\text{effective}} \) into the equation:
\[ \frac{3}{4} g = g - \omega^2 R \]
- Rearranging gives:
\[ \omega^2 R = g - \frac{3}{4} g = \frac{1}{4} g \]

4. **Solving for Angular Velocity \( \omega \)**:
- Now, we can express \( \omega^2 \) as:
\[ \omega^2 = \frac{g}{4R} \]
- Taking the square root gives:
\[ \omega = \sqrt{\frac{g}{4R}} \]

5. **Substituting Known Values**:
- Given \( g = 10 \, \text{m/s}^2 \) and \( R = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \):
\[ \omega = \sqrt{\frac{10}{4 \times 6400 \times 10^3}} \]
- Simplifying:
\[ \omega = \sqrt{\frac{10}{25600 \times 10^3}} = \sqrt{\frac{1}{2560 \times 10^3}} = \frac{1}{\sqrt{2560 \times 10^3}} \]

6. **Calculating the Value**:
- We can simplify \( \sqrt{2560} \) and \( 10^3 \):
\[ \sqrt{2560} \approx 50.6 \quad (\text{since } 50^2 = 2500 \text{ and } 51^2 = 2601) \]
- Thus:
\[ \omega \approx \frac{1}{50.6 \times 31.62} \approx 0.0625 \, \text{rad/s} \]

7. **Final Answer**:
- Converting to a more suitable form:
\[ \omega \approx 0.625 \times 10^{-3} \, \text{rad/s} \]