Find the imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero. (take `g = 10 m//s^(2)` for the acceleration due to gravity, if the earth were at rest and radius of earth equal to 6400 km and `phi = 60^(@)`)

Options

A. `1.25 xx 10^(-3) "rad s"^(-1)`
B. `2.50 xx 10^(-3) " rad s"^(-1)`
C. `3.75 xx 10^(-3) " rad s"^(-1)`
D. `5 xx 10^(-3) " rad s"^(-1)`

Explanation

To find the imaginary angular velocity of the Earth for which the effective acceleration due to gravity at the equator becomes zero, we can follow these steps:

### Step 1: Understanding the Effective Acceleration Due to Gravity
The effective acceleration due to gravity at a point on the Earth's surface is given by the formula:
\[ g' = g - \omega^2 r \cos^2 \phi \]
where:
- \( g' \) is the effective acceleration due to gravity,
- \( g \) is the acceleration due to gravity (10 m/s²),
- \( \omega \) is the angular velocity,
- \( r \) is the radius of the Earth,
- \( \phi \) is the latitude.

### Step 2: Setting the Effective Gravity to Zero
We want to find the angular velocity \( \omega \) such that \( g' = 0 \):
\[ 0 = g - \omega^2 r \cos^2 \phi \]
Rearranging this gives:
\[ \omega^2 r \cos^2 \phi = g \]

### Step 3: Solving for Angular Velocity
Now, we can solve for \( \omega \):
\[ \omega^2 = \frac{g}{r \cos^2 \phi} \]
Taking the square root gives:
\[ \omega = \sqrt{\frac{g}{r \cos^2 \phi}} \]

### Step 4: Substituting Values
Given:
- \( g = 10 \, \text{m/s}^2 \)
- \( r = 6400 \, \text{km} = 6400 \times 1000 \, \text{m} = 6.4 \times 10^6 \, \text{m} \)
- \( \phi = 60^\circ \) (thus \( \cos 60^\circ = \frac{1}{2} \))

Substituting these values into the equation:
\[ \omega = \sqrt{\frac{10}{(6.4 \times 10^6) \left(\frac{1}{2}\right)^2}} = \sqrt{\frac{10}{(6.4 \times 10^6) \cdot \frac{1}{4}}} \]
This simplifies to:
\[ \omega = \sqrt{\frac{10 \cdot 4}{6.4 \times 10^6}} = \sqrt{\frac{40}{6.4 \times 10^6}} = \sqrt{\frac{40}{6400000}} = \sqrt{\frac{1}{160000}} = \frac{1}{400} \]

### Step 5: Final Calculation
Calculating the value:
\[ \omega = \frac{1}{400} \, \text{radians/second} = 2.5 \times 10^{-3} \, \text{radians/second} \]

### Conclusion
The imaginary angular velocity of the Earth for which the effective acceleration due to gravity at the equator shall be zero is:
\[ \omega = 2.5 \times 10^{-3} \, \text{radians/second} \]

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