The angular velocity of the earth with which it has to rotate so that the acceleration due to gravity on `60^@` latitude becomes zero is

Options

A. `2.5 xx 10^(-3) rad s^(-1)`
B. `5.0 xx 10^(-1) rad s^(-1)`
C. `100 rad s^(-1)`
D. `7.8 xx 10^(-2) rad s^(-1)`

Answer

To find the angular velocity of the Earth at which the acceleration due to gravity at a latitude of \(60^\circ\) becomes zero, we can follow these steps:

### Step 1: Understand the relationship between gravity, rotation, and latitude
The effective acceleration due to gravity \(g'\) at a latitude \(\theta\) is given by the equation:
\[ g' = g - R \omega^2 \cos^2 \theta \]
where:
- \(g\) is the acceleration due to gravity at the equator (approximately \(9.8 \, \text{m/s}^2\)),
- \(R\) is the radius of the Earth,
- \(\omega\) is the angular velocity of the Earth,
- \(\theta\) is the latitude.

### Step 2: Set the effective gravity to zero
To find the angular velocity at which \(g'\) becomes zero, set the equation to zero:
\[ 0 = g - R \omega^2 \cos^2 \theta \]
This implies:
\[ g = R \omega^2 \cos^2 \theta \]

### Step 3: Substitute the latitude value
For \(\theta = 60^\circ\):
\[ \cos 60^\circ = \frac{1}{2} \]
Thus,
\[ \cos^2 60^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

### Step 4: Rewrite the equation
Substituting \(\cos^2 60^\circ\) into the equation gives:
\[ g = R \omega^2 \cdot \frac{1}{4} \]
Rearranging this, we find:
\[ \omega^2 = \frac{4g}{R} \]

### Step 5: Take the square root
Taking the square root of both sides, we have:
\[ \omega = \sqrt{\frac{4g}{R}} = 2\sqrt{\frac{g}{R}} \]

### Step 6: Substitute known values
Using \(g = 9.8 \, \text{m/s}^2\) and the radius of the Earth \(R \approx 6.4 \times 10^6 \, \text{m}\):
\[ \omega = 2\sqrt{\frac{9.8}{6.4 \times 10^6}} \]

### Step 7: Calculate the value
Calculating the value inside the square root:
\[ \frac{9.8}{6.4 \times 10^6} \approx 1.53125 \times 10^{-6} \]
Now, taking the square root:
\[ \sqrt{1.53125 \times 10^{-6}} \approx 0.001237 \]
Thus,
\[ \omega \approx 2 \times 0.001237 \approx 0.002474 \, \text{radians/second} \]

### Step 8: Final result
Therefore, the angular velocity of the Earth at which the acceleration due to gravity at \(60^\circ\) latitude becomes zero is approximately:
\[ \omega \approx 2.5 \times 10^{-3} \, \text{radians/second} \]