The angular velocity of the earth about its polar axis so that the weight of the body at the equator will be zero is

Options

A. `1.25xx10^(-3) "rad s"^(-1)`
B. 2 rad `s^(-1)`
C. 1 rad `s^(-1)`
D. 0

Answer

To find the angular velocity of the Earth about its polar axis such that the weight of a body at the equator becomes zero, we can follow these steps:

### Step 1: Understand the Forces Acting on the Body
At the equator, a body experiences two forces: the gravitational force acting downwards and the centrifugal force due to the Earth's rotation acting outwards. For the weight to be zero, these two forces must balance each other.

### Step 2: Write the Equation for Effective Weight
The effective weight (or apparent weight) of the body at the equator can be expressed as:
\[ g' = g - R \omega^2 \]
where:
- \( g' \) is the effective weight (which we want to be zero),
- \( g \) is the acceleration due to gravity at the poles (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( R \) is the radius of the Earth (approximately \( 6.4 \times 10^6 \, \text{m} \)),
- \( \omega \) is the angular velocity of the Earth.

### Step 3: Set the Effective Weight to Zero
To find the angular velocity where the effective weight is zero, we set:
\[ 0 = g - R \omega^2 \]
Rearranging gives:
\[ R \omega^2 = g \]

### Step 4: Solve for Angular Velocity
Now, we can solve for \( \omega \):
\[ \omega^2 = \frac{g}{R} \]
Taking the square root:
\[ \omega = \sqrt{\frac{g}{R}} \]

### Step 5: Substitute Known Values
Substituting the known values:
- \( g = 9.81 \, \text{m/s}^2 \)
- \( R = 6.4 \times 10^6 \, \text{m} \)

We get:
\[ \omega = \sqrt{\frac{9.81}{6.4 \times 10^6}} \]

### Step 6: Calculate the Value
Calculating the above expression:
\[ \omega = \sqrt{\frac{9.81}{6.4 \times 10^6}} \approx \sqrt{1.53 \times 10^{-6}} \approx 1.24 \times 10^{-3} \, \text{rad/s} \]

### Final Answer
Thus, the angular velocity of the Earth about its polar axis so that the weight of a body at the equator will be zero is approximately:
\[ \omega \approx 1.25 \times 10^{-3} \, \text{rad/s} \]

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