What is the angular velocity of a body on'the surface of the earth at the equator ? Also find its linear velocity. Given radius of the earth is 6400 km. Period of rotation of the earth = 24 hours.

Explanation

To find the angular velocity and linear velocity of a body on the surface of the Earth at the equator, we can follow these steps:

### Step 1: Calculate the Angular Velocity (ω)
The angular velocity (ω) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
where \(T\) is the period of rotation.

Given that the period of rotation of the Earth is 24 hours, we need to convert this into seconds:
\[ T = 24 \text{ hours} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} = 86400 \text{ seconds} \]

Now we can substitute \(T\) into the formula for angular velocity:
\[ \omega = \frac{2\pi}{86400 \text{ seconds}} \approx 7.272 \times 10^{-5} \text{ radians/second} \]

### Step 2: Calculate the Linear Velocity (v)
The linear velocity (v) can be calculated using the formula:
\[ v = R \cdot \omega \]
where \(R\) is the radius of the Earth.

Given that the radius of the Earth is 6400 km, we convert this to meters:
\[ R = 6400 \text{ km} = 6400 \times 10^3 \text{ meters} = 6.4 \times 10^6 \text{ meters} \]

Now we can substitute \(R\) and \(\omega\) into the formula for linear velocity:
\[ v = (6.4 \times 10^6 \text{ meters}) \cdot (7.272 \times 10^{-5} \text{ radians/second}) \approx 465 \text{ meters/second} \]

### Final Results
- Angular Velocity (ω) = \(7.272 \times 10^{-5} \text{ radians/second}\)
- Linear Velocity (v) = \(465 \text{ meters/second}\)

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