Find the acceleration of a particle placed on the surface of the earth at the equator due to earth's rotation. The diameter of earth =12800 km and it takes 24 hour for the earth to complete one revolution about its axis.

Explanation

To find the acceleration of a particle placed on the surface of the Earth at the equator due to Earth's rotation, we can follow these steps:

### Step 1: Calculate the Radius of the Earth
Given the diameter of the Earth is 12800 km, we can find the radius (r) using the formula:
\[ r = \frac{\text{Diameter}}{2} = \frac{12800 \text{ km}}{2} = 6400 \text{ km} = 6400 \times 10^3 \text{ m} = 6.4 \times 10^6 \text{ m} \]

### Step 2: Calculate the Velocity of a Point on the Equator
The velocity (v) of a point on the equator can be calculated using the formula:
\[ v = \frac{2 \pi r}{T} \]
where \(T\) is the time period for one complete revolution (24 hours). First, convert 24 hours into seconds:
\[ T = 24 \text{ hours} \times 3600 \text{ seconds/hour} = 86400 \text{ seconds} \]
Now, substituting the values:
\[ v = \frac{2 \pi (6.4 \times 10^6 \text{ m})}{86400 \text{ s}} \]
Calculating this gives:
\[ v \approx \frac{40.212 \times 10^6 \text{ m}}{86400 \text{ s}} \approx 464.6 \text{ m/s} \]

### Step 3: Calculate the Centripetal Acceleration
The centripetal acceleration (a_c) can be calculated using the formula:
\[ a_c = \frac{v^2}{r} \]
Substituting the values:
\[ a_c = \frac{(464.6 \text{ m/s})^2}{6.4 \times 10^6 \text{ m}} \]
Calculating this gives:
\[ a_c \approx \frac{216,000.16 \text{ m}^2/\text{s}^2}{6.4 \times 10^6 \text{ m}} \approx 0.0337 \text{ m/s}^2 \]

### Step 4: Conclusion
The acceleration of a particle placed on the surface of the Earth at the equator due to Earth's rotation is approximately:
\[ a_c \approx 0.0337 \text{ m/s}^2 \]

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