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

Options

A. `465 "m s"^(-1)`
B. `165 "m s"^(-1)`
C. `665 "m s"^(-1)`
D. `410 "m s"^(-1)`

Answer

To find the linear velocity of a body on the surface of the Earth at the equator, we can use the formula for linear velocity \( v \):

\[ v = \frac{2 \pi r}{T} \]

where:
- \( r \) is the radius of the Earth,
- \( T \) is the period of rotation of the Earth.

### Step 1: Identify the given values
- Radius of the Earth, \( r = 6400 \) km
- Period of rotation, \( T = 24 \) hours

### Step 2: Convert the radius from kilometers to meters
Since we want the final answer in meters per second, we need to convert the radius from kilometers to meters:

\[ r = 6400 \text{ km} = 6400 \times 1000 \text{ m} = 6400000 \text{ m} \]

### Step 3: Convert the period from hours to seconds
Next, we convert the period of rotation from hours to seconds:

\[ T = 24 \text{ hours} = 24 \times 3600 \text{ seconds} = 86400 \text{ seconds} \]

### Step 4: Substitute the values into the velocity formula
Now, we can substitute the values of \( r \) and \( T \) into the formula for linear velocity:

\[ v = \frac{2 \pi (6400000 \text{ m})}{86400 \text{ s}} \]

### Step 5: Calculate the linear velocity
Calculating the above expression:

1. Calculate \( 2 \pi \):
\[ 2 \pi \approx 6.2832 \]

2. Now substitute and calculate:
\[ v = \frac{6.2832 \times 6400000}{86400} \]
\[ v \approx \frac{40212352}{86400} \approx 465.1 \text{ m/s} \]

### Final Answer
The linear velocity of a body on the surface of the Earth at the equator is approximately \( 465 \text{ m/s} \).

---