An earth satellite is revolving in a circular orbit of radius 7000km. If its period of revolution is 5500 sec, its orbital velocity is

Options

A. 1 `"kms"^(-1)`
B. 8 `"kms"^(-1)`
C. 2 `"kms"^(-1)`
D. 2.2 `"kms"^(-1)`

Answer

To find the orbital velocity of an Earth satellite revolving in a circular orbit, we can use the formula for orbital velocity, which is given by:

\[ v_0 = \frac{2 \pi r}{T} \]

Where:
- \( v_0 \) is the orbital velocity,
- \( r \) is the radius of the orbit,
- \( T \) is the period of revolution.

### Step 1: Convert the radius into meters
The radius given is 7000 km. We need to convert this into meters for consistency in SI units.

\[ r = 7000 \text{ km} = 7000 \times 1000 \text{ m} = 7,000,000 \text{ m} \]

### Step 2: Use the period of revolution
The period of revolution \( T \) is given as 5500 seconds.

### Step 3: Substitute the values into the formula
Now we can substitute the values of \( r \) and \( T \) into the orbital velocity formula:

\[ v_0 = \frac{2 \pi (7,000,000 \text{ m})}{5500 \text{ s}} \]

### Step 4: Calculate the orbital velocity
Now we can perform the calculations:

1. Calculate \( 2 \pi \):

\[ 2 \pi \approx 6.2832 \]

2. Multiply by the radius:

\[ 2 \pi r = 6.2832 \times 7,000,000 \approx 43982240 \text{ m} \]

3. Now divide by the period \( T \):

\[ v_0 = \frac{43982240 \text{ m}}{5500 \text{ s}} \approx 7996.86 \text{ m/s} \]

### Step 5: Round the result
We can round this to approximately:

\[ v_0 \approx 8000 \text{ m/s} \]

### Final Answer
The orbital velocity of the satellite is approximately \( 8000 \text{ m/s} \).

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