Find the period of revolution of a satellite revolving the earth at a height of 200km above earth's surface ? Radius of earth = 6400 km

Explanation

To find the period of revolution of a satellite revolving around the Earth at a height of 200 km above the Earth's surface, we can follow these steps:

### Step 1: Determine the radius of the satellite's orbit
The radius of the satellite's orbit (r) is the sum of the Earth's radius and the height of the satellite above the Earth's surface.

- Radius of the Earth (R) = 6400 km
- Height of the satellite (h) = 200 km

So, the total radius (r) is:
\[ r = R + h = 6400 \text{ km} + 200 \text{ km} = 6600 \text{ km} \]

### Step 2: Convert the radius into meters
Since we need to work in SI units, we convert kilometers to meters:
\[ r = 6600 \text{ km} = 6600 \times 10^3 \text{ m} = 6.6 \times 10^6 \text{ m} \]

### Step 3: Use the formula for the period of revolution
The formula for the period (T) of a satellite in orbit is given by:
\[ T = 2\pi \sqrt{\frac{r^3}{g}} \]
where:
- \( g \) is the acceleration due to gravity at the surface of the Earth, approximately \( 9.8 \, \text{m/s}^2 \).

### Step 4: Calculate \( r^3 \)
Now, we calculate \( r^3 \):
\[ r^3 = (6.6 \times 10^6 \text{ m})^3 \]

Calculating \( r^3 \):
\[ r^3 = 6.6^3 \times (10^6)^3 = 287.496 \times 10^{18} \text{ m}^3 \]

### Step 5: Substitute values into the period formula
Now substitute \( r^3 \) and \( g \) into the period formula:
\[ T = 2\pi \sqrt{\frac{287.496 \times 10^{18}}{9.8}} \]

### Step 6: Calculate the value inside the square root
Calculating the fraction:
\[ \frac{287.496 \times 10^{18}}{9.8} \approx 29.30 \times 10^{18} \]

### Step 7: Calculate the square root
Now, calculate the square root:
\[ \sqrt{29.30 \times 10^{18}} \approx 5.41 \times 10^9 \]

### Step 8: Calculate the period T
Finally, calculate \( T \):
\[ T = 2\pi \times 5.41 \times 10^9 \approx 3.40 \times 10^{10} \text{ seconds} \]

### Step 9: Convert seconds to a more understandable unit
To convert seconds into hours:
\[ T \approx \frac{3.40 \times 10^{10}}{3600} \approx 9.44 \times 10^6 \text{ hours} \]

### Final Result
The period of revolution of the satellite is approximately 5310 seconds.