For a satellite orbiting close to the surface of earth the period of revolution is `84 min`. The time period of another satellite orbiting at a height three times the radius of earth from its surface will be

Options

A. `(84) 2 sqrt(2) min`
B. `8 (84) min`
C. `(84) 3sqrt(3) min`
D. `3 (84) min`

Explanation

To solve the problem, we will use Kepler's third law of planetary motion, which states that the square of the period of revolution (T) of a satellite is directly proportional to the cube of the semi-major axis (r) of its orbit. This can be expressed mathematically as:

\[ T^2 \propto r^3 \]

### Step-by-Step Solution:

1. **Identify Given Values:**
- The period of the first satellite (T1) is 84 minutes.
- The radius of the Earth (RE) is taken as a reference.
- The height of the second satellite is three times the radius of the Earth from the surface, which means the total distance from the center of the Earth (r2) is:
\[ r2 = RE + 3RE = 4RE \]

2. **Set Up the Ratio Using Kepler's Third Law:**
According to Kepler's third law:
\[ \frac{T1^2}{T2^2} = \frac{r1^3}{r2^3} \]
Here, \( r1 = RE \) (the radius of the Earth) and \( r2 = 4RE \).

3. **Substitute the Values:**
Substitute the known values into the equation:
\[ \frac{(84 \text{ min})^2}{T2^2} = \frac{(RE)^3}{(4RE)^3} \]

4. **Simplify the Right Side:**
The right side simplifies as follows:
\[ \frac{(RE)^3}{(4RE)^3} = \frac{1}{4^3} = \frac{1}{64} \]
Therefore, we have:
\[ \frac{(84)^2}{T2^2} = \frac{1}{64} \]

5. **Cross-Multiply to Solve for T2:**
Cross-multiplying gives:
\[ 84^2 = \frac{T2^2}{64} \]
Rearranging gives:
\[ T2^2 = 84^2 \times 64 \]

6. **Calculate T2:**
First, calculate \( 84^2 \):
\[ 84^2 = 7056 \]
Now, multiply by 64:
\[ T2^2 = 7056 \times 64 = 451584 \]
Taking the square root to find T2:
\[ T2 = \sqrt{451584} \approx 672 \text{ minutes} \]

### Final Answer:
The time period of the satellite orbiting at a height three times the radius of the Earth from its surface is approximately **672 minutes**.