The period of revolution of an earth's satellite close to the surface of the earth is `60` minute. The period of another the earth's satellite in an orbit at a distance of three times earth's radius from its surface will be (in minutes)

Options

A. `90`
B. `90xxsqrt(8)`
C. `270`
D. `480`

Explanation

To solve the problem, we will use Kepler's Third Law of planetary motion, which relates the period of revolution of a satellite to its distance from the center of the Earth.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- The period of the first satellite (T1) close to the Earth's surface is 60 minutes.
- The distance of the second satellite from the Earth's surface is three times the Earth's radius (R). Therefore, the total distance from the center of the Earth for the second satellite (R2) is:
\[ R2 = R + 3R = 4R \]

2. **Apply Kepler's Third Law**:
Kepler's Third Law states that the square of the period of revolution of a satellite is directly proportional to the cube of the semi-major axis (or radius) of its orbit:
\[ T^2 \propto R^3 \]
This can be expressed as:
\[ T^2 = k \cdot R^3 \]
where \( k \) is a constant.

3. **Set Up the Equations**:
For the first satellite:
\[ T_1^2 = k \cdot R^3 \]
For the second satellite:
\[ T_2^2 = k \cdot (4R)^3 \]

4. **Calculate the Radius for the Second Satellite**:
Expanding the equation for the second satellite:
\[ T_2^2 = k \cdot (4R)^3 = k \cdot 64R^3 \]

5. **Relate the Two Periods**:
Now we can relate \( T_2^2 \) to \( T_1^2 \):
\[ T_2^2 = 64 \cdot (k \cdot R^3) = 64 \cdot T_1^2 \]

6. **Substitute the Value of \( T_1 \)**:
Since \( T_1 = 60 \) minutes:
\[ T_2^2 = 64 \cdot (60)^2 \]

7. **Calculate \( T_2 \)**:
First, calculate \( (60)^2 \):
\[ (60)^2 = 3600 \]
Now substitute:
\[ T_2^2 = 64 \cdot 3600 \]
Calculate \( 64 \cdot 3600 \):
\[ T_2^2 = 230400 \]
Now take the square root to find \( T_2 \):
\[ T_2 = \sqrt{230400} = 480 \text{ minutes} \]

### Final Answer:
The period of the second satellite is **480 minutes**.