The period of revolution of an earth satellite close to surface of earth is 90min. The time period of aother satellite in an orbit at a distance of three times the radius of earth from its surface will be

Options

A. `90sqrt(8)min`
B. `360min`
C. `720min`
D. `270min`

Answer

To solve the problem, we need to find the time period of a satellite in an orbit at a distance of three times the radius of the Earth from its surface, given that the time period of a satellite close to the Earth's surface is 90 minutes.

### Step-by-Step Solution:

1. **Understanding the relationship between time period and radius**:
According to Kepler's third law of planetary motion, the square of the time period \( T \) of a satellite is directly proportional to the cube of the semi-major axis \( r \) of its orbit:
\[ T^2 \propto r^3 \]
This can be expressed as:
\[ T^2 = k \cdot r^3 \]
where \( k \) is a constant.

2. **Identifying the parameters**:
Let:
- \( T_1 \) = time period of the first satellite (90 minutes)
- \( r_1 \) = radius of the Earth \( R_e \) (for the satellite close to the surface)
- \( T_2 \) = time period of the second satellite (unknown)
- \( r_2 \) = distance from the center of the Earth for the second satellite, which is \( R_e + 3R_e = 4R_e \).

3. **Setting up the ratio of time periods**:
Using the relationship derived from Kepler's law, we can write:
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} \]
Substituting the known values:
\[ \frac{T_1^2}{T_2^2} = \frac{R_e^3}{(4R_e)^3} \]

4. **Simplifying the equation**:
The expression simplifies to:
\[ \frac{T_1^2}{T_2^2} = \frac{R_e^3}{64R_e^3} = \frac{1}{64} \]
This implies:
\[ T_2^2 = 64T_1^2 \]

5. **Substituting the known time period**:
Now substituting \( T_1 = 90 \) minutes:
\[ T_2^2 = 64 \times (90)^2 \]
\[ T_2 = 90 \times \sqrt{64} = 90 \times 8 = 720 \text{ minutes} \]

6. **Final Answer**:
The time period of the satellite at a distance of three times the radius of the Earth from its surface is:
\[ T_2 = 720 \text{ minutes} \]