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 of finding the time period of a satellite in an orbit at a distance of three times the radius of the Earth from its surface, we can use Kepler's Third Law of planetary motion. Here’s a step-by-step solution:

### Step 1: Understand Kepler's Third Law
Kepler's Third Law 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. Mathematically, this can be expressed as:
\[ T^2 \propto r^3 \]
This means:
\[ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} \]

### Step 2: Identify Given Values
From the problem, we know:
- The period of the first satellite (close to the Earth's surface) \( T_e = 90 \) minutes.
- The radius of the Earth \( R_e \).
- The distance of the second satellite from the Earth's surface is \( 3R_e \), so the total distance from the center of the Earth is \( R_e + 3R_e = 4R_e \).

### Step 3: Set Up the Ratio
Using the values:
- For the first satellite, the radius \( r_1 = R_e \).
- For the second satellite, the radius \( r_2 = 4R_e \).

Now we can set up the ratio using Kepler's Third Law:
\[ \frac{T_e^2}{T_o^2} = \frac{R_e^3}{(4R_e)^3} \]

### Step 4: Simplify the Ratio
Calculating the right side:
\[ \frac{R_e^3}{(4R_e)^3} = \frac{R_e^3}{64R_e^3} = \frac{1}{64} \]
Thus, we have:
\[ \frac{T_e^2}{T_o^2} = \frac{1}{64} \]

### Step 5: Solve for \( T_o \)
Now, rearranging the equation gives:
\[ T_o^2 = 64 T_e^2 \]
Taking the square root:
\[ T_o = 8 T_e \]
Substituting \( T_e = 90 \) minutes:
\[ T_o = 8 \times 90 = 720 \text{ minutes} \]

### Conclusion
The time period of the satellite in an orbit at a distance of three times the radius of the Earth from its surface is **720 minutes**.

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