An earth satellite X is revolving around earth in an orbit whose radius is one - fourth the radius of orbit of a communication satellite . Time period of revolution of X is .

Options

A. 3 hrs
B. 6 hrs
C. 4 days
D. 72 days

Answer

To solve the problem of finding the time period of satellite X, we will use Kepler's Third Law of planetary motion, which relates the time period of revolution of a satellite to the radius of its orbit.

### Step-by-step Solution:

1. **Identify the Given Information:**
- Let \( R_1 \) be the radius of the orbit of the communication satellite.
- Let \( R_2 \) be the radius of the orbit of satellite X.
- According to the problem, \( R_2 = \frac{1}{4} R_1 \).

2. **Apply Kepler's Third Law:**
- Kepler's Third Law states that the square of the time period \( T \) of a satellite is directly proportional to the cube of the radius \( r \) of its orbit:
\[ T^2 \propto r^3 \]
- This can be expressed as:
\[ \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3} \]
- Here, \( T_1 \) is the time period of the communication satellite and \( T_2 \) is the time period of satellite X.

3. **Substitute the Values:**
- Substitute \( R_2 = \frac{1}{4} R_1 \) into the equation:
\[ \frac{T_1^2}{T_2^2} = \frac{R_1^3}{\left(\frac{1}{4} R_1\right)^3} \]
- Simplifying the right side:
\[ \frac{T_1^2}{T_2^2} = \frac{R_1^3}{\frac{1}{64} R_1^3} = 64 \]

4. **Relate the Time Periods:**
- From the equation \( \frac{T_1^2}{T_2^2} = 64 \), we can take the square root of both sides:
\[ \frac{T_1}{T_2} = 8 \]
- Rearranging gives:
\[ T_2 = \frac{T_1}{8} \]

5. **Substitute the Known Time Period:**
- We know that the time period of the communication satellite \( T_1 = 24 \) hours.
- Therefore:
\[ T_2 = \frac{24 \text{ hours}}{8} = 3 \text{ hours} \]

6. **Conclusion:**
- The time period of satellite X is \( 3 \) hours.

### Final Answer:
The time period of satellite X is **3 hours**.