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 revolution of satellite X, we will use Kepler's Third Law of planetary motion, which states that 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.

### Step-by-Step Solution:

1. **Define the Variables**:
- Let the radius of the orbit of the communication satellite be \( R_1 = R \).
- The radius of the orbit of satellite X is given as \( R_2 = \frac{1}{4} R_1 = \frac{1}{4} R \).

2. **Apply Kepler's Third Law**:
- According to Kepler's Third Law, we have:
\[ \frac{T_2^2}{T_1^2} = \frac{R_2^3}{R_1^3} \]
- 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 \( R_2 \) and \( R_1 \):
\[ \frac{T_2^2}{T_1^2} = \frac{\left(\frac{1}{4} R\right)^3}{R^3} \]
- This simplifies to:
\[ \frac{T_2^2}{T_1^2} = \frac{\frac{1}{64} R^3}{R^3} = \frac{1}{64} \]

4. **Take the Square Root**:
- Taking the square root of both sides gives:
\[ \frac{T_2}{T_1} = \frac{1}{8} \]

5. **Find the Time Period of Satellite X**:
- If the time period of the communication satellite \( T_1 \) is 24 hours, then:
\[ T_2 = \frac{T_1}{8} = \frac{24 \text{ hours}}{8} = 3 \text{ hours} \]

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

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