Orbital radius of a satellite S of earth is four times that s communication satellite C. Period of revolution of S is :-

Options

A. 4 days
B. 8 days
C. 16 days
D. 32 days

Answer

To solve the problem, we need to determine the period of revolution of satellite S given that its orbital radius is four times that of communication satellite C. We will use Kepler's Third Law of planetary motion, which 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.

### Step-by-Step Solution:

1. **Define the Variables:**
Let:
- \( r_C \) = radius of communication satellite C
- \( r_S \) = radius of satellite S
- \( T_C \) = period of revolution of satellite C
- \( T_S \) = period of revolution of satellite S

According to the problem, we have:
\[ r_S = 4r_C \]

2. **Apply Kepler's Third Law:**
According to Kepler's Third Law:
\[ \frac{T_S^2}{T_C^2} = \frac{r_S^3}{r_C^3} \]

3. **Substitute the Value of \( r_S \):**
Substitute \( r_S = 4r_C \) into the equation:
\[ \frac{T_S^2}{T_C^2} = \frac{(4r_C)^3}{r_C^3} \]

4. **Simplify the Equation:**
Simplifying the right side:
\[ \frac{T_S^2}{T_C^2} = \frac{64r_C^3}{r_C^3} = 64 \]

5. **Take the Square Root:**
Taking the square root of both sides gives:
\[ \frac{T_S}{T_C} = \sqrt{64} = 8 \]

6. **Express \( T_S \) in Terms of \( T_C \):**
Thus, we can express the period of satellite S as:
\[ T_S = 8T_C \]

7. **Conclusion:**
If the period of the communication satellite \( T_C \) is given, the period of satellite S will be 8 times that period. If we assume \( T_C \) is 1 day, then:
\[ T_S = 8 \text{ days} \]

### Final Answer:
The period of revolution of satellite S is 8 days.

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