The semi -major axis of the orbit of Saturn is approximately nine time of earth. the time period of revolution of Saturn is approximately equal to

Options

A. 81 years
B. 27 years
C. 729 years
D. 9 years

Answer

To solve the problem, we will use Kepler's Third Law of planetary motion, which states that the square of the time period of revolution (T) of a planet is directly proportional to the cube of the semi-major axis (r) of its orbit. The law can be mathematically expressed as:

\[ T^2 \propto r^3 \]

### Step-by-Step Solution:

1. **Identify the semi-major axis of Saturn's orbit relative to Earth:**
- Given that the semi-major axis of Saturn's orbit is approximately 9 times that of Earth's orbit, we can denote the semi-major axis of Earth as \( r_E \) and that of Saturn as \( r_S = 9r_E \).

2. **Apply Kepler's Third Law:**
- According to Kepler's Third Law, we can write the relationship for the time periods:
\[ \frac{T_S^2}{T_E^2} = \frac{r_S^3}{r_E^3} \]
where \( T_S \) is the time period of Saturn and \( T_E \) is the time period of Earth.

3. **Substitute the values:**
- Since \( r_S = 9r_E \), we can substitute this into the equation:
\[ \frac{T_S^2}{T_E^2} = \frac{(9r_E)^3}{(r_E)^3} \]
- This simplifies to:
\[ \frac{T_S^2}{T_E^2} = \frac{729r_E^3}{r_E^3} = 729 \]

4. **Solve for \( T_S \):**
- Taking the square root of both sides gives:
\[ \frac{T_S}{T_E} = \sqrt{729} = 27 \]
- Therefore, we have:
\[ T_S = 27T_E \]

5. **Determine the time period of Saturn:**
- Since the time period of Earth \( T_E \) is 1 year, we find:
\[ T_S = 27 \times 1 \text{ year} = 27 \text{ years} \]

### Final Answer:
The time period of revolution of Saturn is approximately **27 years**.