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 find the time period of revolution of Saturn based on the given information, we can use Kepler's Third Law of Planetary Motion. Here’s a step-by-step solution:

### Step-by-Step Solution:

1. **Understanding Kepler's Third Law**:
Kepler's Third Law states that the square of the time period (T) of a planet's orbit is directly proportional to the cube of the semi-major axis (a) of its orbit. Mathematically, this can be expressed as:
\[ T^2 \propto a^3 \]
or
\[ \frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3} \]

2. **Identifying the Semi-Major Axis**:
According to the problem, the semi-major axis of Saturn's orbit is approximately 9 times that of Earth's orbit. If we denote the semi-major axis of Earth as \( a_E \), then for Saturn:
\[ a_S = 9 a_E \]

3. **Setting Up the Ratio**:
Let \( T_E \) be the time period of Earth (which is 1 year), and \( T_S \) be the time period of Saturn. Using Kepler's Third Law:
\[ \frac{T_S^2}{T_E^2} = \frac{a_S^3}{a_E^3} \]
Substituting the values we have:
\[ \frac{T_S^2}{(1 \text{ year})^2} = \frac{(9 a_E)^3}{(a_E)^3} \]

4. **Simplifying the Equation**:
This simplifies to:
\[ \frac{T_S^2}{1} = 9^3 \]
\[ T_S^2 = 729 \]

5. **Finding the Time Period of Saturn**:
Taking the square root of both sides gives:
\[ T_S = \sqrt{729} = 27 \text{ years} \]

Thus, the time period of revolution of Saturn is approximately **27 years**.