A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is `1.5xx10^(8)` away from the sun?

Options

A. `1.4xx10^(6)` km
B. `1.4xx10^(7)` km
C. `1.4xx10^(8)` km
D. `1.4xx10^(9)` km

Explanation

To solve the problem of finding the distance of Saturn from the Sun, we can use Kepler's Third Law of planetary motion, which states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (r) of its orbit around the Sun.

### Step-by-Step Solution:

1. **Understand Kepler's Third Law**:
According to Kepler's Third Law, we have:
\[ \frac{T_{saturn}^2}{T_{earth}^2} = \frac{r_{saturn}^3}{r_{earth}^3} \]
where \( T \) is the orbital period and \( r \) is the distance from the Sun.

2. **Identify the Given Values**:
- The period of Saturn (\( T_{saturn} \)) is 29.5 times that of Earth (\( T_{earth} \)).
- The distance of Earth from the Sun (\( r_{earth} \)) is \( 1.5 \times 10^8 \) km.

3. **Express the Periods**:
Let \( T_{earth} = T \). Then:
\[ T_{saturn} = 29.5 T \]

4. **Substitute into Kepler's Law**:
Substitute \( T_{saturn} \) and \( T_{earth} \) into Kepler's equation:
\[ \frac{(29.5 T)^2}{T^2} = \frac{r_{saturn}^3}{(1.5 \times 10^8)^3} \]

5. **Simplify the Equation**:
The \( T^2 \) cancels out:
\[ 29.5^2 = \frac{r_{saturn}^3}{(1.5 \times 10^8)^3} \]

6. **Calculate \( 29.5^2 \)**:
\[ 29.5^2 = 870.25 \]

7. **Rearrange to Find \( r_{saturn}^3 \)**:
\[ r_{saturn}^3 = 870.25 \times (1.5 \times 10^8)^3 \]

8. **Calculate \( (1.5 \times 10^8)^3 \)**:
\[ (1.5 \times 10^8)^3 = 3.375 \times 10^{24} \]

9. **Substitute Back**:
\[ r_{saturn}^3 = 870.25 \times 3.375 \times 10^{24} \]

10. **Calculate \( r_{saturn}^3 \)**:
\[ r_{saturn}^3 \approx 2.943 \times 10^{27} \]

11. **Find \( r_{saturn} \)**:
\[ r_{saturn} = (2.943 \times 10^{27})^{1/3} \]

12. **Calculate the Cube Root**:
\[ r_{saturn} \approx 1.43 \times 10^9 \text{ km} \]

### Final Answer:
The distance of Saturn from the Sun is approximately \( 1.43 \times 10^9 \) km.

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