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 find the distance of Saturn from the Sun given that a Saturn year is 29.5 times an Earth year and the distance of Earth from the Sun is \(1.5 \times 10^8\) km, we can use Kepler's Third Law of planetary motion.

### Step-by-step Solution:

1. **Identify the relationship from Kepler's Third Law**:
Kepler's Third Law 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. This can be expressed as:
\[ \frac{T_{saturn}^2}{T_{earth}^2} = \frac{r_{saturn}^3}{r_{earth}^3} \]

2. **Define the known values**:
- Let \(T_{earth} = 1\) year (the period of Earth).
- Thus, \(T_{saturn} = 29.5\) years.
- The distance of Earth from the Sun, \(r_{earth} = 1.5 \times 10^8\) km.

3. **Substitute the known values into the equation**:
\[ \frac{(29.5)^2}{(1)^2} = \frac{r_{saturn}^3}{(1.5 \times 10^8)^3} \]

4. **Calculate \(T_{saturn}^2\)**:
\[ T_{saturn}^2 = (29.5)^2 = 870.25 \]

5. **Rearranging the equation**:
\[ r_{saturn}^3 = 870.25 \times (1.5 \times 10^8)^3 \]

6. **Calculate \((1.5 \times 10^8)^3\)**:
\[ (1.5)^3 = 3.375 \quad \text{and} \quad (10^8)^3 = 10^{24} \]
Thus,
\[ (1.5 \times 10^8)^3 = 3.375 \times 10^{24} \]

7. **Substituting back into the equation**:
\[ r_{saturn}^3 = 870.25 \times 3.375 \times 10^{24} \]

8. **Calculating the product**:
\[ r_{saturn}^3 = 2945.57875 \times 10^{24} \]

9. **Taking the cube root to find \(r_{saturn}\)**:
\[ r_{saturn} = (2945.57875 \times 10^{24})^{1/3} \]

10. **Calculating the cube root**:
- The cube root of \(2945.57875\) is approximately \(14.2\).
- Thus,
\[ r_{saturn} \approx 14.2 \times 10^{8} \text{ km} = 1.42 \times 10^{9} \text{ km} \]

### Final Answer:
The distance of Saturn from the Sun is approximately \(1.42 \times 10^9\) km.