A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is `1.50 xx 10^(8) `km away from the sum?

Explanation

To find the distance of Saturn from the Sun using Kepler's Third Law of planetary motion, we can follow these steps:

### Step-by-Step Solution:

1. **Understand 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:
\[ \frac{T_1^2}{r_1^3} = \frac{T_2^2}{r_2^3} \]
where \(T_1\) and \(r_1\) are the period and distance of Earth, and \(T_2\) and \(r_2\) are the period and distance of Saturn.

2. **Identify the known values**:
- The orbital period of Earth (\(T_e\)) is 1 year.
- The distance of Earth from the Sun (\(r_e\)) is \(1.50 \times 10^8\) km.
- The orbital period of Saturn (\(T_s\)) is 29.5 times that of Earth, so:
\[ T_s = 29.5 \times T_e \]

3. **Set up the equation**:
Using Kepler's Third Law, we can write:
\[ \frac{T_e^2}{r_e^3} = \frac{T_s^2}{r_s^3} \]
Rearranging this gives:
\[ r_s^3 = r_e^3 \cdot \left(\frac{T_s^2}{T_e^2}\right) \]

4. **Substitute the known values**:
- Substitute \(T_s = 29.5 \times T_e\) into the equation:
\[ r_s^3 = r_e^3 \cdot \left(\frac{(29.5 \cdot T_e)^2}{T_e^2}\right) \]
This simplifies to:
\[ r_s^3 = r_e^3 \cdot (29.5^2) \]

5. **Calculate \(r_e^3\)**:
\[ r_e^3 = (1.50 \times 10^8 \text{ km})^3 = 3.375 \times 10^{24} \text{ km}^3 \]

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

7. **Calculate \(r_s^3\)**:
\[ r_s^3 = 3.375 \times 10^{24} \text{ km}^3 \cdot 870.25 = 2.933 \times 10^{27} \text{ km}^3 \]

8. **Find \(r_s\)**:
To find \(r_s\), take the cube root:
\[ r_s = \sqrt[3]{2.933 \times 10^{27}} \approx 1.432 \times 10^9 \text{ km} \]

### Final Answer:
The distance of Saturn from the Sun is approximately:
\[ r_s \approx 1.432 \times 10^9 \text{ km} \]