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The problem provided involves Kepler's Third Law of Planetary Motion, also known as the Law of Harmonies. The law states that the square of the orbital period \( T \) of a planet is proportional to the cube of the semi-major axis \( R \) of its orbit, which is the average distance from the planet to the sun. Mathematically:

\[
T^2 \propto R^3 \quad \text{or} \quad \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}
\]

### Given:
- The small planet is 14 times as far from the Sun as the Earth's distance, which means \( R_{\text{planet}} = 14 \times R_{\text{Earth}} \).
- Earth's distance from the Sun is \( R_{\text{Earth}} = 1.5 \times 10^{11} \) m.
- The constant provided is \( \frac{T}{R^{3/2}} = 2.97 \times 10^{-19} \, \text{s/m}^{3/2} \).

To find the orbital period \( T_{\text{planet}} \) of the planet, we use:

\[
T = \left( \frac{R^{3/2}}{K} \right)
\]

Where \( K = 2.97 \times 10^{-19} \, \text{s/m}^{3/2} \).

### Calculation:

1. For the small planet, the distance from the sun is:
\[
R_{\text{planet}} = 14 \times R_{\text{Earth}} = 14 \times 1.5 \times 10^{11} \, \text{m} = 2.1 \times 10^{12} \, \text{m}
\]

2. Now, plug the value of \( R_{\text{planet}} \) into the formula for \( T \):
\[
T_{\text{planet}} = \frac{(2.1 \times 10^{12})^{3/2}}{2.97 \times 10^{-19}}
\]

3. First, calculate \( (2.1 \times 10^{12})^{3/2} \):
\[
(2.1 \times 10^{12})^{3/2} = (2.1)^{3/2} \times (10^{12})^{3/2} = 3.041 \times 10^{18}
\]

4. Now calculate the orbital period \( T_{\text{planet}} \):
\[
T_{\text{planet}} = \frac{3.041 \times 10^{18}}{2.97 \times 10^{-19}} = 1.02 \times 10^{37} \, \text{seconds}
\]

Now that we have the period in seconds, we can convert it into years.

5. Convert seconds to years:
\[
T_{\text{planet}} = \frac{1.02 \times 10^{37}}{3.154 \times 10^{7}} \, \text{years} \approx 3.24 \times 10^{29} \, \text{years}
\]

Thus, the orbital period of the planet is approximately \( 3.24 \times 10^{29} \) years.

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Would you like further clarification or additional steps? Here are five related questions you might want to explore:

1. How does Kepler's Third Law help in understanding the solar system's structure?
2. Can this method be applied to predict the motion of moons around planets?
3. How do the gravitational constant and mass of the Sun influence planetary orbits?
4. What other applications does Kepler’s Third Law have in astrophysics?
5. How does the discovery of exoplanets use Kepler's laws?

**Tip:** Remember that for large astronomical calculations, it's important to keep track of units and use scientific notation to manage very large or small numbers.

Suppose a small planet is discovered that is 14 times as far from the sun as the Earth's distance from the sun (1.5 × 10^11 m). Use Kepler's law of harmonics to predict the orbital period of such a planet. Given T/R^(3/2) = 2.97 × 10^-19 s/m^(3/2).

Learn how to apply Kepler's Third Law of Planetary Motion to predict the orbital period of a small planet located 14 times farther from the sun than Earth. This problem involves key concepts like proportional relationships between orbital periods and distances, and scientific notation.

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