1. **Stating the problem:** We are given Kepler's third law formula: $$T = \sqrt{kR^3}$$ where $T$ is the orbital period in years, $R$ is the average orbital radius in km, and $k$ is a constant depending on the star's mass. We need to: (a) Calculate $k$ given Earth's orbit: $T=1$ year, $R=1.49 \times 10^8$ km. (b) Use $k$ to find Neptune's orbital period $T$ given $R=4.46 \times 10^9$ km. 2. **Formula and rules:** Kepler's third law states: $$T = \sqrt{kR^3}$$ Squaring both sides gives: $$T^2 = kR^3$$ This allows us to solve for $k$: $$k = \frac{T^2}{R^3}$$ 3. **Part (a) Calculate $k$ for Earth:** Given: $$T = 1$$ $$R = 1.49 \times 10^8$$ Calculate: $$k = \frac{1^2}{(1.49 \times 10^8)^3} = \frac{1}{(1.49)^3 \times (10^8)^3} = \frac{1}{3.311 \times 10^{24}} = 3.02 \times 10^{-25}$$ 4. **Part (b) Calculate $T$ for Neptune:** Given: $$R = 4.46 \times 10^9$$ Using $k = 3.02 \times 10^{-25}$ from part (a), Calculate $T$: $$T = \sqrt{kR^3} = \sqrt{3.02 \times 10^{-25} \times (4.46 \times 10^9)^3}$$ Calculate $R^3$: $$(4.46)^3 = 88.7$$ $$(10^9)^3 = 10^{27}$$ So, $$R^3 = 88.7 \times 10^{27} = 8.87 \times 10^{28}$$ Calculate inside the square root: $$3.02 \times 10^{-25} \times 8.87 \times 10^{28} = 26.78 \times 10^{3} = 2.678 \times 10^{4}$$ Finally, $$T = \sqrt{2.678 \times 10^{4}} = \sqrt{2.678} \times \sqrt{10^{4}} = 1.636 \times 100 = 163.6$$ Rounded to the nearest tenth: $$T = 163.6 \text{ years}$$ **Final answers:** - $k = 3.02 \times 10^{-25}$ - Neptune's orbital period $T = 163.6$ years