1. **State the problem:** We are given Kepler's third law which states that the square of the period $P$ of a planet is proportional to the cube of its average distance $d$ from the sun. 2. **Formula:** The law can be written as: $$P^2 \propto d^3$$ which means $$P^2 = k d^3$$ for some constant $k$. 3. **Find the constant $k$ using Earth's data:** Given Earth has $P = 365$ days and $d = 93,000,000$ miles, $$365^2 = k (93000000)^3$$ 4. **Express $P$ as a function of $d$:** Taking square root, $$P = \sqrt{k} d^{3/2}$$ Using Earth's data, $$365 = \sqrt{k} (93000000)^{3/2}$$ So, $$\sqrt{k} = \frac{365}{(93000000)^{3/2}}$$ 5. **Rewrite $P(d)$:** $$P(d) = 365 \left( \frac{d}{93000000} \right)^{3/2}$$ 6. **Calculate Pluto's period:** Given Pluto's distance $d = 3600000000$ miles, $$P(3600000000) = 365 \left( \frac{3600000000}{93000000} \right)^{3/2}$$ 7. **Simplify inside the parentheses:** $$\frac{3600000000}{93000000} = \frac{3600 \times 10^6}{93 \times 10^6} = \frac{3600}{93} \approx 38.7097$$ 8. **Calculate the power:** $$38.7097^{3/2} = (38.7097)^{1.5} = \sqrt{38.7097^3}$$ Calculate $38.7097^3$: $$38.7097^3 = 38.7097 \times 38.7097 \times 38.7097 \approx 57994.5$$ Then, $$\sqrt{57994.5} \approx 240.83$$ 9. **Calculate Pluto's period:** $$P(3600000000) = 365 \times 240.83 = 87899.95$$ 10. **Round to nearest day:** $$P \approx 87900$$ days **Final answer:** Pluto's orbital period is approximately 87900 days.