Introduction
This problem uses Kepler’s Third Law, which relates the orbital period of a satellite to the radius of its orbit. Specifically, for satellites orbiting the same planet (Saturn in this case), the square of the orbital period is proportional to the cube of the orbital radius:
(T₁/T₂)² = (R₁/R₂)³
Given Data
- Orbital radius of Titan, R₁ = 1.22 × 10⁹ m
- Orbital period of Titan, T₁ = 15.95 days
- Orbital radius of Hyperion, R₂ = 1.48 × 10⁹ m
- We are to find: T₂ (orbital period of Hyperion)
Step 1: Use Kepler’s Third Law
(T₁/T₂)² = (R₁/R₂)³
So, T₂ = T₁ × (R₂/R₁)3/2
Step 2: Plug in the Values
R₂ / R₁ = (1.48 / 1.22) = 1.2131
(R₂/R₁)3/2 = (1.2131)1.5 ≈ 1.336
T₂ = 15.95 × 1.336 ≈ 21.32 days
Final Answer
Estimated orbital period of Hyperion: 21.32 days
Conclusion
Using Kepler’s Third Law, we find that Hyperion, which orbits farther from Saturn than Titan, takes approximately 21.32 days to complete one orbit. This law is very useful in astronomy to relate orbits of celestial bodies around the same central mass.