Question

1. The largest moon of Saturn, Titian has a mean orbital radius of 9 1.22×10 m . The orbital period of Titan is 15.95 days. Hyperion, another moon of Saturn, orbits at a mean radius of 9 1.48×10 m . Use Kepler’s third law of planetary motion to predict the orbital period of Hyperion in hours.

Answer 1

According to Kepler's third law, $\dfrac{T^2}{a^3} = \dfrac{4\pi^2}{GM}$ . Therefore, for two satellites of the same planet we'll obtain

$\dfrac{T_1^2}{a_1^3} =\dfrac{T_2^2}{a_2^3}$ or $\left(\dfrac{T_2}{T_1}\right)^2 = \left(\dfrac{a_2}{a_1}\right)^3$ . In such form we may write periods in arbitrary units, so in days

$\left(\dfrac{T_2}{15.95}\right)^2 = \left(\dfrac{1.48\cdot10^9}{1.22\cdot10^9}\right)^3 , \\ T_2 = 1.336T_1 =21.31^d = 511.5^h$