The mass of Saturn is 5.69 x 1026 kg.
(a) Find the period of its moon Mimas, whose mean orbital radius is 1.86 x 108 m.
(b) Find the mean orbital radius of its moon Titan, whose period is 1.38 x 106 s.
The Correct Answer and Explanation is :
We can solve this problem using Kepler’s Third Law, which relates the orbital period of a moon or planet to its orbital radius and the mass of the central object (in this case, Saturn). Kepler’s Third Law is given by the equation:
[
T^2 = \frac{4 \pi^2 r^3}{GM}
]
Where:
- ( T ) is the orbital period,
- ( r ) is the orbital radius,
- ( G ) is the gravitational constant (( G = 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} )),
- ( M ) is the mass of Saturn.
We can rearrange the equation to solve for the unknowns in each part of the problem.
Part (a): Find the period of Mimas
Given:
- Mass of Saturn, ( M = 5.69 \times 10^{26} \, \text{kg} ),
- Orbital radius of Mimas, ( r = 1.86 \times 10^8 \, \text{m} ).
We need to find the period ( T ) of Mimas. Using Kepler’s Third Law:
[
T^2 = \frac{4 \pi^2 r^3}{GM}
]
Substitute the known values:
[
T^2 = \frac{4 \pi^2 (1.86 \times 10^8)^3}{(6.67430 \times 10^{-11}) (5.69 \times 10^{26})}
]
After calculating this, we find that the period ( T ) is approximately 0.944 days or about 22.7 hours.
Part (b): Find the orbital radius of Titan
Given:
- Period of Titan, ( T = 1.38 \times 10^6 \, \text{s} ),
- Mass of Saturn, ( M = 5.69 \times 10^{26} \, \text{kg} ).
We can rearrange Kepler’s Third Law to solve for ( r ):
[
r^3 = \frac{GMT^2}{4 \pi^2}
]
Substitute the known values:
[
r^3 = \frac{(6.67430 \times 10^{-11})(5.69 \times 10^{26})(1.38 \times 10^6)^2}{4 \pi^2}
]
Solving for ( r ), we get an orbital radius of approximately ( 1.22 \times 10^9 \, \text{m} ).
Explanation
Kepler’s Third Law provides a relationship between the orbital period and the orbital radius, with the mass of the central object playing a crucial role in the orbital dynamics of moons and planets. For part (a), we calculated the orbital period of Mimas based on its orbital radius, and for part (b), we used Titan’s orbital period to determine its mean orbital radius. These calculations are essential for understanding the gravitational interactions that govern the orbits of moons and the influence of Saturn’s massive gravity on its moons.