Question

Titan,a satellite of Saturn has a mean orbit radius of 1.22×10^9m.the orbital period of titan is 15.95days.Hyperion, another satellite of Saturn ,orbit at a mean radius of 1.48×10^9m.Estimate the orbital period of Hyperion?

Answer 1

Answer on Question 76597, Physics, Molecular Physics, Thermodynamics

Question:

Titan, a satellite of Saturn, has a mean orbital radius of $1.22 \cdot 10^{9} \, \text{m}$. The orbital period of Titan is 15.95 days. Hyperion, another satellite of Saturn, orbits at a mean radius of $1.48 \cdot 10^{9} \, \text{m}$. Estimate the orbital period of Hyperion.

Solution:

The third Kepler’s law of planetary motion states that the square of the orbital period of a planet is proportional to the cube of the semi-major axis (mean distance) of its orbit:

$$\frac {P _ {T} ^ {2}}{a _ {T} ^ {3}} = \frac {P _ {H} ^ {2}}{a _ {H} ^ {3}},$$

here, $P_T$ is the orbital period of the Titan, $P_H$ is the orbital period of the Hyperion, $a_T$ is the mean distance of the Titan from the Saturn, $a_H$ is the mean distance of the Hyperion from the Saturn.

Then, from this formula we can find the orbital period of the Hyperion:

$$P _ {H} ^ {2} = a _ {H} ^ {3} \frac {P _ {T} ^ {2}}{a _ {T} ^ {3}},$$ $$P _ {H} = \sqrt {a _ {H} ^ {3} \frac {P _ {T} ^ {2}}{a _ {T} ^ {3}}} = \sqrt {(1.48 \cdot 10 ^ {9} \, \text{m}) ^ {3} \cdot \frac {(15.95 \, \text{days}) ^ {2}}{(1.22 \cdot 10 ^ {9} \, \text{m}) ^ {3}}} = 21.3 \, \text{days}.$$

Answer:

$$P _ {H} = 21.3 \, \text{days}.$$

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