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Chapter 5: Problem 33

Io, one of Jupiter's satellites, has an orbital period of 1.77 d. Europa, another of Jupiter's satellites, has an orbital period of about 3.54 d. Both moons have nearly circular orbits. Use Kepler's third law to find the distance of each satellite from Jupiter's center. Jupiter's mass is $1.9 \times 10^{27} \mathrm{kg}$

Short Answer

Expert verified
Answer: The approximate distance of Io from Jupiter's center is \(4.22 \times 10^8\) m, and the approximate distance of Europa from Jupiter's center is \(6.70 \times 10^8\) m.

Step by step solution

01

Recall and understand Kepler's Third Law

Kepler's Third Law states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (a) of its orbit around the Sun, i.e., \(T^2 \propto a^3\). Since both satellites have nearly circular orbits, we can take the distance from Jupiter's center as the semi-major axis. For each satellite, Kepler's Third Law can be written as: \begin{align} \frac{T_1^2}{a_1^3} = \frac{T_2^2}{a_2^3} \end{align} where T1 and T2 are the orbital periods of Io and Europa, and a1 and a2 are the distances from Jupiter's center.
02

Apply Kepler's Third Law to Io

Given the orbital period of Io (T1): 1.77 d, we can find its distance from Jupiter's center (a1). For this, we can convert the orbital period into seconds and use Jupiter's mass (M) to find the gravitational constant (G) using the formula: \begin{align} T_1^2 = \frac{4 \pi^2}{GM} a_1^3 \end{align} Plugging in the given values, we get: \begin{align} (1.77 \times 24 \times 3600)^2 = \frac{4 \pi^2}{6.674 \times 10^{-11}\times1.9 \times 10^{27}} a_1^3 \end{align} Solving for a1, we find: \begin{align} a_1 \approx 4.22 \times 10^8 \text{ m} \end{align} So the distance of Io from Jupiter's center is approximately \(4.22 \times 10^8\) m.
03

Apply Kepler's Third Law to Europa

Similarly, given the orbital period of Europa (T2): 3.54 d, we can find its distance from Jupiter's center (a2) using the same formula: \begin{align} T_2^2 = \frac{4 \pi^2}{GM} a_2^3 \end{align} Plugging in the given values, we get: \begin{align} (3.54 \times 24 \times 3600)^2 = \frac{4 \pi^2}{6.674 \times 10^{-11}\times1.9 \times 10^{27}} a_2^3 \end{align} Solving for a2, we find: \begin{align} a_2 \approx 6.70 \times 10^8 \text{ m} \end{align} So the distance of Europa from Jupiter's center is approximately \(6.70 \times 10^8\) m.

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