When Will I See You Again?

In an SF story I was working on it was important to know how often two of Jupiter's moons made their closest approach to Ganymede, the third of the four large Jovian moons first spotted by Galileo. In the case of the pair Europa and Ganymede, the answer is quite easy, because of the orbital resonance, or lockstep, of three of the Galilean moons: for every Ganymede orbit around Jupiter, Europa makes exactly two orbits. Thus, the pair makes its closest approach once each Ganymede orbit, or every 7.1546 days.

However, the outermost Galilean moon, Callisto, doesn't participate in the resonance, orbiting Jupiter every 16.689 days. How, then, to figure out how often Callisto and Ganymede make their closest approach?

Here's a diagram with a starting position. Jupiter is in the center, and the two moons are at their closest approach to each other. The outer one takes, let's say, 16 days to complete one orbit, and the inner one 7 days.

After the inner moon has completed one orbit (7 days), it's back where it started from, but the outer moon has had time to complete only 7/16 of an orbit.

The inner moon can now cover that 7/16 of an orbit, but the outer moon will have moved 7/16 * 7/16 of an orbit, or (7/16)².

The speedy inner moon can cross this distance of (7/16)², but the outer moon will have moved a little more ... and so it goes on and on. This is reminiscent of Zeno's paradox of Achilles and the Tortoise.

It's time for a little algebra. Let's say that a represents the time taken for the inner, faster moon to make one orbit, and b represents the time for the outer moon to do so. O represents the number of orbits of the inner moon before the close approach recurs. Our repetition of ever-smaller 7/16 steps can be described by the infinite series,

O = 1 + a/b + (a/b)² + (a/b)³ + …

to simplify, let's represent a/b with r (for the ratio a/b).

O = 1 + r + r² + r³ + …

How can we reduce this series to a simple formula? Here's the trick. Multiply both sides of the equation by r, which yields

r x O = r + r² + r³ + r⁴…

Now subtract the new equation from the original, and you have

O - r x O = 1

pull O out,

O x (1 - r) = 1

and finally divide both sides by (1-r),

O = 1 / (1 - r)

Plugging in 7/16 for r, the number of orbits is 1.777, which at 7 days per orbit is 12.4 days. Using the more precise measures for Callisto and Ganymede, the final result is 1.75 Ganymede orbits, or 12.5 days.