### Jupiter's Moons & Kepler's 3rd Law of Planetary Motion If you have a telescope, by all means aim it at the king planet to view its four major moons. In their outward order from Jupiter, these moons are Io, Europa, Ganymede and Callisto. To know the placement of these moons in your sky for a certain date and time, refer to this handy-dandy Jupiter's Moons Javascript Utility, courtesy of Sky & Telescope magazine.

Using Kepler's 3rd Law to figure Planet Distances/Orbital Periods

First of all, let's use Kepler's 3rd law of planetary motion to determine a planet's distance from the sun or a planet's orbital period. If you know a planet's distance, you can then compute its orbital period. Or if you know the planet's orbital period, you can compute its mean distance from the Sun. For example, let's say a hypothetical planet revolves around the Sun at 4 times the Earth's distance. We can compute this planet's orbital period from Kepler's famous equation: D3 = P2, whereby D = Distance and P = Orbital Period.
 1) D3 (Distance x Distance x Distance) = P2 (Orbital period x Orbital period) 2) 4 x 4 x 4 = P2 3) 64 = P2 4) 8 = Orbital Period of planet in Earth-years

Now let's suppose that a planet revolves around the Sun in 27 Earth-years. How far distant is this planet from the Sun? The answer:
 1) D3 = P2 2) D3 = 27 x 27 3) D3 = 729 4) Distance = 9 times the Earth's Distance from the Sun

Using Kepler's 3rd Law for Jupiter's Moons

Although Kepler's 3rd Law is called the 3rd Law of Planetary Motion, this formula can be applied to the moons of Jupiter as well. The 3 innermost Galilean moons - Io, Europa and Ganymede - are locked into an 4:2:1 orbital resonance. For every 4 times that Io revolves around Jupiter, Europa revolves 2 times and Ganymede revolves once.
Given that Io's mean distance from Jupiter is 262,000 miles, we can compute the distances of Jupiter's other moons. We will compute Europa's mean distance from Jupiter, then leave it to the inquiring reader to compute Ganymede's distance:
 1) D3 = P2 2) D3 = 2 x 2 3) D3 = 4 4) Distance of Europa = 1.5874 times Io's Distance from Jupiter

Tale of 2 Moons: Io and Earth's Moon

Amazing enough, Jupiter's moon Io and the Earth's Moon are roughly the same distance from their parent planets. Io's mean distance from Jupiter is 262,000 miles, and the Moon's mean distance from Earth is 239,000 miles.
Why does Io revolve around Jupiter in only 1.769 days, while the Moon takes a whopping 27.322 days to revolve around the Earth? Shouldn't the orbital periods of both moons be similar since the distances of the two moons from their parent planets are similar? If anything, shouldn't Io's orbital period be a bit longer since Io is farther out from the center its parent planet?
The answer would be yes in both instances, if both Earth and Jupiter had the same mass (weight). However, Jupiter has 318 times the Earth's mass. Or another way of putting it: Earth has 1/318 Jupiter's mass. The more massive the planet, the faster the moon's orbital period. The less massive the planet, the slower the moon's orbital period.

If the Earth's Mass Equaled that of Jupiter, What Would the Moon's Orbital Period Be?

To find out what the Moon's orbital period would be if the Earth equaled Jupiter in mass (318 Earth-masses), take the Moon's orbital period (27.322 days) and divide it by the square root of Jupiter's mass (the square root of 318 = 17.83).

27.322 days divided by 17.83 = 1.53 days

And to find out what Io's orbital period would be if Jupiter equaled Earth's mass (1/318), take Io's orbital period (1.769 days) and divide it by the square root of 1/318. The square root of 1/318 = 1/17.83 = 0.056.

1.769 days divided by 0.056 = 31.55 days

copyright 2009 by Bruce McClure

 Kepler's Third Law of Planetary Motion