Mean Motion Resonance
Examples of second-order resonances would be those in which the orbital period differences were separated by two, such as 3:1 or 5:3, and so forth. The higher the order of the resonance (the greater the difference between the two orbital periods), the less stable the orbits will be over time.
Mean motion resonances can arise between three or more bodies; Jupiter's moons Io, Europa, and Ganymede enjoy a 1:2:4 mean motion resonance among them, such that Io orbits twice per each orbit of Europa and four times per each orbit of Ganymede, while Europa orbits once per two orbits of Io and twice per each orbit of Ganymede .
What about bodies that don't have such neat ratios between their orbits?
In that case, we must to be able to determine the time between any two instances of any particular relationship between two orbiting bodies; for instance, when they both appear “lined up” in the sky, or when they are on opposite sides of the sky, etc. The interval between two instances of these sorts of alignments is their synodic period.
Note: The equations boxed in red above were *incorrect* in previous publications of this site. The equations on the right side were always correct. My apologies if this has caused anyone difficulties in designing planetary or moon orbital systems.
If the synodic period (S) is known and the ratio between Q and P is known (represented by R), then P or Q can be found.
Note: The values for S, P, and Q are always in the same units; e.g., if P and Q are expressed in days, then S is expressed in days.
Note that the fewer decimal places used, the more rounding will cause the values to come out slightly different when doing the verification of the values of P and Q.
Using the same synodic period of 31.415 days, and an orbital ratio of R = 0.75, our orbital periods come out to:
A fictional Example
Let us arbitrarily define that location on their respective orbits as 0°.
Calculation then shows:
Assuming (huge assumption) that the two moons orbit in exactly the same plane, the inner moon will eclipse the outer one every ~22.59 days; at the very least, they will appear in conjunction (“stacked up”) in the sky regularly on this interval, though the stars behind them will be different each time, for the reasons explained next.
Dividing the synodic period (S) we just calculated by the orbital period of the inner moon (P),
Dividing synodic period (S) by the orbital period of the outer moon (Q),
Note that the decimal fractions are identical: 0.9272; this is not a coincidence, and is an important value in the calculations that follow.
Multiplying this shared fractional part of each orbit (0.9272), by the number of degrees in a circle, (360°), yields 333.7920°. This tells us that although the two moons have reached close approach again, they are both (360° - 333.7920° = 26.208°) “short” of the location of the starting point, which we previously defined as 0°. The outer moon has only come 333.7920° around one of its orbits, and the inner moon has completed one full orbit and has only come 333.7920° of a second orbit.
At the next synodic close approach, the outer moon will have again completed 0.9272 orbits, and the inner 1.9272 orbits, and again they are 26.208° “short” of the point where they were last in closest proximity, or 52.416˚ "short" of where they were both at 0°. So, each time they achieve closest approach, they are 26.208° “short” of where they met the previous time. This continues, with each closest approach occurring 26.208° “ahead” of the location of the previous one.
Note that at the 14th synodic conjunction, the line does not fall precisely on the original 0° location, but 6.912˚ “over”:
This interval—between occasions when this orientation happens in exactly the same position relative to an external frame of reference—is found by another method: calculating the least common multiple (LCM) of the two orbital periods. This interval can be much longer than the synodic period.
The least common multiple of their orbital periods is the smallest number into which both orbital periods multiply an even number of times. For instance, the least common multiple of 3 and 5 is fifteen, because this is the least (smallest) number into which they both multiply evenly: 3 goes into 15 a total of exactly 5 times, and 5 goes into 15 a total of exactly 3 times.
One very important thing to note is that LCM can only be determined for integers. If the orbital periods being considered are decimal fractions, they must be converted to integers before finding the LCM. The simplest way to do this is to multiply both periods by the power of 10 necessary to convert the smaller of them into an integer—in other words, the power of ten that will leave the smaller interval with no decimal places.
In this example, multiplying 11.72 by 100 yields 1172; we also must multiply 24.36 by 100 to obtain 2436. Now, we find the LCM of 1172 and 2436, using the equation:
To find the GCD, we first find the factors of the two numbers (an online tool is immensely helpful, here; or use something like Excel™ to jump straight to calculating the LCM), and then we find the largest (greatest) number that appears in both sets of factors.
The factors of 1172 are:
Earlier, we realized that once every ~22.59 days, the two moons would appear in conjunction; now we know that once every ~19.5413 years, their conjunction will occur with exactly (or very nearly) the same stars behind them as when the cycle started.
To summarize: We’ve shown that the outer moon completes 0.9272 orbits in the time it takes the inner moon to compete 1.9272 orbits. We’ve also shown that this point (measuring counter‑clockwise) is 26.208° “short” of the point at which they were last at closest approach, and there will be different stars behind them when they conjunct at this point. After another 0.9272 outer orbits and 1.9272 inner orbits, they are once again at closest approach, again 26.208° “short” of the last point of close alignment … but after 19.5413 years' worth of these "short" conjunctions, this shortage “subtracts out” until the two meet at closest approach once again at the original 0° point where the cycle began.
For instance, Mercury has a Spin-Orbit Resonance of 3:2, such that its orbital period of 87.969 days is exactly 1.5 times its rotational period of 58.646 days; it rotates on its axis three times for every two times it orbits the Sun.
Earth's Moon, on the other hand has a spin-orbit resonance of 1:1, meaning its rotational period is exactly the same as its orbital period. This special case of spin-orbit resonance is called tidal locking.
- The closer two bodies orbit one another, the more likely a tidal lock is to occur in a realistic amount of time;
- A large secondary will lock faster than a small secondary at the same orbital distance; and,
- Secondaries with significantly smaller masses than their primaries will lock long before their primaries do (witness the Earth and Moon).
Below is the equation for calculating tidal lock given at Wikipedia :