## Mean Motion Resonance

When two bodies orbit a third body in common, there will often be

Examples of

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 [1].

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

*resonances*in their orbits. If the inner body orbits exactly twice for each single orbit of the outer body, the two are said to be in*mean motion resonance*, and the relationship of their orbits is 2:1. This is a*first-order resonance*, because the difference between the two values is only 1 unit (2 - 1 = 1). Other examples of first-order resonances would be 6:5, 13:12, etc.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 [1].

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*.## Synodic Period

Below are two related sets of equations that allow calculation of the synodic period of two bodies. Both sets produce exactly the same results. I, personally, prefer the right-hand set, simply because the results don’t have to be inverted to be useful.

Note that if the synodic period (

However....

Thus, where:

*S*) is known, either*P*or*Q*must also be known/chosen in order to calculate the other value. There is no way that knowing*S*alone can tell what*P*and*Q*are.However....

*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.Thus, where:

Then:

Note: The values forS,P, andQare always in the same units; e.g., ifPandQare expressed in days, thenSis expressed in days.

These don't require that either orbital period be known, only that their ratio be known, so let's say we knew we wanted the syndic period to be 31.415 days, and the ratio between the orbits to be 0.382 days. Then, the two orbital periods would work out to be:

And we can double-check by putting these values back into one of the equations above:

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 ofPandQ.

Note also that if

Using the same synodic period of 31.415 days, and an orbital ratio of

*R*is set to a decimal fraction which is the result of the division of two integers, then the orbits will be in*mean motion resonance*, which for this value of*R*would be a*first-order resonance*of*4:3*.Using the same synodic period of 31.415 days, and an orbital ratio of

*R = 0.75*, our orbital periods come out to:Dividing the longer period by the shorter period:

... a mean motion resonance of 4:3, even though the orbital periods themselves are not integer values. Refer to the beginning of this blog for a refresher on mean motion resonances.

## A fictional Example

Imagine two moons orbiting a planet; the inner with an orbital period of 11.72 days and the outer one 24.36 days. Clearly, these two orbital periods are not in a simple mean motion resonance, but the two moons will have a periodicity in their orbital dance.

Let's start with them at the closest approach between them. At this point, they will also be in conjunction with respect to their host planet (they will appear "lined up" or "stacked up" in the sky as seen from the planet).

Let us arbitrarily define that location on their respective orbits as 0°.

Calculation then shows:

Let us arbitrarily define that location on their respective orbits as 0°.

Calculation then shows:

They will be closest to one another in their orbits every 22.5869 days.

Assuming (

Dividing the synodic period (

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*),... tells us that the inner moon has completed one full orbit and 0.9272 of a second orbit.

Dividing synodic period (

Dividing synodic period (

*S*) by the orbital period of the outer moon (*Q*),… tells us that the outer moon has only completed 0.9272 of one orbit.

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”:

*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”:

So, how long will it take until they meet up again at our originally defined 0° point?

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

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:

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:

The greatest common divisor is the largest number that divides both 1172 and 2436 evenly.

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:

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:

[1, 2,

**4**, 293, 586, 1172]… and the factors of 2436 are:

[1, 2, 3,

**4**, 6, 7, 12, 14, 21, 28, 29, 42, 58, 84, 87, 116, 174, 203, 348, 406, 609, 812, 1218, 2436]As shown by the bolding, the largest number that appears in both lists is

**4**; thus,… and, because we multiplied the orbital periods by 100 to calculate the least common multiple, we divide the result by 100 to get our final answer:

If we divide this result by the number of days per year:

… we learn that it takes 19.5413 years for the moons to once again achieve close alignment at 0° (or the exact same place in their orbit relative to their primary). This is approximately 19 years, 28 weeks, 1 day, 17 hours, 2 minutes, and 8.877 seconds.

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.

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.## Spin-Orbit Resonance

Spin-Orbit Resonance occurs when the relationship between the

*rotational*period of a body and its*orbital*period can be expressed in a simple ratio, such as 3:2, which would mean that the body rotates on its axis exactly three times for every two orbits around its primary.*Note that this is different from Mean Motion Resonance, which concerns the orbits of two or more bodies around a central mass.*

This is not to say that the value of either the rotational period or the orbital period is necessarily a neat, whole number; only that the

For instance, Mercury has a Spin-Orbit Resonance of 3:2, such that its orbital period of 87.969 days is

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

*relationship*between the two can be expressed as the*ratio*of two whole numbers.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*.## Tidal Locking

Tidal locking is an extremely complex subject, involving a plethora of characteristics of the two bodies involved, at least some of which are poorly understood, at least for some objects.

Below is the equation for calculating tidal lock given at Wikipedia [2]:

**In general:**- 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 [2]:

The above equation makes some radical assumptions, and the timescale returned may be off by orders of magnitude due to this, so use the equation at your own risk.