I am writing a science fantasy book that takes place on the planet Luyt (inspired by Luyten B/Gliese 273b). Because the orbital period of the planet is about 18.6 days and I have chosen to make the Luyt-Luyten's Star Spin-Orbit Resonance 4:1, there isn't a great natural way to mark days for the humanoids that have been transplanted here.

I want to have a series of satellites that mark time for the inhabitants of Luyt. The year on Luyt is 336 days, 4 x 84 seasons. Each season is comprised of three 24 day months followed be a 12 day festival.

To mark this time there are four groups of satellites (inner, darred, rudden, and outer). There are 3 to 4 satellites in each group. Each group revolves around Luyt at a different distance and a different velocity.

Lunar Group Orbital Period in Luyten days Range of the Orbital radius (km)
Inner 4 0 - 1 x 10^7
Darred 12 1 x 10^7 - 2 x 10^7
Rudden 24 2 x 10^7 - 3 x 10^7
Outer 84 3 x 10^7 - 4 x 10^7

Model: https://www.geogebra.org/m/ptwk37u9

Essentially, I am trying to make a calendar clock with 14 hands out of a planet and satellites. If this calendar is not perfectly accurate, that would actually add to the story (as festivals could be variable length by minutes, hours, or even a day or two to adjust to the placement of the satellites). i.e.: Sommer-fest ends on the syzgy of the moons Merieke and Onadar.

My background is in Mathematics and Computer Science, not Physics or Astronomy. My blundering research leads me to believe that the placement of satellites around a planet is much more complicated than can be accounted for by far future "hand-wavy" science.


  1. Can a planet as close to it's sun as Luyten B is to Luyten's star have any satellites?
  2. Can a series of 14 satellites have anything even close to this relationship (i.e.: make a calendar clock)?
  • 1
    $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$
    – Community Bot
    Mar 9 at 15:58
  • 1
    $\begingroup$ I confess to being slightly confused, are you saying that the orbital period is not dealt with as a "year"? $\endgroup$ Mar 9 at 16:17
  • $\begingroup$ A year on Luyt (used to state the date by those that live there) does not correspond to the orbital period of the planet around the star (because it is based on Luyten B which is reported by NASA to have an orbital period of 28.6 earth days). $\endgroup$ Mar 9 at 16:39
  • $\begingroup$ You need more information to try to work this out. You could try to put the satellites into position, but their orbits may not be stable. Also since you are so close to the star, the orbits will be also be perturbed by the pressure from the light and radiation. They are far enough out that they will be hard to see. $\endgroup$
    – UVphoton
    Mar 9 at 17:42

2 Answers 2


Short Answer:

Your desired orbits for your moons present some problems, but I suggest some possible solutions.

Long Asnwer in Five Parts:

You may want to consider where you want your story to be on the scale of science fiction hardness.


Part One: Orbital problems.

The rings of satellites would have to orbit within the Hill sphere of your planet, the zone where satellites can have stable orbits.

According to Wikipedia:

The Hill sphere for Earth thus extends out to about 1.5 million km (0.01 AU). The Moon's orbit, at a distance of 0.384 million km from Earth, is comfortably within the gravitational sphere of influence of Earth and it is therefore not at risk of being pulled into an independent orbit around the Sun. All stable satellites of the Earth (those within the Earth's Hill sphere) must have an orbital period shorter than seven months.


The large size of Earth's Hill sphere is somewhat favorable for your satellite set up.

However, the more massive a star is, the more luminousit is and the father away its habitable zone is, and the less massive a star is, the less luminous it is and the closer its habitable zone is. And the relationship between the mass of a star and its luminosity (and thus the distance at which a planet can orbit in the habitable zone), is not linear.

A small difference in the mass of stars creates a larger difference in their luminosities. So planets which orbit stars of different luminosity at the distances where they receive the same amounts of radiation from their stars will not experience equal gravitational and tidal foreces from their stars.

Planets orbiting dim red dwarf stars in their habitable zone will be orbiting where the gravity of the itheir stars is much stronger than the Sun's gravitational pull on Earth. That is why it is usally believed that most or all planets that close to red dwarf stars will have become tidally locked to their stars, though your planet seems to have a rotational period/orbital period resonance instead.

As the Wikipedia article explains, the size of a planet's Hill sphere depends on the mass, of the star, the mass of the planet, and the semi-major axis of the planet's orbit.

According to the simplified equation used for low eccentrity orbits, the radius of a planet's hill sphere will be approximately equal to the semimajor axis of the planet's orbit multipled by a number which is the cube root of the ratio of the planet's mass to three times the star's mass.

Luyten b has s masss of 2.89 plus or minus 0.26) that of Earth - 2.63 to 3.15 the mass of Earth. Having a higher mass than Earth's is good for having a large Hill Sphere.

Luyten's Star has a mass of 0.26 times the mass of the Sun. The star having a lower mass than the Sun is good for the planet having a large Hill sphere.

The orbit of Luyten b has a semi-major axis of 0.09110 AU. Having such a small semi-major axis is bad for the planet having a large Hill sphere.

According to my rough calculations, the Hill sphere of Luyten b should equal approximately 0.0211 to 0.0223 of the semi-major axis of its orbit around Luyten's Star. Thus the Hill spere of Luyten b should have a radius of approximately 0.0019222 to 0.0020215 AU, or approximately 287,557.0271 to 302,412.0956 kilometers.

I found an online Hill sphere calculator:


It gives the Hill sphere radius of Luyten b as approximately 265,362.19277423 to 281,810.59421183 kilometers depending on the mass entered for Luyten b, a bit smaller than my calcuations. If the mass of Luyten b is exactly 2.89 that of Earth, the radius of the Hill sphere would be 273,833.39572197 kilometers.

The Hill sphere is only an approximation, and other forces (such as radiation pressure or the Yarkovsky effect) can eventually perturb an object out of the sphere. This third object should also be of small enough mass that it introduces no additional complications through its own gravity. Detailed numerical calculations show that orbits at or just within the Hill sphere are not stable in the long term; it appears that stable satellite orbits exist only inside 1/2 to 1/3 of the Hill radius. The region of stability for retrograde orbits at a large distance from the primary is larger than the region for prograde orbits at a large distance from the primary. This was thought to explain the preponderance of retrograde moons around Jupiter; however, Saturn has a more even mix of retrograde/prograde moons so the reasons are more complicated.3


So if Luyten b has a mass of 2.63 Earth mass the outer edge of its region of true stability would be between 88,454.063 and 132,681.09 kilometers.

So if Luyten b has a mass of 2.89 Earth mass the outer edge of its region of true stability would be between 91,277.796 and 136,916.69 kilometers.

So if Luyten b has a mass of 3.15 Earth mass the outer edge of its region of true stability would be between 93,936.863 and 140,905.29 kilometers.

Using this orbital period calculator, and using the mass range of Luyten b, and setting the mass of a satellite as 0.1 that of the moon:


If Luyten b has a mass of 2.63 Earth, a satellite at the edge of its region of true stability would have an orbital period of 1.86785 to 3.43147 Earth days.

If Luyten b has a mass of 2.89 Earth, a satellite at the edge of its region of true stability would have an orbital period of 1.86789 to 3.43154 Earth days.

If Luyten b has a mass of 3.15 Earth, a satellite at the edge of its region of true stability would have an orbital period of 1.86793 to 3.43160 Earth days.

So the outermost satellites of Luyten b in long term stable orbits would have orbital periods of 1.86785 earth days or 44.8284 Earth hours to 3.43160 Earth days or 93.653282.3584 Earth hours.

There is another calculation to make. Heller and Barnes, in "Exomoon habitability constrained by illumination and tidal heating", 2013:


On page 20, say:

The longest possible length of a satellite’s day compatible with Hill stability has been shown to be about P)p/9, P)p being the planet’s orbital period about the star (Kipping, 2009a).

Since you want the orbital period your planet to be about 18.6 Earth days, and since Luyten b has an orbital period of about 18.650 Earth days, the longest orbital period of a moon in a stable orbit around it would be about 18.650 divided by 9, or 2.072222 Earth days.

Their source is:

Kipping, D.M. (2009a) Transit timing effects due to an exomoon. Mon Not R Astron Soc 392:181–189.

And here is a link if you want to follow what Kipping says:


I believe that you want there to be 4 days of your planet in one orbital period. If th orbitalperiod is 18.650 Earth days, the planetary days will be about 4.6625 Earth days long, which is about 2.25 times as long as the maximum stable orbital period of about 2.072222 Earth days days. And you want the rings of satellites around your planet to have orbital periods of 4, 12,24, and 84 Luyten days, which would equal 18.65, 55.95, 111.9, 391.656 Earth days.

All of which are several times, or tens of times, or hundreds of times as long as the maximum periods of long term stable satellites around Luyten b or any world with a similar mass in a similar orbit around a similar star.

Part Two: Trojan Moons.

There is one bright side: You desire 14 satellites in 4 groups, which makes 2 groups with 3 satellites and 2 groups with 4 satellites.

And a group of three satellites sharing the same orbit is perfectly possible. There could be a large moon orbiting your planet, with two smaller moons in two of its Lagrange points, the L4 and L5 points, 60 degrees ahead of and 60 degrees behind the large moon in its orbit. That is called a trojan orbit.

So each of those three moons would be seen in the same relative position at regular intervals equal to 1/6 th of the orbital period at that distance, followed by no moons for the other 4/6 th of the orbital period.

And there actually is an example of a planet in our solar system which has two different moons that each have two smaller moons in trojan positions. Tethys has 2 trojan moons ahead and behind it, and Dione has two trojan moons in its orbit.

But there is a problem. Caculating the stabiity of trojan orbits is said to be complex, but there is a rule of thumb about the relative masses of the objects for stable orbits.

As a rule of thumb, the system is likely to be long-lived if m1 > 100m2 > 10,000m3 (in which m1, m2, and m3 are the masses of the star, planet, and trojan).


In this case m1 is your planet with 2.63 to 3.15 the mass of Earth, m2 is the large moon in orbit with less than 0.0263 to 0.0315 the mass of Earth, and m3 would be one of its trojans with mass less than 0.000000263 to 0.00000315 the mass of Earth.

If one of the small trojan moons had less than one millionth the mass of your planet, and if it had the same density, it would have one millionth the volume of your planet and thus one hundreth of its diameter.

Luyten b is likely to have less than 1.5 the radius of Earth, and so hypothetical trojan moons of Luyten b would be likely to have a diameter of 0.015 the dddiameter of Earth or less - a diameter of 191.13 kilometers or less. So they would appear as bright dots of light or as small extended objects in its sky.

Another big problem with having 2 of your sets of moons being larger moons with 2 trojans each is that trojan objects do not stay exactly in the L4 and L5 positions. Instead they tend to wander away from them and back again. Thus you could not count on the trojan moons always being exactly 60 degrees ahead of or behind the larger moons. That would not be very good for calendars!

Part Three: Rings of Moons.

sean Raymond's PLanetplanet blog has section The ULitmate Solar system, where he tries designing scientifically plausible solar systems with as many habitable worlds as possible.


In the ultimate engineered solar system he says that a paper by Smith and Lissauer shows that equally spaced and equally mass planets can share an orbit around a star. They calculated stable orbits for 7 to 42 planets sharing an orbit.



So you could put a ring of 7 to 42 equally massed and equally spaced moons around your planet, separated by 8.5741 to 51.4285 degrees. Assuming that the total mass of the ring should be less than about 10 percent of the mass of your planet, each moon could have no more than about 0.00238 to 0.01428 the mass of your planet, and thus no more than about 0.0062594 to 0.0449 the mass of Earth.

Assuming that each moon had the same average density as Earth, their volume would be proportional to their mass relative to Earth. They could have diameters no more than 0.18429 to 0.3555 that of Earth, or no more than 2,348 to 4,529 kilometers. So their maximum possible diameters could be similar to that of the Moon, and they would probably be only a fraction as distant as the Moon is from Earth, so they might be as impressive or even more so than the Moon seen from Earth.

So if you can figure out a way to use a ring of 7 to 42 moons sharing the same orbit equally spaced along it in your calendar, this may be useful to you.

And of course it would be extremely improbable for a planet to naturally acquire or form a ring of equally spaced and equally massed moons. Theodds agains that happening naturally would be astronomical. So it would be very probably that any such ring of moons around a planet wwould have been created artifically by an advanced civilization for some purpose.

Part Four: Synodic Periods.

Did you know that astronomical objects, like the planet Earth, have different types of days, and different types of months, and different types of years, and that the differences betweent the lengths of those periods is important for constructing a calendar?

A sideral day is the natural rotation period of a world, the time it takes to rotate 360 degrees of arc with respect to the distant background stars. The sidereal day of Earth is about 23 hours, 56 minutes and 4 seconds long, or about 86,164 seconds long. There are about 366.25 sidereal days in an Earth year.

So during each sidereal day the Earth moves about 0.98 degree in its orbit around the Sun. So when a sidereal day is completed, and a point on the surface which pointed directly at the Sun at noon is now pointed at the exact same direction relative to the stars, the direction to the Sun has moved by almost one degree, and Earth has to turn for about 4 minutes and 56 seconds more until that point on the surface is pointed directly at the Sun again.

So the solar day, the amount of time it takes for the Sun to more 360 degrees as seen from a point on the surface, equals 24 hours, and there are about 365.25 solar days in an Earth year.


Apparently there are five types of months of different lengths in astronomy:

The following types of months are mainly of significance in astronomy, most of them (but not the distinction between sidereal and tropical months) first recognized in Babylonian lunar astronomy.

The sidereal month is defined as the Moon's orbital period in a non-rotating frame of reference (which on average is equal to its rotation period in the same frame). It is about 27.32166 days (27 days, 7 hours, 43 minutes, 11.6 seconds). It is closely equal to the time it takes the Moon to pass twice a "fixed" star (different stars give different results because all have a very small proper motion and are not really fixed in position).

A synodic month is the most familiar lunar cycle, defined as the time interval between two consecutive occurrences of a particular phase (such as new moon or full moon) as seen by an observer on Earth. The mean length of the synodic month is 29.53059 days (29 days, 12 hours, 44 minutes, 2.8 seconds). Due to the eccentricity of the lunar orbit around Earth (and to a lesser degree, the Earth's elliptical orbit around the Sun), the length of a synodic month can vary by up to seven hours.

The tropical month is the average time for the Moon to pass twice through the same equinox point of the sky. It is 27.32158 days, very slightly shorter than the sidereal month (27.32166) days, because of precession of the equinoxes.

An anomalistic month is the average time the Moon takes to go from perigee to perigee—the point in the Moon's orbit when it is closest to Earth. An anomalistic month is about 27.55455 days on average.

The draconic month, draconitic month, or nodal month is the period in which the Moon returns to the same node of its orbit; the nodes are the two points where the Moon's orbit crosses the plane of the Earth's orbit. Its duration is about 27.21222 days on average


there are more than 13 sidereal months in an Earth year but about 12.3 synodic months in that period.

Wikipedia lists 10 different types of astronomical years:


And these can be usefull in constructing a calendar which uses a "year" that is much longer than the actual orbital period of your planet.

The period it takes a moon to orbit a planet once depends on the mass of the planet and the distance of the moon from the planet. The rotation rates of planets vary greatly, but every planet has a distance at which a moon orbiting above the equator would be in a geostationary orbit and would always seem to have the same position in the sky, though the stars would move behind it.

A moon orbiting closer than that geosynchronous orbit would have a sidereal period shorter than a day, and a moon orbiting above the geosychronous orbit would have a sideral period longer than a sidereal day. The farther inside or outside the geosychronous orbit the moon orbited, the greater would be the difference bwweenits orbit and a day - and thus the shorter would be the difference between its synodic period and a day.

For example, if the moon took 100 days to compete one orbit, it would take the world only a little more than one day to get ahead of the moon and then catch up with it again.

But if the moon took only 0.99 days to complete one orbit, each day the moon wouldd would get 3.6 degrees ahead of the turn world, and after about 100 days the moon would catch up to the turning world and would be in the same position in the sky again.

And if one moon takes 2 days to orbit the planet, and another moon takes 2.1 days to orbit the planet, the inner moon will orbit 180 degrees in 1 day and the outer moon will orbit 171.42857 degrees in 1 day. Each day the inner moon will get 8.57143 days ahead of the inner moon, until after about 41.9999 days, or 42 days if people round up, the inner moon will over take the outer moon and they will be in their same relative positons as when they started.

So if you want to create a calendar period or "year" which is many planetary orbital periods long and equals 336 Earth, or 336 Luyten b days, you can calculate some relationships between the planetary rotation and an orbital period of a moon, or the orbital periods of two or more moons, which has that length.

Part Five: Artificial Satellites.

In Star Trek: The Original Series The Enterprise ws often said to be "orbiting" a planet. But when the ship lost pwoer, the orbit immediately started to decay, unlike a realistic orbit. So possibly those alleged "orbits" involved the starship using its engine power to constantly hold itself up while circling at a speed much less than the orbital speed at its distance.

So possibly the satellites that the settlers on the planet use for their calendar are spacecraft of some kind, which trave in circles around the planet at speeds different from the orbital speeds at their respective distances from the planet.

Those spcef craft could travel around the planet low and be very large, and so be visible from the surface. Or they could "orbit" very high, and shine very bright lights down at the planet, so those lights can be seen from the surface.

And possibly those spacecraft were built by the colonists to serve as factors in their calendar. Or possibly they were left behind by an advanced civilization that the colonists know nothing about.


There are a number of problems with your moons, but I have suggested some possible ways to get around them.


You can answer this question fairly easily yourself, if you have a mathematics background. First, you need to determine the size of your planet's Hill Sphere, using the formulas in that article. You need to decide on your planet's mass to do that, because we have only a vague idea of that from astronomy.

That gives you the maximum distance from your planet that satellites can reach without coming loose from the planet and going into orbit around the star. You can then calculate the orbital period for a circular orbit of that radius using the formula in that article. I suspect you'll find that your planet can have satellites, but they can't have orbital periods nearly as long as you want.

However, there's a different problem with this piece of worldbuilding. Having so many satellites with periods that are an exact number of local days screams "this is artificial." It will be obvious to the reader that these satellites were carefully placed by someone or something with considerable power. If that isn't the case in your story, the reader will have a major red herring to cope with; if it is, you're betraying the existence of those powers as soon as the reader finds out about the orbital periods.


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