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I'm designing a fantasy world on a tidally-locked moon of a gas giant. Because it's primarily a medieval fantasy set on the outward-facing side of the home moon, I'd like to keep units of time similar to Earth and use astronomical events to help shape the units of time. There are a ton of great questions about these types of worlds here, but to me the concept of the earth month (about 30 earth days) is the hardest to map. Because the month was inspired by Earth's moon's phases, is it plausible to have another moon serve this purpose, or is there another astronomical/geological event that could take its place?

Details:

  • The gas giant is smaller and closer to the sun than Jupiter. Its orbit only takes around 4 earthyears, so seasons would be about 1 earthyear each and would be a good substitute for years: He's seen 20 seasons...
  • Days/Months are the same unit of time, as home moon will orbit the gas giant and do 1 rotation in around 25 hours, marking a 'day' (though the inward facing side would always get a few hour eclipse per day)
  • There are other outer moons that the outward-facing side of home moon can see
  • Home Moon is far enough away from the gas giant to have its magnetosphere shield it from most radiation. Tidal heating causes it to be (Removed as this doesn't make sense as pointed out) It is slightly more volcanically active than Earth
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    $\begingroup$ I know it doesn't make a difference to the current question but... _Tidal heating_ causes it to ... Ummm... tides forces strong enough to contribute with heat to the core of a tidally locked moon? By what mechanism? $\endgroup$ Oct 12 at 10:48
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    $\begingroup$ @AdrianColomitchi Io does just fine ;-) $\endgroup$ Oct 12 at 14:45
  • $\begingroup$ @StarfishPrime that was sorta my point, nudging the OP to check his assumptions. With just one other moon in the system, I fear that his Home-moon will need an elongated elliptical orbit to have just a wee more tidal heating than Earth. Io has other 3 around to tug her heart in all directions. Maybe he likes the idea of an elliptical orbit (with the precession that comes with it), maybe he doesn't. $\endgroup$ Oct 12 at 14:52
  • $\begingroup$ @AdrianColomitchi "With just one other moon in the system" disagrees with the OP's "There are other outer moons that the outward-facing side of home moon can see" $\endgroup$
    – PcMan
    Oct 12 at 17:01
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    $\begingroup$ Whenever you look at a moon (including our own), you should always be seeing a ball in space, which is lit up from whatever direction the local sun happens to be. Any moon or planet that you can see with the Sun behind you will look full, and any that you can see with the Sun behind it will look new. So the only time phases don't happen is if you're always near the sun, looking at something that is always far from it. That gives you an Apollo-eye view where nothing is ever shadow. (Well, OK, you and the moon could also orbit the sun with the same period so the sunlight never moves...) $\endgroup$ Oct 12 at 20:10
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A second answer 10-14-2021

Using a Month Based on the Synodic Period of the Two moons.

I suggested that one possible basis for a month equivalent of the habitable moon would be the synodic period of an outer moon with the inner and habitable moon. I also noted that the more similar the orbital periods of the two moons were, the longer their synodic period would be.

Your question requires a day of the tidally locked habitable moon, and thus an oritabl period, of 25 Earth hours.

Suppose that you want the month, based on the synodic period of the two moons, to be 28, 29, or 39 of the habitable moon's 25 hour days. That would make them about 700 hours or 29.1666 Earth days, 725 hours or 30.208333 Earth days, or 750 hours or 31.25 Earth days, long.

According to my rough calculations, for the synodic period & month to be 28 of the habitable moon's days long, the outer moon would have to have an orbital period about 1.0370369 times as long as the habitable moon, or 1.0370369 of the habitable moon's days, or 25.925922 Earth hours.

According to my rough calculations, for the synodic period & month to be 29 of the habitable moon's days long, the outer moon would have to have an orbital period times as 1.0357143 long as the habitable moon, or 1.0357143 of the habitable moon's days, or 25.892857 Earth hours.

According to my rough calculations, for the synodic period & month to be 30 of the habitable moon's days long, the outer moon would have to have an orbital period 1.0344824 times as long as the habitable moon, or 1.0344824 of the habitable moon's days, or 25.86206 Earth hours.

So if the basis of the month of the habitable moon is the synodic period of the two moons, and if you want to make the month used on the habitable moon 28,29, or 30 of the days of the habitable moon, the outer moon will have to orbit with an almost identical orbit and an almost identical orbital period. Thus it should look very large as seen from the habitable moon when it is in opposition and a lot smaller when it is farther away in its orbit.

Using a Month Based on the Phases of the Outer Moon.

I have reconsidered using the phases of an outer moon as the basis for a month of the habitable moon. If the outer moon orbits several times as far from the planet as the habitable moon, and has an orbital period several times as long as the habitable moon, it might work out.

Now try using the orbital relationship between Earth and Saturn as a model for the orbital relationship between the habitable moon and the outer moon. Earth is 1 AU from the Sun, and Saturn is 9.5388 AU from the Sun. Earth has an orbital period 1 Earth year long, and Saturn has an orbital period 29.4577 Earth years long, and their synodic period is 378.09 Earth days, or about 1.035154 Earth years.

So if the habitable moon orbited the giant planet with a semi-major axis of 1 unit, the outer moon would orbit the giant planet with a semi-major axis of 9.5388 units. If the orbital period and thus day of the tidally locked habitable moon was 25 Earth hours long, the orbital period of the outer moon would be 29.4577 times that of the habitable moon, or 736.4425 Earth hours, or 30.685104 Earth days and the synodic period of the two moons would be 25.87885 hours.

So the synodic period of the two moons would be 25.87885 hours, and the orbital period wouid be 736.4425 hours, which is 28.457311 times the synodic period. So during each synodic period the outer moon would move 12.650527 degrees in its orbit.

Suppose that there is a time when the outer Moon is full as seen from the planet, because the star, the planet, and the out moon are arranged in a straight line. And suppose that at this time the habitable moon is also in that line, between the star the planet, and the outer moon, and the outer moon also appears full from the habitable moon.

If a cycle of oppositions starts with a perfect line up, the positions of the two moons will be in the worst viewing position during the next full outer moon. nextpossiblthe

After 736.4425 Earth hours the star, the planet, and the outer moon will once again be be lined up and the outer moon will appear full as seen from the planet. And the inner, habitable moon will have travelled 29.4577 times around the planet and will be 164.772 degrees ahead of fhe outer moon, or 195.228 degreees behind it, and almost opposite to the outer moon.

Going back in time a bit to when the outer moon has travelled exactly 28 times the synodic period, or for 724.6078 hours, it will have travelled for 28.98432 times around the planet, and thus will be 354.35232 degress ahead of the line between the star, the planet, and the outer moon, or 5.64768 degrees behind it. So the line between the planet and the two moons will be only 5.64 degreees from the line between the star, the planet, and the outer moon. So the outer moon should appear almost exactly full when seen from the habitable moon.

And of course in most months the 28th synodic period will happen when the two moons are lined up closer than 5.63 degrees to the line between the star, the planet, and the outer moon,and so the outer moon will look even fuller duirng most of the 28th oppositions in a month.

Thus making the average month in the habitable moon's calendar 29 days, with an occasional 28 day month, should work well.

But there is a problem. The year of the planet is supposed to be 4 Earth years, or about 1,461 Earth days. Thus the planet would travel about 0.2464065 degree per Earth day, or 0.0102669 degree per Earth hour. The sidereal orbital period of the outer moon is 736.4425 Earth hours, so the planet and its moons will travel about 7.5609815 degrees along the planet's orbit dur that period. So the outer moon will have to travel another 7.5609815 degrees along its orbit to line up with the star again, at a speed of 0.4888365 degrees per hour, thus taking 15.467301 hours.

So the synodic month of the outer moon, as seen from the planet, would be 751.9098 hours, which would be 30.076392 of the habitable moon's days, and 29.054992 synodic periods of the two moons. So a month of 30 of the habitable moon's days would be about 28.981195 synodic periods of the two moons. Thus they would have to occassionally have a 31 day month, I think.

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  • $\begingroup$ The synodic period of an outer moon seems like it would most likely be noticed first by ancient astronomers, especially if it looks very large at opposition - larger than earth's moon? Thank you for the extremely helpful answer. $\endgroup$
    – xeg0
    Oct 19 at 11:29
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Short answer: yes, they will.

Long answer: event though each moon is tidally locked, each of them has a different orbit around the gas giant, thus their relative position with respect to the central star and the planet will change over time.

Changing the relative position will change also the point of view and the lightened part which is visible, and this will result in different phases being visible.

For a visual reference, look at this animation of the orbital resonance of Jupiter's moon Europa

enter image description here

you can see how the relative angle of view changes at any moment.

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Home Moon is far enough away from the gas giant to have its magnetosphere shield it from most radiation. Tidal heating causes it to be slightly more volcanically active than earth

I'm not sure you can have all three of "24hr orbital period", "tidal heating" and "far enough away to be safe from radiation belts".

I'm not even sure that you'd get a substantial magnetosphere on a moon at all, even with tidal heating. However, a combination of a thick atmosphere and being well within the parent planet's magnetosphere should provide some protection from solar radiation, and it should be possible to handwave a sufficiently benign planetary magnetosphere so that local radiation belts are not an issue.

Days/Months are the same unit of time

There's been another question about calendars on off-Earth worlds, and month-analogs seem to be a common theme. I'm not really sure why they're needed... it seems unnecessary, especially without some cultural link to Earth (or some other world with unambiguous lunar cycles). Fellow moons won't generate substantial tides, nor will they make much different to ambient light at night time... an important thing when it comes to hunting or fishing.

Seems like only astronomers or astrologers would care very much, but maybe that's enough to drive creation of a suitable calendar.

But on to the meat of your question:

Would Moons of a gas giant appear to phase when looking at them from another, tidally-locked Moon?

Io and Europa seen by Juno as it orbits Jupiter

(image from jpl.nasa.gov, credit CC-BY Roman Tkachenko)

Behold, half-moons! Io is the larger one, closer to Jupiter, and Europa is the further one. Not clear how far away Juno was when it took the image, but there's unambiguous phasing here and I'm pretty certain that an observer on Io could (probably quite briefly, given the weather) see phases of Europa during its orbit.

On, say, a Saturn-sized gas giant, a 1-day orbital period has a radius of about 190000km, and a 30-day orbital period has a radius of about 1860000km. That gives a closest encounter distance of more than 1.6 million kilometres. At that distance, a moon the size of Titan has an angular diameter of about 10 minutes of arc... that's about a third of the diameter of the full moon. This view would only be possible from the hemisphere of the moon facing away from the gas giant.

On the planet-facing of your moon, the Saturn-like world would have an angular diameter of 37 degrees, which would be unambiguously spectacular. As the 30-day moon disappears over the limb of of the gas giant it would be little over 2 million kilometres away, giving it an angular diameter of 8.6 minutes of arc... smaller, but not so much that the phase wouldn't be visible. This view would only be possible from the hemisphere of the moon facing towards the gas giant.

Note that having a really big distant moon and a really big close-in moon is a little unlikely, but for a Saturn-like parent planet not beyond the realms of possibility.

it's primarily a medieval fantasy set on the outward-facing side of the home moon

Missing out on the best view, daily eclipses, bright planet-lit twilight instead of true night time, greater protection from cosmic radiation and meteorites?

Well, to each their own. Astronomy, at least the sort that looks at stars and planets other than the one that Home Moon orbits would certainly be easier from the boring side, and the phases of moons would be more obvious.

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  1. Observers all over Home Moon will see all other satellites, not only those observers on the far side.

  2. Since Home Moon is tidally locked there will be no tidal heating; to understand why, consider a coat hanging on a hook in the wall: does Earth's gravitational force do any work on the coat? Does it warm up the coat? No work means no energy to be converted into heat.

  3. The rapid movement of the other satellites across the sky will be much more easily noticeable than their phases.

  4. The gas giant itself will have very visible and impressive phases.

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  • $\begingroup$ "Since Home Moon is tidally locked there will be no tidal heating" - it may, but I don't think the OP would like that. Because significant tidal heating (to alter the volcanism) will happen with a quite elongated elliptic orbit (equiv hang your coat on a bungee jumping cord) $\endgroup$ Oct 12 at 11:46
  • $\begingroup$ "The gas giant itself will have very visible and impressive phases." A pity that only the proximal face of the tidal locked Moon will see them. Good as a tourist attraction, I think. $\endgroup$ Oct 12 at 11:48
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    $\begingroup$ Io is tidally locked, but still undergoes tidal heating as orbital resonances with other large moons act to de-circularize its orbit. $\endgroup$ Oct 12 at 14:49
  • $\begingroup$ Europa and Enceladus are also tidally locked, and undergoing tidal heating for similar reasons. $\endgroup$
    – notovny
    Oct 12 at 19:13
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I am busy with other stuff right now, so this is a short answer.

If the other moons have a large enough angular diameter as seen from the viewing moon to appear as objects and not mere points of lights, they will certainly show phases.

As the viewing moon and the viewed moons orbit around the planet, the angles between them and between each and the planet will constantly change.

As the viewing moon and the viewed moons orbit around the planet, the planet will also be orbiting around the star. Thus the angles between the moons and the planet will also change relative to the star.

at any one time about one hemisphere of the planet will be facing the star and illuminated and the other hemisphere will be dark. And the same goes for each of the moons.

But when the star, a moon that orbits outside the viewing moon, the viewing moon, and the planet are all lined upin that order, the outer moon will show its dark side to the viewing moon, like the new moon seen from Earth. And the farther away the objects get from being in a stright line, the thicker the illuminated cresent of the outer moon will look from the viewing moon.

In the exact opposite situation from the first one, when the star, the planet, the viewing moon, and the outer moon are all lined up in that order, the outer moon will look full from the viewing moon. And the farther away the objects get from being in a stright line, the thinner the illuminated side of the outer moon will appear from the viewing moon.

It may be very rare for all four objects, the star, the planet, the viewing moon, and the outer moon, to all be lined up in either of those orders. It will probably take many revolutions of the two moons around the planet, as the planet orbits the star, for them to be lined up two successive times.

And if there are two or more outer moons which get close enough to show their phases during at least part of their orbits, it will rare for any one of them to be lined up exactly with the star, the planet, and the viewing moon.

And line ups of five objects would be much rarer than line ups of four objects, and line ups of six objects would be much rarer than line ups of five objects.

In fact, if you add any more outer moons which can pass close enough to sometimes be seen as orbs from the viewing moon, the time elapsed between successive moments when the planet, the star, the viewing moons, and the viewed moons, are all lined up at once could be longer than the time the star spends in its main sequence phase.

Added 10-13-2021

Long Answer.

A science fiction writer should decide where they want their story to be be in the Mohs Scale of Science Fiction Hardness.

https://tvtropes.org/pmwiki/pmwiki.php/Main/MohsScaleOfScienceFictionHardness

And they should decide how they might feel if 1 reader out of 10, or 1 out of 100, or 1 out of 1,000, can see that elements in their story are mathematically possible. Some writers might consider that highly humiliating.

YOu say that your world is a fantasy world, so maybe you won't care how realistic it is except for the fantasy. But if you want to be realistic in some ways you should continue reading.

My answer, and other answers, have already shown that it is possible for people on the outer hemisphere of a tidally locked habitable moon see phases of moons with orbits outside of the orbit of their moon, if, repeat if, those outer moons are large and/or close enough to show visible discs during at least parts of their orbits around the planet.

You desire the system to have the following characteristics.

The gas giant is smaller and closer to the sun than Jupiter. Its orbit only takes around 4 earthyears, so seasons would be about 1 earthyear each and would be a good substitute for years: He's seen 20 seasons... Days/Months are the same unit of time, as home moon will orbit the gas giant and do 1 rotation in around 25 hours, marking a 'day' (though the inward facing side would always get a few hour eclipse per day)

There are other outer moons that the outward-facing side of home moon can see

Home Moon is far enough away from the gas giant to have its magnetosphere shield it from most radiation. Tidal heating causes it to be slightly more volcanically active than earth.

So you want specific orbital periods for both the moon around the planet and the planet around the star.

The orbital speed of one object around another depends on the masses of the two objects, and the distance between them. The time it takes on object to orbit around another object depends on the orbital speed and the total circumference of the orbit, and of course the orbital circumference depends on the radius of the orbit, the distance between the objects.

So if a writer just chooses random values for those factors, the odds are high that the factors won't work with each other but will be contradictory.

The answer by user177107 to this question:

https://astronomy.stackexchange.com/questions/40746/how-would-the-characteristics-of-a-habitable-planet-change-with-stars-of-differe/40758#40758

Has a table giving the characteristics of some spectral classes of stars. One of those characterisits is what I call the Earth Equivalent Distance, the orbital radius where a planet would receive the same amount of radiation from its stars that Earth receives from the Sun. It also gives the orbital speed and the length of the year of a planet orbiting in the EED.

The lengths of orbital periods (or years) of planets orbiting the EED of a star varies from 3.82 Earth days for an M8V star to 1,018.01 Earth days for a F2V star and up to 2,526.01 Earth days for an A2V star.

Since an Earth year is about 365.25 Earth days long, a year 2,526.01 Earth days long would be about 6.915 Earth years long, with (astronomical) seasons about 1.728 Earth years long.

A year about 4 Earth years long would be about 1,461 Earth days long.

A planet orbiting ain the EED of an A8V star would have a year about 1.505.21 Earth days long.

So presumably your planet and its moon could orbit in or near the EED of a star closer to an A8V than to a F2V to have a year about 4 Earth years long.

But writers of science fiction, if not necessarily writers of fantasy, should know that it took Earth billions of years after it formed to produce an oxygen rich atmopshere that multicelled animals and humans can breathe.

As far as I know the main scientific discussion of the factors making a world habitable for humans in particular, and not for liquid water using lifeforms in general, is Habitable Planets for Man, Stephen H. Dole, 1964.

https://www.rand.org/content/dam/rand/pubs/commercial_books/2007/RAND_CB179-1.pdf

Dole decided that a planet would have to be at least 3 billion years old to develop an oxygen rich atmosphere breathable for humans.

Because of calculations of the life spans of various types of stars, Dole calculated that stars would have to be F2V class stars, or of lower mass in spectral classes F, G, and K, to have planets habitable for humans.

So spectral class A stars would be ruled out for having habitable planets. If an F2V class star would be the most massive star that could possibly be old enough to have a planet habitable for humans, a planet habitable fro Humans orbiting at the EED of a star would have a year that would be 1,018.01 Earth days long - or shorter.

A year 1,018.01 Earth days long would be about 2.787 Earth years long. Itwouldhave astronomical seasons about 254.5025 Earth days or 0.696 of an Earth year long.

That is a bit shorter than you want.

But a planet doesn't to orbit exactly at the EED of its star to be habitable for humans.

Each star has a circumstellar habitable zone around it where a planet would receive enough radiation from its star to be warm enough for liquid water on its surface, and thus be habitable for some types of liquid water using life. Humans would not be very comfortable on planets with average temperatures in the lower and higher parts of that temperature range.

So to find the inner and/or outer limits of the habitable zone of a star you just take the inner and/or outer limits of the habitable zone of the Sun and multiply or divide according to the relative luminosity of the other star compared to the Sun.

Except that the inner and outer edges of the Sun's habitable zone aren't known with certainty.

This list of estimates shows a pretty wide variety:

https://en.wikipedia.org/wiki/Circumstellar_habitable_zone#Solar_System_estimates

The planet Earth orbits the Sun at a distance of 1 AU which is the Sun's EED. The planet Mars orbits the Sun at a distance of 1.523 AU and has a year 686.980 Earth days long, or about 1.8808 Earth years long.

If a planet in the EED of an F2V class star would orbit at a distance of 2.236 AU and have a year 1,018.01 Earth days or 2.787 Earth years long, a planet in the Mars equivalent orbit would orbit at a distance of 3.405 AU and have a year 1,914.67 Earth days or 5.242 Earth years long.

So If a planet at a Mars equivalent distance can be warm enough to be habitable for humans, it might have a day at least 4 Earth years long if it orbits a luminous enough star, and a planet somewhat closer than a Mars equivalent orbit around an F2V star, and thus probably warmer, could have an year about 4 Earth years long.

So I think that a planet naturally habitable for humans with a year 4 Earth years long would be just within the bounds of scientific possibility.

And of course a science fiction writer could claim that the planet orbits in the EED of a class A star, and a statistically rare one-in-a-million chain of events caused the moon to become habitable in much less time than Dole's minimum. Or maybe that an advanced civilization terraformed the young moon to become habitable. Or maybe the moon would be too cold for humans except that tidal interactions with the planet and other large moons caused enough tidal heating to keep the moon habitably warm.

Your question does say:

Tidal heating causes it to be slightly more volcanically active than earth.

And a fantasy writer might say that the gods created that solar system exactly the way it is yesterday, and the character's memories of their lives before yesterday are fake memories created by the gods. Or maybe the gods aritrarily decided the orbital distance and and speed of the planet and the legnth of its year without caring that the figures didn't add up according to he laws of science.

The Length of the Moon's day

I should note that a planet, like Earth for example, has at least two different year lengths, and at least two different day lengths.

A sidereal year (UK: /saɪˈdɪəriəl/, US: /saɪˈdɪriəl, sə-/; from Latin sidus "asterism, star"; also sidereal orbital period) is the time taken by the Earth to orbit the Sun once with respect to the fixed stars. Hence, it is also the time taken for the Sun to return to the same position with respect to the fixed stars after apparently travelling once around the ecliptic. It equals 365.256 363 004 Ephemeris days for the J2000.0 epoch.1

The sidereal year differs from the solar year, "the period of time required for the ecliptic longitude of the Sun to increase 360 degrees",2 due to the precession of the equinoxes. The sidereal year is 20 min 24.5 s longer than the mean tropical year at J2000.0 (365.242 190 402 ephemeris days).1

Before the discovery of the precession of the equinoxes by Hipparchus in the Hellenistic period, the difference between sidereal and tropical year was unknown.[citation needed] For naked-eye observation, the shift of the constellations relative to the equinoxes only becomes apparent over centuries or "ages", and pre-modern calendars such as Hesiod's Works and Days would give the times of the year for sowing, harvest, and so on by reference to the first visibility of stars, effectively using the sidereal year.[citation needed] The South and Southeast Asian solar New Year, based on Indic influences, is traditionally reckoned by the Sun's entry into Aries and thus the sidereal year, but is also supposed to align with the spring equinox and have relevance to the harvesting and planting season and thus the tropical year.[citation needed] As these have grown apart, in some countries and cultures the date has been fixed according to the tropical year while in others the astronomical calculation and sidereal year is still used.[citation needed]

https://en.wikipedia.org/wiki/Sidereal_year

A tropical year (also known as a solar year) is the time that the Sun takes to return to the same position in the cycle of seasons, as seen from Earth; for example, the time from vernal equinox to vernal equinox, or from summer solstice to summer solstice. This differs from the time it takes Earth to complete one full orbit around the Sun as measured with respect to the fixed stars (the sidereal year) by about 20 minutes because of the precession of the equinoxes

https://en.wikipedia.org/wiki/Tropical_year

Sidereal time is based on the period Earth takes to rotate relative to the distant stars ("The fixed Stars").

A sidereal day is approximately 86164.0905 seconds (23 h 56 min 4.0905 s or 23.9344696 h).

Because Earth orbits the Sun once a year, the sidereal time at any given place and time will gain about four minutes against local civil time, every 24 hours, until, after a year has passed, one additional sidereal "day" has elapsed compared to the number of solar days that have gone by.

https://en.wikipedia.org/wiki/Sidereal_time

As the planet Earth orbits the Sun, it covers 360 degrees of arc during one year of about 365.25 days. So the Earth travels through about 0.985 degreees along its orbit every day.

So after the planet Earth rotates exactly 360 degrees with respect to the distant stars, a line between the center of the Earth and a spot on the surface of the Earth that pointed directly at the Sun and the stars directly behind the Sun will once again point directly at those stars. But the Sun will not longer be directly in front of those stars, because the Earth will have moved about 0.985 degrees along its orbit, and the direction to the Sun will have changed by about 0.985 degrees.

So the lengths of a sidereal day and a solar day must be at least slightly different.

If a planet has a prograde rotation, rotating in the same direction as it orbits around its star, the solar day will be longer than the sidereal day.

And some writers might want to figure out the exact lengths of the sidereal and tropical years, and the sidereal and solar days, of their fictional planet, and make a calendar that is quite accurate for that planet.

And other writers might be content with a calendar which is only roughly accurate. And possibly some writers might want to write about a boy apprenticed to an astronomer, who finds it too complicated to think about and runs away to become an apprentice to a wandering mercenary instead.

and of course some people may Have noticed that a calendar accurate for a planet might not be accurate for a giant habitable moon that orbits the planet.

I note that you want the year of the planet to be about four Earth years or about 1,461 Earth days long, and you want the day of your tidally locked moon to be about 25 Earth hours long. So your moon's years will be about 35,064 Earth hours, or about1,402.56 of your moon's days.

So in each of your moon's days, the planet, and the moon along with it, will travel about 0.2566 degrees along their orbit around the star. So the difference between the sidereal and solar (or "stellar") days of your moon should be much smaller than the difference between Earth's sidereal and solar days.

There is the complicating factor that the moon will be orbiting around the planet and so will sometimes be trvelling faster than the planet and sometimes slower than the planet, sometiems be aheaed of the planet and sometimes behind the planet as it orbits. Since the day is so short compared to the Moon's day, and the year so long, that will be a much smaller factor than it would be on Earth's Moon.

What length of day would be good for a hypothtetical habitable moon orbiting a giant planet? Stephen H. Dole, in Habitable planets for Man considered the question of the day length of planets habitable for humans.

He believed that a minimum possible day length would be about 2 to 3 Earth hours (0.08333 to 0.125 Earth days) because of the rapid rotation affecting planetary stability and about 96 Earth hours (4 Earth days) due to the daytime getting too hot and the nighttime getting too cold, and plants dying for lack of light in the long nights.

Rene Heller and Roy Barnes, in "Exomoon Habitabiity Constrained by Illumination and Tidal Heating" mentioned possible orbital periods and thus day lengths for hypothetical habitable exomoons.

https://faculty.washington.edu/rkb9/publications/hb13.pdf

The synchronized rotation periods of putative Earthmass exomoons around giant planets could be in the same range as the orbital periods of the Galilean moons around Jupiter (1.7–16.7 d) and as Titan’s orbital period around Saturn (&16 d) (NASA/JPL planetary satellite ephemerides)4

Clearly they haven't calculated any hard limits, but just found known examples of the orbital periods of large, tidally locked moons of giant planets.

They also point out that an exomoon that orbits too close to the planet will receive too much energy from the starlight from the star, reflected starlight from the planet, the planet's own internal heat radiating into space, and tidal heating, and will suffer a runaway greenhouse effect.

Rene Heller and Jorge Zuluaga, in "Magnetic Shielding of Exommons Beyond the Circumplanetary Habitable Edge" discuss the orbits of habitable exomoons. The habitable edge the distance from a planet that an otherwise suitable moon would have to exceed in order to avoid excessive energy and a runaway greenhouse effect.

https://arxiv.org/pdf/1309.0811.pdf

They consider the case of potentially habitable exomoons with the mass of Mars. They would probably be too small to generate their own magnetic fields, and so the stellar wind from their stars might gradually strip the atmsphere from them. Thus they would need to orbit within the magnetic fields of their planets to preserve their atmospheres for geological time periods.

Planetary magnetic fields might not concern hypothetical Earth mass Exomoons of giant exoplanets, which might generate their own magnetic fields to protect their atmospheres from the stellar wind. Of course such Earth mass exomoons would probably generate more powerful magnetic fields if they spin faster, and the speed of spinning of exomoons tidally locked to their planets would depend on orbital periods around the planet.

According to the calculations of Heller and Zuluaga, many sizes of giant planets would not extend their magnetic fields far enough to envelop exomoons orbiting beyond the habitable edge, and exooons orbiting inside the habitable edge would suffer runaway greenhouse effects.

Their conclusion seems to be that a habitable exomoon can orbit at distances between about 5 and 20 planetary radii.

And the angular diameter of the planet as seen from the habitable moon would depend on its distance in planetary radii and diameters from the moon, so if any of your natives on the far side of fhe moon explore far enough for the planet to be visible that will determine its angular size.

The more massive a planet (or other object) is, the faster will be the necessary orbital speed at a specific distance. So the more masive a planet is, the shorter will be the orbital period of an object orbiting it at a specific distance.

According to this orbital calculator:

https://www.satsig.net/orbit-research/orbit-height-and-speed.htm

An object orbiting jupiter 224,750 kilometers above the surface of Jupiter, would have an orbital period of 25.002 hours or 1.041 Earth days. Jupiter has a radius of 71,492 kilometers, so an object 224,750 kilometers above the surface of jupiter would be an orbital semi-major axis of 296,242 kilometers.

The moon Thebe has a semi-major axis of 221,889 kilometers and an orbital period of 0.6778 Earth days or 16 plus hours, while Io has a semi-major axis of 421,700 kilometers and an orbital period of 1.7691 Earth days. So that seems to be consistent with a moon with a semi-major axis at 296,242 kilometers having an orbital period 1.041 Earth days long.

Since jupiter has an equatorial radius of 71,492 kilometers, its habitable edge should be at about 357,460 kilometers or 285,968 kilometers above the surface (where a moon would have an orbital period of 33.140 hours), which is farther than the semi-major axis of 296,242 kilometers for a moon with an orbital period of 25 Earth hours. So a moon with an orbital period of 25 Earth hours would supposedly suffer a runaway greenhouse effect.

Saturn has an equatorial radius of 60,268 kilometers so it's habitable edge should be at 301,340 kilometern or 241,072 kilometers above he surface. At that distance the orbital period would be 46.877 hours.

Note that Saturn is much less massive than Jupiter, and its orbit at the habitable edge has a longer orbital period than that of jupiter.

Uranus is less massive than Neptune, but has a greater radius, 25,559 kilometers. So an orbit at its habitable edge should should have a semi-major axis of 127,795 kilometers, and be 102,236 kilometers above the surface. the orbital period at that distance would be 33.125 hours, very similar to that of the much more massive and larger Jupiter.

Neptune has a lower mass than Saturn. Neptune's radius is 24,764 kilometers, so an orbit at Neptune's habitable edge would have a semi-major axis of 123,820 kilometers, 99,056 kilometers above the surface. With that orbit, the orbital period would be 29.03 hours.

So the order in orbital periods at the habitable edge is Neptune, 3rd in mass, 4th in radius, 29.03 hours, Uranus, 4th in radius and 3rd in mass, 33.125 hours, Jupiter, 1st in mass and radius, 33.140 hurs, and Saturn, 2nd in mass and radius, 46.877 hours.

I note that Neptune has the greatest density (1.638 gm cm2), followed by Jupiter (1.326), Uranus (1.27), and Saturn (0.687).

So I am not certain what factors to modify to get a 25 hour orbit outside the habitable edge.

[Added 10-18-2021. I forgot to write: About getting an orbit more than 5 times the radius of the planet while the moon's day is only 25 hours long, the most massive planets are up to about 13 times the mass of jupiter, but have only slightly larger raddii, or even smaller radii than Jupiter, due to becoming more dense instead of wider with increased mass. The more massive a planet is, the farther away a moon would have to orbit to have a 25 hour day. So a planet with several times the mass of Jupiter, but a radius no larger, could have a moon with a 25 hour day beyond the habitable edge.]

The Month Equivlaent

Since the habitable world is a moon, and its day is equal to its orbital period around the planet, there is no obvious choice for a month on the moon. With the Moon tidally locked to the planet, the planet would always be i in the same spot in the sky as seem from any side of the hemisphere facting it. And since the inhabitants of the moon live on the far side in your story, they won't ever the the planet anyway.

A logical choice would be the orbital period of another, outer moon.

Note the complication, that the Moon has two different months as see from Earth. It has a sidereal month and a synodic month.

The period of the Moon's orbit as defined with respect to the celestial sphere of apparently fixed stars (the International Celestial Reference Frame; ICRF) is known as a sidereal month because it is the time it takes the Moon to return to a similar position among the stars (Latin: sidera): 27.321661 days (27 d 7 h 43 min 11.6 s).5 This type of month has been observed among cultures in the Middle East, India, and China in the following way: they divided the sky into 27 or 28 lunar mansions, one for each day of the month, identified by the prominent star(s) in them.

https://en.wikipedia.org/wiki/Lunar_month#Sidereal_month

The synodic month (Greek: συνοδικός, romanized: synodikós, meaning "pertaining to a synod, i.e., a meeting"; in this case, of the Sun and the Moon), also lunation, is the average period of the Moon's orbit with respect to the line joining the Sun and Earth: 29 d 12 h 44 min and 2.9 s. This is the period of the lunar phases, because the Moon's appearance depends on the position of the Moon with respect to the Sun as seen from the Earth.

While the Moon is orbiting the Earth, the Earth is progressing in its orbit around the Sun. After completing a sidereal month, the Moon must move a little further to reach the new position having the same angular distance from the Sun, appearing to move with respect to the stars since the previous month. Therefore, the synodic month takes 2.2 days longer than the sidereal month. Thus, about 13.37 sidereal months, but about 12.37 synodic months, occur in a Gregorian year.

https://en.wikipedia.org/wiki/Lunar_month#Synodic_month

Ancient people noticed the synodic month befor they noticed the sidereal month.

And of course the synodic month of your hypothetical outer moon wuld be the cycle of its phases as seen from the surface of the giant planet. The habitable moon would usually see the outer moon from a different angle than the planet sees it. Thus the outer moon would have a different cycle of phases than its synodic period as seen from the planet.

Thus it may be preferable to use the synodic period of the orbits of the two moons as the basis of the month, ignoring the phases of the outer moon.

The synodic period of two planets orbiitng a star, or two moons orbiting a planet, is the time it takes for the three bodies to return to the same relativ position. It could be the time it takes between two successive oppositions of the outer moon as seen fromthe habitable moon. In astronomy opposition is when an (outer) planet is on the exact opposite side of the sky from the Sun, as seen from Earth.

Mars has an orbital period of 1.88089 Earth years and a synodic period of 774.96 days, Ceres has an orbital period 4.604 Earth years and a synodic period of 466.6 days, Jupiter has an orbital period of 11.86223 Earth years and a synodic period of 398.88 days, Saturn has an orbital period of 29.477 Earth years and a synodic period of 378.09 days, Uranus has an orital period of 84.03 Earth years and a synodic period of 369.66 days, Neptune has an orbital period of 164.793 Earth years and a synodic period of 367.48 days, and Pluto has an orbital period of 247.689Earth years and a synodic period of 366.72 days.

So the closer the orbital period of an outer planet is to Earth's orbital period, the longer the synodic period will be.

In order to make the synodic period of the outer moon as seen from the inner habitable moon as long as possible, the difference between their orbital periods around the planet should be as small as possible. Andif the outer moon's orbit is only a little wider than that of the habitable moon, it will get a lot closer in each opposition than it normally is, thus looking much larger and brighter than it does during most of the synodic period.

And I think that is all I have to write at the moment.

10-14-2021. Continued in another answer:

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    $\begingroup$ Voted +1, it touches some issues with "inhabitable" moons of gas giants, that I find interesting. By the way, " busy with other stuff right now, so this is a short answer." I really wonder what's going to happen to WB when you would have some time off ? $\endgroup$
    – Goodies
    Oct 15 at 10:08
  • $\begingroup$ @Goodies Thnak you for your kind words. The short answer written on Oct. 12 only consists of the part before "Added 10-13-2021". Today I put in the words: "Long Answer" beneath "Added 10-13-2021", so people will be able to see where the short answer ends and the long answer begins. My long answer did get long enough that I had to continue it in a scecond answer. $\endgroup$ Oct 18 at 21:31
  • $\begingroup$ I have edited the answer to correct a couple of links which should work now. $\endgroup$ Oct 18 at 21:56
  • $\begingroup$ @Goodies About getting an orbit more than 5 times the radius of the planet where the day is only 25 hours long, the most massive planets are up to about 13 times the mass of jupiter, but have only slightly larger raddii, or even smaller radii than Jupiter, due to becoming more dense instead of wider with increased mass. The more massive a planet is , the farther away a moon would have to orbit to have a 25 hour day. So a planet with several times the mass of Jupiter, but a radius no larger, could have a moon with a 25 hour day beyond the habitable edge. $\endgroup$ Oct 18 at 22:02
  • $\begingroup$ Excellent and extremely helpful stats. About the 25 orbital period within the habitable range; if the requirement for tidal heating was removed, could that help change any of the calculations? $\endgroup$
    – xeg0
    Oct 19 at 11:17

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