What is the maximum orbital time for my moon around my planet?

Question

I fear the math involved is beyond my capabilities on this one. I have what I consider to be more than a Layman's understanding of the physics involved, and I believe I can follow the math well enough to spot glaring errors, but practical application, combined with the actual crunching of the numbers is more than I can do, in this case.

Here are the known quantities to start with:

The Planet:

1. Gas Giant (in any layman's definition of that term, even if scientific terms might call it something else like ice giants or brown dwarf or gas dwarf, etc)

2. Maximum mass must be small enough that no layman might mistake it for a star. Other than that, its mass can be adjusted as needed, so long as it can plausibly remain a Gas Giant, with the given mass, for 70 million years (not so small it's gas gets blown away by solar wind faster than that).

3. Radius/diameter must be great enough that the planet would appear, at a minimum, at least as big as Earth's moon when viewed from the surface of its moon (1/2 degree angular size), but has no maximum angular size.

4. Density/composition can be anything scientifically plausible as long as the above mass and radius/diameter limits are met, and it could still be called a gas giant.

The moon:

1. Is rocky/metallic (solid surface, not primarily ice, not gas-giant-like, not water or liquid surface. if it matters)

2. The diameter of the moon cannot exceed 6000 KM (radius of 3000 KM), and would preferably be closer to 5000 KM if other parameters can be met without increasing it further.

3. Surface gravity on the moon must be within a range of 75% - 125% of Earth's gravity.

4. Composition/density can be hand-waved, to some extent, to accomplish the gravity requirement within such a small size (I think [correct me if I'm wrong] this is going to be something in the range of a mostly osmium/platinum core, which I know isn't going to be particularly plausible. But on this one point, only, I don't really care as long as it could be made of something 'stable-ish' on our known periodic table, e.g. no neutronium)

5. Distance from the planet is whatever distance yields the greatest orbital time while keeping the planet close enough to appear as big as earth's moon

I'm guessing the answer is going to involve a brown dwarf (for max mass, and therefore max size for the visibility requirement and max gravity allowing a stronger pull from so far away) with a maximum density moon also of maximum size (again to grant a strong enough pull from so far away) as far as possible from each other to remain within the visibility requirement.

However, I can also see how I might be mistaken, as a more massive planet might crush itself smaller under its own weight, making it harder to stay visible from far enough away to extend the orbit time.

How long can I make this orbit take, within those parameters? And how do I accomplish that?

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EDIT: Either my original question wasn't clear, or I seriously underestimated the importance of another factor, to the point that I omitted it completely. So here I'll address both:

First, I suspect some may think I was asking about the time it takes for the planet+moon pair to orbit their star. I'm not. I'm asking, instead, about the time it takes for the moon to orbit it's planet. I assumed that the influence of the star would be negligible for this, so I did not provide details.

Next, if I'm wrong, and the star is that important, then here are the star's requirements:

1. color: Sol-like (a human tourist to this moon might notice the color difference when arriving on the moon, but would adjust and stop noticing after a day or three)

2. Goldilocks zone: distance from the star should be such that stellar radiation should be a significant factor, but not necessarily the only factor (Tidal forces by the planet, a higher radioactive composition, excessive heat from moon formation, etc., can also be factors but should be kept to a minimum wherever possible) in keeping the moon at a survivable temperature for humans if other life support features (atmosphere, gravity, etc) are also present.

3. Stability: Any scientifically plausible type of star that does not vary drastically enough, during a period of 500 million years, to adversely affect any life already on an otherwise habitable planet or moon in its goldilocks zone.

4. Mass, radius, density, composition, distance from the planet, etc.: can all be adjusted as needed, as long as the color and goldilocks requirements are met. But bonus points if its angular size, viewed from the moon, appears the same size or larger than the planet does.

Summary, in layman's terms: I want the moon to take as long as possible to orbit the planet, but the planet and star should appear at least as big, in the sky, as earth's moon and sun. How long can I make that orbit take?

Answer 1

Hill sphere will define the limit of how distant a moon can be from a planet. Its formula is:

$$r_H \approx a_p(1-e)\sqrt[3]{\frac m{3M}}$$

Where $a_p$ is planet's semimajor axis, e is planet's orbit eccentricity, m is planet's mass and M is star's mass.

For a moon's orbital period, formula is:

$$T = 2\pi\sqrt{\frac {a_m^3}\mu}$$

Where T is orbital period, $a_m$ is moon's orbit semimajor axis and $\mu$ is Gm - the standard gravitational parameter

For round orbits (zero eccentricity), Hill's formula becomes

$$r_H \approx a_p\sqrt[3]{\frac m{3M}}$$

Combining two formulas ($r_H$ is $a_m$), we get:

$$T = 2\pi\sqrt{\frac {a_p^3}{3GM}}$$

Let's substitute with Sun and Jupiter values:

$a_p = 7.78 \times 10^{11} m$

$G = 6.674 \times 10^{-11}\frac {m^3}{kg \times s^2}$

$M = 1.989 \times 10^{30} kg$

$$T_{max} = 2.16 \times 10^8 s$$

or about 6.85 years (maximum)

This is the maximum possible orbital period for a Jupiter's moon. Note that Jupiter (or other gas giant's) mass is irrelevant to the final result. Practically stable orbits are found to be within 1/2 to 1/3 radius of Hill's sphere. Assuming $a_m = r_H / 2$:

$$T = 2\pi\sqrt{\frac {a_p^3}{24GM}}$$

so, realistically

$$ T = 7.63 \times 10^7 s$$

or about 2.42 years

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Now let's see about host planet's visible size. The formula for angular diameter is:

$$\delta = 2 arcsin(\frac d{2D})$$

where d is planetary diameter and D is the distance. Substituting:

$d = 1.4 \times 10^8 m$ (Jupiter's diameter)

$D = 2.65 \times 10^{10} m$ ($a_m$, realistic moon's orbit's size)

we get

$$\delta = 0.302$$

Earth's Moon visible size is about 0.5 degrees. Our moon is just a little bit too far! So, angular size becomes a limiting factor. As @Ash mentioned, gas giants are unlikely to get larger than Jupiter without becoming stars.

Let's reverse the formula:

$$ D = \frac{d}{2 sin(\frac{\delta}2)}$$

For $\delta = 0.5$ degrees, this yields

$D = 1.6 \times 10^{10}$, or 16 million km (double @Ash's estimate)

plugging this number to the orbital period formula, we get:

$$T = 3.57 \times 10^7 s$$

or 413 days or 1.13 years

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Next, let's see if our planet/moon fits into a Goldilocks zone

For Sun, reasonable high end estimate (without designing any exotic planetary atmosphere) is about 2.4 AU. Jupiter's orbit is 5.2 AU, which is definitely too far. Our angular requirement had put the moon on 16 million km orbit - compared to the Hill sphere diameter of 53.1 million km. Let's see how close our Jupiter can be to Sun so that moon's orbit would not exceed 1/2 of the radius of Hill sphere, while host planet's visible size stays at 0.5 degrees.

$$a_p = 2 \times a_m \sqrt[3]{\frac {3M}m}$$

Which gives us

$a_{goldilocks} = 4.69 \times 10^{11} m$, or 3.13 AU. Another limiting factor!

Calculating moon's orbital period for host planet orbiting Sun at 2.4 AU gives us

$$T = 2.395 \times 10^7 s$$ or 277 days, or 0.76 years

Answer 2

Short answer:

It seems quite possible for hypothetical habitable exomoons to have days as long as two Earth weeks. Day lengths of several Earth months or years seem to be less plausible.

Long answer:

Alexander's answer is pretty good as far as it goes.

But according to my rough calculations, a planet orbiting at 2.4 AU from the Sun would have a year about 3.7180 Earth years, or 1,358.0228 Earth days, long, and its hypothetical moon could have a month/day no longer than about 150.8914 Earth days long, not the 277 days that Alexander calculates. There is another complicating factor which Alexander did not allow for in his calculations.

There have been a lot of other questions about habitable moons of gas giant planets in the habitable zones of stars, and it is a good idea to refer to those questions and answers to see if they have any useful information, as I state in my answer to this question:

How long will it take to discover they live on a moon and not on a planet? 1

And I gave links to two earlier questions about habitable exomoons.

The article "Exomoon Habitability Constrained by Illumination and Tidal heating" by Rene Heller and Roy Barnes Astrobiology, January 2013, discusses factors affecting the habitability of exomoons.

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3549631/2

And it suggests that the longest possible day for a hypothetical habitable exomoon would be less than, for example, a single Earth year long.

It is assumed that the vast majority of habitable exomoons would be tidally locked to their primaries, rotating at the same rate as they orbited those planets, and thus keeping one side facing the planet at all times and other side facing away at all times. Thus the moon's month, or orbital period around the planet, should be same length as it's day, the time the moon takes to rotate through 360 degrees.

Thus I tend to call it the month/day of the moon, since as the moon orbits and the planet and also rotates it will rotate in relation to the star or sun in the solar system and thus the sun will rise and set and a spot on the surface of the moon will experience a period of daylight and a period of night during the moon's orbital period around the planet.

On moons, however, tides from the star are mostly negligible compared to the tidal drag from the planet. Thus, in most cases exomoons will be tidally locked to their host planet rather than to the star (Dole, 1964; Gonzalez, 2005; Henning et al., 2009; Kaltenegger, 2010; Kipping et al., 2010) so that (i.) a satellite's rotation period will equal its orbital period about the planet, (ii.) a moon will orbit the planet in its equatorial plane (due to the Kozai mechanism and tidal evolution, Porter and Grundy, 2011), and (iii.) a moon's rotation axis will be perpendicular to its orbit about the planet. A combination of (ii.) and (iii.) will cause the satellite to have the same obliquity with respect to the circumstellar orbit as the planet.

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3549631/2

The longer the month/day is, the hotter the day side will get, and the colder the night side will get. The longest possible day or night for a planet would be if that planet was tidally locked to its sun, and thus had eternal day on the near side and eternal night on the far side of the planet.

There is a fear that a planet tidally locked to its sun would lose its atmosphere and water because the hot air and water vapor from the day side would flow to the night side and condense and freeze until everything was frozen on the night side.

If that was the case, a planet that had a sufficiently long day period would have almost all of its water and atmosphere frozen on the night side during the long night. Only the water and atmosphere that was melted and sublimated at dawn would exist as a thin atmosphere that sublimated at the same rate as it froze out.

On the other hand, it is possible that the circulation of air and water between the light and the dark sides will transfer enough heat to the dark side to keep the air and water unfrozen.

This pessimism has been tempered by research. Studies by Robert Haberle and Manoj Joshi of NASA's Ames Research Center in California have shown that a planet's atmosphere (assuming it included greenhouse gases CO2 and H2O) need only be 100 mbs, or 10% of Earth's atmosphere, for the star's heat to be effectively carried to the night side.[74] This is well within the levels required for photosynthesis, though water would still remain frozen on the dark side in some of their models. Martin Heath of Greenwich Community College, has shown that seawater, too, could be effectively circulated without freezing solid if the ocean basins were deep enough to allow free flow beneath the night side's ice cap. Further research—including a consideration of the amount of photosynthetically active radiation—suggested that tidally locked planets in red dwarf systems might at least be habitable for higher plants.[75]

https://en.wikipedia.org/wiki/Planetary_habitability#Other_factors_limiting_habitability3

So at the present time it seems possible that even a tidally locked planet could be habitable, and thus there doesn't seem to be any known limit based on freezing out the atmosphere to how long the day and night of a habitable exomoon could last, which is good for your desire to have it as long as possible.

According to "Exomoon Habitability Constrained by Illumination and Tidal heating"

The synchronized rotation periods of putative Earth-mass 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. The longest possible length of a satellite's day compatible with Hill stability has been shown to be about Pp/9, Pp being the planet's orbital period about the star (Kipping, 2009a)

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3549631/2

So they estimate that a habitable exomoon might have a month/day as long as maybe 17.0 Earth days. But what is really important is:

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

The source, Kipping 2009a, seems to be:

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

https://arxiv.org/abs/0810.22434

According to Alexander's answer, if the exomoon was orbiting a planet orbiting a star as massive as Sol (the sun) at a distance of 2.4 AU, assumed to be the other limit of the habitable zone, the habitable moon could have a month/day of 277 Earth days or 0.76 Earth years.

If the year of the planet has to be at least nine month/days of the moon long in order for the moon to have a stable orbit, the year of the planet would have to be at least 2,493 Earth days, or 6.825462 Earth years.

There are many different scientific estimates of the habitable zone of the Sun, or of a star that is exactly like the Sun. Some estimates give the some a very narrow habitable zone and other estimates give it a very broad habitable zone.

Since you are interested in the longest possible month/day of your moon, and thus the longest possible year for the planet orbiting it's star, lets calculate it for various outer edges of the Sun's habitable zone.

If the hypothetical moon's planet orbits a star exactly like the Sun at a distance of exactly one AU, the planet will have a year exactly one Earth year long, and the longest possible length of a month/day of a habitable moon of that planet would be one ninth of an Earth year, or about 40.5833 Earth days.

According to this paper:

Hart, M. H. (1979). "Habitable zones about main sequence stars". Icarus. 37: 351–357. Bibcode:1979Icar...37..351H. doi:10.1016/0019-1035(79)90141-6.

The outer edge of the Sun's habitable zone is only 1.01 AU from the Sun. According to my rough calculations, a planet orbiting at that distance would have a year about 1.01503 Earth years, or 370.7424 Earth days long, and its hypothetical moon could have a month/day no longer than about 41.1196 Earth days long.

According to this article:

Vladilo, Giovanni; Murante, Giuseppe; Silva, Laura; Provenzale, Antonello; Ferri, Gaia; Ragazzini, Gregorio (March 2013). "The habitable zone of Earth-like planets with different levels of atmospheric pressure". The Astrophysical Journal. 767 (1): 65–?. arXiv:1302.4566. Bibcode:2013ApJ...767...65V. doi:10.1088/0004-637X/767/1/65.

The outer edge of the Sun's habitable zone is only 1.18 AU from the Sun. According to my rough calculations, a planet orbiting at that distance would have a year about 1.2818 Earth years, or 468.1803 Earth days long, and its hypothetical moon could have a month/day no longer than about 52.0200 Earth days long.

According to this article:

Kasting, James F.; Whitmire, Daniel P.; Reynolds, Ray T. (January 1993). "Habitable Zones around Main Sequence Stars". Icarus. 101 (1): 108–118.

The outer edge of the Sun's habitable zone is 1.37 AU from the Sun. According to my rough calculations, a planet orbiting at that distance would have a year about 1.6035 Earth years, or 585.6943 Earth days long, and its hypothetical moon could have a month/day no longer than about 65.0771 Earth days long.

According to this article:

Kopparapu, Ravi Kumar (2013). "A revised estimate of the occurrence rate of terrestrial planets in the habitable zones around kepler m-dwarfs". The Astrophysical Journal Letters. 767 (1): L8. arXiv:1303.2649. Bibcode:2013ApJ...767L...8K. doi:10.1088/2041-8205/767/1/L8.

The outer edge of the Sun's habitable zone is 1.68 AU from the Sun. According to my rough calculations, a planet orbiting at that distance would have a year about 2.1775 Earth years, or 795.3423 Earth days, long, and its hypothetical moon could have a month/day no longer than about 88.3713 Earth days long.

According to this article:

Spiegel, D. S.; Raymond, S. N.; Dressing, C. D.; Scharf, C. A.; Mitchell, J. L. (2010). "Generalized Milankovitch Cycles and Long-Term Climatic Habitability". The Astrophysical Journal. 721 (2): 1308–1318. arXiv:1002.4877. Bibcode:2010ApJ...721.1308S. doi:10.1088/0004-637X/721/2/1308.

http://iopscience.iop.org/article/10.1088/0004-637X/721/2/1308/meta5

The outer edge of the Sun's habitable zone is 2.00 AU from the Sun. According to my rough calculations, a planet orbiting at that distance would have a year about 2.8284 Earth years, or 1,033.0829 Earth days, long, and its hypothetical moon could have a month/day no longer than about 114.7869 Earth days long.

According to this article:

Ramirez, Ramses; Kaltenegger, Lisa (2017). "A Volcanic Hydrogen Habitable Zone". The Astrophysical Journal Letters. 837: L4. arXiv:1702.08618 [astro-ph.EP]. Bibcode:2017ApJ...837L...4R. doi:10.3847/2041-

http://adsabs.harvard.edu/abs/2017ApJ...837L...4R6

The outer edge of the Sun's habitable zone is 2.4 AU from the Sun. According to my rough calculations, a planet orbiting at that distance would have a year about 3.7180 Earth years, or 1,358.0228 Earth days, long, and its hypothetical moon could have a month/day no longer than about 150.8914 Earth days long. That is the same distance from the Sun that Alexander used to calculate a month/day of 277 Earth days.

However, this seems to involve atmospheric hydrogen concentrations of 1 % to 50 %, which do not seem compatible with an oxygen rich atmospheres suitable for humans.

According to this article:

Fogg, M. J. (1992). "An Estimate of the Prevalence of Biocompatible and Habitable Planets". Journal of the British Interplanetary Society. 45 (1): 3–12. Bibcode:1992JBIS...45....3F. PMID 11539465.

The outer edge of the Sun's habitable zone is 3.00 AU from the Sun. According to my rough calculations, a planet orbiting at that distance would have a year about 5.1961 Earth years, or 1,897.8946 Earth days, long, and its hypothetical moon could have a month/day no longer than about 210.8771 Earth days long.

According to this article:

Pierrehumbert, Raymond; Gaidos, Eric (2011). "Hydrogen Greenhouse Planets Beyond the Habitable Zone". The Astrophysical Journal Letters. 734: L13. arXiv:1105.0021 [astro-ph.EP]. Bibcode:2011ApJ...734L..13P. doi:10.1088/2041-8205/734/1/L13. Cite uses deprecated parameter |class= (help)

http://adsabs.harvard.edu/abs/2011ApJ...734L..13P7

The outer edge of the Sun's habitable zone is 10 AU from the Sun. According to my rough calculations, a planet orbiting at that distance would have a year about 31.6227 Earth years, or 11,550.218 Earth days, long, and its hypothetical moon could have a month/day no longer than about 1,283.3575 Earth days long.

But this last calculation involves planets with significant amounts of hydrogen in their atmospheres, equal to or greater than Earth's total atmospheric pressure, which would not be consistent with a breathable oxygen rich atmosphere for humans.

https://en.wikipedia.org/wiki/Circumstellar_habitable_zone8

Another way to change the possible length of the year of the planet and thus of the month/day of the moon, is to change the mass and thus the luminosity of the star in the system.

A relatively small change in the mass of the star can produce a much larger change in the luminosity, and thus in the distance of the habitable zone, and thus in the length of the years of planets in the habitable zone, and thus in the maximum possible length of the month/days of moons orbiting those planets.

And a relatively small change in the mass of the star can produce a much larger change in the rate at which it uses up is nuclear fuel and thus the time it depends on the main sequence stage of its life before becoming a red giant star and then a white dwarf star.

And if you want your hypothetical moon to have multi celled lifeforms, or an oxygen rich atmosphere breathable for humans, or intelligent natives, or most of the other things which are usually needed to make a world interesting in science fiction, you will want it to be billions of years old and thus you will need the moon's star to be of a spectral type capable of remaining on the main sequence for several billion years.

Answer 3

All the worlds actually plotted (that I've seen the stats for anyway and I looked pretty hard at this a couple of months ago) that are more massive than Jupiter, including small Brown Dwarf Stars, have been found to have a higher average density and thus a smaller radius. So Jupiter would appear to be as large as Gas Giants actually get in nature, purely in terms of radius.

Taking that as a base the orbital distance, for a 1/2 degree angular size, is pretty much 8 million Kilometres, well within Jupiter's Hill Sphere (for an approximately Sol mass star) of 53 million Kilometres. The Earth, or something substantially similar, would orbit it in 145 days and 19 hours.

Here are the tools I used to get the answers:

Jovian Statistics
Angular Size Calculator
Orbit Calculator

All stars have a goldilocks zone so all you need is a yellow star, something with a G spectral classification.