First of all, this website is a dream come true. I'm a total sucker for this stuff.

Secondly, I love astronomy and worldbuilding, and lately, I've been thinking of a hypothetical situation, and was wondering if such a thing was actually possible.

A planet orbits a gas giant. Let's say (for reference's sake) the gas giant is Jupiter, and the planet is Earth. Earth is tidally locked to the gas giant, and has a rotational period of exactly one year. However, let's say it also orbits around Jupiter in exactly one year as well, which means that the planet is cooked on its surface for a while and then vanishes into total freezing darkness behind Jupiter every 6 months, give or take.

This itself is doable. However, would it be possible if Earth was still located rather close to Jupiter, in a way for Jupiter to still dominate the sky (yes, I know that's not Jupiter)?

Of course, it depends on where you're standing on hypothetical Earth to see hypothetical Jupiter, but in this case, we're standing in the right place to expect a giant planet to cover most of our sky. But gas giants' moons orbit around them incredibly fast because they're so close and they have such a strong gravitational pull, which creates a problem, because the ideal orbit here is a slow trip around.

Would other factors be able to slow down the planet's course around this imaginary gas giant and create this situation, such as other moons or planets, size of the planets in question, density of either planey, rings around the gas giant, etc., without the planet being pulled into this gas giant?

We can ignore the more technical things, like the radiation belt around gas giants and whatnot. The most basic question is if an orbit such as this is feasible. I'm asking for a world that's not Jupiter or Earth, so anything goes if it makes it possible- as long as a big gas giant has a moon.

I've ran this through my head a lot, searched for answers, and even bought and messed around in Universe Simulator 2 a little (but it's more fun to blow up planets), and still haven't been able to come up with much of a solution. I'd hope that with the right blend of factors, this would be possible, but I'm not too sure on how this could play out.

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    $\begingroup$ If it orbits gas giant, it's a moon, not a planet. $\endgroup$
    – Mołot
    Apr 24, 2017 at 4:54
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    $\begingroup$ kingledion has a point with "A 1 year long orbit around Jupiter does not work", but technically, the orbital period certainly could be said to be the year. So, are you asking how to make the orbital period of the moon in question an Earth year (365 days, give or take depending on exactly which definition you are using)? Please try to avoid using Earth-centric terms when discussing other orbiting bodies without at least specifying them as such. Note that this is a real-world problem for Mars missions; it has its own Wikipedia article! $\endgroup$
    – user
    Apr 24, 2017 at 5:53
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    $\begingroup$ Definitely related: Habitable moon of a gas giant: working out the sizes and distances $\endgroup$
    – user
    Apr 24, 2017 at 11:19
  • $\begingroup$ This question from 2 weeks ago also features «tidally locked to the gas giant, with an orbiting period of nearly 24h in order to have an Earth-like day/night cycle.» which is also postulating orbit periods. $\endgroup$
    – JDługosz
    Apr 25, 2017 at 6:14
  • $\begingroup$ @Mołot While true, it could still be a habitable moon with Earth-like features. I'm sure if life arose there they'd call the-thing-they-live-on a planet before finding out otherwise. Same reason the Sun was differentiated from the stars before we learned better. $\endgroup$ Apr 25, 2017 at 14:49

5 Answers 5


A 1 year long orbit around Jupiter does not work

Orbital mechanics is actually pretty straightforward mathematically. There are rigidly defined formulas controlling what can and cannot happen.

The formula for distance of the less massive body (Earth) from the more massive body (Jupiter) is given by

$$a = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3}$$

where $GM$ is the standard gravitational parameter of Jupiter ($1.27\times10^{17}\text{ m}^3\text{s}^{-2}$); and T is the desired orbital period (1 year = $3.15\times10^{7}\text{ s}$). Plug those numbers in and we get about 15 million km.

First off, this is not going to work for your desired 'closeness' to Jupiter. Jupiter has a radius of about 70 000 km. Using simple trigonometry, an object that is 140 000 km across at a distance of 15 million km occupies

$$\arctan\left(\frac{140 000}{15000000}\right) = 0.00933 \text{ radians}$$

of arc, equal to about 32 minutes of arc. By comparison, the moon is from 29 to 34 minutes of arc; Jupiter will appear about the same size in the sky than our moon appears to us, in this situation.

Secondly, the existing Galilean moons range from 0.42 million km (Io) to 1.89 million km (Callisto). Large moons do not exist that far from planets, at least not in our solar system. I can offer you this chart for the most distant moon from one of our Gas Giants with a mass with order of magnitude X.

X  Moon                Distance 
18 Sycorax (Uranus)  12 179 000 km
19 Nereid (Neptune)   5 513 818 km
20 Iapetus (Saturn)   3 560 820 km
21 Iapetus (Saturn)   3 560 820 km
22 Callisto (Jupiter) 1 882 709 km
23 Callisto (Jupiter) 1 882 709 km

As you can see, you just don't get large moons that far out. Jupiter's largest moon at 15 million km or greater is about 60 km across. The moral of this story is that it is likely an object as massive as the Earth (50 times again more massive than any Moon) would not stay stable in an orbit that far from a gas giant in a busy solar system.


You can use that orbital period equation (and others in the Wikipedia link) to help determine how to make your planets orbital characteristics more like what you want.

I would advise trying to make the Gas giant even bigger, there are planets out there 10 times the size of Jupiter. Just be sure to keep the mass of the gas giant under ~0.08 solar masses, which is the point at which the giant could ignite into a star itself (Jupiter itself is just under 0.001 solar masses).

EDIT - As @Tradeylouish points out in the comments, even if you made the 'Jupiter' bigger, the increase in mass would cause the distance required for an object to be in a 1 year orbit to increase proportionally; the result would be that the 'Jupiter' would stay approximately the same size in the sky.

I was suggesting that the 'Jupiter' be made bigger to help clear the space around it, allowing it to hold onto a satellite at such a long distance. However, this will not help your planet appear massive in the sky.

@Tradeylouish's suggestion about density is the way to go if you want the planet to be huge in the sky (though it won't necessarily help keep the satellite in a far away orbit). Density of gas giants a can be pretty low apparently; check out TrES-4b (no relation) which has about the mass of Jupiter but a density of 200 kg/m$^2$...about the same as balsa wood.

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    $\begingroup$ What is Callisto doing twice in your list of moons? $\endgroup$
    – user
    Apr 24, 2017 at 5:55
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    $\begingroup$ I believe your angular diameter calculation is off by a factor of 2, as it only covers one half of Jupiter. Also, making the gas giant bigger gives it more mass, increasing its gravitational pull and hence increasing the radius at which the moon must orbit in order to maintain a 1-year period. Assuming that the central body maintains a constant density when made larger, it should actually appear almost exactly the same size in the moon's sky. For the planet to appear larger, its density must be decreased so that the moon orbits closer given some planet size. $\endgroup$ Apr 24, 2017 at 10:30
  • $\begingroup$ @MichaelKjörling Callisto is both the farthest moon with mass > 1e22 kg and the farthest moon with mass > 1e23 kg. $\endgroup$
    – kingledion
    Apr 24, 2017 at 12:21
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    $\begingroup$ @Tradeylouish Thanks for the comments, I've made some corrections and edits. $\endgroup$
    – kingledion
    Apr 24, 2017 at 12:34
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    $\begingroup$ equal to about 32 minutes of arc. By comparison, the moon is from 29 to 34 minutes of arc; Jupiter will appear about the same size in the sky than our moon appears to us, in this situation. Neat! I haven't seen anyone do that math before (all the artistic representations just replace the moon with the other object, keeping it at the same distance, giving a middle finger to gravity). $\endgroup$ Apr 25, 2017 at 14:47


An orbit is elliptic (instead of linear as Newton's First Law predicts) because the planet is attracting the orbitter1.

So, we have only the force of the planet applied to the body that orbits around it. Why doesn't that object just fall?

Because the pull of the planet2 provides an acceleration; that is, a change in the speed of the body. Instead of flying straight, that acceleration makes it change the direction of its speed, curving its path.

Now, we have 3 options:

  • The body moves so fast that the pull of the planet does not change its speed enough to keep into the current orbit: the body switches to a higher orbit or simply escapes the planet.

  • The body moves so slow that the pull of the planet changes its speed so much that it cannot keep the orbit; the body orbits are each time closer to the planet.

These two points above work as follow: if a body moves apart from the planet it orbits, it loses energy (that goes to compensate the gravitational energy) and slows down. And, at the same time, a wider orbit means that, even if the gravitational pull is slower, it keeps affecting the orbitting body for more time (as the revolution time increases). When the body moves towards the planet it is just the opposite, it accelerates until it finds a lower orbit that fits its new speed (or it crashes).

  • The body moves reaches an equilibrium speed. The pull of the planet changes its speed enough to keep it in orbit, but without changing it. In a perfect, circular orbit, that could be describe as the body always having the same magnitude of speed (the same km/s) while the direction of the speed is continuously changing by the same amount, enough to follow the orbit.

So, it is not as if you can slow the orbitting body somehow. The issue is that, as soon as you slow it, by whatever the means, it begins falling towards the planet, because the only thing that keeps it in orbit is its current speed3.

TL;DR The factors that determine a orbit are the speed of the orbitting body and the mass of the body around which it orbits. The only factor that could be changed would be the gas giant mass, but then you have the issue of how making a non-massive-gas-giant (hint, gas has a tendency to run away unless there is lot of attraction from its planet).

1 So Newton's First Law is not of application there. Did you thought I was going to say that Newton's First Law does not work?

2Or of any force.

3Technically an orbitting body is always falling -the free fall term means that- towards the body they orbit, but constantly missing it.

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    $\begingroup$ While this answer explains the orbital mechanics of a body orbiting a planet, it doesn't answer the question posed. Namely, can an Earth mass moon orbit a gas giant planet in one year (the 365 day year)? Your answer can be improved by dealing with that matter. $\endgroup$
    – a4android
    Apr 24, 2017 at 11:00

I found a useful article "Exomoon Habitability Constrained by Illumination and Tidal Heating" (Kipping, 2009a).

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. So, if the moon somehow managed to orbit around a giant planet with a period of one Earth year, the giant planet's orbital period around their sun would have to be at least nine earth years.

If the moon receives as much radiation from its star as the Earth does from the Sun, and orbits that sun with a period of at least nine years, its star would probably have to be so massive and luminous that it would not remain on the main sequence long enough for the moon to become habitable for humans, develop multi celled lifeforms, or have a native intelligent species. Unless super advanced aliens terraformed the moon and made it habitable and gave it advanced life forms.

A very dim star would have its habitable zone very close to it and its tidal forces would make any planet orbiting in the habitable zone tidally locked with one side always facing toward its star in eternal day and the other side always facing away from its star in eternal night.

In our solar system astronomers believed that Mercury was tidally locked by the Sun, with one side in eternal day and hellish heat, and the other side freezing in eternal night and cold. But in 1964 it was discovered that Mercury is tidally locked, but not as strongly as a 1:1 resonance. Mercury has a 3:2 resonance. The orbital period or year of Mercury is 87.969 Earth days. The sidereal day or rotational period of Mercury relative to the stars is 58.646 Earth days. Thus there are three Mercurian sidereal days in two Mercurian years. But a solar day, the time between two successive sunrises or sunsets on a spot on the surface of Mercury, is two Mercurian years long, or about 175.938 Earth days.

Some astronomers thought that Venus might also be tidally locked in a 1:1 resonance, and some old science fiction stories were set on such a Venus. That is not the case but Venus does have a odd relation between its year length and day length. The orbital period or year of Venus is 224.701 Earth days. The sidereal day or rotational period of Venus is 243.025 Earth Days, longer than the year. All planets in our solar orbit the Sun in a counter-clockwise direction as seen from above the Earth's North pole. Most planets also rotate in a counter-clockwise or prograde direction.

If Venus did that its solar day, the time between two successive sunrises at the same spot on its surface, would be several Venus years long and thus more than an Earth year long. But Venus rotates in the opposite direction, clockwise as seen from above Earth's north pole, or retrograde. This makes the length of a solar day on Venus "only" 116.75 Earth days, less than that of Mercury. Nobody knows what gave Venus its long sidereal day and retrograde rotation. A common theory is a giant impact billions of years ago.

If the long day and retrograde rotation of Venus have the same cause, it should be rare for a planet to have a long sidereal day like Venus without also having the retrograde rotation, which would make the solar day shorter than the sidereal day. But if the long day and retrograde rotation of Venus have two different and independent causes, it should be much more common for a planet to have a long sidereal day like Venus without also having the retrograde rotation, and thus solar days as long as an Earth year would be far more common.

From what I have heard, a planet with days and nights much more than a few Earth days long would suffer from extremes of heat and cold during the days and nights. Lifeforms could flourish only during comparatively short periods near sunrise and sunset. They would have to go into some sort of suspended animation or die and leave protected seeds and eggs, twice each year-long day.

Added 04-25-2017. Or lifeforms could move with the sunrise and sunset. With an equatorial circumference of about 25,000 miles and a day about 365.25 Earth days long, animals would have to move at an average speed of 68.446 miles per day or 2.851 miles per hour at the equator. At higher latitudes where the planet's circumference was much less, they could move slower. Where the circumference was only 2,500 miles they would need an average speed of 6.8446 miles per day or 0.2851 miles per hour.

  • $\begingroup$ I don't think you are using the term resonance correctly. I understand two planets to be in resonance when they have a certain ratio of orbital periods. You appear to be using it to refer to the ratio of day to year length. Also, the whole last part of your post doesn't seem to be relevant (the parts about Mercury and Venus). The first 4 paragraphs seems like a good and efficient answer. $\endgroup$
    – kingledion
    Apr 25, 2017 at 15:09

Diamagnetic moon.

The problem with this problem is gravity. The orbital math posted above by @kingledion is inflexible. To get a satellite as close as you want and as slow as you want the net attractive force needs to be less. A lighter satellite cannot do it - as Galileo demonstrated dropping the balls from the leaning tower of Pisa. The attractive force is M1*M2 and when M1 is immense M2 does not matter much. You could make Jupiter lighter but what fun is that?

What is required is a force which could oppose gravity such that the net attractive force on the moon was lower. I can think of 2 which might work: electrical repulsion and diamagnetism. Diamagnetism is magnetic repulsion: the effect that allows certain nonparamagnetic items (like frogs) to be levitated in a strong magnetic field. Achieving levitation means the force of gravity is completely opposed.

Jupiter is a good candidate for this because it has a very strong magnetic field. If one starts by accepting diamagnetic repulsion can be strong enough to oppose gravitational attraction on this scale, then for a diamagnetic satellite one could assert that the magnetic repulsion opposed gravity to whatever degree desired. An arbitrarily weak or strong attractive net force would allow your satellite to orbit at whatever distance you like.

This assumes that the magnetic field around Jupiter is uniform but if it is anything like Earth's it is not. I could imagine a satellite which sped up and slowed down / moved higher and lower as it traversed the irregularities of the magnetic field and the net attractive force waxed and waned.

  • $\begingroup$ This is a cool idea, but I had to reality-check your reality-check. Assuming Jupiter and Earth with 2million km between them, I calculate the force of gravity at 1.9e23 N. To get 10% of this force with Lorentz force in a magnetic field (given orbital speed of 10 km/s), you need $q\hat{v}\times B$ (where $\hat{v}$ is the unit vector of velocity) to equal 1.9e19 Coulomb*Amp/meter. First, I don't see any mechanism to get a planet charged that way, and second I don't see magnetic fields getting to that order of magnitude (maybe a pulsar?). $\endgroup$
    – kingledion
    Apr 25, 2017 at 15:07
  • $\begingroup$ When a person busts out the Coulombs I know I am licked. Hopefully someone handy with the physics will sally in and rescue this concept. $\endgroup$
    – Willk
    Apr 25, 2017 at 20:38
  • $\begingroup$ I am struggling along. I do not think diamagnetism is the same thing you are talking about. You are talking about an induced magnetic field related to charge. Diamagnetism is an intrinsic property: the levitating frog is not charged nor magnetic. The idea is for the diamagnetic moon to be magnetically levitated on Jupiters immense magnetic field. $\endgroup$
    – Willk
    Apr 25, 2017 at 21:21

It might be possible to replace Jupiter with more of a cloud, rather than legitimate gas planet. Some sort of star maybe with very low density. I'm not sure if such objects exist, but it might provide the large object in the sky you're looking for.

As another answer gave, your distance will have to be 15 million km if your masses are the same (there's no real reason to change the masses). It's hard to tell what you mean by "dominate the sky", but let's go with 5 degrees (0.08727 radians).

So, in order for your planet to be 5 degrees large at a distance of 15 million km, you need a diameter of roughly tan(0.08727)*15,000,000 = 1,312,000 km. With a mass of 1.898e27kg, your density is roughly 6.7 kg/m^3. In other words, this would not be considered a planet, and I'm not sure if such a cloud could form without it forming together. Possibly your planet has just formed out of the the wreckage of some star.

  • $\begingroup$ This doesn't really provide an answer to the question. The post is tagged with science-based. Could you add more information on how such a cloud-like Jupiter would work, and what distances would be involved in the orbit, etc. $\endgroup$
    – kingledion
    Apr 25, 2017 at 14:59

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