Astronomers place planet Earth under what they call a "Goldilocks Zone"--a spot in the solar system where conditions are ideal for liquid water to form, thus making the creation of life possible.

In Norse mythology, there are nine separate worlds--Midgard, Asgard, Vanaheimr, Jotunheimr, Alfheimr, Hel, Nidavellir, Niflheim and Museplheim--all of which are inhabited by creatures of one form or another.

If all this were to be realistically put on a linear plane like our solar system, what kinds of astronomical conditions are required for all nine of them to be placed in a Goldilocks Zone?

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    $\begingroup$ @nigel222 how about lagrangian points? wouldn't that help with stability? $\endgroup$
    – njzk2
    Oct 5, 2015 at 18:10
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    $\begingroup$ Does it have to be nine different "Earths"? Canonically the realms have vastly different terrain and climate - Niflheim could be a frozen planet ala Neptune, while Muspelheim could be a hot plant like Mercury or Venus. $\endgroup$
    – Nathan
    Oct 5, 2015 at 23:01
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    $\begingroup$ For what it's worth: I tried this configuration with Universe sandbox and it was perfectly stable. 9 earths aligned on the same plane between 130m and 180 million km from the Sun. The planets have an average temperature between 40 and -5 Celsius. $\endgroup$
    – Vincent
    Oct 5, 2015 at 23:10
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    $\begingroup$ @JohnWDailey universesandbox.com $\endgroup$
    – barbecue
    Oct 6, 2015 at 1:03
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    $\begingroup$ @NateKerkhofs : was it confirmed that the 9 words in Norse mythology are 9 separate planets? They might be 9 continents on the same planet, or 9 different states of existence, "alternate universes", whatever. $\endgroup$
    – vsz
    Oct 6, 2015 at 6:08

12 Answers 12


You can have 36 habitable planets/moons as follows (all shamelessly taken from here):

  • We can fit six stable orbits into the habitable zone. Each orbit has two sets of binary Earths. These are Earth-sized planets with Earth-sized moons. Each binary planet is in a Trojan (co-orbital) configuration with another binary planet, separated by 60 degrees on their orbit around the star.
  • Now let’s include gas giant planets. We can fit the orbits of four gas giants in the habitable zone (in 3:2 resonances). Each of those can have up to five potentially habitable moons. Plus, the orbit of each gas giant can also fit an Earth-sized planet both 60 degrees in front and 60 degrees behind the giant planet’s orbit (on Trojan orbits).
  • Let’s add it up. One gas giant per orbit. Five large moons per gas giant. Plus, two binary Earths per orbit. That makes 9 habitable worlds per orbit. We have four orbits in the habitable zone. That makes 36 habitable worlds in this system!

enter image description here

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    $\begingroup$ I get the sense that even a small comet could destabilize the system. There's no way this is stable even on a million year time-scale. The gas giants would destabilize the Trojans and send each other spiralling into the star! $\endgroup$ Oct 5, 2015 at 19:48
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    $\begingroup$ The trojan points are stable, its the gas giants mass that makes them so. Note that in order to accommodate co-orbitals, the main orbits have had to be spaced out more than would otherwise be needed for orbital stability. Anyway, the OP only need 9 earths, so can take his pick from these 36 options, and no doubt reduce the systems mean time before failure to something acceptable. $\endgroup$
    – rumguff
    Oct 5, 2015 at 20:21
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    $\begingroup$ Well, they may be moderately stable with regards to their own orbiting giant, but the giants in neighboring orbits would quickly pull them out of the narrow Trojan orbit (remember, we're in the habitable zone, so orbits are packed close). $\endgroup$ Oct 5, 2015 at 20:30
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    $\begingroup$ If I lived in a place like that, I would be worried about sneezing :-p $\endgroup$
    – SJuan76
    Oct 5, 2015 at 20:35
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    $\begingroup$ No m'lud, I can't prove it nor did I have any intention of doing so, especially since it ain't my own work (but also because I wouldn't know where to start). I hoped my admission of plagarism would spare me from such strictures. My defence shall be simply that its really cool. $\endgroup$
    – rumguff
    Oct 5, 2015 at 22:22

Well, if you think about it, there are 2 main things that make a planet habitable, in the sense of being able to sustain a liquid ocean of water:

1) Distance from the system's star.
2) Density of greenhouse gases in the atmosphere

enter image description here

For instance, we normally think of Earth as solidly in the habitable zone, but the average temperature of the moon is around -20 C. Without greenhouse effects, Earth's oceans would be frozen. You can see how you can extrapolate from here.

Nine might be a bit on the high side, but in theory, if Jupiter formed further out (so Mars could feed on a bit more material), and Venus lost most of its carbon in some neat collision and so got to hold on to its water instead of being a greenhouse hell, there's no reason why there shouldn't be 3 Earth-like planets in our system instead of one.

So imagine a series of planets, spaced just far apart that they don't interfere with each other gravitationally overmuch, where the density of greenhouse gases increases with the distance from the star. you might need to also tweak the sizes a bit (so they won't all be identical), and as I said, 9 might be a bit of stretch, but in theory I see no reason why it should not be possible.

There's a lot more to it (near-solar planets tend to lose their volatiles on formation, while places past the ice-line tend to accumulate hydrogen and turn into giants; greenhouse gases tend to have a limited life-span with CH4 getting blasted into pieces and CO2 being weathered away and recirculated via mantle convection, which needs moving tectonic plates, which only Earth has, possibly because of its oceans, etc -- see what I mean, it gets really complicated fast), but let's not get into it now, shall we?


Use Complex Orbits

While in our solar system, planets are generally alone in their orbit, or perhaps orbited by one or more smaller bodies, this need not be the case. Specifically, you can easily have two planets that are about the same size forming a binary pair, which orbit around their shared center of gravity the same way the Earth and the Moon do. This binary pair could then be orbited a bit further away by a third planet, so long as that planet was close enough to the binary pair to fall within its gravity well and be pulled away by the gravity of the star. While uncommon in planets, this sort of system is very common in stars, with star systems having been found with up to seven stars. A glob of seven planets orbiting one another in a way that is stable with regards to the gravity of the star they all orbit is highly unlikely, but three should be feasible.

enter image description here
HD 98800, a system comprised of two pairs of binary stars. Planets could orbit one another the same way.

Within the orbit of the Earth, if you placed one trinary planet system where Earth is and one at the L4 Lagrange point, you would already have six planets within the orbit of the Earth. If we can fit two orbits in the Goldilocks zone of a star, that would give us space for as many as twelve planets.

  • $\begingroup$ Would they still be habitable, though? $\endgroup$
    – HDE 226868
    Oct 5, 2015 at 22:47
  • $\begingroup$ Provided they're all in the habitable zone, yes. Tides, however, will be intense and complex in something like a trinary star system. $\endgroup$
    – ckersch
    Oct 6, 2015 at 2:01
  • $\begingroup$ I'm just concerned about fitting all these orbits into a habitable zone. $\endgroup$
    – HDE 226868
    Oct 6, 2015 at 2:54
  • $\begingroup$ Although the L4 point is stable for small objects, I think something large enough to count as a habitable planet would preclude that stability. $\endgroup$ Oct 7, 2015 at 8:26

It's realistically possible. But liquid water alone doesn't mean earthlike life ... Presented below is a diagram of a planetary system containing 4 planets. The central star is roughly the same size as our sun.

star with four closely-packed orbits indicated

Planet 1 (innermost). Distance from sun is 45 Mmi (million miles) which is slightly more than Mercury's distance from our sun. Atmosphere is thin, consisting mainly of nitrogen (70%) and oxygen (30%). Greenhouse gases do not exist at all. Planet is mostly water (80%). Land is present as a series of islands. Axial tilt is less than 10°. Rotation speed is high and a day-night cycle finishes in 13 hours. This planet can have liquid water on its surface due to large water body that regulates the temperature. The core is dead so that there are no live volcanoes. Also, since the planet revolves so fast, sunlight does not fall at the same location so long to make it scorched.

Planet 2. Distance from sun is 70 Mmi. Atmosphere is slightly thinner than Earth's, consisting mainly of nitrogen and oxygen. Greenhouses gases are ~4% of total atmospheric weight. Axial tilt is <10° and a day-night cycle is 17 hours. Surface is 75% water and 25% land. This planet is basically a Venus with far less dense atmosphere and a large water content.

Planet 3. Earth. At 95 Mmi from parent star.

Planet 4. Distance from sun 115 Mmi. Atmosphere contains 10% greenhouse gases. Day-night cycle completes in 30 hours. Surface is 60% water and 40% land. This planet can have liquid water due to its high greenhouse gases which absorb sun's energy well. 15 hours of daylight assure that enough sunlight falls on a single place to warm it up.

Planet 5. Distance from sun is 135 Mmi. Atmosphere contains 20% greenhouse gases by weight. Surface water content is 45%. Day-night cycle lasts 40 hours. Crust contains a high graphite (carbon) content. This planet can also have liquid water due to high greenhouse phenomenon that traps most of the heat it receives. High graphite content assures that the crush, too, absorbs heat well.

Beyond that distance the planets will have to have a very active core, really dense atmosphere with increasingly high greenhouse gases ratio and longer day-night cycles (very unlikely) to have any hope of having liquid water on them.

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    $\begingroup$ I question the habitability of the innermost planet—it seems like the sheer amount of radiation falling on it (4.4x Earth's by area) will cause the water to boil away quickly—with all that water vapor acting as a greenhouse gas, the temperature could easily get high enough for it to escape. $\endgroup$
    – lirtosiast
    Oct 6, 2015 at 1:21
  • $\begingroup$ That is possible indeed, but not guaranteed. Due to the extremely fast spinning rate (which is, by the way, very unlikely in practice, but could happen to be in theory), sunlight (including UV) would not have enough time to boil away the water it falls on. Plus, the high oxygen content would ensure a very thick layer of ozone shield. $\endgroup$ Oct 6, 2015 at 9:57
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    $\begingroup$ Hm... water does not need to heat to 100 deg. C to evaporate, and in a thin atmosphere, it doesn't need that much to boil. A "dead" core means no magnetic field to shield from hard radiation. I think your innermost planet will be just as dead as mercury. ;) $\endgroup$
    – DevSolar
    Oct 7, 2015 at 9:09
  • $\begingroup$ Nnnope! A dead core means the core isn't molten and tectonically active. As long as the planet is spinning and has a ferromagnetic metals based core, a magnetic field will be present. $\endgroup$ Oct 7, 2015 at 13:22

The simplest complex orbit would be a complex horseshoe orbit.

Each planet would have to be roughly the same size, due to how many different large bodies there will be, and there can be no other large bodies in the system (besides the parent star, or course).

And, importantly, this is not a natural system. Something will have to have engineered this orbit intentionally. It is possible to see two similarly sized planets in a mutual horseshoe orbit in a system formed naturally, but there will have to be no other large bodies in that system.

How it works:

How fast a planet completes an orbit depends on how close it is to its star. The closer, the shorter its orbit (the faster it completes the orbit). Mercury takes 88 Earth days to complete one of its years.

The more energy you put into an object's orbit, the higher its orbit goes. Seemingly paradoxically, this slows the object down. (Really, it just has more distance to travel, so it only seems that the object is slowed down.)

Thus, we can use gravity to help things stay balanced out on geologic time scales (hundreds of millions of years, enough time for humans to evolve from the rat-like mammals that survived the Dinosaur extinction at least 3 times over), and hope that we don't lose the ability to arbitrarily move planets, which was necessary to get them in this configuration in the first place.

The way it works:

Each planet will have a leading and a following planet, and will start out in ballance. Nothing ever stays in balance forever, though. Eventually, one planet will move closer to one of its companions, and further from the other, though this shouldn't be perceptible for at least 10 million years.

Let's name three of the planets Alice, Bob, and Charlie, with planet Bob in the middle and Alice being the lead planet (Both Bob and Charlie are orbiting towards Alice.)

Bob starts moving closer to Alice. Because gravity gets more intense between the two, and weaker from Bob to Charlie, Bob gains orbital speed, Alice loses orbital speed, and Charlie loses orbital speed.

Because of the gain in energy, Bob's orbit moves outward, slowing its year. Alice's and Charlie's orbits both move inwards, speeding up their years.

This causes Bob to start moving away from Alice and towards Charlie, where the situations will soon reverse; Charlie starts stealing Bob's energy, slowing Bob, and causing Alice and Charlie to speed up.

This also kicks off imbalances in the rest of the planets in the system, and soon, over the next million years or so, every planet has a slightly perceptible wobble.

Assuming a Sun sized star and an Earth orbit, that puts the orbital radius at 939,953,595 kilometers. Each planet would start out about 100,000,000 km from each other, which is twice the closest distance that Venus gets to Earth.

From the surface of one of these planets, the two closest companion planets would be plainly visible, always as bright as Venus at her brightest (which holds the distinction as the 4th brightest object in the sky, recently usurped by the International Space Station, and formerly third only to the Moon and the Sun). These two closest planets would always rise and set at the same time of day, rather than changing by season like the stars do, or in more complex patterns like our own system's planets do. We would not be able to see either of the two closest planets at midnight.

The next two distant planets will be visible as well, though not as bright, and setting earlier/rising later than the closest two.

The next two would be barely visible just before dawn and just after dusk, and may need a trained eye to spot them.

The final two would be drowned out by the glare of the star. Once a society develops calculus, their existence will immediately be inferred. They will be quickly photographed within two decades of unmanned spaceflight starting.

If such an orbital system were allowed to destabilize (say, the original system architects all went extinct and nobody on any of those planets re-developed the tech necessary), then things get interesting.

The distances between the planets are huge... It will take hundreds of millions of years for things to destabilize enough for any real danger to start for the inhabitants; perhaps even a billion years.

It will take a couple of decades for each cycle of planet oscillation to complete, but the effect, from the perspective of Planet Bob starting about 3 cycles before the ultimate disaster is:

As Bob approaches Alice, it will do so from Alice's day side. Bob will see Alice grow brighter and brighter, and about a month before closest approach, stop being just a point and become a sphere barely discernable from the naked eye. Alice grows larger until the day of closest approach, when it will be about 60 times larger than our moon (it will be 1/4th as close as our moon, and 4 times the Moon's apparent size). There will be earthquakes. Everyone will be prepared for the earthquakes, because the earthquakes will have been happening on closest approach for the entirety of the society's written memory.

Then, Alice disappears into Bob's day side, obscured by the star.

20ish years later, and the predicted coming of Charlie happens, as gradually building earthquakes herald that planet's appearance out of the glare of the sun, to retreat to a point much the way that Alice appeared from a point 20 years ago. Thus completes the first of the last three cycles.

Then, on Alice's next approach, the earthquakes tear apart the crust. The entire planet, heated by the tidal stresses of the recent near misses, becomes a molten hunk. The surface does manage to cool down before its next approach from Charlie, but Charlie's appearance only stirs the fresh, brittle, thin crust up and the surface quickly becomes molten, ending the penultimate cycle.

Finally, Alice's last approach tears the mantle and crust away from both planets, leaving super-heated rock spinning quickly away from both planets, bombarding the other planet, distributing the debris beyond each other's reach. Bob is now a much lighter, but much denser, core... hot, molten, with a crystal chunk of iron... and slightly more speed than it should. The ultimate cycle ends with Bob delivering a glancing blow against Charlie, adding more mass to what we should, by all rights, call Charlie's daughter planet, because neither Bob nor Charlie exist any more. Alice, at this time, is being flung out of the solar system from its near miss of its more distant companion, which will quite possibly find its way into the star in another couple million years... Charlie's descendent will probably impact its more distant companion as well, and has the greatest chance of surviving the ensuing chaos, though at this point nothing is certain.

  • $\begingroup$ Quite a creative setup. I'm not sure that earthquakes would happen - the idea that tidal forces can cause them isn't fully accepted yet - and it fails the stability criterion, but it's a clever solution. $\endgroup$
    – HDE 226868
    Oct 6, 2015 at 0:15
  • $\begingroup$ @HDE226868: Approaching the Roche Limit would cause earthquakes. The ground would be, quite literally, tearing itself apart. -- Under slow conditions (which, if they're mutually orbiting the same star in the same orbit, applies), planets don't quite collide, they just tear each other apart. $\endgroup$
    – Ghedipunk
    Oct 6, 2015 at 1:46
  • $\begingroup$ This and this have some further information on horseshoe orbits, and pretty pictures. They're about near-Earth asteroids, and don't include multiple bodies at once, but the concept should be similar. $\endgroup$
    – MichaelS
    Oct 6, 2015 at 1:57

I just skimmed all the answers and saw they all predated the recent NASA announcement, so I'm assuming this was not mentioned elsewhere...

NASA recently announced the discovery of 7 approximately earth-sized planets orbiting the same star, multiple of them being in the "goldilocks zone" calculation. Remember also that the "goldilocks zone" is only a very rough approximation; though unlikely, it is possible that most of these planets are habitable.

You can find one (of the many) articles about this here.

NASA's Spitzer Space Telescope has revealed the first known system of seven Earth-size planets around a single star. Three of these planets are firmly located in the habitable zone, the area around the parent star where a rocky planet is most likely to have liquid water.

  • $\begingroup$ That was weird. While I was typing my answer, I was getting the notification that more answers were posted to this question while I was working on mine. Supposedly there were "11 new answers to this question" posted by the time I submitted, though I see no such thing here. $\endgroup$
    – Loduwijk
    Mar 31, 2017 at 18:30
  • $\begingroup$ TRAPPIST is too small to make a proper analogy. $\endgroup$ Mar 31, 2017 at 19:40
  • $\begingroup$ TRAPPIST does not completely answer the OP's question, but it does provide the real, hard reality that is already halfway there (sort of). Complete answer for 9 planets? No. Good analogy? Yes. It also helps that the scientists talking about those 7 planets are very vague about the details, which makes sense since they don't have them. We assume those planets are tidally locked, water cooked off them, etc. But we don't know that. It could be that 4+ of them are habitable by us. Not likely, but not insignificant odds either. Since this is fiction, world-building its easy to run further with it $\endgroup$
    – Loduwijk
    Mar 31, 2017 at 21:04
  • $\begingroup$ @JohnWDailey And second, I could have made a long, fuller answer, but plenty of others have already done a much better job of the fictional side and the quasi-science (30+ planets in perfect configuration) side and have explained it somewhat, so I did not go deep. Shouldn't be the accepted one, no. And I don't mind if someone uses my info to make another answer better. If you said "That's 7, not 9, and we don't know if any are habitable" I'd agree. I just thought this page was begging for the closest real world approximation that we have, and I think it complements the other answers perfect. $\endgroup$
    – Loduwijk
    Mar 31, 2017 at 21:09

A couple of largish stars in a far binary orbit (ie far enough out to have minor effects on the other star's goldilocks zone). Now place a 3rd star even farther out and in orbit around both (like Alpha Centaury except with bigger stars and more spread out).

Place a super earth in the warmer end of the goldilocks zone for each star. Place a "moon" around each that are large enough to hold their own atmosphere.

Now place a smaller planet (earth-like) towards the outer edge of the goldilocks zone.

This will give you 9 planets to play with.

  • $\begingroup$ This does depend on what you meant by "linear" plane. All of these planets are in the same system, but not around the same star. $\endgroup$ Oct 5, 2015 at 17:19

Realistically I'd say no chance. With an artificial setup, maybe two gas giants orbiting in the habitable zone. Binary pairs of planets in L4 and L5 points of the gas giants and the 9th a moon of a gas giant. Throw in some orbital resonances and it might be stable. Invoke superhuman powers to stabilise it if someone with more maths than myself proves not!


To add to Rumguff's excellent answer:

In our own system Jupiter has 4 moons (the Galilean moons) of about the same size as our own moon, and saturn has one (Titan.)

At some time in the future, the sun will have expanded and it's likely that all five of these bodies will have "goldilocks" temperatures at once (Saturn is about twice the distance from the sun as Jupiter, so Titan's going to be a lot colder than the others.)

The one big caveat is that these five bodies have surface gravity similar to our own moon, which is insufficient to hold onto a decent atmosphere. High atmospheric pressure has virtually no effect on the freezing point of water, but it raises the boiling point, and if there isn't enough atmospheric pressure, liquid water is simply impossible. Titan is currently the only non-planet in our solar system that has an appreciable atmosphere, with 1.5 times the surface pressure of Earth, but this will surely partly evaporate when the sun expands and warms the solar system.

So, we probably need a system with bigger moons, which probably means bigger gas giants and a bigger star. That said, our own moon is much larger in comparison to Earth than the other moons of the solar system in comparison to their respective planets. So there are processes that can form large moons.

Apart from that, the only other caveat to having 5 goldilocks bodies in our solar system in the distant future, is that Io has intense volcanic activity due to tidal warming from Jupiter and the other moons, which will probably render it always unliveable.


Bending the question to give 9 separate worlds of a similar surface area and climate to our own; I would suggest a partially complete Dyson Sphere. Instead of a filled in sphere (a Dyson Shell) it could be 9 main habitable areas with supporting structures and a rainbow road transit system between them. That might make it more like a Dyson Swarm or Dyson Bubble, but the overall idea stands.

As a suggested step in the evolution of a civilisation it could be that the Avguard were in the process of constructing such a thing when they were required to interact with the lower technology nordic civilisation. I am not sure how much more complex it would be to add a planet to such a construct. Perhaps 8 of the realms are part of the Dyson thing but the norse one is a free planet elsewhere in the habitable zone.


See here for a simple explanation of how tightly you can pack planets into the habitable zone: https://planetplanet.net/2014/05/21/building-the-ultimate-solar-system-part-3-choosing-the-planets-orbits/

And here is a followup with a couple of ways to squeeze in extra worlds, in the form of planets on Trojan orbits as well as moons: https://planetplanet.net/2014/05/21/building-the-ultimate-solar-system-part-3-choosing-the-planets-orbits/

Finally, regarding Trojan worlds, I want to reassure you that Trojan planets are: 1) A natural outcome of planet formation. I have run thousands of simulations of planet formation, and Trojans are inevitable. Exactly how common they are depends on some of the assumptions we make in our models. But they are a simple consequence of physics and they must exist 2) Dynamically stable. A system with two planets in a Trojan configuration and no other planets will generally be stable forever. Many sets of Trojans can exist in the same system, although their combined mass must be taken into account in terms of the spacing of adjacent sets of many planets.

The orbits of planets packed into the habitable zone of our chosen star, with co-orbitals (Trojan planets). Each orbit is occupied by two planets separated by 60 degrees. The planets are either 0.1, 1 or 10 times Earth’s mass. The shaded area represents the habitable zone, which extends from about 0.2 to 0.4 Astronomical Units (AU; 1 AU is the Earth-Sun distance) for our chosen star. The number of pairs of co-orbital planets that can be packed into the habitable zone is 9, 6, and 2 for planets with 0.1, 1, or 10 times Earth’s mass, respectively.  From: https://planetplanet.net/2014/05/22/building-the-ultimate-solar-system-part-4-two-ninja-moves-moons-and-co-orbital-planets/


If they were all on the same orbit, they could all be in the same "zone". Impossible to achieve naturally and it would require some sort of magic to stabilize (https://www.quora.com/Can-two-planets-share-the-same-orbit) but it could be done. The star would have to be pretty big to support so many of them without gravitationally induced collisions.

Or you could put them on differently oriented orbits roughly the same distance from the sun and assume they were either in different dimensional states such that their gravitational fields didn't interfere with each other and they never collided.

Bottom line, there is no natural way it could be achieved, but with magic (or sufficiently advanced science) anything is possible.


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