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I'm imagining a system with a star and something like 6-8 planets. The planets' orbits are (relatively) close to each other, and all share an orbital period that is exactly the same. I.e. the length of a year would be exactly the same on all planets. I've even gone so far as to imagine a system where the planets are all in a line, with a gravity elevator linking each planet to its neighbor(s). Something like this:

(star)                        A-----B-----C-----D-----F-----G

I understand that they would have to be offset a bit (or in slightly different orbital planes) in order to not perpetually eclipse each other. Assume that technology exists allowing a species to exactly place a planet into the desired orbit, and even make routine corrections (though I'd prefer to not have to if possible). In other words, you can almost treat each planet as a giant spaceship as long as it would not need to use any thrust 99% of the time.

In this scenario the planets would all be roughly Earth-sized and have somewhat Earth-like climates, though probably the innermost would be hotter and the outermost colder.

From what I understand, a planet orbiting at x (average) distance from the star has a specific range of velocities it must adhere to--if it is too slow it would crash into the star, and if it is too fast it would escape the system altogether. I also know that the innermost planets would travel slower and the outer ones much faster in order to make one revolution in the same period. Finally, I'm guessing the distance between each planet would vary throughout their orbits since the orbits would be elliptical (so the gravity elevators would be long and flexible). But I don't understand the math enough to do the calculations.

So specifically I'd like to know:

  1. Is is possible for such a system to exist?

  2. If so, are there are limitations/constraints on the length of a year in this system, or on the type of star, distance from the star, etc?

  3. How far apart would the planets have to be in order for the gravity of each (assuming Earth-like mass) to not pull its' neighbors out of orbit?

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    $\begingroup$ I honestly fail to see why people downvoted this question. Sure, everyone is free to use their downvotes as they please, but to me, this is one of the better questions I've seen on the site today. The premise is clear; there isn't a load of irrelevant material; and it's complete in that it provides most if not all of the pertinent information right off the bat. It's even appropriately tagged! Sure, it asks three subquestions, but all three are very closely related. Yes, the answer is obvious if you know orbital mechanics; but really, that's part of what we're here to help people with. $\endgroup$ – a CVn Nov 16 '17 at 18:33
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    $\begingroup$ @jamesqf It bugs me when people judge a question based on what they already know is the answer. Questions and answers should be judged separately and on their own merit. $\endgroup$ – corsiKa Nov 17 '17 at 16:14
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    $\begingroup$ @ corsiKa: Well, that's you :-) OTOH, it bugs me when people ask questions that seem to display complete ignorance of physics. I mean, what can you write as a meaningful answer, except "No, the universe doesn't work like that"? $\endgroup$ – jamesqf Nov 17 '17 at 19:08
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    $\begingroup$ @jamesqf How is it ignorant of physics to ask "What does physics say about this set of conditions?" Seems that is exactly the opposite of ignorance, but rather a very healthy respect that physics has the answer, OP just doesn't know what that answer is. $\endgroup$ – corsiKa Nov 19 '17 at 7:55
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    $\begingroup$ @jamesqf A basic understanding of physics would be knowing Newton's laws or understanding that friction and gravity are forces. Kepler's laws would not fall under a basic understanding of physics. Sorry... $\endgroup$ – corsiKa Nov 19 '17 at 18:15

14 Answers 14

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Is it possible for such a system to exist?

I'm sorry, but no. At least not according to orbital mechanics as currently understood.

Kepler's third law of planetary motion is one of the old workhorses of orbital mechanics, and applies in this case. As translated and summarized by Wikipedia, it states that:

The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

or, mathematically,

$$ P^2 \propto a^3 $$

or, stated differently, there exists some constant $k$ such that

$$ P^2 = k a^3 $$

The semi-major axis is one of the defining parameters of an ellipse. (Put simply, the semi-major axis is the longer radius of the ellipse.) Since orbits are ellipses (also), this applies.

Consequently, the least you change the distance at which the planet orbits from the star, the orbital period will change, however little. If the orbital periods are different, then the planets will drift apart over time, however slowly. (They will occasionally line up, assuming that the orbits are themselves stable and closed. A variant of this occured for some of the planets in our solar system in the 1970s-1980s, giving our solar system grand tour taken by the Voyager 1 and Voyager 2 probes.) Ergo, the system you describe cannot exist when all planets are in the same plane.

If the planets are in different planes (technically, have different inclinations relative to the solar system ecliptic, which in this case could probably conveniently be defined as the equator plane of the star), then the distance between the planets will change as they move through their orbits. You can visualize this by considering two planets, orbiting the same star with the same velocity but at different inclinations; if you trace their orbital trajectories, you will see that the distance between the two planets varies throughout their orbits. Any kind of rigid construction attaching them to one another would interfere with their movement and either cause them to crash into each other, tear the structure apart, or tear the structure from one or both of the planets involved. Either way, having the planets in different planes is not an option either.

So, sorry, no, you can't have what you want.

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    $\begingroup$ That's not entirely true. Kepler's Laws only apply to an approximate two-body system, where perturbations from other bodies are negligible. This system explicitly violates that constraint, so a more detailed analysis is required. Such a system certainly could not form naturally, and would not be in a stable configuration, but the question assumes planetary-scale station-keeping is possible, and it's clear that a steady-state mathematical solution for such a system does exist. $\endgroup$ – Logan R. Kearsley Nov 17 '17 at 5:34
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    $\begingroup$ Fully agree with the physics, but it might be interesting to note that it is possible to have these 5 planets take the same orbit. I.e. instead of being at different distances from the sun, they are simply in different position of the same orbit, basically a conga line. If the orbit is circular, then a space elevator should be possible (due to no relative drift between the planets) $\endgroup$ – Flater Nov 17 '17 at 8:58
  • $\begingroup$ @LoganR.Kearsley: If the planets are sufficiently perturbing eachother, then they would have to orbit eachother in order to continually "miss" eachother due to their interactions. Which is essentially the same thing as our Earth and the Moon; which renders the "five planets" system more like a "one planet and four moons" system $\endgroup$ – Flater Nov 17 '17 at 9:01
  • $\begingroup$ @Flater (I'll have to consider Logan's input, but) the Earth and the Moon are probably not a very good example in this case, as the distance between the two bodies varies between approximately 356 Mm and 407 Mm. I know you said circular orbit, but having a zero-eccentricity orbit is not exactly trivial. Though I suppose if we posit the ability to position planets in any orbit desired and keep them in the proper place, then pretty much anything is probably possible. Whether it'd be reasonable is another matter... $\endgroup$ – a CVn Nov 17 '17 at 9:05
  • $\begingroup$ @MichaelKjörling: You're conflating two separate comments. The circular orbit has nothing to do with the moon setup. The circular orbit would be necessary to create a space elevator between 5 planets that all orbit the sun in the same orbit (but do not orbit eachother, e.g. they're in a train/conga line formation). If the orbit were not circular, the planets would librate (relative to eachother), making a space elevator nigh impossible. The shape of the orbit of the "moons" is irrelevant, at least for the comment I made about them. $\endgroup$ – Flater Nov 17 '17 at 9:08
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The only way such an arrangement could exist is having external planet being the heaviest and the others each in L1 point of next outer one.

Unfortunately such an arrangement is not stable, so it would need corrections to keep alignment and avoid planets to drift out. This kind of correction would be quite small, if you do it before planet/spaceship drifts too far from its ideal position.

Any (sufficiently strong) physical link between planets (as OP seems to indicate talking about "gravity link", which I interpret as a "space elevator" thing, but I could be very wrong) would contribute to stabilize the system (a really linked system would assume the required configuration in any orbit due to tidal forces, but required tensile stress rapidly becomes unmanageable, even in the SF context, as soon as planets are not in the proposed configuration). It is unclear how such a linked system may work without tidal locked planets.

Note: Lagrangian points are computed on a 3-body system, your multi-body system would need some adjustment, but any alien race able to move planets shouldn't have problems working out details ;)

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  • $\begingroup$ Ooh, yes, that seems like a plausible scenario. Two planets is clearly possible this way. Do you happen to know if a chain of L1 points, like this, is feasible with occasional thrust to compensate for the instability? (yes, I'm aware that this amounts to asking for an analysis of a n-body problem for n > 3) $\endgroup$ – Ethan Kaminski Nov 17 '17 at 8:48
  • $\begingroup$ @EthanKaminski: no, I don't have the data available and, sincerely I don't have the time to invest in a reasonable simulation (as inferred by last paragraph in my Answer). Note: a "simple" orbital simulation wouldn't suffice because orbits are unstable. I would expect instability to be more (bu not much more) evident than in 3-body case. This might be mitigated with an external planet being much heavier (Jovian) than the ones in its L1 (and L1 of planet in L1 ...). $\endgroup$ – ZioByte Nov 17 '17 at 10:15
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    $\begingroup$ Three body problem was still considered unsolvable last time I checked... $\endgroup$ – Mołot Nov 17 '17 at 18:10
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    $\begingroup$ @Mołot: I said "computed", not "solved". 3-body problem does not have an analytical solution, but there are tons of numeric integrations. Add to that 3-body problem does not have (analytical) solution for the "general case", but it is readily solvable in several "special cases" including, but not limited to, the Lagrangian Points which which have very specific requirements (e.g.: circular orbits). $\endgroup$ – ZioByte Nov 17 '17 at 22:49
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    $\begingroup$ I came to post this. From a physics perspective, setting a body in the L1 point is the only way to have a synchronous orbit for two bodies orbiting a third at different distances. That said, chaining L1 points seems possible intuitively, but orbital mechanics is far from intuitive. Also, as noted, L1 points are not stable and require stationkeeping. $\endgroup$ – dotancohen Nov 19 '17 at 13:20
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TL;DR: Nothing like you describe is possible with current technology (or anything likely in the near future), but several configurations are possible using technology well within the realm of science fiction.

As others have mentioned, the linear arrangement you mentioned is not possible (unless the connections between the planets are exremely strong and rigid). As others have also mentioned, a Klemperer rosette would allow the planets to maintain fixed relative distances, but the orbits would not be stable to small perturbations. The Klemperer rosette also has the possible disadvantage that the distance between the planets is similar to the distance between the planets and the central star (i.e. quite large).

There is a another configuration worth considering, although it doesn't solve all of these problems: the proposed orbital configuration for the Laser Interferometer Space Antenna (LISA): LISA orbit diagram

In the proposed LISA mission, three satellites orbit the sun with roughly the same orbital period as the Earth. Their orbital planes are oriented at slightly different angles and their orbital phases are synchronized so that the three sattelites orbit the sun while maintaining a fixed relative distance. As seen in the picture, the constellation of sattelites also rotates in the plane defined by the three satellites. Note that the satellites are not attached to each other in any way (the lines on the diagram indicate laser beams used to measure the stretching of space).

While this keeps the planets relatively close together, I believe* it suffers from the same instability problem as the Klemperer rosette, so while it works for satellites which are small and far away from each other, the gravitational effects between the planets in a LISA-like configuration would be considerable, and would lead to significant instability. The interplanetary gravitational forces would also cause the planets to be non-periodic, although not in a way that would neccesarily present any practical problems.**

So, to answer your questions:

  1. It is not possible for a system like the one you describe to exist without some repeated adjustments of the orbits, or extremely strong connections between the planets. However, some configurations require more energy to maintain (or stronger connections) than others. The Klemperer rosette and the LISA configuration both require orbital adjustments only to fix any drift away from their initial configuration, so in principle this could be done with a relatively small amount of energy or relatively weak interplanetary connections.***

  2. According to Kepler's third law, the length of year, distance of the planet from the star, and mass of the star cannot be varied independently, although there is no restriction on any one of these parameters independently. In principle**** you can pick whatever you want for any two of these parameters, but your choice of the first two determines the third parameter. This is still true for the Klemperer rosette and the LISA configuration, although the exact relationship between the parameters would be different.

    In the Klemperer rosette, the gravitational forces between the planets would shorten the length of year for any given mass of star and radius of orbit, although the effect would be very small unless the planets were very massive, or the star very small.

    In the LISA configuration, the time it takes for the planets to orbit the star would be about the same as you would expect from Kepler's third law. Depending on the rotational axes of the planets themselves, the seasons could change in very complicated ways since the season depends on the orientation of the planet's rotational axis with respect to the incoming starlight. This in turn depends on the orientation of the rotational axis relative to the orbital plane of the planetary constellation, as well as how the constellation is rotated in its own plane. Since the rotation of the constellation has a different period than the orbit of the constellation around the sun**, the angle between a planet's rotational axis and the incoming starlight could change in quite complicated ways over cycles of many (orbital) years.

  3. The gravitational force between two objects is proportional to $1/r^2$, where $r$ is the distance between the objects. This means that planets will feel each other's gravitational forces no matter how far away they are from each other. In the case of the Klemperer rosette and the LISA configuration, these forces will affect the orbits (see the previous two paragraphs), but these effects are not catastrophic. The bigger problem is that these configurations are unstable, so the gravitational effects of everything else in the universe will be catastrophic in the long run. It is these effects that you have to counteract using orbital adjustments or strong interplanetary connections. Luckly, most of the rest of the universe is very far away, so these effects are small and, with the right sci-fi technology not terribly hard to fix.

In short, if technology capable of significantly influencing the orbits of planets is avaiable, the Klemperer rosette or the LISA configuration are both feasible.

* While I am a physicist, orbital mechanics is not my area of expertise, and I haven't done any calculations related to this post, so take what I say with a grain of salt.

** The planetary constellation would rotate around its center of mass faster that it would due to the relative canting of the orbital planes. After one orbit around the central star, the constellation would be in the same place, but rotated differently than it was at the same time in the previous year.

*** Of course adjusting a planet's orbit by any measureable amount is still very hard, so the amount of energy required would be enormous compared to anything feasible with our current technology.

**** There are practical limits on the distance to the star and the mass of the start however. The planets should not be inside the star, or so far away that they are close to other stars. If the mass of the star is too small it will just be a planet or cloud of dust and gas, if it is too large it will turn into a black hole, making things complicated nearby.

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  • $\begingroup$ The LISA thing is really cool, but doesn't the reorientation of the orbit (in order to always face the sun, instead of its natural tendency to not direct itself towards the sun) rely on _continual (minor) thrust adjustments? A planet is considerably heavier than a satellite in terms of changing its velocity vector... $\endgroup$ – Flater Nov 17 '17 at 9:06
  • $\begingroup$ @Flater: My understanding is that the LISA satellite constellation naturally follows the orbits described with no thrust needed for orbital adjustments. To put it another way, the orange path in the diagram above is a standard elliptical orbit, as are the paths followd by the other 2 satellites. The satellites orbit the sun, not each other, but the orbits are carefully chosen so that the satellites stay in this rotating constellation. $\endgroup$ – Z2h-A6n Nov 17 '17 at 15:16
  • $\begingroup$ That said, your comment made me do a bit more research: $\endgroup$ – Z2h-A6n Nov 17 '17 at 15:16
  • $\begingroup$ I inferred the dotted line to imply an orbit around Earth (and the orange line to be a calculated position at given intervals), my mistake :) It seemed like the goal was to ensure that the satellites orbit the Earth in such a way that they always have sunlight (for their solar panels), which you can achieve by having the orbit align itself according to the Sun's position relative to Earth (hence needing to continually correct it). $\endgroup$ – Flater Nov 17 '17 at 15:21
  • $\begingroup$ Whoops, I posted my last comment incomplete. It should have said: "That said, your comment made me do a bit more research: Apparently the satellites do move relative to each other, but only at a maximum relative speed of 13 m/s. It seems that there is a fair amount of ongoing research regarding the details of the LISA orbits, as would be expected for a proposed space mission costing 250M Euro." $\endgroup$ – Z2h-A6n Nov 17 '17 at 15:26
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This problem is easiest to solve in a co-rotating frame, where you sum together the sun's gravity and the centripetal force to produce an effective potential with a flat spot which tells you where a planet can sit. If the flat spot is a minimum, the system is stable; if it's a maximum, it's unstable, but can be stabilized with appropriate application of external restoring forces (planet-scale station-keeping thrusters of some sort!).

In this case, of course, you won't just have the sun's gravity to worry about--you'll be summing up the gravities of all the different planets, which gives you a system of multiple equations to solve, but the basic idea is still the same. Fix the orbital periods to all be the same (exactly what the value is doesn't matter, we'll just make it a variable that we can solve for), work in a co-rotating frame to turn it into a one-dimensional problem, and solve for the distances that balance forces so the planets don't move inward or outward.

We can also significantly simplify things by assuming that all the planets have the same mass.

The centrifugal acceleration of each planet is given, in terms of orbital period, by $4\pi^2\frac{r}{T^2}$, where $T$ is the orbital period. We're fixing the orbital period as constant for all the planets, so we can gather all the constant terms together and write the centrifugal acceleration as $\alpha r$--i.e., linear in radius.

The gravitatonal acceleration will be the sum of contributions from the sun and all the other planets. Since the sun has a different mass from each planet, and it's gravity will always be in the same direction for every planet, it will be convenient to separate out its contribution. So, the net radial acceleration of each planet can be written as

$$a_p = -\frac{GM_S}{r_p^2} + \alpha r_p + GM_P\sum_i \frac{sgn(r_i-r_p)}{(r_i-r_p)^2}$$

If there are exactly 6 planets, that expands out to, e.g.,

$$a_1 = -\frac{GM_S}{r_1^2} + \alpha r_1 + GM_P(\frac{sgn(r_2-r_1)}{(r_2-r_1)^2} + \frac{sgn(r_3-r_1)}{(r_3-r_1)^2} + \frac{sgn(r_4-r_1)}{(r_4-r_1)^2} + \frac{sgn(r_5-r_1)}{(r_5-r_1)^2} + \frac{sgn(r_6-r_1)}{(r_6-r_1)^2})$$

(Note that the sign of the planet's radius minus itself is zero, so the self-interaction term falls out when you expand the summation.)

Now, we want the radial accelerations to all be zero. So, we can write out the complete system of equations (with signs resolved assuming they are ordered from 1 to 6 moving outwards, and terms re-ordered to make opportunities for cancellation more obvious) as follows:

$$-\frac{GM_S}{r_1^2} + \alpha r_1 + GM_P[(r_1-r_2)^{-2} + (r_1-r_3)^{-2} + (r_1-r_4)^{-2} + (r_1-r_5)^{-2} + (r_1-r_6)^{-2}] = 0$$

$$-\frac{GM_S}{r_2^2} + \alpha r_2 + GM_P[-(r_1-r_2)^{-2} + (r_2-r_3)^{-2} + (r_2-r_4)^{-2} + (r_2-r_5)^{-2} + (r_2-r_6)^{-2}] = 0$$

$$-\frac{GM_S}{r_3^2} + \alpha r_3 + GM_P[-(r_1-r_3)^{-2} - (r_2-r_3)^{-2} + (r_3-r_4)^{-2} + (r_3-r_5)^{-2} + (r_3-r_6)^{-2}] = 0$$

$$-\frac{GM_S}{r_4^2} + \alpha r_4 + GM_P[-(r_1-r_4)^{-2} - (r_2-r_4)^{-2} - (r_3-r_4)^{-2} + (r_4-r_5)^{-2} + (r_4-r_6)^{-2}] = 0$$

$$-\frac{GM_S}{r_5^2} + \alpha r_5 + GM_P[-(r_1-r_5)^{-2} - (r_2-r_5)^{-2} - (r_3-r_5)^{-2} - (r_4-r_5)^{-2} + (r_5-r_6)^{-2}] = 0$$

$$-\frac{GM_S}{r_6^2} + \alpha r_6 + GM_P[-(r_1-r_6)^{-2} - (r_2-r_6)^{-2} - (r_3-r_6)^{-2} - (r_4-r_6)^{-2} - (r_5-r_6)^{-2}] = 0$$

Now, you've got 6 equations and 6 unknowns (the radii for each planet), which you can go ahead and solve in terms of the orbital period, solar mass, and planetary mass. The same applies for any number of planets.

Once you've done that, you can try varying the radii by small amounts to calculate how much station-keeping force you'll need for each planet.

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  • $\begingroup$ Out of curiosity, do you have any (maybe order-of-magnitude) ideas of what the radii might be, say in relation to $r_1$? $\endgroup$ – HDE 226868 Nov 18 '17 at 3:13
  • $\begingroup$ @HDE226868 Having not bothered to actually solve the system... it'll depend on the solar / planetary mass ratio, but my guess would be that it would be pretty close to placing each subsequent planet at the L2 point of the sun & the center of gravity of the collection of previous planets, contracted a little bit due to reduction in the sun's effective gravity by the influence of the outer planets. Given 1 solar mass, and planets of 1 Earth mass, that would suggest $r_2$ should be about 1% larger than $r_1$, $r_3$ about 1.505% larger, $r_4$ about 2.01% larger, and so forth. Very approximately! $\endgroup$ – Logan R. Kearsley Nov 18 '17 at 4:09
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Given sufficiently different eccentricities planets with the same year are possible, but they won't stay a constant distance apart. The more eccentric ones spend more of their year in the cold outer dark.

If the orbits are such that the semi-major axes are collinear, then I'm pretty sure that the resonance would mess things up in a hurry. Spacing the axes around the circle would give you a solution called a klemplerer rosette

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

This is a known unstable situation.

If planets were in pairs, with large eccentricities in opposite directions, and opposite in phase (one is near the star when the other is far) and each pair with wildly different incinations to the ecliptic I think you would have a system that is at least short term stable.

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Short answer: still impossible.

Given the other answers, the closest thing to what you want I could think of was an Earth-like planet captured near a much bigger planet's Lagrangian point L4 or L5. Actually, you can get both and a few large moons orbiting the central giant.

Bear in mind that L1..L3 points are unstable, so no more than 3 planets and the central one probably uninhabitable due to immense gravity. (You may need something like Jupiter to stabilize the system, but I can't tell for sure what the limits are).

As for your third question, gravitational pull of planets on each other is tiny, however, over long time anything unstable eventually gets kicked out of orbit by resonance.

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    $\begingroup$ Even if their position can be defined in the OP's proposed system, L4 and L5 won't result in the situation the OP is after. I know you imply this by your statement that this might be "the closest thing to what you want", but you might possibly want to emphasize that even more. $\endgroup$ – a CVn Nov 16 '17 at 18:53
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Yes... if the order of the planets at any time is flexible.

If the inter-planetary tethers are strong enough (insanely strong), you could have the 6-planet system spinning about its center of mass, and that spinning system would have a single orbit.

If the axis of rotation was normal to the orbital plane, all planets would have the same length of day, which would be the period of rotation of the planetary chain and which is arbitrary above the minimum required to prevent gravitational collapse.

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  • $\begingroup$ The orbit you describe isn't stable. I believe 2 planets orbiting their own center of mass is a stable arrangement, but as soon as you add more then any small perturbation will bring one pair closer to each other than the rest and they will start to fall towards each other. I believe this would require a lot more effort to maintain than the 1% correction asked for in the question. $\endgroup$ – Jules Nov 19 '17 at 10:40
  • $\begingroup$ @Jules You could be correct. I envisaged the spin to be "high", so the outward centripetal force is "large", hopefully keeping things taught and stable $\endgroup$ – Bohemian Nov 19 '17 at 19:53
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First of all, see Kearsley's answer for the general formulas that will provide you with all possible solutions (he has my upvote for that). This answer gives a simple configuration that includes six planets and should be reasonably stable (= require only minimal amounts of corrections):

You have

  • one sun

  • two big planets
    (a bit heavier than earth but less dense and thus bigger in size so that surface gravity is not too much larger than 1g)

  • four small planets
    (a bit lighter than earth but more dense and thus smaller in size so that surface gravity is not too much smaller than 1g)

The two big planets orbit the sun in the same orbit, but on opposite sides (they are in each other's L3 point). The four smaller/lighter planets occupy the L1 and L2 points of the larger planets. Thus, all planets and the sun are in a single straight line like this:

 p1-----P2-----p3---------S---------p4-----P5-----p6

L2P2   L3P5   L1P2                 L1P5   L3P2   L2P5

Since all planets are Lagrange Points relative to each other, their orbits are relatively stable, however they will still require corrections from time to time since it's just the L1,2,3 points.


If you want all planets to be on the same side of the sun, you must use hirarchical Lagrange Points: Once you place a planet in a Langrange point, two new Langrange Points appear, one between the two planets, and on the other side of the smaller planet. So, adding a third, smaller class of orbiting objects I'll call moons, you'd get a configuration like this:

S---------------m1-----p1-----m2-----P2-----m3-----p3-----m4

               L2p1   L1P2   L1p1          L1p3   L2P2   L2p3

This gives you a total of seven orbital bodies in a single line on one side of the sun.

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As you can see from your picture, they are not at the very same distance from the central star.

This alone tells you that the alignment you in have in the picture won't last. Planet A will have a different orbital speed than all the others, and the same holds for all the other planets.

Moreover, if they get too close each other, they will end up either crashing one on another or with some of them being ejected out of the system.

What you can have (but it is rather different than what you state) is a big central planet with a lot of moons around it, a la Jupiter. But they won't be constantly aligned.

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The diagram you gave with the six planets in a line is not possible because of the huge forces involved. If the technology existed to tie planets together like that the tidal forces would make them not usable as planets. Their shape would be distorted and holding them together to maintain spheres would require extra technology.

I suggest six planets of equal size and mass surrounding the sun spaced 60 degrees apart sharing the same exactly circular orbit. This is like a Klemperer rosette as mentioned in other answers. They key in this case is having the massive sun at the center instead of empty space. That makes it so that each planet has two adjacent planets in its L4 and L5 points making this configuration relatively stable.

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Although other answers are right, it is possible to have several bodies orbiting a star with the same period while keeping close to each other: usually they are called satellites (all of them except for one). Furthermore, if two bodies are of similar sizes, they stop being a planet and a satellite and they became a double planet.

However, if you need distance between planets to be fixed in order to place an space elevator between them, that could be done only in a pair of tidally locked bodies - that is, a tidally locked double planet.

In short, the system on your question is not possible, but you may have an smaller version of it with a double planet.

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Maybe...

But not really if you just have one star. If you had a binary star system where the two stars had equal mass and the distance between the stars was just right to get just the right amount of sunlight, then the region directly between those two stars might be a place your system might work.

At the center of mass you would have zone where the gravity of the two stars nearly cancels out. But the equilibrium is unstable. If a planet drifted to one side or the other then the gravity from the star on that side would be a little stronger and make the situation worse.

The gravitational potential would look a little like the image below.

enter image description here

The y axis represents the gravitational potential energy, the x axis the position and the two stars are located at -4 and +4.

Usually, we would want to place a habitation in a situation that is dynamically stable. If a system is dynamically stable, then small changes in the system just result in oscillations.

If a system is dynamically unstable, then small changes produce large results. This is like a ball sitting on top of a hill. It will sit there if undisturbed, but any small push will eventually lead to the ball rolling down the hill.

If you set things up just right, the additional forces from the stars as you moved either way from the center could balance out the force of gravity from the other planets.

I think an electromagnetic tether could be used to overcome the dynamic instability of the system. But this tether would have to be something incredible.

A gigantic solar sail might be a good way to augment the tether, especially is the forces from each star would normally balance out. Moving either way would usually cause the net force on the solar sail to be a restoring force, and so we might make the equilibrium a dynamically stable one.

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1: Probably in the case of 2 or more stars rotating on eachothers axis and permanently maintaining the gravitational pull at a stable output. With 1 star highly unlikely for sustained amount of time (like others before me stated)

2&3: Related to the gravitational balance needed to do nr1 , they can even orbit dwarfs stars or black holes as long as the orbital equilibrium is maintained constantly. But the inevitable alterations in every star/black hole will eventually break the orbits down)

If you want to go scify rout go for 3 star solar system (2 big stars and 1 dwarf )the planets orbit the dwarf that is locked in place by the orbit of the 2 big ones. But this solution is a big NO if you ask "is this possible ?" just giving you a possible solution that unless you know the precise math behind it normal people won't notice it at first glance.

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Their might be a way too get something like what you want.

If you use the elevator to tether the planets together(assuming specfic strength ) then they will want to travel in the same orbit. (The lower planets will be pulled up and the higher planets down)

Now if you have the engines to counter this then they can stay in the seperate orbits. This will cause the non middle planets to have a sideways acceleration though.

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