I've been doing a little research (including reading Howard Curtis' Orbital Mechanics for Engineering Students) because I want to build a solar system with a planet that has a true three-dimensional orbit. Regrettably, I'm not far enough into Curtis' book to answer this question, if it is answerable from it.

I understand that orbits occur in three dimensions, however, the examples I've found so far are all 2D in the plane of the elliptic. In other words, the orbit is actually two-dimensional, it simply can be rotated as a flat plane in lots of different directions.

Question: Is it possible to set up a series of stars and/or planets such that one planet has an actual three-dimensional orbit, meaning the orbit is not flat (two-dimensional) in the plane of the elliptic?

For illustration purposes only

Here's our solar system showing Neptune's normal orbit. Notice that both Neptune and Pluto can be thought of as 2D orbits in the plane of their elliptic. It only becomes 3D when you compare the two orbits.

enter image description here

And here' a modified mock-up showing Neptune's orbit as a 3D orbit (gravity well causing this behavior not shown). Neptune's orbit is no longer on a flat plane.

enter image description here


If you're tempted to edit my question and change every instance of "elliptic" to "ecliptic..." please don't do it:

It is incorrect as "ecliptic" refers the the solar plane, not the orbital plane of an individual planet. Pluto does not orbit on the ecliptic plane, but it does orbit on its own elliptic plane.

The gist of my question can be summarized as follows: if you press all of the orbits to the ecliptic plane without modifying the basic shape of the orbit itself, what you get in our solar system is a 2D orbital example (the ecliptic and elliptic planes would coincide). If you flatten my example of Neptune, above, it's 3D because part of the orbit sweeps out of the ecliptic plane.

In other words, the orbit I'm interested in does not inhabit a single elliptic plane and could not be contained within the ecliptic plane if pushed into it.

  • Comments are not for extended discussion; this conversation has been moved to chat. – L.Dutch Aug 21 at 6:33

14 Answers 14

up vote 103 down vote accepted

There are no known orbits of this kind, but they aren't proven to be impossible.

You need to have 3 objects in order to have a 3d orbit, and it's known that the general 3-body problem is chaotic and difficult to work with. However, researchers do explore restricted 3-body problems. In these problems, we assume that one object is negligable in mass to the others (mathematically, we treat it as massless). That means two of the objects orbit in a predictable 2-body configuration, and the 3rd smaller object flits around between them.

There are two classes of problems you are looking at: 3D CR3BP and 3D ER3BP. Yes, they gave the names. The 3D part is what you are interested in: the solutions aren't planar. The CR and ER refer to a circular obit restriction and an elliptical orbit restriction, referring to how the large two objects orbit each other. 3BP references the fact that it's a 3 body problem. So you're looking for non-planar solutions with either circular or elliptical orbits for the 2 main bodies in a 3 body problem. *Phew**

First the good news. A shout out to b.Lorenz and their answer, because it suggests there's some possibility of it working. The paper cited explores a 3D CR3BP problem, specifically the Earth/Moon system. It also has some glorious looking pictures showing 3d movement resonating around these two massive objects, suggesting interesting parking orbits.

Resonant orbits

However, there's some bad news.If you note, these orbits get "pinched" in some places. None of the orbits found in this paper were stable. They decayed in dozens or hundreds of orbital periods. For a planet, you need something that remains stable for millions or even billions of periods!

Indeed, it appears we have not yet found a stable 3D ER3BP orbit, as of this paper published in August 2015:

A few families of periodic orbits in the elliptic Hill problem were discovered by Ichtiaroglou [1981], who found all of them to be highly unstable. In more recent work, Voyatzis et al. [2012] calculated a large set of families of periodic orbits in the elliptic Hill problem, determined their stability and applied the results to motions of a satellite around a planet.

Reading the Voyatzis et al. paper, we find that they did indeed find stable ER3BP solutions, but they were concerned with the planar ER3BP problem, not the 3D one.

Indeed, it appears that has been enough of a search for 3D ER3BP solutions that the paper pays a substantial amount of attention to periodic ejection-collision orbits in the 3D ER3BP space. Many papers have indeed found such orbits, but by their nature, they cause the small object to pass so close to one of the massive ones that there will either be a collision or the small object will be ejected out of the system.

So overall, we've found nothing which prevents us from finding a stable 3D ER3BP solution. However, in the current art, such a solution has yet to be found.

The paper I cited does mention that some of the solutions used to approach planar ER3BP problems show promise for continuation into the 3d problem. But, as of publication, the authors had not yet seen this done in any journal.

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    You stated that your source is a paper about Sun/Earth systems, however the diagrams you provided clearly indicate that they show Earth/Moon systems. Is there a terminology disconnect, or perhaps just a simple mistake? – Kamil Drakari Aug 20 at 13:52
  • @KamilDrakari It is my understanding that the moon is the "massless" object in these 3-body problems, and the diagrams are simply shown with Earth as the fixed central point, omitting the sun (since 2-body orbits are well-known). – R. Barrett Aug 20 at 14:33
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    @R.Barrett I haven't read the paper certainly, but the Abstract provided mentions "Earth-Moon system" 3 times (plus once in the title), and "Sun-Earth system" not at all. Even if the intent was to have the Moon act as the "massless" 3rd object after the Sun and Earth, I see nothing to indicate that "Sun-Earth system" is the correct way to explain what the paper is about. – Kamil Drakari Aug 20 at 14:42
  • @KamilDrakari You are absolutely right. That was a mistake on my part. I have edited it. Good catch! – Cort Ammon Aug 20 at 15:00
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    The impression I get from reading the papers is that 3D orbits around the L4 and L5 points are mostly ignored as being uninteresting, with the implication that they're stable and have been known to be so for a long time. – Mark Aug 20 at 20:19

It is indeed possible.

This is for Earth-Moon system, but shows that a small object can orbit two others in a 3d way. And since the third (the spaceship considered) is so small, the other two's motion is nonchaotic.

You would need luck to get a planet on such an orbit, but as you see, the article is quite above KSP-level astrogation, thus fulfilling your desire for pain.in the ass navigation.

As Justin Thyme's insightful answer pointed out, you would need probably more than a bit of luck, as the almost-planar structure of the solar system is the result of the angular momentum of the gas cloud of which it was formed.

Perhaps a giant space station with station-keeping engines would be a better candidate for these orbits, as it would have reason to be there, and means to fight orbital instability and remain there.

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    Note, though, that -- according to your citation -- the orbits are unstable. – RonJohn Aug 19 at 17:32
  • Oh yes. They were intended by the researchers as short-term parking orbits for spacecrafts. – b.Lorenz Aug 19 at 17:35
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    There's no need to tack on "EDIT" everywhere, the system tracks the edit history for those who are curious. It just makes it difficult to read for everyone else. – pipe Aug 19 at 20:32

Try the Kozai mechanism

The basis for the Kozai mechanism is that in a binary system, a third orbiting body - with a much lower mass than the other two - has a quantity that doesn't change in time: $$L_z=\sqrt{1-e^2}\cos i$$ where $e$ is the orbital eccentricity and $i$ is the orbital inclination - the amount it tilts in comparison to a reference plane. As the planet orbits the main body - in this case, a star - the second body perturbs it such that the planet's eccentricity and inclination oscillate in time, with a characteristic period. Both orbital elements are changing, but $L_z$ remains constant. Therefore, even though the planet's orbit is at a given point in time Keplerian, it's actually following a much more complex path in three dimensions.

There are two scenarios where this is commonly considered:

  • A Jupiter-like planet perturbing minor bodies, such as asteroids.
  • A brown dwarf perturbing the orbit of a planet orbiting a companion star.

In your case, I think adding a brown dwarf would be a good idea. If the system is largely compact with the exception of one planet, and the brown dwarf is far enough away from the star, the inner planets wouldn't be affected, and their orbits would be stable and Keplerian. The farther-out planet, with the greater period, would be far more likely to be perturbed by the brown dwarf, and would indeed execute a strange three-dimensional orbit.

Here's an example of how eccentricity changes with time, from Takeda & Rasio (2005):

Eccentricity plots from Tekda & Rasio
Figure 2, Takeda & Rasio. The two curves show two different scenarios, illustrating how the Kozai timescale can be large or small.

If you know $L_z$, you can calculate $i(t)$ from the values of $e(t)$ for any given $t$.

Example: TRAPPIST-1

The TRAPPIST-1 system is an appealing example. It is extremely compact - all seven planets have semi-major axes less than 0.07 AU. They're held stable by resonances. What if we were to put another planet in orbit at, say, 2 AU, and to add a brown dwarf companion at 8 AU?

The mass of TRAPPIST-1 is $0.089M_{\odot}$; a brown dwarf might have a mass of $0.02M_{\odot}$. Kepler's third law tells us that the planet's orbit is 9.48 years. The period of Kozai oscillations is, in the case where the brown dwarf is much more massive than the planet, $$P_{\text{Koz}}=P_p\frac{M_*}{M_b}\left(\frac{a_b}{a_p}\right)^3(1-e_p^2)^{3/2}$$ where the subscripts $p$ and $b$ refer to the planet and brown dwarf. Say the initial eccentricity is $0.5$. For us, this turns out to be $185P$, or 1785 years. In other words, noticeable inclination changes could happen in less than two millennia, which is fairly small on astronomical timescales.

The oscillations will happen quicker if

  1. $M_*/M_b$ is small, which means a low-mass star (like a red dwarf) is a safe bet for the primary, or
  2. $a_b/a_p$ isn't too large.

If we substituted the Sun in for TRAPPIST-1, the effects would be much different, and the period of oscillation would increase by more than an order of magnitude, producing less drastic changes.

The maximum eccentricity is a function of the initial inclination of the planet: $$e_{\text{max}}\simeq\sqrt{1-(5/3)\cos^2(i_0)}$$ If, say, $i_0=75^{\circ}$, we find that $e_{\text{max}}=0.94$, which is enormous. We then calculate that if $e_0=0.5$ and $i_0=75^{\circ}$, $L_z=0.058$. We can then graph $e(t)$ and $i(t)$:

Eccentricity plot

Inclination plot

This assumes that the eccentricity varies sinusoidally; it's likely that the period motions are more complicated than that.

Lissajous orbits and Halo orbits seem to be relevant here.

Lyapunov orbits around a Lagrangian point are curved paths that lie entirely in the plane of the two primary bodies. In contrast, Lissajous orbits include components in this plane and perpendicular to it, and follow a Lissajous curve. Halo orbits also include components perpendicular to the plane, but they are periodic, while Lissajous orbits are not.

As others have mentioned, the "third body" in these three-body problems seems to have negligible mass.

These particular orbits don't conform exactly to the example in your image, but they do demonstrate that orbits don't necessarily have to take place in a 2D plane.

I would suggest that the solution would involve another significant gravity source outside of the elliptical. I am thinking perhaps FOUR suns forming a 3-D triangle, with the planets weaving between them. Or perhaps a major planet rotating around a sun in an orbit on the z-x or x-y axis, with other major planets orbiting on the x-y planar axis. Thus, the smaller planet would seem to wobble as the influence of each major planet came closer to it. I suspect such a system would need a computer simulation to model.

I am thinking along the lines of, after the planetary system were formed, a galactic wanderer came in to the system high above the elliptic, and was captured by the star in some orbit not on the elliptic.

That is, suppose Pluto were an absolutely massive planet. As it orbited off the elliptical, it would deform the orbit of the other smaller planets in some strange wobbly way.

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    That's actually a really cool question. Can a tetrahedral system of bodies (i.e. suns) be stable? Seems to me like they'd have some wildly complex orbits about the common barycenter to give each the angular momentum necessary to counteract gravity. – Dubukay Aug 19 at 17:58
  • @Dubukay I know it is not the same forces, but I was thinking in terms of tetrahedral molecules. – Justin Thyme Aug 19 at 18:41

I am not an expert on orbital mechanics, but I can suggest a possible situation in which a star system has some three dimensional orbits.

I can point out that from one point of view all the planets, asteroids, comets, etc. in our solar system have 3 D orbits. They orbit around the Sun, each orbit being apparently 2D, while the Sun orbits around the center of the Galaxy in a 200,000,000 year or so long orbit. And the plane where the Sun orbits and the plane of the planetary orbits are very much non co planer. So instead of moving in closed elliptical orbits, the planets, etc. in our solar system actually movie in orbits that are open helices from the point of view of the galaxy.

And there is another factor that actually makes the orbits of all bodies in the Solar System slightly 3 D. The planets, asteroids, comets, etc., don't all orbit in the same plane. They orbit in planes that are very closely aligned but still slightly tilted in various directions compared to other orbital planes.

And because the orbit of another planet is tilted slightly with respect to Earth's orbit, half the time that other planet will seems to be In a direction that could be arbitrarily called "above" Earth's orbital plane, and half the time that other planet will seem to be in a direction that could be arbitrarily called "below" Earth's orbital plane.

And because Earth and that other planet are attracting each other with gravity all the time, they are constantly "pulling" on the plane of each other's orbit and tilting the orbits a little more toward each other. But since their relative positions are constantly changing as they orbit the sun separately the relative directions of that pull are constantly changing.

And because the mass of Jupiter is much greater than the pass of all other planets combined, the ever changing direction and distance to Jupiter is the main factor that determines how much force in pulling on Earth and in which direction, but the changing directions and distances to the other planets modify the strength and direction of that force.

And so the tilt of Earth's orbital plane compared to the equatorial plane of the Sun is constantly changing very slightly in one direction or another.

Now image a solar system with two stars, and planets orbiting each star, instead of orbiting both stars. When planets orbit one star in a binary system they are said to have "S-type" orbits instead of "P-Type" or "circumbinary" orbits. In which planes will the planets orbit? The planets of each star in "S-type" orbits will orbit in planes tilted only slightly to the equatorial plans of their stars, since the vast mass of the equatorial bulges of the stars will gradually pull the planetary orbital planes closer and closer to the equatorial planes of the stars.

So can two stars in a binary system have different equatorial planes? If the two stars formed in the same collapsing protostar nebula then they are likely to have very similar equatorial planes which should also be very similar to the orbital plane in which they revolve around each other. But possibly various factors can make the three planes varying by much more than a few degrees.

Stars tend to form in vast collapsing nebulae that contain many smaller collapsing protostar nebulae that form the individual stars. And after all the stars are formed they make up an open star cluster, which gradually dissipates over hundreds of millions of years as the gravity of other stars pull the cluster apart. And while the new stars are orbiting each other in a relatively dense star cluster for tens and hundreds of millions of years there are any opportunities for gravitational interactions to make two stars capture each other and form a binary star.

In any case it seems that it might be possible for the orbital plane that two stars in a binary orbit around each other in to be very highly tilted compared to the orbital plane of any planets in S-type orbits around either of the stars. In some cases the orbit of the other star could be tilted approximately 90 degrees compared to the orbit of a planet around its star.

And thus the gravity of the other star, many times more massive than Jupiter, should pull the orbital plane of a planet "up" on one side and thus "down" on the other, since it has to stay centered on its own sun. But since the years of the planet as it orbits its star close in will be many times shorter than the orbital period of the other star several times farther out, when the planet moves to the other side fr its orbit the other star's pull will now pull the other side of the orbital plane "up".

Thus I picture the orbital plane of the planet wobbling from side to side, going in one direction for half a planetary year and then going in the other direction for the other half of the planetary year.

Any other planets that orbit the same star as this planet will also experience wobbling of their orbital planes, but to greater or lesser degrees, the lesser degrees closer to their star, and the greater degrees farther from their star.

And I think that it might be possible, under some conditions, for some or all of the planets that orbit Star A to have their orbital planes do a full circle around Star A at about 90 degrees to the equatorial plane of Star A. Thus each planet might have an orbit around star A lasting one planetary year in the planet's orbital plane, and also a much longer orbit of its orbital plane as that orbital plane rotates around Star A at an angle of about 90 degrees.

And to me that seems like it might be considered a three dimensional orbit.

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    Yes, I think a binary star system is the answer here. Possibly you could have a difference in size between the two stars as well to set up some interesting scenarios. – Tim B Aug 19 at 19:20

You could use a binary planet. Then the planets would describe a 2D orbit around a common center of gravity, and this would too describe a 2D orbit around the primary.

The resulting orbit could be fully three-dimensional.

I think it cannot happen.

When you have a single attractor, the force it exerts, combined with the motion of the planet, defines a plane, which is the ecliptic.

When you have multiple attractors, you can have two different situations:

  1. the planet is far enough from them to be simply attracted by the center of mass of the attractors, ending up in the single attractor case. Think something like the Earth in the system Sun-Jupiter-Earth
  2. the planet is closer to one of the attractors so that the others can be neglected, but in this case we would again be in the single attractor case. Think something like the Moon in the system Sun-Jupiter-Earth-Moon

Worrying about the attractors being out of the plane of the ecliptic is not needed. For the solar system the plane of the ecliptic is tilted with respect to the galactical plane. Still the orbits we see are contained in a plane.

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    So what of the case in where Pluto were a very massive planet, captured by the system? – Justin Thyme Aug 19 at 18:43
  • I think you only consider two extremity, but there's plenty of room between them to explore. If the planet is close enough to attractor A to orbit it, but attractor B is not very-very far, it can still significantly perturbate the planet's movement. If the orbital planes are not aligned, these will be in 3D. The same goes for case 1. There will always be perturbations, the question is only how severe. Whether this setup may result in a stable system is another question. – Neinstein Aug 19 at 19:05
  • @Neinstein, the moon orbital plane is tilted with respect to the plane where the Earth orbits the Sun. There are perturbation, but not a significantly 3D orbit. – L.Dutch Aug 19 at 19:21

While it seems next to impossible to have a stable orbit that deviates significantly from a planar ellipse, some orbits are truly 3D in a different way: They precess in interesting ways.

Around the earth, there is one especially important class of these precessing orbits, the Sun-synchronous orbits. These orbits (which happen to pass near earth's poles) are arranged in such a way that the plane of their orbit rotates slowly around the earth's axis due to the equatorial bulge of the earth. This rotation is exactly as fast as earth's rotation around the sun, so the satellite always crosses the equator at the same local time.

Satellites that need a lot of power (= sun light on their collectors) just love the Sun-synchronous orbits. They allow them to avoid earth's shadow entirely. True 24/7 power by an orbit that traces across an entire sphere (relative to earth) over the course of a year.

In the coordinate system of the sun the motion can be quite complex.


What are the prerequisites?

  • You actually only need two celestial bodies, a big one, and a tiny one (compared to the big one). More won't hurt, as long as your small body stays close to the big one.
  • The big body must have a bulge. This is easiest to achieve with some smart rotation. Earth's rotation is enough to provide enough bulge for a full precession within a year. Spin the big body faster, and you enable faster precession.
  • The orbit must have a significant inclination relative to the bulge of the big body.

As far as rotational speed goes, the gas giant Jupiter turns around its axis once every ten hours. So that's what's totally believable for a celestial body. It's enough to give Jupiter a visible vertical squeeze, and it should be easy to find some fast precessing orbits around this planet.

The problem with a fast precessing orbit is, that it cannot result from planet formation from an accretion disk. You need a significant inclination to get meaningful precession, and that's what an accretion disk does not provide. So, your small body must have been caught by the big body to explain the inclination.

This also means that the small body will move through the equatorial plane at very significant speeds. If there's anything in its way, you get a nice big explosion. In the long run, these collisions would change the parameters of the orbit to make it more equatorial. So, there must be no remnants of an accretion disk remaining around the big body. (Sorry, Jupiter won't do with its wonderful rings...)


So, take a fast spinning gas giant like Jupiter, subtract rings, add a moon that was caught from interstellar space which happens to have entered a highly inclined orbit. This moon's orbit will precess around the gas giant in a matter of months, giving you your true 3D orbit.

Yes, a binary pair of planets can have a crown-shaped orbit.

enter image description here

Imagine two planets of equal mass $m=m'$. They mutually orbit each other in a circle of radius $r$, centered on their mutual center of mass (barycenter), 180° out of phase with each other. Let's call this circle the "wheel".

This binary pair orbits a star of mass $M>>m$ at a radius $R$. The "axle" of the "wheel" runs along the radius to the star $R$. In other words, the plane of the "wheel" is inclined 90° to the orbit around the star.

If $R>>r$, then the orbits should be stable. The gravitational force radial to the star is essentially only that of the star; thus, each planet orbits the star as if the other planet is absent. The gravitational force tangential to the star is essentially that of only the other planet; thus, the planets orbit each other as if the star was not there.

The path of each planet is a trochoid wrapped around a cylinder. This makes a 3-dimensional "crown" shape. It is not planar, although in order for $R>>r$, it would likely be fairly "flat".

Pluto and its "moon" Charon are close examples to such a system. Their barycenter is outside of either body, and their orbit round each other is inclined about 120° to Pluto's orbit around the Sun.

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    Looks like your planets rotate their orbital plane so its always tangential to their orbit around the star. In this universe the orbital plane should not rotate (much) to conserve angular momentum. I guess, the tidal forces of the star may tilt it slowly, I guess till it is aligned with the ecliptic. – darsie Aug 21 at 8:45
  • As darsie said: There is nothing that can induce the precession of the binary planet in this setup. Look at my answer for a similar setup that actually works. The difference is, that you need a flattened planet (due to rotation) that's orbited by a much smaller satellite with a tilted orbit. With that, the equatorial bulge of the big planet can provide the angular force necessary to create the precession, which an equal sized binary other can't. – cmaster Aug 21 at 21:14

The orbit of the Moon (or the Earth) around the Sun is 3D as the Earth-Moon system has 5.145° inclination to the ecliptic.

Tidal forces caused the Moon to become tidally locked to Earth (=the Moon always points one side towards Earth). Similarly also Earth will become tidally locked to the Moon (this already happened with Pluto and Charon). This will also change Earth's equator to match the Earth-Moon orbit. Since angular momentum is conserved this change in Earths axis has to change the Earth-Moon orbital axis to keep the overall angular momentum constant. So even in a two body system the orbit can be 3D, although since we need to deal with real bodies that can flex, bulge, break and have friction, the gravity of all the particles (atoms) excerted on each other is relevant and their electromagnetic interactions (chemical bonds, repulsion, attraction) which give these bodies their consistency (hard rocks, oceans) over a gas.

IIRC the Sun will go red giant before this finishes and possibly destroy the Earth-Moon system.

Tidal locking

Angular momentum

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    Citations would go a long way to backing this answer up, though it is an interesting spin on orbital mechanics! – GOATNine Aug 20 at 19:54

I got a very loose idea to get your "impossible to nagivate" world that's not immediately disprovable. Set your fiction in a globular cluster. The stars in a globular cluster have box orbits around the center of the cluster. The thing about box orbits is they don't really repeat and you quickly get a mess.

You would want several inhabitable worlds (gotta make em G stars but as big as you dare with your planets as close as you dare without becoming implausible in that direction) and no outer planets at all. Every few years you get a system-system encounter where another star passes 10-20 of your AU (measured by your home planet of course) away. Getting from one inhabitable planet to another is quite the challenge as the energy and time efficient methods probably involve slingshots off other (quite probably larger) stars.

Make sure when setting up your worlds you give them all really strong magnetic fields. The radiation is going to be something else.

This phenomenon is observed in the case of the moons of Jupiter - Janus and Epimetheus.

Source: http://www.planetary.org/blogs/emily-lakdawalla/2006/janus-epimetheus-swap.html

Two moons swap their orbits, so these orbits are 3D.

One example comes to mind: There are several earth observing satellites that are in sun-synchronous orbits. See https://en.wikipedia.org/wiki/Sun-synchronous_orbit The orbit is nearly polar, and is chosen so that it precesses due to the non-spherical earth.

This probably isn't useful to you: For earth, the typical altitude is about 600-800 km, and orbital inclination is about 98 degrees. For a planet the star would have to be seriously oblate. 2 hour orbits tend to make for toasty temperatures around a star (Yes, it would be longer than 2 hours,due to lower density of stars, but still way too close for comfort. Put on your silvered asbestos long johns.)

I suspect that having a jupiter mass inner planet would give the effect of an obese primary. E.g. put a jupiter mass in Mercury's orbit, then orient your planet at 95 degrees to the ecliptic.

Note: You need to explain the unusual inclination. However for story purposes, you can state it as an observation, and claim that the mechanics of how it got that way are still being investigated by the boffins.

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