So we have had multiple questions on having multiple moons and this is another. I did check and didn't find questions that covered this specific setup.

I have an world I am working on, including a similar moon.

In my world I would like to add a second smaller moon that is about half the size of our moon and orbits Earth farther out than our moon. The orbit, if you are looking down at the north pole (the dot), would look something like this. I would also like the outer/smaller moon to orbit at a different rate so that at times it is behind the larger moon and at other times is on the other side of the planet.

enter image description here

Is this setup possible?

  • Is Earth's gravity strong enough to have a second moon further from us than our current moon?
  • Is there a certain size it would have to be?
  • Can an orbit like this be stable? (two moons, one half the size of the Moon and moving at different speeds, relative to Earth)
  • 2
    $\begingroup$ Good question. Would an answer have to take into account the formation of the extra moon? Earth's Moon formed from a giant impact, and capture would require interaction with another body - an unlikely chance. That would seem to indicate another impact of some sort, which could change the planet quite a bit. $\endgroup$ – HDE 226868 Aug 26 '15 at 21:49
  • $\begingroup$ Another interesting feature i have added to my answer - the luminosity of this moon would be very different. Distance affects light intensity, and amount of surface area would also affect amount of light reflected, so this moon would be very much dimmer than our current moon, unless its surface contained some highly reflective material. $\endgroup$ – user11864 Aug 27 '15 at 22:12
  • $\begingroup$ Look up Earth's Hill Sphere, perhaps on a different SE, to see how far out another moon could be, The Earth had other moons, but this big one ran them down as it receeded. In your plans, consider the history of formation, not just the current state. $\endgroup$ – JDługosz Aug 29 '15 at 3:40
  • $\begingroup$ Why such a big moon? A moon the size of Ceres would provide an area about the size of Argentina with a mass of only 1/80 moon masses (as far as I know this is small enough for the L4 or L5, but you can put it further away in orbital resonance) but the surface gravity would be a bit low at 0.029g although enough to keep objects from floating away. $\endgroup$ – k-l Feb 12 '16 at 0:12
  • 4
    $\begingroup$ I'm just thinking of Minmus. I've been playing to much Kerbal Space Program. $\endgroup$ – Amziraro Feb 13 '16 at 14:19

Note: If anyone can double-check my numbers and result at the end of this answer, that would be much appreciated.

This looks like a question that can be broken down as per your bullet points, so I think I'll do it like that.

Is earth's gravity strong enough to have a second moon further from us than our current moon?

As JDługosz pointed out, the answer to this depends on whether or not the moon is inside Earth's Hill sphere, the region in which it can hold satellites in stable orbits. The general formula for a body of mass $m$ orbiting a body of mass $M$ with a semi-major axis of $a$ is $$r\approx a\sqrt[3]{\frac{M}{3m}}$$ For Earth, this is[1] $$r_{\oplus}\approx a_{\oplus}\sqrt[3]{\frac{M_{\odot}}{3m_{\oplus}}}=1.496\times10^{9}\text{ meters}$$ This is roughly one tenth of the distance to the Sun (and four times the Moon's orbital radius). It doesn't take into account influences from the other planets, but that shouldn't be an issue. After all, Venus and Mars are both much further than 0.1 AU away from Earth, even at their closest passes.

Regarding your request for different angular velocities: Two bodies in circular orbits with different radii will always move at different angular and tangential velocities, so you're fine there.

Is there a certain size it would have to be?

So long as the mass of the extra moon is much less than that of the planet, there should be no ill effects on either Earth or the Moon. Any non-zero mass will perturb the orbit of the Earth a bit, so there's no cutoff line. You just have to specify what limit is okay with you.

It's interesting to think of the effects of the moon on the Moon. I have something to say on the subject - something that I've been meaning to write about for a while - but I'll make that a separate section.

Can an orbit like this be stable? (two moons, one half the size of the moon and moving at different speeds, relative to earth)

Well, yes, it can - again, so long as the moon is far enough away from the Moon. The places where the setup are unstable can be found easily - again, this will come later. The short answer, though, is that this should be perfectly safe. Just use Newton's law of gravitation: $$F=G\frac{m_{\text{Moon}}\frac{1}{2}m_{\text{Moon}}}{(r_{\text{Moon}}-r_{\text{moon}})^2}$$ This gives you the force at closest approach - which should be minimal, for large enough values of $r_{\text{moon}}$.

Here's the section on stability - the stability of the Moon's orbit - which I alluded to earlier. I'm going to base it strongly off of these lecture notes.

Here, we use the disturbing function, which has the same magnitude as the gravitational potential from the perturbing1 body - in this case, the extra moon. The general form of it is $$\mathcal{R}=-\frac{Gm_{12}m_3}{R}+\frac{Gm_2m_3}{|\mathbf{R}-\alpha_1\mathbf{r}|}+\frac{Gm_1m_3}{|\mathbf{R}+\alpha_2\mathbf{r}|}\tag{1}$$ where the subscripts $_1$, $_2$, and $_3$ refer to Earth, the Moon, and the moon, respectively, $m_{ij}=m_i+m_j$, $\alpha_i=m_i/m_{12}$.

To analyze this, we have to expand it - basically, write it in another form using summations and functions called spherical harmonics. For those interested, the lecture notes provide a full derivation, but to make it short, the intermediate expansion is $$\mathcal{R}=G\mu_im_3\sum_{l=2}^{\infty}\sum_{m=-l}^l\left(\frac{4\pi}{2l+1}\right)\mathcal{M}_l\left(\frac{r^l}{R^{l+1}}\right)Y_{lm}(\theta,\varphi)Y_{lm}^*(\Theta,\psi)\tag{2}$$ where $$\mathcal{M}_l=\frac{m_1^{l-1}+(-1)^lm_2^{l-1}}{m_{12}^{l-1}}$$ and $Y_{ab}(\beta,\gamma)$ is a spherical harmonic.

Why do we care? Good question. The disturbing function allows us to identify orbital resonances2, which can help a system stay stable.

To do this, we have the expand the disturbing function again, this time as a Fourier expansion. We get $$\mathcal{R}=G\mu_im_3\sum_{l=2}^{\infty}\sum_{m=-l}^{l}c_{lm}M_l\left(r^le^{imf_i}\right)\left(\frac{e^{-imf_o}}{R^{l+1}}\right)e^{im(\varpi_i-\varpi_o)}\tag{3}$$ where $$c_{lm}=\frac{(l-m)!}{(l+m)!}[P_l^m(0)]^2$$ where $P_l^m(x)$ is an associated Legendre polynomial and $f$ and $\varpi$ are orbital elements.

This can all be expanded further by way of . . . am I boring you? Well, I'll skip to the good bit. I just wanted to emphasize that this one little function contains a lot of information.

We eventually arrive at a function $\phi_{n'nm}$ that is called the resonance angle, given by $$\phi_{n'nm}=n'M_i-nM_o+m(\varpi_i-\varpi_o)\tag{4}$$ $M$ is the mean anomaly and $\varpi$ is the argument of periapsis.

We can find the rate of change of this with respect to time, $\dot{\phi}$, and then write a function $\mathcal{E}$ as $$\mathcal{E}=\frac{1}{2}\dot{\phi}^2-\omega^2(\cos\phi+1)$$ There are three possible classes of values of $\mathcal{E}$:

  1. $\mathcal{E}>0$: The system is circulating.
  2. $\mathcal{E}<0$: The system is librating.
  3. $\mathcal{E}=0$: The system is at an unstable equilibrium (think of a pendulum pointed straight up).

We can then write down another equation where $$\ddot{\phi}\propto\sin\phi$$ The associated $\mathcal{E}$ tells us if the system is stable.

There's an explicit stability algorithm by the author of the lecture notes in a related paper (see Mardling (2008) using a different expansion. It should be able to give you a yes-or-no answer for resonant stability for given orbits of the Moon and the moon. I'll see if I can apply it here.

The precise algorithm is

(1) Identify which $[n : 1](2)$ resonance the system is near and calculate the distance $\delta_{\sigma n}$ from that resonance: $\delta_{\sigma n} =\sigma-n$, where $n = \lfloor\sigma\rfloor$. (the nearest integer for which $n \leq \sigma$);
(2) Take the associated resonance angle to be zero rather than the definition (2.18) (see discussion below): $\phi_{2n1} = 0$;
(3) Calculate the induced eccentricity from (5.1) and (if $m_1 = m_2$) the maximum octopole eccentricity from (5.3). Determine $e_i = \text{max}[e^{(ind)}_i, e^{(oct)}_i]$ for use in $s^{(22)}_1 (e_i)$;
(4) Calculate $\mathcal{A}_{2n1}$ from (3.6);
(5) Calculate $\mathcal{E}_{2n1}$ and $\mathcal{E}_{2 \ n+1\ 1}$: deem the system unstable if $\mathcal{E}_{2n1} < 0$ and $\mathcal{E}_{2 \ n+1\ 1} < 0$.

I've run some numbers, and gone through a resonance or two. Mardling focused on $[n:1](2)$ resonances, which are important here. I found that for a $[2:1](2)$ resonance, the system is unstable at most eccentricities, but it seems - judging from some of the graphs given - that at higher resonances, the system is perfectly stable.

I ran through a certain combination of values to test for stability; it would be awesome of someone could check them. Here they are, with intermediate values (the subscripts $_i$ and $_o$ refer to the inner and outer satellites, the Moon and the moon).

The given data are (I've chosen the last two): $$m_1=m_{\text{Earth}}=5.9722\times10^{24}\text {kg}=81.285 m_2$$ $$m_2=7.3462\times10^{22}=1m_2$$ $$m_3\equiv\frac{1}{2}m_2$$ $$e_i(0)=0.0549$$ $$e_o(0)=0.01$$ $$a_o=2.75a_i$$ The intermediate values are: $$\sigma=4.5466$$ $$n=\lfloor\sigma\rfloor=4$$ $$\delta\sigma_4=\sigma-4=0.5466$$ $$n=4,n'=1,m=2$$ $$e_i^{(eq)}=0.02206$$ $$A=1.4891$$ $$e_i^{(oct)}=0.09902$$ $$l=l_{min}=2$$ $$s_{224}=0.44833$$ $$F_4^{(22)}(e_o)=2.712\times10^{-6}$$ $$f_4^{(22)}(e_o)=2.6513\times10^{-6}$$ $$\beta_4=5.6513\times10^{-8}$$ $$e_i^{(ind)}=0.0549$$ $$e_i=\text{max}[e_i^{(ind)},e_i^{(oct)}]=e_i^{(oct)}=0.09902$$ $$s_4^{(22)}(e_i)=-0.16443$$ $$\mathcal{M}_2=1$$ $$M_i^{(2)}=0.006114$$ $$M_o^{(2)}=0.0119568$$ $$\sigma_4=1$$ $$c_{(22)}=\frac{3}{8}$$ $$\mathcal{A}_{241}=3.9059\times10^{-8}$$ $$\delta\sigma_{41}=\sigma-n/n'=0.5466$$ $$\bar{\mathcal{E}}_{241}=0.1493857019$$ This is greater than zero, so the system is stable.

Update, 28 October 2016

I've been going through and checking some of these numbers again, in the process of trying to write an algorithm around this, and I'm fairly certain that there are some errors, potentially including (and thus starting with) $s_4^{(22)}(e_i)$. This may mean that the system is unstable, as kingledion's results show.

I've been running some simulations of my own, using the European Space Agency's open source ORSA software. I've only done four or five runs over shot periods of time, but they're starting to fall into two categories: Either the secondary moon is slowly ejected, or the two go into separate elliptical orbits, with periodic changes in eccentricity and semi-major axes. For the sake of accuracy, I've only run the simulations over one year, but it seems like there may be stable setups and there may be unstable setups.

Additionally, I haven't included the Sun in this, so I don't know how it could change any of this.

1 In this case, I'm talking about perturbing the Moon.
2 Specifically, orbit-orbit resonances.

| improve this answer | |
  • $\begingroup$ Could you add a table of the acceptable distance / mass ranges? $\endgroup$ – nu everest Feb 23 '16 at 1:32
  • $\begingroup$ @nueverest That would require a lot of computations. Some of the terms (e.g. $e_i^{(oct)}$) are defined using piecewise functions, meaning that it's not quite as simple as plotting a stability boundary as a smooth analytical function. The author of the paper does provide some graphs of $\sigma$-vs.$e_o$ (distance-vs.-eccentricity, instead of distance-vs.-mass) as a result of theoretical computations; would that be good? $\endgroup$ – HDE 226868 Feb 23 '16 at 1:39
  • $\begingroup$ I was thinking in terms of attaching a propulsion system to an asteroid. According to this en.wikipedia.org/wiki/Comet_nucleus Halley's Comet has a mass of 3E14 kg. We could use that to limit the objects mass. $\endgroup$ – nu everest Feb 23 '16 at 1:54
  • $\begingroup$ @nueverest Hm. Well, regarding the comet, I highly doubt that would perturb the system significantly. It's ~100,000,000 times less massive than the Moon. $\endgroup$ – HDE 226868 Feb 23 '16 at 1:56
  • 1
    $\begingroup$ Well I don't think I have to familiarity with orbital mechanics to parse out that paper and try to run a simulation myself, but I did put your given data into rebound. Rebound had the second moon escaping earth's orbit and getting into its own orbit around the sun somewhat outside of the orbit of earth. I ran the simulation for an hour or so, but presumably if I had more computing power the second moon would end up crashing into the earth, or the moon, or both. I can post my simulation setup in a separate answer if you want. $\endgroup$ – kingledion Oct 27 '16 at 2:16

No, that setup is not stable

There do not appear to be any stable orbits for another moon, of any size, outside the orbit of Luna

I ran a simulation in rebound to see what would happen with this moon setup. I used a grid of different behaviors to check how the second moon would react in different scenarios.

Here was my setup in rebound:

import rebound
from math import sqrt

for m_selene in [3/2, 1, 1/2, 1/4, 1/10, 1/100, 1/1000]:
    for a_selene in [3/2, 2, 3, 5, 10]:
        for e_selene in [0, 0.01, 0.02, 0.05, 0.1, 0.25]:
            sim = rebound.Simulation()
            sim.integrator = 'whfast'
            sim.units = ('AU', 'days', 'Msun')

            sim.add(m=0.000003004, a=1, e=.016709)
            sim.add(primary=sim.particles[1], m=0.000000037, a=0.00257, e=.0549)
            sim.add(primary=sim.particles[1], m=0.000000037*m_selene, a=0.00257*a_selene, e=e_selene)

            earth_luna = sqrt((sim.particles[2].x - sim.particles[1].x)**2 + (sim.particles[2].y - sim.particles[1].y)**2)
            earth_selene = sqrt((sim.particles[3].x - sim.particles[1].x)**2 + (sim.particles[3].y - sim.particles[1].y)**2)
            print(m_selene, a_selene, e_selene, "{0:.2f}".format(earth_luna), "{0:.2f}".format(earth_selene))

First I ran the simulation for just the earth-moon system to ensure it was stable, and it was. Then I ran it again with all objects (Sun, Earth, Luna — the real moon, Selene — a new outer moon).

I modified the above program to find 'breakpoints' where different behavior was observed. This is what I got:

  • Let m_selene be the mass of Selene as a proportion of Luna's mass (so 0.5 means Selene is half the size of the moon).
  • Let a_selene be the semi-major axis of Selene as a proportion of Luna's semi-major axis (so 2 means twice as far from the Earth as Luna).
  • Let e_selene be the eccentricity of Selene.

Then there are four scenarios.

  • Luna is ejected from the system, and Selene becomes the new moon. This occurs only in rare circumstances when Selene is large and close to Luna (m_selene > 1.25, a_selene < 1.75) or when Selene is medium size but with a highly eccentric orbit (m_selene > 0.35, 1.25 < a_selene < 2.25, e_selene > 0.075).

  • Luna and Selene are both ejected from the system. This is likely when Luna and Selene are about the same size and close to each other (m_selene > 0.35, a_selene < 2.25). This happens in both highly eccentric and non-eccentric orbits of Selene. It can also happen rarely for smaller masses of Selene (as low as m_selene > .125) but not for larger orbits.

  • Luna and Selene are (somehow?!?!) ejected from the solar system. I don't see how this is possible, but it occurs in the case of m_selene = 0.5, a_selene = 1.5, and the eccentricities of selene and the moon are about the same.

  • The most common situation is that Selene is ejected from the system and Luna remains. This happens in every (actually all but one) case where m_selene < 0.35 and a_selene > 2.25, and rarely when Selene is larger or closer.

So what you do not notice is any situation where Selene and Luna are both stable in orbit around earth. So this is not an exhaustive numerical solution, nor is it an analytical proof, but from a grid search of a numerical solution, I cannot find any stable orbits for a second smaller moon outside the orbit of Luna.

Edit: In response to @Molot's request for orbits inside of Luna, it turns out that most orbits inside Luna's are stable for at least a short time. The grid search only integrates 10 years, so that doesn't prove long term stability, but I ran a simulation for:

  • m_selene = 0.5
  • a_selene = 0.5
  • e_selene = 0.02

and it was stable for 100,000 years. The moon moved slightly outwards over the first 1000 years so that Luna and Selene achieved a resonance, but then the two were stable for as long as the simulation was running.

I can't promise stability for billions of years, and to be truly accurate I'd have to account for Jupiter and Saturn (at least) and the early solar system planet migrations, but it seems plausible that Selene could exist in an orbit inside that of Luna.

| improve this answer | |
  • $\begingroup$ How hard would it be to look inside orbit of Luna, instead? $\endgroup$ – Mołot Oct 29 '16 at 22:47
  • 2
    $\begingroup$ @Molot Not too hard, I'll give that a go and post the results. $\endgroup$ – kingledion Oct 29 '16 at 23:29
  • $\begingroup$ Awesome. Thanks for running these. Putting the small moon inside the orbit of the other could work just as well for what I am trying to do. $\endgroup$ – James Oct 31 '16 at 13:45
  • $\begingroup$ Very helpful for building intuition about how systems like this work in general. $\endgroup$ – ohwilleke Nov 3 '16 at 7:48

(Cliff notes version for Hard-science)

Yes, the Earth could have multiple moons. Mars has two, though granted they are tiny.

The big issue you have is that the orbit determines the speed at which one body orbits another. So the farther away you are, the slower/longer the orbit. So in this case the smaller moon would be slower than the big moon, which would seem to pass it by now and then.

The other issue would be that the orbits would need to be far enough apart that their gravity doesn't mess each other up: Epimetheus and Janus around Saturn switch orbits periodically, because their orbits are too close together.

| improve this answer | |
  • $\begingroup$ I just made an edit, I guess I don't care if it is faster or slower relative to an earth viewer, just that they are different. $\endgroup$ – James Aug 26 '15 at 21:38
  • 1
    $\begingroup$ +1 for very interesting (and on point) reference to Epimetheus and Janus. $\endgroup$ – user11864 Aug 27 '15 at 21:52
  • 1
    $\begingroup$ At twice the distance from the earth, its velocity would be 1.4 times slower (velocity squared = (G * m(e))/r, where G is gravitational constant, m(e) is mass of earth, and r is radius of orbit). The mass of the object does not matter, its orbital speed is dictated purely by its distance from earth. $\endgroup$ – user11864 Aug 27 '15 at 21:56
  • 6
    $\begingroup$ This is not a hard-science answer. $\endgroup$ – JDługosz Aug 29 '15 at 3:41
  • 1
    $\begingroup$ Sorry man, you got outscienced by @HDE 226868 $\endgroup$ – James Feb 12 '16 at 20:15

The moon's orbit is about 380 thousand kilometers from the earth. The moon is about 1/100th the mass of the earth and it's diameter is about 1/3.5 that of the earth (1/3.66... but close enough). The moon is less dense than the earth.

Suppose you had a second moon twice as far out as our current moon, and say 1/2 the diameter of our moon. Say that it is the same density as our moon. It's mass would be 1/8th that of our moon just by the difference in volume, which would make it 1/800 the mass of earth. When aligned with our moon, the new moon's pull on our current moon, being at the same distance, would only be about 0.125% that of the earth. It would cause a pretty good wobble on the moon's orbit, but nothing out of the realm of possibility.

Our moon is about 100 times less massive than earth, but, when aligned, would be at half the distance from this new moon. So our Moon's pull on this new moon would be about 4% that of the earth. The wobble in the orbit of the new moon would be significantly more pronounced, but again nothing that the right orbit could not account for.

This new moon would be 1/8 the mass of our moon, and twice as far, making its pull on earth 1/32 that of our moon, or about 3% that of the moon. I do not think this would yield any significant atmospheric effects, but again, I am no astrophysicist. Mind you, this smaller moon's orbit would also change in speed near the alignment. As it approached alignment, it would speed up, and as it passed alignment it would slow back down. Overall, at twice the distance from earth, it would move about 1.4 times slower than our current moon. But it's angular speed, which is what we would "see" would be about 2.8 times slower than the moon (the square root of 8 for anyone who cares). Also, light intensity changes inversely proportional to the square of the distance, so at twice the distance, the same object would be 1/4 as bright, but this moon would be much smaller (1/4 the surface area) and possibly made of different material, so even if it was quite reflective, it would be somewhere near one tenth as bright.

Finally, as others have said, other planets in our solar system have multiple moons. I am no astrophysicist (just a physics nut) but it would seem there is nothing to preclude your scenario. So yes, this is plausible, and the orbit would be stable. At those distances, the orbits would have some interesting characteristics, but I am fairly sure it would be something for astronomers, not very visually apparent to the average earthman observer.

| improve this answer | |
  • 2
    $\begingroup$ Sources, please. $\endgroup$ – HDE 226868 Aug 27 '15 at 20:25
  • $\begingroup$ @HDE226868 my own calculations based on F(g) = G*m1*m2/r^2, m=density x vol, and vol = c x r^3, and rotational acceleration = w^2 x r. All formulas any physics student will know by heart. The distances and mass of the earth and moon are easily found on Wikipedia or one of a hundred sources. $\endgroup$ – user11864 Aug 27 '15 at 20:39
  • $\begingroup$ Also, @bowlturner, in his post, which I liked very much, points out Saturn's moons Epimetheus and Janus. Their orbital separation is in the order of 50 km, over orbits of 91,000 kilometers. That is why their effect on each other is much more pronounced. When you are talking about orbital separation as big as the orbit of the inner moon, the effects will be nowhere near that. $\endgroup$ – user11864 Aug 27 '15 at 20:44
  • 1
    $\begingroup$ "This new moon would be 1/800 the mass of our moon," — No, 1/8 of the mass of the moon, according to your assumptions above. $\endgroup$ – celtschk Aug 28 '15 at 20:01
  • $\begingroup$ @celtschk - you are correct. I edited accordingly. That was a bit of a brain-hiccup. Sorry about that. $\endgroup$ – user11864 Aug 28 '15 at 21:42

Here's an example that's weird & wonderful. Usual disclaimers about numerical integration apply, of course. Mass is ~21% of the mass of the moon, which works out to a diameter of ~60% of the moon, assuming equal densities (a probably-bad assumption).

Some thoughts:

  • ~21% of the mass of the Moon.
  • In an (retrograde!) 2:3 resonance with the moon. Well, sort of. It may be one of those weird resonances that's close to 2:3.
  • It'll occasionally be eclipsed by the Moon, but not very regularly, due to the inclination. (Also, the eclipses will be fairly short, due to the retrograde orbit.)
  • It's stable over as long as I care to run it for (~100 years with the accurate integrator, >10,000 with the fast integrator and default step size (0.001 day), >512k with the fast integrator and step size of 1/1000th of a year). On the other hand, I'm not a theory guy - this not actually be stable at longer timescales.
  • I'm just simulating Earth / Moon / Selene / Sun. I'm not simulating e.g. Jupiter, which may (read: will) affect things.
    • Correction: I've tried simulating Mars + Jupiter also, at least a few years, but I haven't tried doing any long-term simulations with them. This resonance seems surprisingly robust considering how "wobbly" it is.
  • Tides will be interesting.

    import time
    import rebound
    from math import sqrt
    sim = rebound.Simulation()
    sim.integrator = 'hermes'
    sim.units = ('AU', 'days', 'Msun')
    sim.add(primary=sim.particles[1], m=0.000000037*0.21564912733016417, a=0.00257*1.8967736522524086, e=0.08825717827598856, inc=2.695021633949315, Omega=5.385750562430302, omega=0.42668650997546287, f=0.7633635278610188)
    for orbit in sim.calculate_orbits():
    sim.ri_whfast.safe_mode = 0
    sim.ri_whfast.corrector = 11
    earth_luna = sqrt((sim.particles[2].x - sim.particles[1].x)**2 + (sim.particles[2].y - sim.particles[1].y)**2)
    earth_selene = sqrt((sim.particles[3].x - sim.particles[1].x)**2 + (sim.particles[3].y - sim.particles[1].y)**2)
    print("{0:.8f}".format(earth_luna), "{0:.8f}".format(earth_selene))

(With an answer of)

    0.00249790 0.00466839

I found this via a repeated random-start local search over the search space for orbits that took the longest to escape; I was unable to improve beyond this using this method because I have yet to have had this orbit escape. Perhaps I'll switch back to "minimize max distance from Earth - min distance from Earth over a set timeperiod". Though that tends to end up with "boring" orbits.

enter image description here

This is the Earth-Moon-Selene system after 1 year, sidereal coordinates centered on the barycenter of said system (hence why Earth moves slightly), coordinates are in AU. In the original, 1 frame == 1 / 10th of a day; I am unsure as to if this persists.

| improve this answer | |
  • $\begingroup$ What are the "usual disclaimers about numerical integration"? $\endgroup$ – Kate Bertelsen Apr 27 '18 at 17:10
  • $\begingroup$ @KeithB - N-body systems tend to exhibit chaotic behavior. There's a big difference between "seems to be stable" and "is stable, and I didn't do much of any analysis as to what the timescale of said chaos is. Or the magnitude of any numerical inaccuracies, for that matter. Take this as 'a good possibility it is possible', not 'proof it is possible'. $\endgroup$ – TLW Apr 29 '18 at 22:09

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