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Is it possible for a spacecraft to achieve an orbit around a celestial body that has no orbital eccentricity at all? The smallest orbital eccentricity of a natural satellite is that of Triton's orbit around Neptune. Triton's orbit is almost perfectly circular despite allegedly being a captured dwarf planet. Is it possible for spacecraft to achieve an even more perfect orbit or a perfect one with no orbital eccentricity? Please don't say geostationary satellites since they don't orbit the Earth relative to its surface. I'm asking for craft that actually revolve around the Earth or another celestial body. One with little mascons, obviously since such strong mascons like on the Moon would make a perfectly circular orbit unlikely.

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    $\begingroup$ I'm a bit unclear why you're excluding geostationary orbits. There's nothing particularly special about them, except that they have their orbital period set to match the planet's rotational period. They're still an orbital path, same as any other. Some orbits are certainly intended to be circular, but of course aren't exactly circular. That's really just down to a 'circle' being a geometrical concept, whereas we live in the real-world, full of non-spherical cows and air resistance. $\endgroup$ – Josh Eller Apr 3 at 16:22
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    $\begingroup$ There are no perfect circles in nature. Anywhere. There are no geometrically perfect objects or world lines in spacetime - man made or otherwise. Anywhere. No spheres, no squares, no parabolas, no sine waves - not perfect. $\endgroup$ – J... Apr 3 at 19:08
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    $\begingroup$ @J... is exactly correct. A perfect circle is a mathematical construct, which requires infinitesimal precision that is impossible to achieve in the physical universe - and it's not because we don't have precise enough tools, it's because the very laws of physics preclude that level of precision. At extremely small scales, it's not even possible to say that the orbit is here and not there, which means we can never have a perfectly defined circle in reality. $\endgroup$ – Nuclear Hoagie Apr 3 at 19:40
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    $\begingroup$ This question needs clarification. "Perfect" isn't very meaningful. To what accuracy are you interested in? Why exclude geostationary orbits? If you can make a circular geostationary orbit, then you can make a circular orbit at some other altitude. $\endgroup$ – Harabeck Apr 3 at 20:19
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    $\begingroup$ @user30007 And who, exactly, is out there with all the digits of pi that verified the perfection of Bondone's circle? Yeah... $\endgroup$ – J... Apr 4 at 11:52
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If you define perfection as absolutely zero eccentricity then perfection is impossible. There will always be eccentricity in an orbit, even if it is very small. Orbits vary due to:

  • Spacecraft system inaccuracy: no spacecraft is perfect, no matter how accurate
  • Changes in the density of the orbited planet
  • Gravitational influence of other celestial bodies: my answer to this question shows that Jupiter's influence on a body in Earth orbit when the Earth and Jupiter are close together is 3.2e-7 m/second squared. That isn't much, but it will have in influence

No orbit can be perfect because the gravity of other celestial bodies will always cause a small amount of eccentricity. If you had a perfectly even sphere far away from any other celestial bodies, say halfway between galaxies, and a spacecraft with an extremely accurate guidance and control system you could get close to perfection. However, gravitational influence will always cause some minute eccentricity even if it is undetectable.

So you could get close to, but never achieve perfection.

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    $\begingroup$ Software should still handle perfectly circular orbits correctly to cope with finite precision. you wouldn't want your software to go haywire just because the readings momentarily show a perfectly circular orbit with undefined ap/pe. I mean, its unlikely that you'll ever measure that, but not impossible (reminds me of the engine software for aircraft that couldn't cope with exactly 70 mph readings). $\endgroup$ – Polygnome Apr 3 at 9:23
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    $\begingroup$ @Polygnome that's why god invented " >=" :-) . The other common disaster is dividing by X and assuming X is never zero. $\endgroup$ – Carl Witthoft Apr 3 at 11:42
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    $\begingroup$ Okay folks please tone it down and restrict comments to the topic. @user30007 I think CW used lower case to avoid referring literally to a deity, but instead to use a common expression in English that is understood to be secular only. However CW's last comment is way out of line and I've flagged it for moderator attention. $\endgroup$ – uhoh Apr 3 at 15:50
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    $\begingroup$ @uhoh Thanks for drawing attention to this. Yes, user30007 and Carl, you both need to refrain from making this personal. user30007, this is a secular site, and we don't police the theological correctness of users' speech. Carl, you didn't have to target the user's personal life to respond here. $\endgroup$ – called2voyage Apr 3 at 16:04
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    $\begingroup$ Might also mention that you can't claim to have achieved any goal with greater precision than your ability to measure. $\endgroup$ – Solomon Slow Apr 3 at 17:39
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You would need a perfect two body system, a perfect central body and a perfect spacecraft. No other planets or stars influencing the orbit.

The central body should be a sphere with constant density or at least spherical symmetry.

You need a perfect measurement of orbital parameters.

The thrusters of the spacecraft needed for circularation of the orbit should be controlled with infinite precision.

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  • $\begingroup$ Spacecraft in an orbit should be in free fall, so no use of thrusters. $\endgroup$ – user35272 Apr 3 at 13:42
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    $\begingroup$ @user30007 Thrusters are needed to correct errors of the orbit to circularize it perfectly. Once the orbit is perfect the thrusters are not used anymore. $\endgroup$ – Uwe Apr 3 at 13:53
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    $\begingroup$ The sphere doesn't have to have a constant density, but it does have to be shperically symmetrical, i.e. at a given radius from the center the density is constant in all directions. Essentially, the sphere can consist of spherically symmetrical shells. $\endgroup$ – Tom Spilker Apr 3 at 17:14
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    $\begingroup$ Technically you'd need an empty universe too, this is another one of those relative to what questions... $\endgroup$ – Magic Octopus Urn Apr 3 at 21:06
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    $\begingroup$ @MagicOctopusUrn You might not need an entirely empty (aside from the two identical bodies) universe. If all the other mass in the universe is concentrated along the line normal to the bodies' orbit plane and through their barycenter, for a specified orbit radius it changes the orbit period, but I don't think it perturbs the circularity of the orbit. But that's a pretty special universe, completely divorced from reality! :) This works for Newtonian physics and special relativity but I haven't pondered the implications of general relativity. But there are other real world complications... $\endgroup$ – Tom Spilker Apr 3 at 21:29
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The answer depends: how accurate is the physics you consider?

Considering the symmetric system postulated in other answers/comments, one with two spherically-symmetric bodies of equal masses, orbiting their barycenter: if you use only Newtonian physics, empty the universe of all masses other than the ones you're focusing on, and make the bodies perfectly rigid (i.e., ignore the physics of non-rigid bodies to maintain their spherical symmetry), then indeed theoretically you could have perfectly circular orbits. You don't have to specify just two bodies because under these assumptions theoretically you could have three or more such objects orbiting the barycenter in circular, coplanar orbits. This geometry was discussed in a science fiction novel some decades ago — Ringworld? I don't have time to look it up.

But the foregoing assumes three things that we know not to be true in the real world. What happens when you replace these ideological models with real-world realities?

Apart from a single case, as soon as you allow other mass in the universe, "perfect" goes out the window. That single case is actually a class of cases, in which 1) all the other mass in the universe is concentrated along the line normal to the bodies' orbit plane and passing through the barycenter, and 2) those masses are either static (non-moving) or moving in such a way that the net gravitational force at the orbit of the bodies in question doesn't change. This works for Newtonian physics and special relativity, but as I said in a comment, I haven't pondered the implications of general relativity. For that matter, case 2 above is difficult to pull off, even in Newtonian physics, without resorting to electric charges to counter what gravity is trying to make them do. Hmm ... most models of the universe don't involve such geometry ... and they're based on data. So much for that single case!

If you allow non-rigid bodies (but keep Newtonian physics and an otherwise-empty universe), unless the bodies rotate at the same angular rate as they orbit (they are locked in terms of relative orientation), then tidal effects perturb the orbits. Assuming two bodies, their mutual tidal forces perturb their shapes into prolate spheroids — like rugby balls. If they rotate at the same angular rate as they orbit, then the long axes of the prolate spheroids line up with each other and there is no non-radial (from the barycenter) component of gravitational force, so a circular orbit is possible. Rotation at a rate different from synchronous with the orbit then drags the long axes of the prolate spheroids away from pointing directly toward the other prolate spheroid's center. This results in a non-radial component of force that either adds orbital energy to the other object or removes energy, depending on whether the angular rotation rate is faster or slower than the orbital rate. That makes the orbit spiral either outward or inward.

A spiraling orbit is not perfectly circular!

Once you abandon Newtonian physics and include general relativity, all possibility of a perfectly circular orbit goes out the window. Even if you have the symmetric two-body system that under Newtonian physics yields circular orbits, under general relativity the system radiates gravitational-wave energy at the expense of the system's orbital energy. The objects spiral inward.

A spiraling orbit is not perfectly circular!

The net conclusion: if you entertain a simple, ideological universe you can contrive situations that appear to give perfectly circular orbits — theoretically. As soon as you open the door to any form of real-world universe, especially real-world physics, all such possibility goes away.

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    $\begingroup$ Yes, the Ringworld (1970) series. One of the later ones (2007), larryniven.fandom.com/wiki/Fleet_of_Worlds, was even titled after (and set in) bodies in a "Kemplerer [sic] rosette". (kinda spoilers for a plot point of older books / short stories, future readers beware) In real life, a Klemperer rosette (wikipedia), note spelling difference, is not exactly what Niven described, but similar). $\endgroup$ – Peter Cordes Apr 4 at 2:03
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    $\begingroup$ Oh good point about gravity waves. We don't even have to invoke quantum mechanical fluctuations and what it even means to measure position, or Is spacetime discrete or continuous? - but those are potential sticking points as well. $\endgroup$ – Peter Cordes Apr 4 at 2:07
  • $\begingroup$ So to sum it up: only orbits of tidally locked bodies can be perfectly circular and only in Newtonian physics? Such as Pluto-Charon? $\endgroup$ – user35272 Apr 4 at 6:06
  • $\begingroup$ @user30007 Yep, but only in a highly contrived universe. The Pluto-Charon system is close — but no cigar. There are other moons in the system that perturb the orbit, and both Pluto and Charon don't have spherically symmetrical mass distributions, so Charon's orbit eccentricity is really small, 0.0002, but not zero. $\endgroup$ – Tom Spilker Apr 4 at 16:35
  • $\begingroup$ @PeterCordes Thanks so much for the reliable reference to the Ringworld series! Also for your observations about the nuances of physics at the other extreme from gravitation and general relativity. $\endgroup$ – Tom Spilker Apr 4 at 16:41

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