Though it is understandable that the sun and the earth may be revolving around a barycenter, but, if so, not only the sun and Jupiter should also be revolving around some barycenter, the same should be true about the other planets as well? So it has to be true that the sun revolves around as many barycentres as the number of the planets we have in our solar system. I am quite confounded — how any object may at all revolve around multiple barycenters? And if an object can't revolve around more than one point, does it not nullify the theory that the sun and the earth revolve around a barycenter?

  • $\begingroup$ No… the main point of there being a "solar" system is that everything else orbits the sun. Do you have a friend who might help provide a better translation for this Question? $\endgroup$ Commented Jan 15, 2021 at 19:46
  • $\begingroup$ Paper That Blames The Sun For Climate Change Was Just Retracted From Major Journal "In the now-retracted paper, Zharkova et al. contended that the motion of the Sun around the barycentres created by the gas giants was enough to alter the distance between Earth and Sun by up to 3 million kilometres (1.85 million miles), over a timeframe of a few hundred years. But, as other scientists were quick to point out on PubPeer, Earth doesn't orbit those barycentres. It orbits the Sun." $\endgroup$ Commented Jan 27, 2021 at 4:05

4 Answers 4


The short answer is no; there is only one barycenter. Yes, you can count the Sun/Jupiter barycenter or the Sun/Saturn barycenter, or whichever barycenter you want, but the net effect of all Solar System bodies is to be considered when you calculate the actual barycenter of the Solar System. (And yes, that would include counting all the small asteroids and moons, even those yet unknown to humans, even though their combined effect is negligible.)

One could see it in such a way that yes, there are many barycenters, but the movement of the bodies is around the “average” barycenter. Somehow. But that’s not a good way to describe the system.

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    $\begingroup$ In summary, dynamics with >2 bodies is very complicated $\endgroup$
    – OrangeDog
    Commented Jan 13, 2021 at 10:35
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    $\begingroup$ More to the point, the barycenter isn't at a fixed place in the system. It's a chaotic system, none of the orbits are true ellipses, and the barycenter moves around due to the combined effects of the positions of all the bodies. $\endgroup$
    – Barmar
    Commented Jan 13, 2021 at 16:26
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    $\begingroup$ @Barmar of course, there is only one fixed place in a system and all other points move around. The barycenter certainly qualifies as one of those places which can be defined as fixed, and I am even inclined to say that it is one of, if not the most natural place to be defined as fixed. $\endgroup$ Commented Jan 14, 2021 at 12:39
  • $\begingroup$ @RenéNyffenegger Of course I realize that space really has no fixed coordinate system. But for the purposes of my comment I was assuming a Newtonian model, not Einsteinian. $\endgroup$
    – Barmar
    Commented Jan 14, 2021 at 15:33
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    $\begingroup$ @paul23 the far away point is static relative to what? $\endgroup$
    – Erbureth
    Commented Jan 15, 2021 at 12:52

The Sun's movement in the Solar System can be thought of as its movement around all the individual pairwise barycenters at once, or as a movement around the Solar System barycenter, which itself is constantly moving.

Suppose Mercury was the only planet. The mutual barycenter of Mercury and the Sun is about 10km from the center of the Sun, which is inside the Sun. The Sun would be orbiting this barycenter inside itself every 88 days.

Now, suppose Mercury and Jupiter were the only planets. The Sun/Jupiter barycenter is located barely outside the Sun (about 1.07 Solar radii or 745,000km). In this two planet system, the Sun would rotate around the Sun/Jupiter barycenter about every 4,333 days, but at the same time, it would be rotating around the Sun/Mercury barycenter every 88 days. The center of mass of the Sun wouldn't quite be tracing out curlicues like a spirograph, but it would be wavering around its orbit of the Sun/Jupiter barycenter due to gravitational perturbations by Mercury.

If we consider the full Solar System, with all the massive bodies, the Sun is orbiting all the individual barycenters as well as the whole barycenter. Here is a picture of the movement of the Sun around the barycenter taken from ProfRob's answer to What does the Sun's orbit within the Solar System look like?. If we were able to "zoom in" enough, we would see the line "wiggle" due to the location of the inner planets.

enter image description here

Of course, this image is just created with the known Solar System masses. What happens if we eventually discover the theorized Planet 9? It could be out at 800 AU away with up to 10 times Earth's mass, giving a barycenter distance from the Sun of as much as 3,592,000 km (over 5 times the Sun's radius. If Planet 9 exists, we will then learn that this whole diagram might really be stretched out and slowly rotating around a barycenter as much as five Solar radii away!!!

Summary: The Sun rotates around the Solar System barycenter, but the barycenter is constantly moving since the planets all have different orbital speeds. The Sun's rotation around the barycenter is a strange wavering curve due to its simultaneous gravitational interaction with the rest of the Solar System bodies.

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    $\begingroup$ Related: Can the paper narrowing Solar System's barycentre to within 100m help find Planet Nine? $\endgroup$
    – ksousa
    Commented Jan 13, 2021 at 22:25
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    $\begingroup$ And it should be pointed out that we can’t infer the presence or absence of “Planet Nine” from the position of the Solar System barycenter, as all bodies in the Solar System orbit it. In other words, we have no way of distinguishing the actual position of the Solar System barycenter from orbits alone. $\endgroup$ Commented Jan 13, 2021 at 23:48
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    $\begingroup$ At what point does it stop being meaningful to talk about a barycenter that includes more and more distant objects? Should we be considering Alpha Centauri and so on? The rest of the stars in the Milky Way? Obviously their effects are negligible given the huge distance, but they're massive and far away. I guess that would mean you're mostly looking at the Sun's motion within the galaxy, with perturbations to that path from motion around the proper solar system's barycenter. $\endgroup$ Commented Jan 15, 2021 at 9:47
  • $\begingroup$ @PeterCordes I think this would make an excellent question for Astronomy Stack Exchange. $\endgroup$
    – Connor Garcia
    Commented Jan 15, 2021 at 18:19
  • $\begingroup$ Part 2 of Ilmari's answer on this question is sufficient for me. It's just an approximation tool, and it works best when you want to group together and simplify multiple nearby objects that are "nearby" wrt. another more distant one. And yes, real interactions from other bodies can be modeled as perturbations. Also comments on that answer address part of the point of where it becomes silly season. $\endgroup$ Commented Jan 15, 2021 at 19:28

The motions of the Sun, the planets and their moons and everything else in the solar system are well described by Newton's laws of motion and gravity (with some minor relativistic corrections needed to fully account e.g. for the perihelion precession of Mercury). These laws make absolutely no reference to a "barycenter" in any form, so the whole concept of a barycenter isn't really needed to describe the solar system. If you want, you can just forget that it even exists!

So why do we care about the barycenter, then? I'd say there are two main reasons:

  1. Newton's first law says that, in the absence of external forces acting on it, an object at rest will stay at rest, and an object in motion will continue moving at the same speed in the same direction. Clearly, that's a very useful law of physics. But wait — what if the object is spinning, or flexing, or even composed of multiple parts only loosely attached to each other? Does the first law still apply, and how do we even measure the velocity of such objects?

    Fortunately, it turns out that Newton's first law does apply to such extended, rotating and possibly non-rigid objects, but only if we measure the velocity from the object's barycenter. The barycenter (also known as the center of mass) of any extended object (including even "objects" like the entire solar system!) always follows Newton's first law, moving at a constant velocity in the absence of external forces, no matter how much the various component parts of the object might spin or wobble around it.

    Thus, for example, if we're numerically simulating the motion of the solar system, it's a good idea to do so in a coordinate system where the velocity of the barycenter of the system is zero — because if we don't, then the entire system, Sun, planets and all, will gradually drift further and further away from its initial coordinate location. (It's also common to choose the location of the barycenter as the origin of the coordinate system, but there's no real reason for that choice except for mathematical convenience.)

  2. Also, for a system consisting of only two massive bodies (e.g. the sun and a planet, or a planet and its moon), approximated as point-like masses, Newton's laws turn out to have an exact mathematical solution, and the solution turns out consist of the two bodies following elliptical (or possibly parabolic or hyperbolic) orbits around their mutual barycenter.

    Now, of course, the real solar system has many more than just two bodies in it. But it turns out that most orbits in it can, at least over short timescales, be approximated with combinations of such elliptical two-body orbits.

    For example, to a first approximation, we can describe the mutual orbits of the Sun, the Earth and the Moon by assuming that a) the Earth and the Moon follow two-body elliptical orbits around their mutual barycenter, b) this combined Earth+Moon system (approximated by a single point mass located at its barycenter) and the Sun each follow two-body orbits around their mutual barycenter, and c) the effects of all other planets and moons don't matter.

    Of course, over time, the orbits in this simplified model will start to deviate from the real ones, both because in reality the Earth+Moon system is not a single point mass, and also because the effects of other planets do matter somewhat in the sufficiently long run. But it's still possible to start with the simple "hierarchical two-body" model and add perturbation terms to refine it and correct for the minor effects that the simple model leaves out.

    More generally, whenever we have a system consisting of two widely separated groups of objects — say, the Sun and its inner planets on one hand, and Jupiter and its moons on the other — we can approximate it quite well just by treating each group as a point mass located at the group's barycenter, and with these two (approximate) point masses following simple two-body orbits around their mutual barycenter. And this approximation will work regardless of how complicated the orbits within each group might be, as long as both groups stay together and separated from each other.

    (Also, to a first-order approximation, the movement of the bodies in each group relative to the group's barycenter is not affected by any bodies outside the group, since — being far away — the gravity of those bodies exerts the same force per mass on each body in the group.)

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    $\begingroup$ @SCSawhney: Yes, every pair of bodies in the solar system has a barycenter. For that matter, so does any non-empty group of objects you care to consider. If you want, you may e.g. calculate approximately where the barycenter of Mars's moon Phobos, former U.S. president Barack Obama and your own left big toe is. (FWIW, it's somewhere fairly close to Phobos, since that's by far the most massive object in this group.) And none of those barycenters have any actual physical effect on anything — they're all just imaginary points in space. $\endgroup$ Commented Jan 14, 2021 at 19:05
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    $\begingroup$ … I suspect you may be confused by statements like "The Moon and the Earth orbit their mutual barycenter." While that's (approximately) true in a certain sense, it doesn't mean that the barycenter itself has any physical effect on the movement of either the Earth or the Moon. It cannot, since it's just an imaginary point in space. $\endgroup$ Commented Jan 14, 2021 at 19:06

Though we may calculate the position of the barycenter of the Sun and the Mercury also in the same manner as we calculate the position of the barycenter of the Earth and the Moon as shown in the following diagram; we can’t calculate the position of the barycenter of the Sun and the Earth in the same manner.

[Method of calculating the position of the barycenter of the Sun and Mercury1] (It may be clarified that though the barycenter of Mercury lies inside the Sun — it has been shown outside the Sun in this diagram only since the diagram has been drawn basically with the intention of showing the theoretical part of the manner in which we calculate “d1” and “d2”.)
Before we calculate the position of the barycenter of Sun and the Earth; we shall have to calculate the position of the barycenter of the Venus, in the following manner.

Method of calculating the position of the barycenter of the Sun, Mercury and the Venus

Since we are going to talk about multiple barycenters — let us designate the barycenter of the Sun and the Mercury as “BC(1)” and designate the “Pair of the Sun and the Mercury” as the subset “SS(1)” of the solar system. If we may call the subset of the Sun, Mercury and the Venus as “SS(2)” and call their barycenter as “BC(2)”; we shall have to calculate d1 of the Venus in the following manner keeping in mind that, though the Sun and the Mercury keep on revolving around BC(1); the entire subset “SS(1)” would revolve around BC(2) since BC(1) happens to be the “Mass Center” of the subset “SS(1)”. d1 of Venus = M(♀) x d2/{M(☉) + M(☿)}, where d2 = (0.728 AU – d1); M(☉) = Mass of the Sun; M(☿) = Mass of the Mercury and M(♀) = Mass of the Venus. The same way, we shall have to calculate d1 of the Earth, as follows.

Method of calculating d1 of the Earth and the other planets

If we designate the barycenter of the Earth as “BC(3)”; the subset SS(2) would have to revolve around BC(3) and the value of d1 of the Earth shall have to be calculated, as follows. d1 = M(♁) x d2 /{M(☉) + M(☿) + M(♀)} where d2 = (1.0 AU – d1) and M(♁) = Mass of the Earth.
And the same way, for all other planets with the following values of d2. (i) d2 = (1.52 AU – d1) to calculate the d1 of the barycenter of SS(3) and the Mars. (ii) d2 = (5.2 AU – d1) to calculate the d1 of the barycenter of SS(5) and the Jupiter. (iii) d2 = (9.58 AU – d1) to calculate the d1 of the barycenter of SS(6) and the Saturn. (iv) d2 = (19.2 AU – d1) to calculate the d1 of the barycenter of SS(6) and the Uranus. (v) d2 = (30.1 AU – d1) to calculate the d1 of the barycenter of the solar system, that is, the barycenter of SS(7) and the Neptune.


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