# Thread: Is the Solar System's Barycenter Inside or Outside the Sun?

1. Originally Posted by George
It gets worse....

Using the radial shift formula in Wiki (r = a x ratio of masses) and making the Earth a unit shift distance for the Sun....

Mercury......0.02
Venus.........0.59
Earth.........1.00
Mars..........0.16
Jupiter...1,652
Saturn.....911
Uranus....280
Neptune...516

I was once pretty good on a see-saw, too.
Convert those into percent of solar radius, and I think it'd make profloater's image make sense

2. Originally Posted by Robert Tulip
I attach again the Fourier Transform of the SSB which illustrates that the phrase ‘wobbling as the gravitational action of the giant planets jerks it around’, while technically correct, has potential to be misleading and produce an incorrect assumption.
My bold. Sure, if someone misinterprets it in that incomplete quote as meaning the barycenter. In the complete final sentence in post #6 it should be clear that I was saying that the Sun is being jerked around by the giant planets.

3. Originally Posted by tony873004
If Planet 9 gets discovered the barycenter will have to be recalculated.
Great point. I woke up this morning thinking the same thing. The predicted path of the Sun about the barycenter would be off by this effect from Planet 9, right? What a great help this would be in locating it.

Check out the effect on the barycenter of Andromeda, or, for that matter, a tennis ball at the edge of the universe.
I won't say sour grapes, but Wow!, what a monster hiccup! Hornblower is correct because the math from grapes' wiki link does indeed match that of Tulip's simple see-saw -- the farther away an object is, the greater the distance the counterbalancing mass must be located. [Simple summation of moments.]

Yesh! How does one make sense of this stuff? Would Feynman suggest that all the other objects in great beyond all cancel each other in some mass isotropic effect somehow? Or, is this one of those equations where objects with Lim --> 0 for angular velocity negates their role on our barycenter?

But, to grapes' point, Planet 9's (@ 700 au) barycenter effect would be 7,030 in the table values I last gave, more than 4x that of Jupiter! Heavens to Betsy! And Betsy ain't in sight!
Last edited by George; 2017-Mar-14 at 03:05 PM.

4. Here is how I would make sense of what George is concerned about. Yes, the overall barycenter is several million miles from the Sun if this reputed Planet 9 is for real, but for the purpose of analyzing the motions of the known planets it is a mathematical curiosity of no physical importance. The gravitational gradient across the expanse of Neptune's orbit is too slight to cause observable perturbations of the major planets; otherwise we would have inferred its existence just as Adams and Leverrier inferred Neptune's existence from observed perturbations of Uranus.

I have seen assertions that our planets are in Keplerian orbits with a common focus at the overall barycenter of the solar system. That is not a realistic picture. For a comet far beyond the major planets, the best fit ellipse will indeed have a focus near the overall barycenter, as the Sun and the planets will act on it roughly in unison. For Mercury, the best fit ellipse will have a focus much closer to the center of the Sun, and this ellipse precesses and jiggles as a result of the gravitational perturbations of the planets. For something like Ceres, with a strong but highly variable perturbation from Jupiter, I would not expect to find a fixed focus anywhere. My educated guess is that the focus of the best-fit reference ellipse is somewhere between the center of the Sun and the Sun-Jupiter barycenter.

5. Let me expand on this with a more down to Earth example. A satellite in low Earth orbit has its best-fit focus at or very near the center of the Earth, nowhere near the Earth/Moon barycenter 3,000 miles away.

6. Originally Posted by Hornblower
Let me expand on this with a more down to Earth example. A satellite in low Earth orbit has its best-fit focus at or very near the center of the Earth, nowhere near the Earth/Moon barycenter 3,000 miles away.
Nor, heaven forfend, anywhere near the Sun/Earth barycenter

7. Originally Posted by Hornblower
Yes, the overall barycenter is several million miles from the Sun if this reputed Planet 9 is for real,...
I think this would indeed be true if we only had one planet in the system -- Planet 9. Since the Sun is orbiting the barycenter many hundreds of times for each would-be orbit of Planet 9, then the see-saw model is inadequate. Orbital motion seems to make the big difference and Andromeda does not waltz. I'm assuming angular momentum presents the better model for finding the barycenter, but perhaps this is too simple as well.

...but for the purpose of analyzing the motions of the known planets it is a mathematical curiosity of no physical importance. The gravitational gradient across the expanse of Neptune's orbit is too slight to cause observable perturbations of the major planets; otherwise we would have inferred its existence just as Adams and Leverrier inferred Neptune's existence from observed perturbations of Uranus.
That make sense because the inverse square law for gravity is still applicable.
Last edited by George; 2017-Mar-14 at 05:45 PM.

8. Originally Posted by George
I think this would indeed be true if we only had one planet in the system -- Planet 9. Since the Sun is orbiting the barycenter many hundreds of times for each would-be orbit of Planet 9, then the see-saw model is inadequate.
Objects orbit the barycenter if they are a long ways out from the other masses, or if there is only one other mass.

After all, the sun would orbit the Jupiter/sun barycenter once per Jupiter orbit, but the Saturn/sun barycenter once per Saturn orbit
Orbital motion seems to make the big difference and Andromeda does not waltz. I'm assuming angular momentum presents the better model for finding the barycenter, but perhaps this is too simple as well.

That make sense because the inverse square law for gravity is still applicable.

9. Originally Posted by George
I think this would indeed be true if we only had one planet in the system -- Planet 9. Since the Sun is orbiting the barycenter many hundreds of times for each would-be orbit of Planet 9, then the see-saw model is inadequate. Orbital motion seems to make the big difference and Andromeda does not waltz. I'm assuming angular momentum presents the better model for finding the barycenter, but perhaps this is too simple as well.

That make sense because the inverse square law for gravity is still applicable.
My bold. No, you are making the concept of the barycenter more complicated than it is. Its position is purely a function of the positions and masses of the bodies. We could create a hypothetical system in which the masses and relative positions at the present time are identical to those of our own solar system, but give them different amounts of angular momentum. The relative positions of the barycenters would be the same, and if stationary in the chosen frames of reference they would remain stationary.

If we look at our known solar system as a tight cluster, with Planet 9 far away proportionately, we could make a rough analogy with a small but heavy seesaw on a fulcrum that in turn is resting on one end of a much longer seesaw, with Planet 9 at the far end and with the beam resting on a fulcrum near the cluster. The latter fulcrum would be at the barycenter of the overall system. Of course, in this thought exercise we make the beams massless. Note that angular momentum does not figure here.

10. Originally Posted by grapes
Objects orbit the barycenter if they are a long ways out from the other masses, or if there is only one other mass.

After all, the sun would orbit the Jupiter/sun barycenter once per Jupiter orbit, but the Saturn/sun barycenter once per Saturn orbit
So it looks like what is happening is that the Sun's motion is a net motion based on all those masses and orbital periods, and we can sum all those barycenters into one. I think Hornblower is saying the same thing, but he knows I, and others, like pictures and I like, ok...... see-saws.

Yet.....

11. Originally Posted by Hornblower
My bold. No, you are making the concept of the barycenter more complicated than it is. Its position is purely a function of the positions and masses of the bodies. We could create a hypothetical system in which the masses and relative positions at the present time are identical to those of our own solar system, but give them different amounts of angular momentum. The relative positions of the barycenters would be the same, and if stationary in the chosen frames of reference they would remain stationary.

If we look at our known solar system as a tight cluster, with Planet 9 far away proportionately, we could make a rough analogy with a small but heavy seesaw on a fulcrum that in turn is resting on one end of a much longer seesaw, with Planet 9 at the far end and with the beam resting on a fulcrum near the cluster. The latter fulcrum would be at the barycenter of the overall system. Of course, in this thought exercise we make the beams massless. Note that angular momentum does not figure here.
That's very helpful. Thanks. If I had some time, I think a graphic of that would look pretty cool.

[Yet....] If we were to add and remove the one see-saw that belongs to Planet 9 (~4x the effect Jupiter has), for example, then there should be (cutting to the chase) a noticeable shift in the barycenter. [I'm imagining using a stellar blink-like comparator for these point shifts.] So what am I now missing?
Last edited by George; 2017-Mar-14 at 07:38 PM.

12. Originally Posted by George
That's very helpful. Thanks. If I had some time, I think a graphic of that would look pretty cool.

[Yet....] If we were to add and remove the one see-saw that belongs to Planet 9 (~4x the effect Jupiter has), for example, then there should be (cutting to the chase) a noticeable shift in the barycenter. [I'm imagining using a stellar blink-like comparator for these point shifts.] So what am I now missing?
What's the mechanism for "add and remove"? Wouldn't you have to know where Planet 9 was? So, that's not going to help you find it!

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Originally Posted by George
... I think a graphic of that would look pretty cool. ...
http://orbitsimulator.com/gravitySim...thPlanet9.html
This simulation includes Planet 9 as defined by the orbital elements on Wikipedia.
At first it looks just like the version that doesn't contain planet 9. But over centuries, you can see this pattern drift. In 10s of thousands of years, it completes a full circuit.

14. Originally Posted by grapes
What's the mechanism for "add and remove"? Wouldn't you have to know where Planet 9 was?
Why wouldn't it help you find it? If we can accurately plot, somehow, the location of the barycenter based on the known objects and compare it to the actual motions to determine the observed barycenter, then the difference would guide as to which direction the Planet 9's "see-saw" is angled, right? Is this beyond precision astronomy? [An absolute aether would be nice.]

[I admit I still don't have a handle on all that has been said, so my question may be mute. But when has the stopped me. ]

15. Originally Posted by tony873004
http://orbitsimulator.com/gravitySim...thPlanet9.html
This simulation includes Planet 9 as defined by the orbital elements on Wikipedia.
At first it looks just like the version that doesn't contain planet 9. But over centuries, you can see this pattern drift. In 10s of thousands of years, it completes a full circuit.
That's cool! But the variation would already have been noticed if anyone was plotting actual vs. known planetary models, right? [Brown (et al) would already be all over this by now, I would think.]

16. Originally Posted by George
That's cool! But the variation would already have been noticed if anyone was plotting actual vs. known planetary models, right? [Brown (et al) would already be all over this by now, I would think.]
Not necessarily. The reputed Planet 9 is so far away in proportion to the orbital radii of even Neptune that the known planets' orbits will drift roughly in unison with the Sun, so that teasing the perturbations out of the overall pattern from within is going to be murderously difficult with only a couple of centuries of data. We might see it if we could observe for half the period of Planet 9, but that would take thousands of years as compared with 40 or so for Uranus when detecting the perturbations that pointed to Neptune.

17. Originally Posted by Hornblower
Not necessarily. The reputed Planet 9 is so far away in proportion to the orbital radii of even Neptune that the known planets' orbits will drift roughly in unison with the Sun, so that teasing the perturbations out of the overall pattern from within is going to be murderously difficult with only a couple of centuries of data. We might see it if we could observe for half the period of Planet 9, but that would take thousands of years as compared with 40 or so for Uranus when detecting the perturbations that pointed to Neptune.
Sorry, disregard what I just said. We should not need a large portion of Planet 9's orbital period, just a large portion of Neptune's period. Observing over the course of Neptune's orbit would be enough if the perturbation is large enough to tease out of observational uncertainties. The problem is that the strength of the perturbation would drop off as the cube of the distance to the perturber, and that is some 10 times Neptune's orbital radius. It would be roughly 1,000 times weaker than what Neptune was doing to Uranus.

18. Originally Posted by George
That's very helpful. Thanks. If I had some time, I think a graphic of that would look pretty cool.

[Yet....] If we were to add and remove the one see-saw that belongs to Planet 9 (~4x the effect Jupiter has), for example, then there should be (cutting to the chase) a noticeable shift in the barycenter. [I'm imagining using a stellar blink-like comparator for these point shifts.] So what am I now missing?
You appear to be misconstruing the concept of the barycenter, but I cannot tell just how you are going astray. When I get a chance I will be happy to post some sketches, drawn to scale on graph paper, to reinforce what I have attempted to present verbally. This definitely is a case in which a picture is worth many words, and I will admit that my technical writing is not always as clear to others as I think it is.

19. Originally Posted by grapes
Convert those into percent of solar radius, and I think it'd make profloater's image make sense
How 'bout actual distance (km)?...

Mercury.......10
Venus.........265
Earth............449
Mars..............74
Jupiter........742,420
Saturn........409,052
Uranus.......125,510
Neptune.....231,767
Pluto.................37
Planet 9...3,157,292
V1 (And)...1.58E+15

Originally Posted by Hornblower
You appear to be misconstruing the concept of the barycenter, but I cannot tell just how you are going astray.
Agreed! I appreciate all construeing efforts. The last two on the list show you why things are going awry. Assuming 400 billion stars in Andromeda, then the Sun-to-barycenter distance would be extremely influenced by some seemingly magical force; the see-saw suddenly has excessive leverage. There must be some constraints on that simple formula (per Wiki) that I'm missing.
Last edited by George; 2017-Mar-15 at 03:11 PM.

20. Originally Posted by George
Agreed! I appreciate all construeing efforts. The last two on the list show you why things are going awry. Assuming 400 billion stars in Andromeda, then the Sun-to-barycenter distance would be extremely influenced by some seemingly magical force; the see-saw suddenly has excessive leverage. There must be some constraints on that simple formula (per Wiki) that I'm missing.
Excessive? I don't understand why you'd say it's excessive? Not that I've checked your calculations, but as I said earlier, even a tennis ball, at great distance, shares a barycenter with the sun that extends a long ways from the center of the sun. The farther away, the farther away.

ETA: The barycenter of the sun and a similarly-massed twin is going to be exactly half-way between them, right? No matter where it is in the universe.

21. Originally Posted by grapes
Check out the effect on the barycenter of Andromeda, or, for that matter, a tennis ball at the edge of the universe.

The barycenter calculation has zero effect on the tides. Rotate a fixed orientation body around a barycenter, you'll see that the centrifictional forces at each point of the body are equal and parallel.
Well, if I consider just Jupiter, it exerts a force just like the moon does on the fluids of Earth. The force is negligible as a lift force but has days to act on the tangents of the fluid sun. The situation is directly comparable

22. Originally Posted by George
Agreed! I appreciate all construeing efforts. The last two on the list show you why things are going awry. Assuming 400 billion stars in Andromeda, then the Sun-to-barycenter distance would be extremely influenced by some seemingly magical force; the see-saw suddenly has excessive leverage. There must be some constraints on that simple formula (per Wiki) that I'm missing.
I don't think you're doing anything wrong in your calculations. I just think you're ascribing some importance to the barycenter that it doesn't really have. It's certainly true that the Andromeda galaxy totally dominates the barycenter of the Sun-Andromeda system, just as you'd expect. But I can't think of any significance to looking at the Sun-Andromeda system as a whole. Especially since considering the barycenter of a system of masses is mostly useful when outside influences can be easily isolated or considered separately. But all the stars in the Milky Way would also have a dramatic effect on the Sun-Andromeda barycenter (and they'll have very different effects on the components of that system).

The reason the barycenter of the solar system is a useful concept is because the gravitational effects of objects outside the solar system is more or less constant on all of the objects in the solar system (i.e., the solar system is in freefall relative to the combined gravitational influence of the rest of the stars of the Milky Way, and the tidal gradient across the solar system from even the nearest stars is insignificant). So you can ignore those outside gravitational effects, consider the barycenter of the solar system to be fixed in place, and reckon the movements of everything around it. It provides a convenient point of reference. But it doesn't have any gravitational attraction of its own or anything like that.

23. Originally Posted by grapes
Excessive? I don't understand why you'd say it's excessive? Not that I've checked your calculations, but as I said earlier, even a tennis ball, at great distance, shares a barycenter with the sun that extends a long ways from the center of the sun. The farther away, the farther away.
I seem stuck on seeing a "the barycenter" point for the Sun, as would be see from afar, where the Sun would be seen actually orbiting given all the objects, and not a two-body only view. [This is why I facetiously added Feynman to help with a net barycenter.]

24. Originally Posted by Grey
... I just think you're ascribing some importance to the barycenter that it doesn't really have. ...
I agree. We had a thread here about 8 years ago (can't find it now), that was similar to this. Turned out the OP thought that the planets all traveled in perfect ellipses around the barycenter ... which is only true in a completely isolated two-body system. Our solar system is way more than two bodies, but if you're only plotting the movements of the Sun and Jupiter, the approximation is good to an extra decimal place. Three or more body systems are chaotic in the long term, and highly accurate orbits need to be calculated continuously. There is no simple model.

25. Originally Posted by Grey
I don't think you're doing anything wrong in your calculations. I just think you're ascribing some importance to the barycenter that it doesn't really have. It's certainly true that the Andromeda galaxy totally dominates the barycenter of the Sun-Andromeda system, just as you'd expect. But I can't think of any significance to looking at the Sun-Andromeda system as a whole. Especially since considering the barycenter of a system of masses is mostly useful when outside influences can be easily isolated or considered separately. But all the stars in the Milky Way would also have a dramatic effect on the Sun-Andromeda barycenter (and they'll have very different effects on the components of that system).
The Sun's position relative to the barycenter is a simple linear relation with distance to the other mass (and linear with mass). This is the same for a see-saw or a lever, the latter maybe being the better analogy to address my impediment. The leverage Andromeda, or especially any quasar, holds in effecting a net(?) barycenter seems untenable. ["Net" being the hiccup point for me, probably.]

The reason the barycenter of the solar system is a useful concept is because the gravitational effects of objects outside the solar system is more or less constant on all of the objects in the solar system (i.e., the solar system is in freefall relative to the combined gravitational influence of the rest of the stars of the Milky Way, and the tidal gradient across the solar system from even the nearest stars is insignificant).
Yet if we understand that gravity is causal to the barycenter's location [again a "net" point and ignoring a 2-body view], how can we reconcile the extreme shift in a barycenter location from extremely distant objects, and the more distant the more leverage, ironically. Even if we have an isotropic and homogenous mass distribution view of the universe (beyond the solar system) to give us a net of zero (simplifying since a fixed shift would count as zero as well) shift in the barycenter -- thus giving the solar system's masses their ability to shift the barycenter -- I would still struggle with why the inverse square law for gravity wouldn't apply and nullify the leverage effect? So it's the idea of giving all that leverage to all those outsiders that seems to be where I am struggling.
Last edited by George; 2017-Mar-15 at 04:49 PM.

26. Originally Posted by George
The Sun's position relative to the barycenter is a simple linear relation with distance to the other mass (and linear with mass). This is the same for a see-saw or a lever, the latter maybe being the better analogy to address my impediment. The leverage Andromeda, or especially any quasar, holds in effecting a net(?) barycenter seems untenable. ["Net" being the hiccup point for me, probably.]

Yet if we understand that gravity is causal to the barycenter's location [again a "net" point and ignoring a 2-body view], how can we reconcile the extreme shift in a barycenter location from extremely distant objects, and the more distant the more leverage, ironically. Even if we have an isotropic and homogenous mass distribution view of the universe (beyond the solar system) to give us a net of zero (simplifying since a fixed shift would count as zero as well) shift in the barycenter -- thus giving the solar system's masses their ability to shift the barycenter -- I would still struggle with why the inverse square law for gravity wouldn't apply and nullify the leverage effect? So it's the idea of giving all that leverage to all those outsiders that seems to be where I am struggling.
Well, again, the barycenter of a collection of objects is just really a fancy name for their center of mass. We generally only use the term for objects that are orbiting each other, because when doing orbital calculations, it's a particularly convenient coordinate system. It's completely true that the center of mass of the Andromeda galaxy plus the Sun is much closer to the center of the Andromeda galaxy than it is to the Sun, just like adding an extra grain of sand on one side of a balanced rock won't change the balance point very much.

Originally Posted by George
I seem stuck on seeing a "the barycenter" point for the Sun, as would be see from afar, where the Sun would be seen actually orbiting given all the objects, and not a two-body only view. [This is why I facetiously added Feynman to help with a net barycenter.]
There is no "the barycenter" for the Sun. There's a barycenter for the solar system, including all the objects contained within it, and that's a useful point of reference for computing the movement of those objects. But if you add Andromeda into the mix, it's not that it moves "the barycenter" of the solar system. It's that you can also calculate the center of mass for a system that consists of the solar system and the Andromeda galaxy. You could use that as the center of a coordinate system for something if you really wanted to, but it's not a particularly useful one, so it doesn't seem like you would be likely to do that.

27. Originally Posted by grapes
Excessive? I don't understand why you'd say it's excessive? Not that I've checked your calculations, but as I said earlier, even a tennis ball, at great distance, shares a barycenter with the sun that extends a long ways from the center of the sun. The farther away, the farther away.

ETA: The barycenter of the sun and a similarly-massed twin is going to be exactly half-way between them, right? No matter where it is in the universe.
Well surely the shell theorem applies to other stars and galaxies, the barycentre is a local effect for the solar system.

28. Originally Posted by Grey
Well, again, the barycenter of a collection of objects is just really a fancy name for their center of mass.
Yep, though "center of gravity" is another expression for it and it is part of my problem. But, while here, perhaps another analogy will help, probably more me than thee, since I'm the one stumbling (partly because I am juggling many work issues, so thanks for everyone's patience).

If we load a machine on a trailer and calculate the axle loads, all we need is the weight of the machine, its c.g. and where we place it on the trailer (ignoring the weight distribution of the unloaded trailer). We can add other machines on the trailer, if it's long enough, and recalculate for each weight, c.g. and placement to determine the axle loads, and a net c.g. for that matter (if only calculating axle loads). We aren't adding the c.g. of anything not on the trailer because they will not have any noticeable influence on the axle loading.

So I am trying to apply similar direct causation from anything and everything that might affect the barycenter we use for the Solar system (though I have loosely called it the Sun's barycenter). I can see the difference between things on the trailer and things off, but it isn't so easy for masses in space; space itself is the trailer deck. If not, why not?

We generally only use the term for objects that are orbiting each other, because when doing orbital calculations, it's a particularly convenient coordinate system. It's completely true that the center of mass of the Andromeda galaxy plus the Sun is much closer to the center of the Andromeda galaxy than it is to the Sun, just like adding an extra grain of sand on one side of a balanced rock won't change the balance point very much.
I can see where we can use a computer program to calculate the barycenter of the solar system if we know the initial conditions and mass/orbit selections. I see nothing wrong in the math, of course, just how to keep the distant, and more leveraged, masses off our "trailer".

One thought, given my magic wand, would be to freeze the universe, go out and measure all the masses in the observable universe (Just to save time, I won't go beyond today's observable edge when I get that far and, nor surprisingly, see more masses. ) and get their distances. Recompute to get the c.g. which will be light years away, given anisotropy.

Hmmm, this would still not give us much of an orbital motion of the Sun about this c.g. What we want are all the masses that have an orbital influence about a point we will call the barycenter, which is the c.g. for that specific group of masses (including the Sun) that are moving (ie orbiting) in a way that directly relate to the orbital motion of the Sun.

You could use that as the center of a coordinate system for something if you really wanted to, but it's not a particularly useful one, so it doesn't seem like you would be likely to do that.
Right, and I think it must be the need to consider relative angular motion in determining a useful barycenter. This is the part I seem to be missing.
Last edited by George; 2017-Mar-15 at 09:22 PM. Reason: grammar

29. If we make the loads on the trailer analogous to the Sun and planets, the forces they exert on the trailer are irrelevant to how they interact among themselves. For interaction among themselves we are concerned with the horizontal gravitational forces they exert on one another. Those forces get weaker as they are moved farther apart, while the leverage the bodies exert on the trailer gets stronger. The latter is because each body's gravitational weight remains the same, while its moment arm gets longer. If we magically eliminate the trailer and let the bodies be in free fall, they will fall in unison without changing their relative positions, and we can ignore the external gravity, as has been said repeatedly.

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George,

Your truck is on a teeter-totter. With no trailer attached, the tractor
needs to be over the fulcrum in order to be balanced. Attach an empty
trailer, and the tractor needs to move forward. Start loading cargo into
the trailer, and it has to keep moving forward farther and farther.

The tractor is the Andromeda galaxy. The trailer and its contents is
the Milky Way. The Solar System is one little tiny package inside the
trailer, but without the trailer. You are trying to find a balance point
between the big, heavy tractor and the tiny little package, while
ignoring the trailer and all the rest of the packages it contains.
Why? What is the significance of that balance point? The answer
is that there is no significance. The Sun is part of the Milky Way.
So if you want to find a balance point between the Sun and the
Andromeda galaxy, you have to include the rest of the Milky Way
on the side of the teeter-totter that has the Sun on it.

-- Jeff, in Minneapolis

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