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

1. Originally Posted by grapes
I think I see what you're asking for, good question.

All sorts of calculations could show it. Maybe the easiest way to see it is through energy (how many times have I said that?)

Force is the derivative of energy .

There's the inverse linear relationship you've been looking for (and, it's proportional to mass).

Conservation of energy then has that center of mass not changing, without some other ("external") force, whatever the motion of the objects.
I stand by what I said in my previous post. The force need not be an inverse square function of the separation. In a thought exercise it could be inverse cube, inverse linear, constant, linear, you name it. Once again, the key point is that the forces on the two bodies are of equal magnitude and opposite direction along the line connecting the bodies. As a result, at any time after they are released their accelerations, and thus their distances covered in any amount of elapsed time, are inversely proportional to their masses. Thus they will meet at the barycenter. The variation in the forces as a function of separation drops out in the analysis.

Originally Posted by George
It's easier than expected. Using R=1/2at2, and T1 = T2 since they, in free-fall, will meet at the barycenter, we get m1r1 = m2r2! Summation of moments (seesaw)from inverse square law and F = ma. I didn't think we could do it.

iphone
I am having a hard time following you algebra, but the one thing I see for sure is that the inverse square law for the magnitude of the accelerations does not appear in this presentation.

Throughout this thread George appears to be seeking a resolution to an initial perception of a conflict between inverse square gravitation and the location of the barycenter. Such a conflict simply does not exist. In all fairness, I can see how this absence of a conflict can be counterintuitive.

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Originally Posted by Hornblower
It is my understanding that in general relativity, a frame of reference in which the Earth is stationary and non-rotating is admissible, but it remains my educated guess that calculations of orbital dynamics in the solar system, not to mention the galaxy, would be horrendously messy and of no practical value.
On the point of whether the calculations would be horrendously messy, Hoyle states:

"The relation of the two pictures is reduced to a mere coordinate transformation, and it is the main tenet of the Einstein theory that any two ways of looking at the world which are related to each other by a coordinate transformation are entirely equivalent from a physical point of view. Moreover, in the Einstein theory the method of calculating the effect of gravitation is changed to a form which applies equally to all such related ways of expressing a problem."

3. Originally Posted by wd40
On the point of whether the calculations would be horrendously messy, Hoyle states:

"The relation of the two pictures is reduced to a mere coordinate transformation, and it is the main tenet of the Einstein theory that any two ways of looking at the world which are related to each other by a coordinate transformation are entirely equivalent from a physical point of view. Moreover, in the Einstein theory the method of calculating the effect of gravitation is changed to a form which applies equally to all such related ways of expressing a problem."
Are you arguing that Hoyle's expression "applies equally" means just as easy to calculate for practical purposes in either coordinate system? If so, can you demonstrate it with some sample calculations, in appropriate mathematical detail?

4. Originally Posted by wd40
On the point of whether the calculations would be horrendously messy, Hoyle states:

"The relation of the two pictures is reduced to a mere coordinate transformation, and it is the main tenet of the Einstein theory that any two ways of looking at the world which are related to each other by a coordinate transformation are entirely equivalent from a physical point of view. Moreover, in the Einstein theory the method of calculating the effect of gravitation is changed to a form which applies equally to all such related ways of expressing a problem."
You've clearly been misled by the phrase "mere coordinate transformation", which says nothing at all about whether a specific coordinate transformation would be horrendously messy or not. Hoyle was describing a way of thinking, not a practical approach.

Grant Hutchison

5. Originally Posted by grapes
I think I see what you're asking for, good question.
Y'all are better at this than I am, but it is much easier than I thought. [And I have been limited to an iphone lately.]

F = mA
A is acceleration; we will use “a” for the semi-major axis of the system later
1) s1 =1/2 * A1 * t2
2) A1 = 2*s1/ t2 = 2*r1/ t2…. Because r1 = s1

F1 = F2….. from the inverse square law (i.e. gravity)

From, F = mA
m1*A1 = m2*A2

Substituting from eq. 2…
m1*(2*r1/ t2) = m2*(2*r2/ t2)

Resolves to…
3) m1*r1 = m2 * r2……. This is the seesaw equation; summation of moments = 0 stuff.

To get it to the common barycenter equation….
4) a = r1 + r2………… where “a” is the semi-major axis.

From eq. 3…
r2 = m1*r1/m2
a = r1 + r2 = r1+ (m1*r1/m2)
a = r1*(1 + m1/m2)

thus…
r1 = a/(1+m1/m2)

As Hornblower has stated, no orbital dynamics are needed to give us a c.g. or barycenter, though as a layperson I suggest the use of "barycenter" be associated with orbital masses. [When I was very little, I still recall seeing a picture (perhaps my first) of an airplane in the air and it was just.... resting there in the air. I don't see two masses in space just sitting there worth determining their barycenter; if they aren't going to fly, let 'em fall.]

The seesaw analogy is helpful but it assigns the force, F, perpendicular to the board (moment arm), whereas here we use F (from gravity) aligned radially with the "board". [The perpendicular F for seesaws is really the same because the F used there is radially with the c.g. of the Earth (gravity). We will ignore the case for a vertical seesaw if for safety only. ]

I had thought that something more than F=ma, presented by Grant, would be needed because I couldn't get the orbital motions out of my head. But they aren't needed to derive the barycenter, after all.
Last edited by George; 2017-Mar-26 at 06:43 PM.

6. Originally Posted by George
Y'all are better at this than I am, but it is much easier than I thought. [And I have been limited to an iphone lately.]

F = mA
A is acceleration; we will use “a” for the semi-major axis of the system later
1) s1 =1/2 * A1 * t2
2) A1 = 2*s1/ t2 = 2*r1/ t2…. Because r1 = s1
What are r1 and s1?
F1 = F2….. from the inverse square law (i.e. gravity)
Or inverse cube, inverse 4th, inverse linear, constant, linear, square, cube, you name it. F1 = F2 because they are equal and opposite central forces in accordance with one of Newton's laws, regardless of how their magnitude varies, if at all, with the separation of the bodies.
From, F = mA
m1*A1 = m2*A2

Substituting from eq. 2…
m1*(2*r1/ t2) = m2*(2*r2/ t2)

Resolves to…
3) m1*r1 = m2 * r2……. This is the seesaw equation; summation of moments = 0 stuff.

To get it to the common barycenter equation….
4) a = r1 + r2………… where “a” is the semi-major axis.

From eq. 3…
r2 = m1*r1/m2
a = r1 + r2 = r1+ (m1*r1/m2)
a = r1*(1 + m1/m2)

thus…
r1 = a/(1+m1/m2)

As Hornblower has stated, no orbital dynamics are needed to give us a c.g. or barycenter, though as a layperson I suggest the use of "barycenter" be associated with orbital masses.
To do so or not is an arbitrary verbal decision that is of no consequence of analyzing it as an exercise in physics.
[When I was very little, I still recall seeing a picture (perhaps my first) of an airplane in the air and it was just.... resting there in the air. I don't see two masses in space just sitting there worth determining their barycenter; if they aren't going to fly, let 'em fall.]

The seesaw analogy is helpful but it assigns the force, F, perpendicular to the board (moment arm), whereas here we use F (from gravity) aligned radially with the "board". [The perpendicular F for seesaws is really the same because the F used there is radially with the c.g. of the Earth (gravity). We will ignore the case for a vertical seesaw if for safety only. ]

I had thought that something more than F=ma, presented by Grant, would be needed because I couldn't get the orbital motions out of my head. But they aren't needed to derive the barycenter, after all.

7. Originally Posted by George
I had thought that something more than F=ma, presented by Grant, would be needed because I couldn't get the orbital motions out of my head. But they aren't needed to derive the barycenter, after all.
George, your claim here that orbital motions aren’t needed to derive the barycentre is sufficient to place the barycentre in space, but not in time.

The movement of the barycentre over time against the sum of mass of the solar system is depicted in detail over several thousand years in diagrams that I have posted in this thread at

#26 https://forum.cosmoquest.org/attachm...3&d=1489653849
- Quantifies power of orbital relations between gas giants as primary causes of distance of sun from barycentre
- Also identifies unknown factors revealed by NASA’s barycentre calculations

#85 https://forum.cosmoquest.org/attachm...7&d=1489914377
- Stable similar flower patterns of barycentre in successive 178.9 year periods, each divided into fifteen stable segments

and

#93 https://forum.cosmoquest.org/attachm...1&d=1490063170
- Wave depiction of stable similar solar radius function of barycentre caused by orbital movement of planets over time

8. Originally Posted by Robert Tulip
George, your claim here that orbital motions aren’t needed to derive the barycentre is sufficient to place the barycentre in space, but not in time.

The movement of the barycentre over time against the sum of mass of the solar system is depicted in detail over several thousand years in diagrams that I have posted in this thread at

#26 https://forum.cosmoquest.org/attachm...3&d=1489653849
- Quantifies power of orbital relations between gas giants as primary causes of distance of sun from barycentre
- Also identifies unknown factors revealed by NASA’s barycentre calculations

#85 https://forum.cosmoquest.org/attachm...7&d=1489914377
- Stable similar flower patterns of barycentre in successive 178.9 year periods, each divided into fifteen stable segments

and

#93 https://forum.cosmoquest.org/attachm...1&d=1490063170
- Wave depiction of stable similar solar radius function of barycentre caused by orbital movement of planets over time
Of course the position of the barycenter as a function of time depends on the motions of the bodies. I am not aware that anyone is saying otherwise. All I am saying is that at any given moment in time the position of the barycenter is a function of the positions and masses of the bodies at that time, whether they are moving or not.

9. Originally Posted by Hornblower
Of course the position of the barycenter as a function of time depends on the motions of the bodies. I am not aware that anyone is saying otherwise. All I am saying is that at any given moment in time the position of the barycenter is a function of the positions and masses of the bodies at that time, whether they are moving or not.
Hornblower, you are missing the point of my post, which responded specifically to George's statement that orbital motions aren’t needed to derive the barycentre. He is using a limited static meaning of 'derive' when a dynamic model is useful.

I have illustrated in the links that you quoted that in fact orbital motions are needed to derive the barycentre as a dynamic moving location, and illustrated exactly how this is done, opening up further scientific questions (eg the unknown factors influencing NASA calculations) that a dynamic model indicates.

The wave function of the barycenter has slowly shifting axes of symmetry caused by the orbital interaction of Jupiter, Saturn and Neptune, illustrated in the attached.

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Originally Posted by grant hutchison
Hoyle was describing a way of thinking, not a practical approach.
Would the ECI be used if it didn’t work?

11. Originally Posted by wd40
Would the ECI be used if it didn’t work?
What is the ECI?

12. Originally Posted by Robert Tulip
Hornblower, you are missing the point of my post, which responded specifically to George's statement that orbital motions aren’t needed to derive the barycentre. He is using a limited static meaning of 'derive' when a dynamic model is useful.

I have illustrated in the links that you quoted that in fact orbital motions are needed to derive the barycentre as a dynamic moving location, and illustrated exactly how this is done, opening up further scientific questions (eg the unknown factors influencing NASA calculations) that a dynamic model indicates.

The wave function of the barycenter has slowly shifting axes of symmetry caused by the orbital interaction of Jupiter, Saturn and Neptune, illustrated in the attached.
I am not missing your point. I just don't think it is all that important for reassuring George and others that there is no conflict between the position of the mutual Sun/Neptune barycenter and the relative weakness of Neptune's gravitational action at that distance, as compared with the corresponding items for Jupiter. George appears to be reassured on that point.

Yes indeed, quantitative analysis of the dynamic model shows us why the planets and the Sun move as they do, and how the Sun traces its loopy path around the overall barycenter. The Fourier transform technique teases out periodic components of that motion that might be missed if we just looked at the raw data.

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Originally Posted by Hornblower
What is the ECI?
"Earth Centered non-rotating Inertial frame"

14. Originally Posted by Hornblower
What are r1 and s1?
r1 is the distance to the barycenter from m1. s1 is typically used in the F=ma equations for distance, so I elected to go with it. So r1 is s1.

Or inverse cube, inverse 4th, inverse linear, constant, linear, square, cube, you name it.
Yes, good point. I should have just used "gravity" but I sometimes enjoy calling it the inverse square law to get the math feel to it, I suppose.

15. Originally Posted by Hornblower
I am not missing your point. I just don't think it is all that important for reassuring George and others that there is no conflict between the position of the mutual Sun/Neptune barycenter and the relative weakness of Neptune's gravitational action at that distance, as compared with the corresponding items for Jupiter. George appears to be reassured on that point.
I think both the static and dynamic presentations are very applicable and quite interesting. The static model makes it easy to see how to calculate the barycenter, just like a seesaw.

But the dynamic model may be even more important. The idea that a distant and somewhat small mass can have such a huge difference in establishing a barycenter is where I became conflicted. But I put the cart before the horse, meaning that unless they are already orbiting one another, the barycenter has little meaning. As I learned, there is no leveraging as can be found on a seesaw.

For instance, if two masses are traveling mostly toward one another, say an interstellar rogue planet coming into town, then their mutual barycenter simply moves closer and closer to the Sun's center -- it has little meaning if it is headed straight for the solar c.g. -- though how often does that ever happen for any object? But if a mass, say hypothetical Planet 9, is already orbiting, then the barycenter has far more meaning. The Sun really does swing way out beyond the Sun-Jupiter-Saturn (etc.) barycenter. That's why I did the graphic. But what must be kept in mind is that this swing (Sun's motion about this barycenter) is, at best, barely noticeable since it is, very roughly, perhaps 15,000 year addition to Tulip's plots. A Sun-quasar barycenter (if we simplify and ignore all other distant masses) is even more superfluous because of... time, and that's pretending they are orbiting one another.
Last edited by George; Yesterday at 03:20 PM.

16. Originally Posted by wd40
"Earth Centered non-rotating Inertial frame"
There ain't no such animal. Since Earth's motion is accelerated, any frame centered on it is going to be non-inertial. GR enables us to use it in principle, but I continue to think it will be horrendously messy

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Originally Posted by Hornblower
There ain't no such animal. Since Earth's motion is accelerated, any frame centered on it is going to be non-inertial. GR enables us to use it in principle, but I continue to think it will be horrendously messy

https://en.wikipedia.org/wiki/Earth-centered_inertial

"for objects in space, the equations of motion that describe orbital motion are simpler in a non-rotating frame such as ECI."

ECI is the only coordinate system used for successful GPS operations.
Last edited by wd40; Yesterday at 06:17 PM.

18. Originally Posted by wd40
https://en.wikipedia.org/wiki/Earth-centered_inertial

"for objects in space, the equations of motion that describe orbital motion are simpler in a non-rotating frame such as ECI."

ECI is the only coordinate system used for successful GPS operations.
Simpler for objects in orbit around the Earth. Which rules out pretty much the entire Universe.

Grant Hutchison

19. Originally Posted by wd40
https://en.wikipedia.org/wiki/Earth-centered_inertial

"for objects in space, the equations of motion that describe orbital motion are simpler in a non-rotating frame such as ECI."

ECI is the only coordinate system used for successful GPS operations.
OK, the creators of the term are fudging the word "inertial". The article acknowledges that such a frame is not truly inertial, and that in some applications the departure from being inertial cannot be ignored.

20. I would assume the of an ECI model would be for Nasa's Earth satellites, for example, but very messy, as mentioned, beyond that.

21. Originally Posted by George
the dynamic model may be even more important. The idea that a distant and somewhat small mass can have such a huge difference in establishing a barycenter is where I became conflicted.
The influence of Neptune is seen in the wave function graph that I recently posted, and in the table below.

Isaac Newton calculated the barycenter solar distance based only on the visible planets, producing a smooth sine wave. I saw his diagram a few years ago, but now I can't find it.

The influence of Neptune changes the smooth Jupiter-Saturn wave into the near-symmetrical 179 year wave seen in the graph. The reason for the symmetry is that the approach of Neptune to the Jupiter-Saturn inferior and superior conjunction points is a mirror image of its movement away from these points, and this same pattern repeats itself with strong stability over several cycles.

I think the departures from symmetry are mainly from the disruptive influence of Uranus, as indicated in the attached tabulation of spectral power, which shows Neptune has more than double the influence of Uranus.

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