Solved

3 or 4 bodies moving in a circle

Posted on 2011-02-24
43
1,178 Views
Last Modified: 2012-05-11
In a plane, I would like to learn how to compute the initial velocities of 4 bodies (all having same mass), starting off in a square (whose sides are S) so that in absence of all external forces, these 4 bodies rotate in the same circle and whose 4 points always maintain a square. So, only internal gravitational forces on the 4 bodies apply. (I suspect that a small variance from the perfect numbers will cause problems, but let's keep this theoretically Newtonian.)

Note: If the math/physics is easier for 3-bodies (in an equilateral triangle) is easier to use than 4 bodies, then I'd be happy to start with that and not be concerned in this question with 4-bodies. If the problem solutions are sufficiently different and help is available on both, then I'll break one of these cases into a separate question.

I'm guessing that the initial velocity should be tangential to the circle that includes the 4 bodies, and that by symmetry, the speeds should be identical. But, even if right, how do I prove that? And how do I come up with the initial speed?

If the speed is less than the required circle speed, then I am guessing that they spiral into each other gaining speed (but total angular momentum should remain constant), and then they may spiral out. (If the initial speed is 0, then they won't spiral out, but just fall into each other.)

I'm guessing the solution to the circle path is the balancing act where the minimum distance to the origin = maximum distance to the origin = initial radius of the circle. So, maybe I need to figure out the min/max relationship to the initial speed, and then set min=max=initial radius to determine the speed.

I found an article on 3-body spiral that used complex numbers and some kind of Hermite transformation that I wasn't familiar with. Now, I did find this link on motion and complex_numbers (and I pretty much understand the material that preceeds this section), but seeing the Hermite stopped me from proceeding. An explanation or link of simpler math/physics to help me setup initial conditions would be very nice. If Hermite is absolutely required for me to proceed, the let me know how to learn all about it. Preferably, other more basic concepts can be used.

I'd like to keep the coordinate system fixed at the origin, if at all possible.

If this topic gets too complex, then I'll break it into multiple questions and/or increase points.
0
Comment
Question by:phoffric
  • 20
  • 11
  • 9
  • +1
43 Comments
 
LVL 27

Expert Comment

by:aburr
ID: 34976674
You seem to have a pretty good grip on the problem. Advanced math is not needed.
Start by drawing a good diagram. Let r be the radius of the circle on which the masses will travel.
The key physics principle is that the masses will move in a circle IF there is a centripetal force applied. That force is the force of gravity. (F)
F = (m * v^2)/r         Gravitation force is F = G mM/r^2
V is perpendicular to radius because you do not want to change that radius
You are correct to appeal to symmetry to say that you can tread each mass the same.
Now get the gravitational force
You have three masses acting on the mass you are discussing.
One, the farthest away and two with equal radial components (The non-radial compontnts cancel out.
F = G (m^2)/(2 * r^2)+ ((2 * G 8* m^2)/S^2)* cos(45 degrees)  which = (m*v^2)/r
You know everything but v so calculate
More help if needed.
0
 
LVL 32

Author Comment

by:phoffric
ID: 34976709
Thanks aburr for your reply. I plan on trying out your suggestions tomorrow, and working out any details left for me to do. I was so confused by the spiral link, I just assumed that some complex math was needed. I'll let you know how I did.
0
 
LVL 37

Expert Comment

by:TommySzalapski
ID: 34987903
Hello gents, nice to see you on the other side of a question, phoffric. Hope I can be of some help. The reason the 3-body spiral problem is so complicated is because the system changes so drastically as the bodies move. If the bodies remain equidistant, then the problem just becomes like one of those geosynchronous orbit problems we saw in first semester physics. So I agree with aburr.

>>then I am guessing that they spiral into each other gaining speed (but total angular momentum should remain constant), and then they may spiral out.

They will not spiral out unless the initial speed is too high. Since the force is inversely proportional to the square of the distance and the velocity is linearly proportional, the longer they spiral in, the more trapped they will be. They will eventually collapse into a single mass, rotating (as you noted) at the same angular velocity as the orbit had been.
0
 
LVL 32

Author Comment

by:phoffric
ID: 34988758
aburr,
That Hermitian tranformation in the OP reference really blindsided me and made me think all multi-body problems were harder than they need be. Your idea seems to make sense.

Could you just double-check in your formula for F that in the first term, the 2 in the denominator is ok; and in the 2nd term that the 8 is ok. If it is ok, I'll try again and try to match your formula.

Also, from your argument, it seems that it doesn't matter whether there are 3, 4, or even 100's of bodies in regular polygon, just as long as you get the geometry right, the formulas will work out.

Tommy,
Yeah, I start getting bleary eyed when the number of bodies > 0.

>> They will not spiral out unless the initial speed is too high.
If the initial speed is just a little off, then won't they spiral out, but then fall back in?

If very high initial speed, then probably some escape possibility exists.

But if initial speed is just a little lower from the V that aburr is computing, then I didn't realize they would spiral into the origin. I was thinking that the PE gets converted to the KE speeding them up so much that at the lower radius, they would be going faster than the just-right velocity to maintain that radius. Actually, I'm a little biased, because I tried a simulation (that could be wrong) and saw them close in and then go out. Do you know of a way to show that the increased speed is still below the just-right speed for a given radius? If you do, and if complicated, I can put this in a different question.

What I find a little confusing is that one of the bodies (call it A) is not really moving around the system's CM (initially, the origin), but rather around the CM of the other 3 bodies (call it CM3). So, just as the moon is constantly falling into the earth, but missing it, I thought that A might be missing the CM-3, but as it gets closer, it wraps around CM-3 fast enough to spiral out.
0
 
LVL 37

Expert Comment

by:TommySzalapski
ID: 34988968
If the initial speed is just a little off, then won't they spiral out, but then fall back in?

I'm assuming we are in a frictionless state, so no. As you already know, the angular velocity never changes. As you get farther away, the speed needed to break free decreases. So even if the velocity was still perpendicular to the radius, they would still go farther out; however, the direction of the velocity is also changing, making them sail away even faster.
Oh, if the initial velocity is not perpendicular to the radius, then you could see that happen. But I don't think that was the issue

I was thinking that the PE gets converted to the KE speeding them up so much that at the lower radius, they would be going faster than the just-right velocity to maintain that radius.

Yes, I need to amend my earlier statement. They would spiral in, but not in a 'circular spiral.' Rolling a marble in a bowl would give you an idea of how they would move. I don't know the name for the 'shape,' but let's call it a decreasing elliptical spiral since the path would be elliptical, but each 'loop' would have a smaller and smaller area.

Do you know of a way to show that the increased speed is still below the just-right speed for a given radius?
Sure. Proof by contradiction. If you could generate enough energy to break out of orbit when you didn't have it to begin with, then you would be creating energy out of thin air. The math would work out to show it analytically, but it would be fairly complex.

What I find a little confusing is that one of the bodies (call it A) is not really moving around the system's CM (initially, the origin), but rather around the CM of the other 3 bodies (call it CM3).

But CM3 is also moving. In fact it moves in a circle concentric with the orbit of A always directly across the center from A. So it would actually 'appear' to orbit around the center anyway. If you start something in a circular orbit, it will remain in a circular orbit.
0
 
LVL 37

Expert Comment

by:TommySzalapski
ID: 34988978
I'm curious what you used as a simulation. A computer program, I guess? Otherwise the frictionless thing might be hard, but marbles in a bowl would actually be similar, (only for the case where they all orbit around the center of course).

I'd write up a simulator myself just because you have me all intrigued, but I'm so behind with the new baby and all that I shouldn't even be spending as much time here as I already have.
0
 
LVL 27

Expert Comment

by:aburr
ID: 34989002
"F = G (m^2)/(2 * r^2)+ ((2 * G 8* m^2)/S^2)* cos(45 degrees)  which = (m*v^2)/r
-
my typing must be pretty bad. what I wrote was

F = G (m^2)/(2*r)^2 + 2*G*(m^2/S^2)*cos 45
-
I have trouble putting in (). In the first term I missed the position and in the second I forgot the shift and missed a key when intending (
0
 
LVL 84

Assisted Solution

by:ozo
ozo earned 200 total points
ID: 34989710
>  I am guessing that they spiral into each other gaining speed
Angular momentum and energy are conserved. so one can spiral in only if another spirals out.
But starting from symmetry, symmetry would have to be preserved too, so they would have to move together
But you can have elliptical orbits, with the radius changing cyclically, or hyperbolic orbits, where they escape to infinity.  
0
 
LVL 32

Author Comment

by:phoffric
ID: 34990409
Thanks aburr for the rewrite. So here is the perfect speed, |Vp|, formula for the 4-body is:
f = G (m²)/(2*r)² + 2*G*(m²/S²)*cos 45 = m |Vp|²/r
I just have to solve for the perfect speed, |Vp|, and under perfect frictionless conditions, the 4 bodies will remain in a pure circle forever. (And Vp is tangent to the circle that includes the 4 bodies in the square.)

Tommy,
>> the angular velocity never changes
For a circular orbit, v = w r (w == angular velocity). If I have two circles with radii, r1 > r2, and if v were the same speed, then w2 > w1, and so angular velocity appears to be changing.

ozo,
>> Angular momentum and energy are conserved. so one can spiral in only if another spirals out.
From above equation, I now know |Vp|, so the 4 bodied square should remain as a rotating equivalent square forever with the rotional speed constant.

If the bodies' initial speed is slightly smaller speed than |Vp|, then all 4 bodies begin to spiral in. Then in this symmetric case, are you confirming that the 4 bodies will continue to spiral in (always maintaining a square at any point in time) with the square's dimension approaching 0 until the bodies concurrently collide?
0
 
LVL 84

Assisted Solution

by:ozo
ozo earned 200 total points
ID: 34990611
If the bodies initial speed is slightly smaller than |Vp|, they do not spiral in.
They would be in an elliptical orbit rather than a circular orbit.

It is probably easiest to conserve energy and angular momentum.
gravitational energy is 4*(G*m²/(2*r) + 2*G*m²/(sqrt(2)*r))
kinetic energy is 4*mV²/2
Total energy is the sum of gravitational potential energy and kinetic energy
When at the highest and lowest points in the orbit, where the radial velocity is 0,
total angular momentum would be 4*mVr

0
 
LVL 32

Author Comment

by:phoffric
ID: 34990970
Just one lasat thing, please, to reconcile with "They will eventually collapse into a single mass,..."
(BTW - I am not worried about whether they bind into a single mass, or just bounce off each other; I'm just trying to understand conceptually whether the shapes' dimensions (square or triangle) will oscillate or just keep closing in towards the origin until they collide with each other.)

So, if the bodies initial speed is much smaller than |Vp|, say 1/2, or 1/10, or even 1/100 |Vp| (but not 0, of course), then initially, at least, the dimensions of the shape start to shrink. Looking at an individual body, its initial descent would appear to be spiraling inwards (or perhaps it is elliptical also). To simplify the scenario, and to try to avoid collisions, let's assume that the 4 bodies are point masses. Will the shape reach a minimum dimension > 0, and then restore itself back to their initial conditions (or perhaps the dimension might even exceed this initial state)?

As in my previous remark to aburr, I assumed that whether there were 3 or 4 bodies (or perhaps even more bodies in a regular polygon), then the high-level remarks would remain the same. If this assumption is incorrect, then please just address one specific case, and I'll ask about the other case(s) in a separate question.


Just looking for high-level qualitative posts at this point. I will investigate more detailed scenarios after better understanding this and hopefully will be able to ask more quantitative questions later.

Although interested, I don't need in this question to know the exact nature of the path of an individual body (e.g., whether it is some kind of spiral, or elliptical). In this regards, I'll may ask in a related question about the path of an individual body after investigating some scenarios. Thanks.
0
 
LVL 37

Expert Comment

by:TommySzalapski
ID: 34991962
Ozo made some good points in http:a#34989710 about conservation of energy that helped me make an important observation. No they will never actually spiral in to the middle and stay there. That was my mistake. Since I usually work with systems that have friction, I forget about all the funny things that happen when no energy is lost. Sorry.

Using Vs to mean aburr's static orbit velocity and Vi to mean the initial
If Vi = 0 and they are somehow immune from collision they will have 'flat' 'orbits' of length 2r
If Vi > 0 and Vi < Vs they will have elliptical orbits
If Vi = Vs they will have circular orbits (which is of course an ellipse)
If Vi > Vs then they will have hyperbolic orbits and fly away forever

If you rotate your system at the same speed as the bodies you will see all of them have 'flat' orbits.
If Vi is not perpendicular to the radius, then you were, or will end up in, one of the above cases. Since escape velocity is irrespective of angles, you can still get Vs and all the above still applies (except the = case which requires perpendicularity)
0
 
LVL 32

Author Comment

by:phoffric
ID: 34992458
Tommy,
   Here's where I saw the "spiraling in, then out" simulation. Take a look at http://faculty.ifmo.ru/butikov/Projects/Collection6.html . I'll write up a simulation one of these days (hopefully next month) and post a text of the data.

  Congratulations on the new baby!
0
 
LVL 37

Expert Comment

by:TommySzalapski
ID: 34992530
Ah, watch the time carefully. It goes for a while and then reverses. You aren't seeing it spiral in and then out, you are seeing it play backward after a bit. It's a cool simulator, but I'd rather see one that I could set the inputs on.
0
 
LVL 32

Author Comment

by:phoffric
ID: 34992685
Due to a java applet error, I can't get the whole scenario, but I think I got enough to illustrate what I perceive as spiraling towards the CM and then spiraling out. Here is the starting point: Start and here is the finish: Finish
I am surprised that the 3 curves don't look similar given the symmetry arguments. One curve seems to be going through the CM of the 3 bodies. But I'll leave that for another question.
0
 
LVL 37

Expert Comment

by:TommySzalapski
ID: 34993427
Oh, that's just what I've been calling an elliptical orbit.
0
 
LVL 32

Author Comment

by:phoffric
ID: 35000348
Follow-on Q: Escape velocity of 4 bodies for those interested. Thanks.
0
 
LVL 84

Assisted Solution

by:ozo
ozo earned 200 total points
ID: 35003223
Looks like it's unstable, despite what Niven implied.
http://en.wikipedia.org/wiki/Klemperer_rosette
(since he made the same mistake about Ringworld, I guess it's not surprising that the Puppeteers's Rosette would be unstable too)
0
 
LVL 32

Author Comment

by:phoffric
ID: 35007749
ozo,
I am not understanding this statement:
>> so one can spiral in only if another spirals out.
Could you elaborate on this for the 3 or 4 body symmetric problem. (I think I saw a spiral galaxy simulation and was surprised to see stars moving from one spiral leg to another from an inner leg to outer, and vice-versa.)

aburr,
Could you please repeat your first post with the corrected formula. For the PAQ, I'd like to get that explanation corrected. Thanks.
0
 
LVL 84

Assisted Solution

by:ozo
ozo earned 200 total points
ID: 35009251
Stars in a spiral galaxy do not move in spirals.
They move in circles (approximately)
The spirals you see in a  spiral galaxy are density waves which move through the stars, not the paths of stars.

But in a galaxy, momentum can be transferred from one star to another, which lowers the orbit of one and raises the orbit of another (and destroys any symmetry that may have existed between those stars, so it's a very different situation)
0
 
LVL 32

Author Comment

by:phoffric
ID: 35009276
Then what did you mean by
>> so one can spiral in only if another spirals out.
Could you elaborate on this for the 3 or 4 body symmetric problem.
0
How your wiki can always stay up-to-date

Quip doubles as a “living” wiki and a project management tool that evolves with your organization. As you finish projects in Quip, the work remains, easily accessible to all team members, new and old.
- Increase transparency
- Onboard new hires faster
- Access from mobile/offline

 
LVL 84

Assisted Solution

by:ozo
ozo earned 200 total points
ID: 35009362
The bodies start out in the same orbit, with the same momentum.
Since momentum is conserved, the only way for one to lose momentum is for another to gain it,
(assuming that the bodies are the only things which exist in the universe)
this would break the symmetry.

Although as noted in the article, Klemperer rosettes are unstable.
so in the real universe, the symmetry will eventually break down, and you would be likely to see
one orbit drop and another orbit rise.
0
 
LVL 32

Author Comment

by:phoffric
ID: 35009474
>> Since momentum is conserved, the only way for one to lose momentum is for another to gain it
Ok, that makes sense. Were you specifically talking about linear or angular momentum, or both?

Now your comment was in response to
>  I am guessing that they spiral into each other gaining speed

I am not seeing the connection between "they spiral into each other gaining speed" (which we agree is correct??) and that one body losing momentum can only occur if another gains momentum. Don't all the bodies gain speed and KE (and therefore linear momentum) at the expense of losing PE?

>> Klemperer rosettes are unstable
Hmm, I will see if I can follow this article better (if not, maybe another question).
0
 
LVL 27

Accepted Solution

by:
aburr earned 200 total points
ID: 35011879
repeat of previous answer (corrected) as requested

You seem to have a pretty good grip on the problem. Advanced math is not needed.
Start by drawing a good diagram. Let r be the radius of the circle on which the masses will travel.
The key physics principle is that the masses will move in a circle IF there is a centripetal force applied. That force is the force of gravity. (F)
F = (m * v^2)/r         Gravitation force is F = G mM/r^2
V is perpendicular to radius because you do not want to change that radius
You are correct to appeal to symmetry to say that you can tread each mass the same.
Now get the gravitational force
You have three masses acting on the mass you are discussing.
One, the farthest away and two with equal radial components (The non-radial components cancel out.
F = G (m^2)/(2*r)^2 + 2*G*(m^2/S^2)*cos 45  which = (m*v^2)/r
You know everything but v so calculate
More help if needed.
0
 
LVL 84

Assisted Solution

by:ozo
ozo earned 200 total points
ID: 35013923
spiraling down to a lower orbit requires losing energy and momentum,
which can only go into the only other bodies in the system, boosting them into a higher orbit.

Maybe you were thinking of elliptical orbits when you said spiral in.
But an elliptical orbit is still a fixed orbit, with a fixed energy and fixed angular momentum.
(intermediate between the energy and momentum of circular orbits at the apoapsis and periapsis distances)
It does move in during half of the orbit, and out during the other half, and perhaps that's what you meant.
0
 
LVL 32

Author Comment

by:phoffric
ID: 35016869
>> spiraling down to a lower orbit requires losing energy and momentum,
Ok, I think I see what you mean. In a lower stable circular orbit, the speed is lower since r is smaller.

Re: elliptical orbits
I've been seeing this as I hit the web pages for this type of problem. I know that an object around a stationary object will move in an elliptical orbit (if it doesn't escape).

I don't understand why we have an elliptical orbit for one body, since it is trying to rotate around the CM of the other bodies; but as the other bodies also move in corresponding symmetry, their CM is accelerating.

I am wondering how to prove under these more complex conditions that each body has an elliptical orbit. This sounds complicated, so if anyone knows how to prove this, I'd be happy to ask another question. Just let me know if there is a proof (hopefully one that deals with basics).
0
 
LVL 32

Author Comment

by:phoffric
ID: 35016938
Take that back. I just use (m v^2/r) as centrifugal force and used gravity force formula and came up with: v^2 = GM/r
So, I think I was wrong. In a lower orbit, the object must at faster speed to maintain stability. So, then in a lower orbit, the linear momentum now appears to be greater than in a higher orbit.
0
 
LVL 37

Assisted Solution

by:TommySzalapski
TommySzalapski earned 100 total points
ID: 35020615
Yes. It would be much faster. You can see this in our own solar system.
But the total energy in the system is acutally less. If an object was orbiting and some external force slowed it down, it's orbit would have a smaller area (either due to having a smaller radius or being less circular).
0
 
LVL 37

Expert Comment

by:TommySzalapski
ID: 35020657
since it is trying to rotate around the CM of the other bodies;
I don't know why this keeps coming up. Are you sure it doesn't orbit around the center of mass of the system? If they only orbited around the other masses, then we should expect to see the earth orbit the moon (centered at the moon's center) while the moon orbits the earth (centered at the earth's center) but they appear to both orbit the center of mass of the whole system.
The mini simulation you linked to also seems to show that.
0
 
LVL 84

Expert Comment

by:ozo
ID: 35022107
> their CM is accelerating.
The CM of the system of bodies is not accelerating.
0
 
LVL 32

Author Comment

by:phoffric
ID: 35023279
>> Are you sure it doesn't orbit around the center of mass of the system?
    From symmetry, I believe that the 4 bodies orbit around the CM of the 4 body system. But..

>>  since it is trying to rotate around the CM of the other bodies; but as the other bodies also move in corresponding symmetry, their CM is accelerating.
>> >> The CM of the system of bodies is not accelerating.

    In the 4 body problem, break this down into two systems: system1 around three bodies, and system2 of the remaining body. System2 has mass M and system 1 has mass 3M. I see this as system2 (one body) rotating around the CM of system1. Even when the 4 bodies are rotating in a stable circle, the CM of system1 is accelerating as it moves in a circle. When the 4 bodies move in a stable orbit (but not a circle, and allegedly an elliptical orbit), then the CM of system1 is accelerating (also possibly in an elliptical orbit).
0
 
LVL 37

Assisted Solution

by:TommySzalapski
TommySzalapski earned 100 total points
ID: 35028355
The CM of the whole system doesn't move at all, ever. Even in the cases where the objects fly off into space, the CM remains static.

The CM of a group of 3 objects will always balance the fourth object across from the CM of the whole system, so it will mimic the single object.

The objects are drawn toward each other, but all the orbits end up being around the CM of the whole system.
0
 
LVL 32

Author Comment

by:phoffric
ID: 35032159
Since my post, http:#35016869 , you have been trying to help me understand something (appreciating the sentiment), which I believe I already understood. I think the discussion arose due to my imprecision in a statement where I used pronouns causing ambiguity.

As this is subsequent discussion is dealing with "their CM is accelerating", let me rephrase my statement removing ambiguities (i.e., pronouns):

I don't understand why we have an elliptical orbit for one body, since the one body is trying to rotate around the CM of the other 3 bodies; but as the other 3 bodies also move in corresponding symmetry, the CM of the other 3 bodies is accelerating.

So, now I hope that we all agree that the CM of the other 3 bodies is accelerating. The question about the path one body takes as being elliptical or not can be deferred to another question. (I'm going to think about paths of binary stars which are both rotating around each other.)

>> Yes. It would be much faster. You can see this in our own solar system.
Right, that is what I realized when writing http:#35016938 where I took back the previous post. In the previous post, I was trying to fit "spiraling down to a lower orbit requires losing energy and momentum," into my model.
0
 
LVL 32

Author Comment

by:phoffric
ID: 35032166
So, only one item is left for clarification.

I am still confused as to what "spiraling down to a lower orbit requires losing energy and momentum" means. As the 4 bodies get closer to the origin, they appear to be gaining linear momentum rather than losing it.
0
 
LVL 37

Assisted Solution

by:TommySzalapski
TommySzalapski earned 100 total points
ID: 35032289
By "spiraling down to a lower orbit" we are talking about the actual total orbit getting smaller (which I have been referring to as the area of the orbit).
In an elliptical orbit the kinetic energy increases (object goes faster) as it gets closer to the center but the potential energy of course gets lower. And PE+KE is naturally constant.
You need to lose energy to make the entire orbit smaller. Does that clear up what we were talking about now?
0
 
LVL 32

Author Comment

by:phoffric
ID: 35032893
Looking back, I see:
>> Angular momentum and energy are conserved.
So, strike my comments on linear momentum as they are not applicable since ozo is talking about angular momentum.

As you reiterated, PE+KE is a constant. The 4-body square oscillates between a min and max area. And so, the total energy is the same at any given point in time whether the square is min area or max area. This is what is confusing me, since I don't see how we are losing energy when we are in the min area state. The KE increases, and the PE decreases so that the energy is constant; so we don't have a total lower energy.


>> spiraling down to a lower orbit requires losing energy and momentum, which can only go into the only other bodies in the system, boosting them into a higher orbit.

Perhaps, there are some semantic misunderstandings here. I hope ozo will be able to further clarify this statement addressing the points that are still confusing me.
0
 
LVL 37

Assisted Solution

by:TommySzalapski
TommySzalapski earned 100 total points
ID: 35033081
When Ozo said "lower orbit", she didn't mean a 'lower' portion of the same orbit she meant an actual different orbit. Like, if the earth decided to "spiral down" and orbit right behind Mercury, it would have to lose some total energy. If some external force changes the energy in the system, then the orbits could change.
0
 
LVL 32

Author Comment

by:phoffric
ID: 35033092
Oh, ok - me dense. As soon as ozo confirms your statement, I'll close this question.
0
 
LVL 84

Assisted Solution

by:ozo
ozo earned 200 total points
ID: 35042181
> The KE increases, and the PE decreases so that the energy is constant;
In a lower orbit, as opposed to a different portion of the same orbit
KE is increased by about half as much as the PE is decreased, so PE+KE is lower.
Within the same orbit, yes, PE+KE remains constant, but I would not call that a "spiral"
(unless you consider each half of an ellipse to be a "spiral")
0
 
LVL 32

Author Comment

by:phoffric
ID: 35045746
From the Klemperer rosette link:

any tangential perturbation causes a body to get closer to one neighbor and farther from another; the gravitational force becomes greater towards the closer neighbor and less for the farther neighbor, thus further pulling the perturbed object towards its closer neighbor, enhancing the perturbation rather than damping it. An inward radial perturbation causes the perturbed body to get closer to all other objects, increasing the force on the object and increasing its orbital velocity---which leads indirectly to a tangential perturbation and the argument above.

I think this means that if running a simulator too long, then even very small numerical errors (e.g., truncation, round-off) will result in a tiny perturbation which eventually lead to the destruction of the stable paths taken by the 4 bodies.
0
 
LVL 84

Assisted Solution

by:ozo
ozo earned 200 total points
ID: 35045788
> even very small numerical errors ... eventually lead to the destruction of the stable paths
that's correct

So if you want to simulate the perfectly symmetrical situation, I'd suggest that you take advantage of the symmetry
and simulate just one body, placing the other 3 by their symmetry.
That way, all your very small numerical errors will be symmetric.
0
 
LVL 32

Author Closing Comment

by:phoffric
ID: 35056601
Thanks to all of you and your patience with the terminology issues of "spiral" vs "elliptical". Reading through this again, I do understand much better all of your posts.

BTW - I looked at some articles on 2-bodies which explain that their rotation is elliptical, and I think that explanation can be applied to this problem in explaining why the paths are elliptical under certain conditions.

Thanks again!
0
 
LVL 32

Author Comment

by:phoffric
ID: 35057519
Related question for anyone interested:
    http://www.experts-exchange.com/Other/Math_Science/Q_26868214.html

Having trouble with consistency with the force computation.
0

Featured Post

What Is Threat Intelligence?

Threat intelligence is often discussed, but rarely understood. Starting with a precise definition, along with clear business goals, is essential.

Join & Write a Comment

Suggested Solutions

Complex Numbers are funny things.  Many people have a basic understanding of them, some a more advanced.  The confusion usually arises when that pesky i (or j for Electrical Engineers) appears and understanding the meaning of a square root of a nega…
Foreword (May 2015) This web page has appeared at Google.  It's definitely worth considering! https://www.google.com/about/careers/students/guide-to-technical-development.html How to Know You are Making a Difference at EE In August, 2013, one …
This video discusses moving either the default database or any database to a new volume.
Access reports are powerful and flexible. Learn how to create a query and then a grouped report using the wizard. Modify the report design after the wizard is done to make it look better. There will be another video to explain how to put the final p…

744 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

14 Experts available now in Live!

Get 1:1 Help Now