Understanding the Relativity-Induced Precession of Mercury's Orbit

[Bjarne]

Can someone explain the cause of the Precession of Mercury’s Orbit in a "language" so simple that my grandmother can understand it?

I mean only the part (43 arc second) that is due to relativity.

 

[Bill_K]

Probably not. Does your grandmother understand the precession in Mercury's orbit that is not due to relativity?

 

[Jonathan Scott]

Can someone explain the cause of the Precession of Mercury’s Orbit in a "language" so simple that my grandmother can understand it?

I mean only the part (43 arc second) that is due to relativity.

Hmmm. Has your grandmother studied General Relativity?

The direct cause is that space is very slightly curved around a large mass, like a strip around the surface of a cone. This is the main practical difference between General Relativity and Newtonian theory.

I think the simplest way to think about it is to imagine going round a large circle on the curved surface of the earth. If you take the strip of the surface of Earth in which you travelled, it cannot be laid flat without leaving a gap, so in going all the way round you have actually turned by less than 360 degrees. (The missing angle is called the "angular deficit" or "angular defect" and on a sphere the total angular deficit for the perimeters of the areas enclosed by any network of paths is always 720 degrees or 4 pi radians).

Similarly, space is effectively slightly "conical" around any massive object. If you follow space all round the sun, then when you pass the direction where you started, you will not yet have turned through 360 degrees, and that will actually occur slightly after the point where you started. Orbits are ellipses which normally effectively repeat every 360 degrees, so this effect, "precession", causes the orbit to repeat at a slightly later position each time round. The closest point of an orbit around the sun is called "perihelion" and can be observed in order to determine the precession rate.

(The perihelion precession of a planet is also affected by other planets, but that effect can be calculated using normal Newtonian gravitational theory in flat space).

 

[tom.stoer]

I'll try to.

There are two cebtral potentials V(r) which - in Newtonian mechanics in three dimensions- have closed, periodic (elliptic) orbits, namely V(r) ~ r² and V(r) ~ 1/r.

In GR one can derive "correction terms" to the usual V(r) = -g/m; V_eff(r) ~ V(r) + L²/2r² effective potential, especially a first correction with v(r) = -gL²/c²r³. So instead of the effective potential V_eff(r) one has to solve for an orbit of the corrected potential V_eff(r)+v(r) - which leads to a small precession of an orbit.

 

[Bill_K]

Similarly, space is effectively slightly "conical" around any massive object. If you follow space all round the sun, then when you pass the direction where you started, you will not yet have turned through 360 degrees, and that will actually occur slightly after the point where you started.

Sorry, that is absolutely, positively, totally incorrect! Not to mention imaginary and misinformative. Space is most certainly *not* conical around a massive object. Where did you ever get that idea? An r=const surface in the Schwarzschild metric has (guess what?) the metric of a perfectly normal sphere of radius r. The distance around it in any direction is 2πr, the surface area is 4πr² and a gyroscope transported all the way around it will continue to point in the same direction.

 

[bcrowell]

The simple way to look at it is to compare with falling through the event horizon of a black hole. Let r be the distance at periastron (perihole?), and let h be the radius of the event horizon.

At large r, you have Newtonian orbits, which are periodic, closed-path trajectories.

For r \le h, you fall through the event horizon and never come back out.

As we consider orbits with r values that get smaller and approach h, we expect that the time spent near periastron should increase. This must be true, since the time blows up to infinity for r \le h. If you spend more time near periastron, it makes sense that you wrap around through a greater angle.

 

[DrGreg]

Image credit: ©Allen McCloud CC BY-SA 3.0

Actually Bill_K and Jonathan Scott are both right but they are talking about different things.

The diagram above represents the geometry of a two-dimensional slice of space around the Sun, with the Sun at the centre. (See Flamm's paraboloid for the technical details.) The region between two concentric circles has approximately conical geometry. Imagine cutting such a region out the diagram with a pair of scissors. To make the region lie flat, you have to make another cut in the radial direction to open it out, and then you will have an annulus with a gap missing. If you now draw an ellipse on your almost-annulus, there will be a gap. This gap represents the precession.

 

[Bjarne]

Hmmm
I have heard if the orbit of Mercury was circular there would not be any anomaly at all.

So this must mean that it cannot be general relativity that is the cause of the anomaly.

As I understand it when time is changing due to gravity and hence ticks slower the further to the Sun we get, then also distances gets larger.

Therefore there will be relative more time closer to the Sun (Mercury’s orbit) but distances is also proportional larger. Therefore there is also more time to complete a orbit that have relative "longer" distance.

The cause should be (according to what I know) that the orbit of Mercury is very elliptical, and this is of course causing different / larger speed by perihelion. And due to that also slower ticking time.

So now we are not speaking GR but SR, - the cause of the Precession of Mercury’s Orbit is therefore due to special relativity, and that time is changing relative much in the orbit of Mercury (e.g.; compared to the orbit of the earth) ..

Is that correct ?

 

[tom.stoer]

No, I would say that it's incorrect.

Some reasonable explanations in the context of GR have already been presented here.

Treating the problem in SR is simply not possible as SR does not allow for a formulation of the gravitational law - that's why Einstein had to invent GR :-)

 

[bcrowell]

I have heard if the orbit of Mercury was circular there would not be any anomaly at all.

If a circular orbit precesses, it doesn't change, so there would be no way to *define* precession.

No, it is not something that can be explained in SR. It requires GR. Looking back through the explanations that have been given, note that they all contain some general-relativistic ingredient.

 

[Jonathan Scott]

Sorry, that is absolutely, positively, totally incorrect! Not to mention imaginary and misinformative. Space is most certainly *not* conical around a massive object. Where did you ever get that idea? An r=const surface in the Schwarzschild metric has (guess what?) the metric of a perfectly normal sphere of radius r. The distance around it in any direction is 2πr, the surface area is 4πr² and a gyroscope transported all the way around it will continue to point in the same direction.

Not in this universe!

The coordinate system is of course flat - that's why we use it. However, local space isn't. This is usually illustrated using Flamm's paraboloid (thanks DrGreg for the picture). If you take a narrow band of Flamm's paraboloid around a given orbit, this illustrates the locally conical property nicely.

A gyroscope does change direction when transported around an orbit; this is called the "geodetic effect" or "de Sitter precession" and is partly due to the static curvature of space and partly due to "Thomas precession". See the Wikipedia entry for http://en.wikipedia.org/wiki/Geodetic_effect" for more information.

 

[bcrowell]

A gyroscope does change direction when transported around an orbit; this is called the "geodetic effect" or "de Sitter precession" and is partly due to the static curvature of space and partly due to "Thomas precession". See the Wikipedia entry for http://en.wikipedia.org/wiki/Geodetic_effect" for more information.

A minor quibble here. This idea of attributing part of the geodetic effect to the Thomas precession is IMO very unnatural, and AFAICT idiosyncratic to Rindler (who is referenced in the WP article). It's especially unnatural because the sign comes out wrong. For more discussion of this: http://www.lightandmatter.com/html_books/genrel/ch06/ch06.html#Section6.2 (at "One will see apparently contradictory..."). MTW, for example, explicitly say that there is no Thomas precession for a satellite.

 

[Jonathan Scott]

A minor quibble here. This idea of attributing part of the geodetic effect to the Thomas precession is IMO very unnatural, and AFAICT idiosyncratic to Rindler (who is referenced in the WP article). It's especially unnatural because the sign comes out wrong. For more discussion of this: http://www.lightandmatter.com/html_books/genrel/ch06/ch06.html#Section6.2 (at "One will see apparently contradictory..."). MTW, for example, explicitly say that there is no Thomas precession for a satellite.

You may be right. I must look that up again.

I used to understand this stuff pretty well about 20 years ago! If you consider the point of view of a gyroscope in free fall, the sun going round it causes Lense-Thirring precession relative to the fixed stars, and I had initially expected that to be equal to the precession as seen from a solar system frame of reference, but I couldn't get them to match. I eventually found, with help from Clifford Will, that I'd missed an effect of the accelerated frame of reference from the gyroscope's point of view, and that the two points of view do come out equal when that extra effect is considered.

I also use complex four-vector algebra (now better known as the "Algebra of Physical Space") for all quick calculations on this sort of thing. They are very good for SR and work in the same way as spinors for calculate Thomas precession, but I find spinors rather unnatural because I prefer a notation which is isotropic in 3D. However, when I want to communicate with people using tensors and similar GR notation I tend to get confused because I'm used to different metric sign conventions and notation.

 

[Jonathan Scott]

I should also point out that to be accurate, my explanation of the relativistic perihelion precession of Mercury as being directly due to the curvature of space is not the whole story - I just thought it should suffice for a "grandmother"!

In terms of the parameterised post-Newtonian (PPN) approximation parameters, the relative amount of perihelion shift for different theories compared with GR is given by the following expression (part of MTW equation 40.18):

<br /> \frac{(2 - \beta + 2 \gamma)}{3}<br /> [/itex]<br /> <br /> For GR, \beta, the parameter describing the non-linear effect of gravitational energy, and \gamma, the parameter describing the curvature of space, are both equal to 1. This means that loosely speaking the curvature of space contributes 2/3 of the perihelion shift and effects relating to non-linearity contribute the other 1/3.

 

[Bjarne]

I read somewhere at the internet that the 43 arc second, is seen from the Earth, seen from Mercury is "should" be 80 arc second.
Is that correct or false?.

 

[DaveC426913]

I read somewhere at the internet that the 43 arc second, is seen from the Earth, seen from Mercury is "should" be 80 arc second.
Is that correct or false?.

I do not understand what you are asking.

Are you asking about diameter of Mercury's disc?

If I read correctly. Mercury's disc is approximately 5 arcseconds at superior conjuction and about 10 to 12 arcseconds at inferior conjunction. That's an order of magnitude smaller than your numbers so I don't know what you're referring to.

 

[edguy99]

Can someone explain the cause of the Precession of Mercury’s Orbit in a "language" so simple that my grandmother can understand it?

I mean only the part (43 arc second) that is due to relativity.

According to Newton, the force of attraction between two objects with mass is inversely proportional to the square of the distance between them. If you modeled gravity using the inverse square law, no 2 bodies in orbit could have a "precession" (all orbits end up being ellipses). The orbit of Mercury can be very accurately measured by lining it up with distant stars and it is NOT an ellipse like it should be if the inverse square law were true. Something in addition to the inverse square law is required.

The orbit of Mercury cannot be accurately predicted over time with the inverse square law calculations. Adding in the first correction with v(r) = -gL²/c²r³ mentioned earlier adds in a factor inversely proportional to the CUBE of the distance between 2 bodies in orbit which is important for very close objects and very heavy objects and will cause "precession" in orbits (including circular orbits - you can tell because the orbit will be a little faster then it would be if calculated using the inverse square law).

 

[Bjarne]

I read it here > http://milesmathis.com/merc.html


Now, if we subtract Mercury's precession from the Earth's, we achieve the apparent precession of Mercury as seen from the Earth. This is the number we want. This gives us a difference of

ΔP = .8 arcsec/yr

Or 80 arc secs per century. Therefore we will see Mercury precess about 80 arc seconds per century, due to curvature of the field alone. This is almost double Einstein's 43, which is enough to disprove his math and postulates. It also means that we will have to re-figure the perturbation total. The number 528 from above cannot be correct, as I said, since that is a Newtonian number, not a Relativistic number.

 

[Bjarne]

Image credit: ©Allen McCloud CC BY-SA 3.0

Actually Bill_K and Jonathan Scott are both right but they are talking about different things.

The diagram above represents the geometry of a two-dimensional slice of space around the Sun, with the Sun at the centre. (See Flamm's paraboloid for the technical details.) The region between two concentric circles has approximately conical geometry. Imagine cutting such a region out the diagram with a pair of scissors. To make the region lie flat, you have to make another cut in the radial direction to open it out, and then you will have an annulus with a gap missing. If you now draw an ellipse on your almost-annulus, there will be a gap. This gap represents the precession.

Still I am not sure what is the cause of the precession.
1.) Is it because of higher speed by perihelion relative how strong the curvature of space is ?
2.) Is it because of inertia, - increasing resistance against motion the further a body approach c ? - I mean when accelerating 100 km/h + 100km/h is not 200 km/h but 199,999...etc..km/h,
3.) or how can the cause / result of it be described.

 

[tom.stoer]

The simplest model derives a general-relativistic 1/r³ correction to the usual 1/r potential which causes the angular deficit.

 

[A.T.]

Can someone explain the cause of the Precession of Mercury’s Orbit in a "language" so simple that my grandmother can understand it?

I mean only the part (43 arc second) that is due to relativity.

Here is the simplest explanation of precession due to spatial curvature that I know of:

http://www.physics.ucla.edu/demoweb/demomanual/modern_physics/principal_of_equivalence_and_general_relativity/curved_space1.gif
http://www.physics.ucla.edu/demoweb/demomanual/modern_physics/principal_of_equivalence_and_general_relativity/curved_space2.gif

From: http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html

However, I was told (and a quick estimation seemed to confirm it) that the purely spatial curvature causes just a small part of the 43 arc second relativistic contribution. While most of it is due to gravitational time dilation (time curvature) and effective potentials:

http://www.bun.kyoto-u.ac.jp/~suchii/eff.potent.html

 

[FtlIsAwesome]

I read it here > http://milesmathis.com/merc.html


Now, if we subtract Mercury's precession from the Earth's, we achieve the apparent precession of Mercury as seen from the Earth. This is the number we want. This gives us a difference of

ΔP = .8 arcsec/yr

Or 80 arc secs per century. Therefore we will see Mercury precess about 80 arc seconds per century, due to curvature of the field alone. This is almost double Einstein's 43, which is enough to disprove his math and postulates. It also means that we will have to re-figure the perturbation total. The number 528 from above cannot be correct, as I said, since that is a Newtonian number, not a Relativistic number.

That site seems to basically be all about saying everybody's "wrong" with science and math.


157. How to Build a Nucleus without the Strong Force. With simple logic and diagrams.

129. A Recalculation of the Roche Limit. Showing that current math is wrong, and how to find a different kind of limit.

175. The Extinction of π. Here I show that the true value of π, defined as the ratio of circumference to diameter, is 4.


So, I think that site is only good for humor.

 

[Bjarne]

So, I think that site is only good for humor.

I think that is correct, but I just wonder; is the 43 arc second deviation per century seen from a Mercury or Earth perspective?

http://www.physics.ucla.edu/demoweb/demomanual/modern_physics/principal_of_equivalence_and_general_relativity/curved_space2.gif

I understand that space is curving (Fig.1+2+3+4+5+6+7).
But the confusion is with references to Fig 8 + 9.
Fig 8 - “Similar Mercury executes its Newtonian ellipse in local space”
Fig 9. ---but the global curvature causes the ellipse to precess.

I understand; that:
Fig 8.; “ Newtonian space is without any curvature” and --
Fig.9 is with curvature.

So what the illustration only say is; That the curvature causes the ellipse to precess, - not WHY.

However, I was told (and a quick estimation seemed to confirm it) that the purely spatial curvature causes just a small part of the 43 arc second relativistic contribution. While most of it is due to gravitational time dilation (time curvature) and effective potentials:

http://www.bun.kyoto-u.ac.jp/~suchii/eff.potent.html
[PLAIN]http://www.bun.kyoto-u.ac.jp/~suchii/eff.potent.jpg

The essence from this link is so far I understand Fig 2 > “precession because of extra dwell time at inner part.

According to classic understanding I don’t think there should be any extra “dwell time” – WHY should that happen due to general relativity?

I still do not understand the cause of that anomaly, (and do not have a mathematical background to do so).
Is there someone what can explain the cause in simple words?

I simply cannot find any description in this thread, that explain the cause of that precision anomaly.

* * *​

I have considered another option; - we know it requires more and more energy to get a diminishing increase in speed.

So each time when Mercury accelerate towards perihelion, the planet need more potential energy to be able to reach that speed it should according to classis orbit mechanics, to overcome the increasing inertia against acceleration (towards perihelion).

But where must that extra energy come from?

As I see it this must mean that Mercury will lose speed, and hence approach the Sun. (Circling inwards to the Sun, in the same way as Illustration 3 above; “Captured and then plunged”. -

But this does not fit with reality, - why not ? .
What prevent that from happen?
Maybe this question do not have anything with the precession anomaly to do, But I would anyway appreciate to get some suggestion why Mercury due to Inertia (by acceleration) not is losing speed / energy. - It must be a fact that it will not reach that speed it "should" according to classis orbit mechanics.

 

[D H]

I think that is correct, but I just wonder; is the 43 arc second deviation per century seen from a Mercury or Earth perspective?

Neither. It is a calculated result.

From the perspective of the Earth, Mercury's observed apsidal precession is 5600 arcseconds/century if the observations are expressed in mean of date coordinates or 574 arcseconds/century if the observations are expressed in J2000 or ICRF coordinates. The difference between these two values results from the 5026 arcseconds/century general precession of the equinox. In either case, the observed precession is not 43 arcseconds/century. Let's call the observed precession 574 arcseconds/century.

Newtonian mechanics explains most but not all of this observed precession: The outer planets cause Mercury's orbit to precess by 531 arcseconds/century -- which is off by 43 arcseconds/century. That 531 arcseconds/century is a calculated result; all that can be observed is the 574 arcseconds/century value.

 

[MikeLizzi]

Here is the simplest explanation of precession due to spatial curvature that I know of:

http://www.physics.ucla.edu/demoweb/demomanual/modern_physics/principal_of_equivalence_and_general_relativity/curved_space1.gif
http://www.physics.ucla.edu/demoweb/demomanual/modern_physics/principal_of_equivalence_and_general_relativity/curved_space2.gif

From: http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html

However, I was told (and a quick estimation seemed to confirm it) that the purely spatial curvature causes just a small part of the 43 arc second relativistic contribution. While most of it is due to gravitational time dilation (time curvature) and effective potentials:

http://www.bun.kyoto-u.ac.jp/~suchii/eff.potent.html

I tried this. The ellipse on my piece of paper didn't precess. What's the trick?

 

[Jonathan Scott]

I tried this. The ellipse on my piece of paper didn't precess. What's the trick?

Draw the ellipse on the flat paper with an extra piece of paper in the gap to complete the shape. If you then close the gap, making the cone, and place the extra piece of paper adjacent to the join so that it still continues the ellipse on one side, it will not quite join up correctly on the other side. You could make it join to the next bit by keeping the extra piece of paper still and turning the cone under it to bring the join to the other side. This represents the precession from one orbit to the next (or at least the component of precession due to space curvature).

 

[MikeLizzi]

I tried this. The ellipse on my piece of paper didn't precess. What's the trick?

I found the trick to the original exercise. Don’t draw the ellipse directly opposite the cutout. That’s the one place where any drawing will not change orientation when the disc is pulled into a cone.

I’m not stating this as an insult. I know no one in this forum invented that exercise. But it tells me nothing about relativity. I am a little better at Paper Mache though.

 

[Jonathan Scott]

I found the trick to the original exercise. Don’t draw the ellipse directly opposite the cutout. That’s the one place where any drawing will not change orientation when the disc is pulled into a cone.

I’m not stating this as an insult. I know no one in this forum invented that exercise. But it tells me nothing about relativity. I am a little better at Paper Mache though.

Ah. If you do that, the drawing will still look similar to an ellipse, but it will have a slight point at the join.

 

[A.T.]

I found the trick to the original exercise. Don’t draw the ellipse directly opposite the cutout. That’s the one place where any drawing will not change orientation when the disc is pulled into a cone.

There is no trick. You can draw the ellipse directly opposite the cutout, just as shown in the picture. The lines will meet then, but not at zero angle. So the object will not continue on the old ellipse but a shifted one. That's precession.

 

[A.T.]

But the confusion is with references to Fig 8 + 9.
Fig 8 - “Similar Mercury executes its Newtonian ellipse in local space”
Fig 9. ---but the global curvature causes the ellipse to precess.

I understand; that:
Fig 8.; “ Newtonian space is without any curvature” and --

By "Newtonian ellipse" they mean the inverse square law, that accelerates the object towards the cone tip. In flat space (called Euclidean not Newtonian) that acceleration law creates a closed ellipse.

Note that in reality the trajectory caused by the time curvature is different but they want to show only the effect of purely spatial curvature on the precession.

Fig.9 is with curvature.

On the cone the curvature exists only if you include the tip. Locally, excluding the tip it is still Euclidean, so the shape of the ellipse doesn't change there. But because it goes around the tip, it doesn't close properly, because there is an angle defect.

Note that in reality the spatial geometry is not cone, and the curvature is everywhere, but the cone approximation is good to illustrate the principle.

So what the illustration only say is; That the curvature causes the ellipse to precess, - not WHY.

It does show WHY. The trajectory is not closed properly. When the object arrives at the starting point, it has a different direction, than last time it was there. So it will not take the same path again.

 

[MikeLizzi]

There is no trick. You can draw the ellipse directly opposite the cutout, just as shown in the picture. The lines will meet then, but not at zero angle. So the object will not continue on the old ellipse but a shifted one. That's precession.

I don't know what you are talking about.

If I draw a vertical ellipse on the upper part of a disc symetric to the y-axis and make a cutout in the lower part of the disc symetric to the y-axis and close the disc into a cone symetric to the y-axis my ellipse is still vertical and symetric to the y axis. I don't get any precession.

 

[Jonathan Scott]

I don't know what you are talking about.

If I draw a vertical ellipse on the upper part of a disc symetric to the y-axis and make a cutout in the lower part of the disc symetric to the y-axis and close the disc into a cone symetric to the y-axis my ellipse is still vertical and symetric to the y axis. I don't get any precession.

It's not an ellipse then - it's got a slight point at the join (at least in theory). If you follow the direction of the line from one side of the join it will diverge slightly from the line on the other side. In practice, you'd have to have a highly eccentric ellipse and a large angle missing to make this visible with ordinary pencil and paper.

 

[A.T.]

There is no trick. You can draw the ellipse directly opposite the cutout, just as shown in the picture. The lines will meet then, but not at zero angle. So the object will not continue on the old ellipse but a shifted one. That's precession.

If I draw a vertical ellipse on the upper part of a disc symetric to the y-axis and make a cutout in the lower part of the disc symetric to the y-axis and close the disc into a cone symetric to the y-axis my ellipse is still vertical and symetric to the y axis.

Being symmetric doesn't make it an ellipse. An ellipse is a smooth shape, but this will have a sharp corner, where you stitch the cone together.

I don't get any precession.

You get it, if you continue the path smoothly over the seam (dashed line in the picture). Eventually you have to make the cut-out-angle bigger and layout the seam area flatten out to see how the path will continue.

But it's easy to show mathematically. Just consider the angles between your original ellipse and your cuts. On the tip side they total less than 180°, so when you weld the cut lines, you don't get a smooth path.

 

[DrGreg]

Here's a very exaggerated version (not to scale) with 120° precession to get the point across.

 

[Bjarne]

OK I begin to understand, we can say that the deformation of space happens along the orbit with different values, and the result is that the angle defects whereby the ellipse cannot close. Simple…. When first the confusion first is gone. Thank’s .

However, I was told (and a quick estimation seemed to confirm it) that the purely spatial curvature causes just a small part of the 43 arc second relativistic contribution. While most of it is due to gravitational time dilation (time curvature) and effective potentials

Time and distance deformation should (as I understand it) not be able to cause the other part of the precession. Because time and distance dilation are (as I understand it) proportional to each other (?).

But what about the; - “relativistic resistances” ?.
It requires more and more energy to get a diminishing increase in speed.

So each time when Mercury accelerate towards perihelion, the planet need more potential energy to be able to reach that speed it should according to classis orbit mechanics, to overcome the increasing relativistic resistances (towards perihelion).
But Mercury do not get that extra speed from anywhere, which mean that each time heading perihelion, - speed must be a little too low, compared to what it must be according to classis orbit mechanics.
Does that too not also have an influence ?

 

[A.T.]

http://www.bun.kyoto-u.ac.jp/~suchii/eff.potent.html
[PLAIN]http://www.bun.kyoto-u.ac.jp/~suchii/eff.potent.jpg

The essence from this link is so far I understand Fig 2 > “precession because of extra dwell time at inner part.

According to classic understanding I don’t think there should be any extra “dwell time” – WHY should that happen due to general relativity?

I still do not understand the cause of that anomaly, (and do not have a mathematical background to do so).
Is there someone what can explain the cause in simple words?

I don't understand it well enough to explain it simply, yet still correctly. But here are more visualizations and explanations:
http://www.fourmilab.ch/gravitation/orbits/

 

[MikeLizzi]

Being symmetric doesn't make it an ellipse. An ellipse is a smooth shape, but this will have a sharp corner, where you stitch the cone together.


You get it, if you continue the path smoothly over the seam (dashed line in the picture). Eventually you have to make the cut-out-angle bigger and layout the seam area flatten out to see how the path will continue.

But it's easy to show mathematically. Just consider the angles between your original ellipse and your cuts. On the tip side they total less than 180°, so when you weld the cut lines, you don't get a smooth path.

Thanks for spending some time with me. I still don't know what you (or Dr Greg) are talking about. I will just assume I am too stupid to understand.

 

[A.T.]

Thanks for spending some time with me. I still don't know what you (or Dr Greg) are talking about. I will just assume I am too stupid to understand.

Dr Greg showed an extreme example where the trajectory precesses by 90° on each orbit. Start on the right end of the red ellipse going CCW, then it goes to the green, orange and finally blue one.

Try it yourself on paper, with different orientations of the initial ellipse. Make sure when you cross the seam, that the path keeps it's local orientation relative of the seam.