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Is there an intuitive reason for why the shape of the orbit at perigee is the mirror image of that at the apogee? It seems strange to me, because the situation is so different at the perigee compared to at the apogee. It is not symmetrical, and yet the orbit is symmetrical.

It's an ellipse with the earth at one focus, a completely unsymmetrical situation.

Edit: Take your pick between relativistic and Newtonian models. I mean, feel free to ignore either model in your explanation. Feel free to pretend Newtonian mechanics is exactly correct.

Edit: What I mean by "unsymmetrical" is that the speed is completely different, and the earth is not in any shape or form near the center of curvature at the apogee.

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    $\begingroup$ I see the problem I think: "It's an ellipse with the earth at one focus, a completely unsymmetrical situation.". Nope, an ellipse is symmetric along its two axes. For one whole orbit (ignoring Kepler's 2.) it doesn't matter in which focus the earth lies (simple two body problem, higher orders are chaotic). $\endgroup$
    – user40414
    May 2 at 12:58
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    $\begingroup$ I don't know how to give an intuitive explanation, even just using Newtonian mechanics. An ellipse (including the circle as a special case) is just about the simplest closed curve, and it seems reasonable to me that a particle moving in a $1/r$ potential field ought to have a fairly simple trajectory. Showing that such trajectories are, in fact, conic curves takes about a page of vector calculus, see Wikipedia $\endgroup$
    – PM 2Ring
    May 2 at 13:37
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    $\begingroup$ I don't think that this question has a good answer in a format that is good for this forum (that does not mean it's not a good question). What is 'intuitive' is a near-canonical example of something which is a matter of opinion: what was intuitive to Richard Feynman might well not be intuitive to me. For what it's worth, the answer for me is 'no': I have to work it out (and it's obviously only even true the simple Newtonian two-body case anyway). This is why I'm going to vote to close it. $\endgroup$
    – user21103
    May 2 at 16:12
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    $\begingroup$ It's obvious to me that the planets should follow an egg-shaped orbit with the Sun towards the 'pointy end' of the egg. (It would be great to have someone update my intuition) $\endgroup$
    – Roger Wood
    May 2 at 17:26
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    $\begingroup$ I disagree strongly with the close votes. We've allowed other "help my intuition" questions before. "What shape do you think a Keplerian orbit should be" is opinion-based and unanswerable; this question is clearly answerable. $\endgroup$ May 3 at 20:01
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If we assume Keplerian/Newtonian mechanics, then we can see a way to rendering the same local curvature of the path at perigee and at apogee (terms for orbiting Earth, of course).

At both points the motion is perpendicular to the applied gravitational force. So, the curvature is given from Newtonian mechanics as $f/(mv^2)$ where $f$ is the magnitude of the applied gravitational force, $m$ is the mass of the orbiter and $v$ is the orbital speed. Then:

  • $f$ will be inversely proportional to $r^2$ from Newton's gravitational law, $r$ being the distance form the 9rbiter to the gravitating body at either extremal point.

  • $m$ is constant in Newtonian mechanics.

  • And ... when the extremal points with the perpendicular motion to gravity are compared, $v$ will be inversely proportional to $r$ from Kepler's equal-areas law.

So ... the dependence on $r$ cancel, making the curvature $f/(mv^2)$ equal at the extremal points.

What goes around just misses coming around

What happens if we recognize that gravity does not exactly follow Newton's model? We still have $v$ inversely proportional to $r$ in the third point above, but no longer is the curvilinear effect of gravity inversely proportional to $r^2$. The gravitational effect is stronger at perigee compared with apogee than the $r^{-2}$ proportionality would imply, so the real, relativistic orbit has extra curvature at perigee.

This extra curvature throws the path forward in the direction of orbital motion, leading to the observed direction of precession in a situation where this can be measured such as Mercury orbiting the Sun.

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    $\begingroup$ This is really beautiful! $\endgroup$
    – uhoh
    May 2 at 21:29
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    $\begingroup$ @Oscar Lanzi My intuition has been reset - thank you! $\endgroup$
    – Roger Wood
    May 3 at 3:33
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    $\begingroup$ @Oscar Lanzi a similar question that equally eludes intuition: "Why does the orbital period depend on the major axis but not the minor axis?". Obviously it will take the more time to go round a fat ellipse than a thin one. $\endgroup$
    – Roger Wood
    May 3 at 22:23
  • $\begingroup$ You have me stumped on that one. I have been for years. $\endgroup$ May 3 at 23:02
  • $\begingroup$ While this is correct physics, it doesn't answer the question. It's not an intuitive reason. $\endgroup$
    – jamesqf
    May 3 at 23:14
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As someone who computes simple orbits for a living, I would say your question has meaning only in the original Keplerian model of projectile motion: gravitational fields of central bodies are point symmetric, barycenters are at the geometric centers, and other bodies are not present. Then the symmetry arises from the straightforward, simple math used to describe Keplerian motion. As soon as the primary becomes oblate (i.e., no longer a point symmetric central force field), other celestial bodies are present, and the gravitational warping of spacetime are included, then "symmetry" is more an analogy than a reality. One might think instead of these other realistic phenomena as perturbations that destroy the original -- but simplistic -- symmetry.

The short answer to your original, basic question is that it's because it's an ellipse. An ellipse is mirror symmetric about any axis of "anti-reflection" that passes through the center of the ellipse (not a focus). In particular, it is plain to see an ellipse is symmetric about the major and minor axes. The difference in speed, however, is entirely due to the central force being at only one focus. The unoccupied focus is "just for looks" -- it serves no physical purpose (that I ever heard of) and is only used for geometric computations.

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    $\begingroup$ Can you add something on how to make the keplerian case more intuitive. $\endgroup$
    – lijat
    May 2 at 16:55
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I think the premise is wrong: in physics "everything is similar" unless proven otherwise. It's one of the principles of physics.

So you say that the orbit (shape? Or did you mean velocity, or something else) should't be "similar": but in what way should't it be similar? You should specify that, what did you expect?

If your premise is: "the velocity at perigee is much higher than the apogee, so how can the curvature be equal and form an ellipse". If that is the real question we can look at Kepler's first law, and/or second law (again based on what other things you consider "logical").

The first law is just a proof using calculus, it's straight forward but a bit lengthy, it starts with the basic equations of motion (or energy), and shows that the resulting motion equals that of the mathematical equation of an ellipse.

It's a bit lengthy to repeat here but here is a small pdf: https://radio.astro.gla.ac.uk/a1dynamics/ellproof.pdf

The second law (it's the first one Kepler noticed from observing the bodies in the sky) is a bit more intuitive and it states that "if" the motion is a conic, the "area" swept by the motion (from a line at the central body) at each point and delta time is equal. This means that the velocity at the apoasplis "must" be smaller than the velocity at the center, Oscar lanzi explained the details a bit of this law, but here is a neat animation from Wikipedia (https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion) regarding this:

enter image description here

The blue area stays equal in size.

So in conclusion: please be more careful on defining "is there intuitive reason" - specify why you find it "unsymmetrical", especially since you already state it is an ellipse.

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  • $\begingroup$ The animation is helpful. What I mean by unsymmetrical is that the speed is completely different, and the earth is not in any shape or form near the radius of curvature at the apogee. $\endgroup$ May 3 at 18:52
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    $\begingroup$ @MatthewChristopherBartsh is IS symmetrical. but not around any 2-d reference system. It is symmetrical on the energy axis. You are just looking at it from the wrong viewpoint. $\endgroup$ May 3 at 22:11
  • $\begingroup$ @MatthewChristopherBartsh - so it's the second law, and you understand that the orbits are an ellipse? Well the "intuitive" reason is indeed angular momentum. The proof is quite a trivial thing and basic calculus: burro.case.edu/Academics/Astr221/Gravity/kep2rev.htm - Then if you follow that the first conclusion you should also understand is conservation of angular momentum. - some system has always (when no work is being done) a constant angular momentum. Thus if the orbit has a higher radius it turns "slower": similarly how ice skaters turn faster if they pull their arms. $\endgroup$
    – paul23
    May 4 at 13:36

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