All trajectories available in space: circle, ellipse, parabola, hyperbola, straight line are parts of cone. Why? Does the spacetime have a conic shape?


5 Answers 5


Not all trajectories in space are conic sections, only those that are a two body problem. One planet in orbit around a star is a two body problem. Only two body problems are solved by a conic section.

The gravitational force of a star is inverse proportional to the square of the distance between the star and the planet. The centrifugal force is inverse proportional to the distance. The balance between gravitational and centrifugal force is therefore a quadratic equation.

All conic sections are described by a quadratic equation. Therefore the solutions of the balance equation are conic sections.

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    $\begingroup$ "All conic sections are described by a quadratic equation." More importantly; all quadratic equations describe conic sections. $\endgroup$
    – user44687
    Dec 7, 2021 at 2:59
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    $\begingroup$ The first paragraph is good, but the rest of the answer is very misleading. "The same is true for the centrifugal force" -- how do you get $1/r^2$ there? The second paragraph suggests that the exponent 2 in the inverse square law of gravity is linked straightforwardly to the exponent 2 in the quadratic equation for an orbit. It is a much more subtle and unique relation. Orbits in an inverse cube force aren't given by cubic equations $\endgroup$
    – nanoman
    Dec 7, 2021 at 5:54
  • $\begingroup$ @nanoman Are there orbits with an imaginary cube force? If so, what equations would describe them? (Ah, I see, Ryan seems to suggest there aren't.) $\endgroup$ Dec 8, 2021 at 7:30

The question of which force laws allow what kinds of orbits received considerable attention in classical mechanics, some of it very recent. The strongest result dates to 1873, when Joseph Bertrand proved that of all possible laws where the attractive force scales as some power of the distance, the only two such forces for which all orbits are closed are $r^{-2}$ and $r^{+1}$ --- the inverse square law and the simple harmonic oscillator.

The constraints under which this holds are

  • the force is central: its magnitude is a function only of radius, and its direction is along the radial position vector, $\mathbf{\vec{F}}=f(r)\mathbf{\vec{r}}/r = f(r)\mathbf{\hat{r}}$
  • the force is conservative: difference in energy depends only on start and end points, not the path between them; implies the force can be written as the negative gradient of some potential energy function, $\mathbf{\vec{F}}=-\nabla V(r)$, so $f(r)= -\partial V \!/ \partial r$
  • the force is proportional to just one power of the radius: $f(r)=ar^{-X}$, not $f(r)=ar^{-X}+br^{-Y}$
  • "orbit" means bounded motion, because the kinetic energy is not great enough to escape
  • "closed" means the motion eventually repeats its track exactly

There are also two weaker related results which can be discovered along the way as part of some derivations. First, $f(r)=ar^{-X}$ has stable circular closed orbits for any $X<3$. Second, for $X=3-Q^2$, where $Q$ is any rational number, it has stable closed orbits that look like this. Integer ripples of a wave quantized on a circle These are standing waves on a circle, as found in electron orbitals in the Bohr model, the jet stream, and the allowed pitch harmonics if you bent an organ pipe into a torus.

The ripples shown in the image are in general greatly exaggerated for the central force problem, where such orbits are only truly closed at first order in the Taylor series approximation of $f(r)$. [However, the wiggles in the atmosphere which we call the jet stream really are that big, as a high-altitude weather map of a sufficiently large region shows.] Of course, the reason physicists are so in love with the simple harmonic oscillator is that $f(r)=-kr$ equals its own one-term Taylor expansion, so one-term Taylor approximation turns everything into a simple harmonic oscillator. That means the ripples can indeed get that large, but only in the other Bertrand-allowed case, that is if the two point masses were joined by an ideal spring (Hooke's law).

As Michael Seifert mentioned in a comment, trajectories should be concave with respect to a center of attractive force, or convex towards the center of a repulsive one. What's going on here with these wiggles, then? The derivation I learned from Goldstein (1980) makes use of the "equivalent one-dimensional problem", which looks solely at the radial component of the motion, but accounts for the conserved angular momentum, and the kinetic energy of rotation it carries, by changing the form of the potential energy function. That is, although for the full three-dimensional Kepler problem, the potential $V(r) = -k/r$, bookkeeping the angular momentum $L$ means that to match the same behavior in the equivalent one-dimensional problem, we have to add an extra term, resulting in $V(r)=-k/r + L^2/2mr^2$. This means that for large enough $r$ (for any given $k$, $m$, and $L$), the potential is attractive, and for small enough $r$, the potential is repulsive. Motion that is precisely circular in 3D here must be translated to the 1D constant point solution $r(t)=r_0$. This $r_0$ is the value at which $V(r)$ takes its minimum, and where it changes from attractive to repulsive. Trajectories which do not escape this transformed equivalent potential are bounded between two circles, and swing back and forth between them while going around with constant $L$.

What this truly looks like in the Kepler problem is what happens when you draw a precessing ellipse: it either slightly undershoots or slightly overshoots the point where it would close, by the same amount every orbit, leading its path to gradually form a figure like this, which has the correct concavity:

many iterations of nonclosing elliptical orbit

The names we give to these two special radii, at least for orbiting the Earth, are the familiar "apogee" and "perigee". Eventually, for most potentials, the whole space will be filled in (in the asymptotic sense of arbitrarily close approach to any point within a computable time); for rational exponents ($X=3-Q^2$) such patterns can theoretically repeat, but aren't stable to any perturbation.

The original paper, entitled "Théorème relatif au mouvement d'un point attiré vers un centre fixe", published in Comptes rendus de L'Académie des Sciences (77) 849-853, is available in French from the Bibliothèque Nationale. A direct translation into English published in 2011 is this PDF.

There are at least a dozen alternate derivations, counting just those cited in "A truly elementary proof of Bertrand's theorem", Chin (2014), plus "An Even Simpler “Truly Elementary” Proof of Bertrand’s Theorem", Galbraith and Williams (2019).

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    $\begingroup$ $F(r) = a r^{\color{red}{-X}}$ has stable closed circular orbits for any $X < 3$, surely? $\endgroup$ Dec 6, 2021 at 19:27
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    $\begingroup$ Also, those wavy orbits you've depicted won't actually be precisely closed. They're perturbations to a circular orbit for which the radial oscillation frequency of the first-order perturbations is commensurate with the orbital frequency, so they're closed to the extent that you view the higher-order perturbations as negligible. But the point of Bertrand's theorem is that when you actually look at the higher-order perturbations, the orbits no longer close exactly. $\endgroup$ Dec 6, 2021 at 19:31
  • $\begingroup$ One last comment: it's worth emphasizing that the amplitude of the radial oscillations in the diagrams is hugely exaggerated. A "real" trajectory for an attractive force should always be concave towards the origin; only if the force is repulsive will you get trajectories that are convex towards the origin like those shown. $\endgroup$ Dec 6, 2021 at 19:37
  • $\begingroup$ @MichaelSeifert oops, good catch on the minus, thank you. my response to the longer comments won't fit here, but i'll expand the answer to touch on those issues as well. $\endgroup$
    – Ryan C
    Dec 6, 2021 at 21:14
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    $\begingroup$ @EricDuminil: Yeah, I phrased that weirdly. By "concave towards the origin" I meant that the center of the osculating circle would lie inside the trajectory, not outside. Which is equivalent to saying that the entire orbit is convex. $\endgroup$ Dec 6, 2021 at 22:28

The existing answers here are either wrong (making it sound like the conic section orbits are easily deduced by simple algebra from the inverse square law of gravity) or overly extensive and technical for what OP asked.

The conic orbits are indeed a remarkable fact. To Kepler they suggested a mysterious cosmic harmony. Newton found that they are merely a contingent and approximate outcome of a deeper principle -- the $1/r^2$ force law. To show this, Newton had to invent calculus -- no small feat.

The reason for conic orbits for the two-body problem is no more or less than that they are the solution (after eliminating the time variable $t$) of the differential equations $$\frac{d^2x}{dt^2} = \frac{{-}kx}{(x^2 + y^2)^{3/2}}, \quad \frac{d^2y}{dt^2} = \frac{{-}ky}{(x^2 + y^2)^{3/2}}.$$ I show these equations not to delve into them, but to make the point that without understanding what they mean, one is not in a position to actually understand why the solutions are conics.

In a broader sense, if we imagine possible universes with different dimensionalities and force laws, the anthropic principle argues that life would arise in a universe where planetary orbits can be stable. Our actual universe is such that this holds -- not that orbits are exactly conic sections (they aren't, due to multi-body perturbations, relativistic effects, etc.), but that they're close enough to be stable. Other force laws tend not to produce stable orbits at all.

So arguably, we may owe our very existence to some quite subtle and lucky math regarding the dynamical consequences of $1/r^2$ forces. For me this inspires awe.

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    $\begingroup$ 1/r^2 for all "flux-like" fields is a consequence from space being 3-dimensional. That probably implies that higher-dimensional universes don't even have condensed matter (no electron orbits, no crystals etc. ...). $\endgroup$ Dec 8, 2021 at 7:37
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    $\begingroup$ @Peter-ReinstateMonica There's a part of the Feynman lectures that veers near this point. The impression that I get is that it's not so clear cut that electric fields must go as 1/r^2 due to the dimensionality. feynmanlectures.caltech.edu/II_04.html $\endgroup$ Dec 8, 2021 at 14:52
  • $\begingroup$ @WaterMolecule If I read it right, he says "yeah, well, the 'bullet' model helps us find the right mathematics and there is no harm in thinking this way, so long as we do not say that the electric field is made out of bullets". But it seems more fundamental than that. Oh, and he mentions that the fact that the circle integral is zero is only dependent on E being radial and spherically symmetric -- 1/r^2 is not a condition. $\endgroup$ Dec 8, 2021 at 16:41
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    $\begingroup$ "without understanding what they mean, one is not in a position to actually understand why the solutions are conics." +1. Step one: learn calculus. The only questions I have left require The Calculus to understand them. $\endgroup$
    – Mazura
    Dec 8, 2021 at 21:24

Conic sections tend to happen when you have $r^2$ terms, as they have such terms in the equations that define them. Here are the equations for some of these shapes. They are conic sections because gravity is the dominate force in space travel, which falls away as $r^2$

  • $y=x^2$
  • $a^2 \cdot y^2+b^2 \cdot x^2 = r^2$
  • $a^2 \cdot y^2-b^2 \cdot x^2 = r^2$

Essentially for gravitational problems that result in a single body being the dominate force, and a secondary body (Called 2 body problems), the second body, when not accelerating, will have to be accelerating as some form of a conic section, because gravity is the only force, operating from a single point, and it is defined by falling away as the square of the distance.

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    $\begingroup$ If it was that simple, one woud expect all trajectories in a force field which falls away as $1/r$ to be straight lines. $\endgroup$
    – Litho
    Dec 6, 2021 at 14:34
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    $\begingroup$ No, it has nothing whatsoever to do with light speed. $\endgroup$
    – PearsonArtPhoto
    Dec 6, 2021 at 17:21
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    $\begingroup$ This is not at all why orbits in a $1/r^2$ force are conics. See my comment on Uwe's answer. You have to solve a nonlinear second-order differential equation for the orbit. You can't just intuit the result of that from some naive algebra. Given a power law force, you can't even assume the orbit is described by a polynomial equation at all. $\endgroup$
    – nanoman
    Dec 7, 2021 at 6:03
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    $\begingroup$ @Peter The differential equations in my answer are what a $1/r^2$ force means -- force is proportional to the second time derivative of position. If someone has a valid method to derive the resulting orbit shape without using calculus, great -- but it's not here. There is no general principle that orbits are polynomial curves involving the same power that's in the denominator of the force law. As Ryan's answer notes, orbits generally aren't closed curves at all. If we didn't already know (from calculus) what the orbits are, the equations in this answer would be just a wild and lucky guess. $\endgroup$
    – nanoman
    Dec 8, 2021 at 8:42
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    $\begingroup$ @Peter And the equations in this answer are misleading anyway: $r$ has to represent a constant for them to describe conic sections, but this isn't the same as $r$ in the force law (which varies over a non-circular orbit). Plus the conic orbits are generally not centered at the center of force ($x$ and $y$ need to be offset) and there's no indication of how one would obtain this. It isn't even an attempt at a convincing derivation. "Invert the force law and apply that function to the body's Cartesian coordinates with some coefficients" is nonsense not justified by the meaning of force. $\endgroup$
    – nanoman
    Dec 8, 2021 at 9:03

The answer from Uwe above is almost "there", but not very accessible. The two-body equation describes the motion of a small mass orbiting a larger body. There is an elegant derivation in Vallado, p.27 starting with crossing the specific angular momentum vector with the two body equation. You do a couple of pages of cool substitutions, exercising your vector identities, and end up with the polar form of a conic section (p.29). That's the expansion of Uwe's comment.

When you get the moon involved you have to restrict the problem to be able to analytically solve it, this is most famously the Circular Restricted 3-Body Problem (CR3BP), see Vallado figure 12-13, p.968. That equation does not result in the polar form, you get stable and unstable gravitational "manifolds" to mess around with, "Highways in Cis-Lunar Space."

Include the Sun and Jupiter and well, have a good day.


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