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Just wanted to know why planetary orbits get to be elliptical? I'm not debating how stable these orbits are (obviously almost completely stable), but why don't planets revolve in the special ellipse known as a circle? Wouldn't the centrifugal/-pedal forces between the planet and Sun make the planet revolve around the Sun in a perfect circle? Is there a 'third force' that attracts planets at a specific point (a hypothetical reasoning for elliptical orbits)? I'm not trying to disprove Kepler over here, but I don't exactly get it.


marked as duplicate by Jim, user10851, ACuriousMind, Bernhard, John Rennie Sep 7 '14 at 15:39

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  • $\begingroup$ Circles are ellipses (with eccentricity 0). $\endgroup$ – NeutronStar Sep 7 '14 at 3:09
  • $\begingroup$ Are you suggesting there is something that there is some force or mechanism to make elliptical orbits circular over long periods of time? $\endgroup$ – Brandon Enright Sep 7 '14 at 3:23
  • $\begingroup$ A planet (lets say Earth) does not orbit around the center of the Sun. It orbits around the center of mass of the Earth-Sun system (which, due to the difference of masses, happens to be very close to the center of the Sun. The Sun moves around that point, too. While I doubt it is enough to explain all of the ellipsis of the Earth orbit, it explains why it cannot be "a perfect circle" around the Sun (unless the orbitying body has a near despreciable mass, v.g. an asteroid). $\endgroup$ – SJuan76 Sep 7 '14 at 11:32
  • $\begingroup$ possible duplicate of Why are orbits elliptical? $\endgroup$ – GOTO 0 Sep 7 '14 at 13:13
  • $\begingroup$ A generalization of this question is why are two body orbits conic sections? $\endgroup$ – ja72 Sep 7 '14 at 13:40

Wouldn't the centrifugal/-pedal forces between the planet and Sun make the planet revolve around the Sun in a perfect circle? Is there a 'third force' that attracts planets at a specific point (a hypothetical reasoning for elliptical orbits)?

I need to rant a bit, not at you, Hassaan Qazi, but at whoever taught you (and who teach so many others) that orbits are a balance of a centrifugal force and a centripetal force. There is no third force, and indeed, there is no centrifugal force. There is but one force, gravitation, and it is a central force rather than a centripetal force.

In particular, there is no centrifugal force if you look at the solar system from the perspective of an inertial frame of reference. From this perspective, the solar system comprises a number of planets, asteroids, and other bodies, each of which orbits the Sun at its own orbital velocity. The only force acting on these bodies is gravity.

Each of these objects more or less follows a conic section: A hyperbola, a parabola, an ellipse, or a circle. Some comets are on a hyperbolic or parabolic trajectory. We see them once, and then they are gone forever. The planets and asteroids follow paths that are very close to elliptical. Note that a circle is just a special case of an ellipse. If some planet did have a circular orbit (but none do), it would still be okay to call that planet's orbit elliptical.

I intentionally wrote "more or less" and "very close to" in the above. Those conic sections only arise in the case of the two body problem. Our solar system comprises many bodies. While the Sun is the most influential body gravitationally, planets are gravitationally attracted to other planets as well as to the Sun. If by some fluke a planet was on a perfectly circular orbit, it wouldn't stay that way for long because one of those other planets would perturb it from that perfectly circular orbit.


Consider a planet near a star at one moment in time. They are separated by a certain distance $r$, and so the planet feels the corresponding $GM_\mathrm{star}/r^2$ acceleration toward the star. That's due to the only force on the system.

But of course velocity is not necessarily in the direction of acceleration; acceleration merely acts to change the velocity over time. So to fully specify what happens to the planet, we need its current velocity as well. Let's assume without loss of generality1 that this velocity is perpendicular to the acceleration right at this moment. What value will the speed $v$ be for a circular orbit? Well, we must have $v^2/r = GM_\mathrm{star}/r^2$, which you can solve to find $v_\mathrm{circ} = \sqrt{GM_\mathrm{star}/r}$

But what if the speed is something other than this exact value? Does that mean the planet is no longer allowed to orbit? No. As long as its total energy per unit mass $v^2/2 - GM_\mathrm{star}/r$ is negative, the planet is bound to the star.

So any velocity in the range $(0, \sqrt{2}v_\mathrm{circ})$ will lead to an orbit, but only the exact value $v_\mathrm{circ}$ leads to a circular orbit. This just goes to show that circular orbits are very special cases. No other forces enter the picture.

1Proof: If the planet is in some repeating orbit, it can't be moving only inward or outward forever. So if its velocity is sometimes not perpendicular to acceleration (which is always radial), it must both approach and recede from the star at different points in the orbit. By the intermediate value theorem, there are points in any orbit, circular or not, in which velocity is exactly perpendicular to acceleration.


Generally, if they start off as circular, and the planet is disturbed in its orbit (by passingt planets or commets), they come to be elliptic.

One should note that stable elliptic orbits are a feature of three dimensions only. They don't work in four or higher dimensions.

  • 2
    $\begingroup$ One could add, that in two and three dimensions each solution of the Kepler problem is given by two parameters: the energy and the angular momentum. Obviously, circular solutions in a plane have only one degree of freedom, the radius. An ellipse, on the other hand, is also characterized by its eccentricity. $\endgroup$ – CuriousOne Sep 7 '14 at 2:26
  • $\begingroup$ Elliptic (or hyperbolic/parabolic) orbits are not (directly) a feature of three dimensions, but of an inverse-square attraction force. They would equally well arise for inverse-square attraction forces in two, four, or higher dimensions. It turns out that, exclusively in three dimensions, inverse-square force fields have an additional property, namely that they are divergence-free (and centrally symmetric divergence-free fields in other dimension would be different, and not have elliptic orbits). But that is a rather separate matter. $\endgroup$ – Marc van Leeuwen Sep 7 '14 at 17:08
  • $\begingroup$ Elliptical orbits are only stable if the radiant force is inverse-square, but the radiant model supposes flux radiates in all directions, and so one only gets inverse-square flux if space is bounded by a hedrix: ie 3D. They are not stable in hyperbilic space, where the surface is exp(r/k), more or less. $\endgroup$ – wendy.krieger Sep 8 '14 at 0:43

When looking at the mathematics of Kepler orbits (a two body problem), elliptical orbits are just as stable and periodic as circular orbits. And an elliptical orbits will not be pulled into an circular orbit. When only considering Newtonian gravity, and there are no perturbations by other bodies, an elliptical orbit will not change its trajectory over time. This is because gravity is a conservative force, so energy will be conserved. The same goes for angular momentum.

A followup question might then be about why all planets in out solar system such low eccentricity. This has to do with the fact that out solar system does not only have two bodies, there are multiple planets. Before the planets formed there probably were a lot more, but most collided with each other which formed the planets. During those collisions momentum would be conserved, meaning that if all those small bodies had randomly distributed eccentric orbits around the sun, that an average would results in a roughly circular orbit. Another way of looking at it is that a planet will have a bigger change to "survive" if its orbit does not get near orbits of other planets, since otherwise they will interact with each other sooner or later, which could lead to a collision or one flinging the other into another orbit. Roughly circular orbits, with different radii, will minimize those interactions, while allowing the most amount of bodies in the system.


why don't planets revolve in the special ellipse known as a circle?

You've answered your own question: the circle is special and thus, highly improbable. In fact, a perfectly circular orbit is essentially impossible and is, in fact, an idealization that can only be approximated.

Think about it - a genuine circular orbit is like a genuine sinusoidal function of time that is a solution to a differential equation - it exists for all time.

There is no such physical orbit - it is an abstraction. Physical orbits are only approximations to the solutions to the idealized differential equations of orbital mechanics.

  • $\begingroup$ OK…….so basically what you are trying to say is that in a perfect world, where no external not-so-negligable forces exist, the orbit would be a perfect circle. The non-circular ellipse forms because of other forces in the equation, a bit like static through an old TV aerial, even when tuned into a proper show. $\endgroup$ – Hassaan Qazi Sep 7 '14 at 2:44
  • $\begingroup$ @HassaanQazi, I'm not even saying that. In the idealized mathematical model, given a particular energy and no dissipation, circular orbits exist but such orbits are circular for all time - they are eternally circular. An orbit that decays or changes with time is not circular. $\endgroup$ – Alfred Centauri Sep 7 '14 at 2:57
  • $\begingroup$ @AlfredCentauri: Likewise, any orbit that has not existed for all time cannot e circular. $\endgroup$ – Pieter Geerkens Sep 7 '14 at 4:59

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