I read about Kepler's Laws and in one of them he mentions that the path of a planet is an ellipse, with the sun as one of its foci (I'm narrowing down this to only our solar system).

However though I'm not experienced in this subject, I had a doubt.

I read in places that the Sun is not stationary. Please correct me if it's not the case.

But if it is the case, then the path of the planets is an ellipse only with respect to the Sun.

So the actual path of a planet observed from, let's say, a point in space, would differ from an ellipse?

Or is this already factored into the law?

Please help me because I'm new to this concept.

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    $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$ – David Z Jul 3 '20 at 0:33
  • $\begingroup$ What is a "point in space"? $\endgroup$ – pericynthion Jul 3 '20 at 2:42
  • $\begingroup$ @pericynthion I mean a reference frame not on the sun or earth. $\endgroup$ – Vamsi Krishna Jul 3 '20 at 5:05
  • $\begingroup$ @VamsiKrishna But your choice of reference frame has to include the relative velocity as well as the position — in fact, that’s generally more important. $\endgroup$ – Mike Scott Jul 3 '20 at 10:01
  • $\begingroup$ @MikeScott Yes I agree , but that would be required if I need an exact path tracing. I was just looking for a rough idea of whether it deviates from an ellipse. $\endgroup$ – Vamsi Krishna Jul 3 '20 at 11:56

This is an interesting question, since it raises the problem of the reference frame where Kepler's laws are true, which is often neglected.

As a consequence of Newton's laws, in the inertial reference frames where the center of mass (c.m.) is fixed (there is a triple infinity of them, differing only with respect to the position of the c.m.) both planet and Sun describe an elliptic motion having the center of mass as one focus of the ellipse. The two ellipses are similar, with a rescaling factor equal to the planet/Sun mass ratio.

In every other inertial frame, the elliptic motion is combined with a uniform translation, therefore, in such systems, no closed orbit exist anymore.

There are two additional reference frames where the orbit is an ellipse. Both are non-inertial. One is the non-inertial reference frame where the Sun is fixed. You correctly noticed that the Sun is non-stationary. But this is true in any inertial frame. If one picks precisely the non-rotating, non-inertial system where the Sun is fixed, it stays forever at the position of one focus of the elliptic orbit of the planet. Similarly, one could sit on the planet without rotations, and in that system the orbit of the Sun would be again an ellipse like the one of the planet, with the planet at one focus position.

In conclusion, there is not the actual path. Shapes and properties of the orbits are not invariant with respect to changes of reference.

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    $\begingroup$ What is a "triple infinity"? $\endgroup$ – Schwern Jul 1 '20 at 19:47
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    $\begingroup$ @Schwern It is an usual way to say that there are three idependent possible displacements of the center of mass. $\endgroup$ – GiorgioP Jul 1 '20 at 21:52
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    $\begingroup$ @Schwern It's a fancy way of referring to the fact that what point one considers the "origin" of one's coordinate system can vary in three dimensions. "Three degrees of freedom" would be a more standard way of referring to it. $\endgroup$ – Acccumulation Jul 2 '20 at 5:06
  • $\begingroup$ I've deleted a number of obsolete comments and/or responses to them. $\endgroup$ – David Z Jul 3 '20 at 0:36

Take a basket ball, how would you map it mathematically? With an equations describing a sphere, where the center of the sphere is the center of mass of the ball, no?

Throw it to the basket, would you still call it a sphere?

The difference with the elliptical trajectory of a planet around the sun is that it is not solid. Still it is a mathematical mapping of the trajectory where the sun is in one of the focuses. The mathematics does not change if the observational reference system changes. The whole ellipse will be describing an additional motion, but the description of sun-planet will be always an ellipse with a sun as the focus. The trajectory of the planet itself will be different for different reference frames but the ellipse mapping will always be there

  • $\begingroup$ By elliptic mapping you are describing an implici change of reference. However, there i no way, orbits are not frame-invanriant. Take a circular orbit. It can be described by the parametric equation $x=R cos( \omega t) ; y = R sin( \omega t)$ in one frame. In a uniformly translating frame the equations could be $x=v t + R cos( \omega t) ; y = R sin( \omega t)$. The first curve is closed and not self-intersecting. The second one is open and self-intersecting. Therefore the shape of the orbit is frame-dependent. This does not exclude that some invariand relation beteween ponts could exist. $\endgroup$ – GiorgioP Jul 1 '20 at 13:40
  • $\begingroup$ In the previous example it is the distance between a point of the trajectory and the origin, in the first case, and a point of the trajectory and the point of coordinates $(vt,0)$, in the second. $\endgroup$ – GiorgioP Jul 1 '20 at 13:41
  • $\begingroup$ @GiorgioP I am trying to give the analogy that the mathematical map is always there when in the correct reference frame, and will be there in a much more complicated way if the reference frame is moving. Think of describing the shape of the basket ball while in motion. It is always very easy to get complicated mathematical relationships out of simple ones by changing frames of reference.. $\endgroup$ – anna v Jul 1 '20 at 15:10
  • $\begingroup$ @annav I thought about it as, at any particular instant in a planet's path it tends to revolve in an elliptical orbit. That's the instantaneous path. But as the focus moves too, the orbit shifts to a different ellipse. Is it a right way to think about it? $\endgroup$ – Vamsi Krishna Jul 1 '20 at 15:59
  • $\begingroup$ @VamsiKrishna if you want to get the mathematics at that reference frame, you can be thinking of a different ellipse. It is a moving ellipse map $\endgroup$ – anna v Jul 2 '20 at 3:25

with the sun as one of its foci

That's probably the source of your confusion. The focus (one of them) is the place of the center of mass of the system (also called barycenter). The elliptical orbit is around that CM of the entire system.

But because usually the star takes up the large part of the mass (well into the high-90%s for single-star systems such as the Sun), then the center of mass ends up very close (often inside) the star, but not at the center of the star. From afar (enough that a pointlike star is a good enough approximation), it resembles that the star is on one of the foci of the ellipse.


This one has different answers depending on which of the sun's motions we consider. Kepler's Law is only true when there are exactly two bodies under consideration. They orbit around their common center of mass.

However, you talk about the solar system, which has several planets. Even if we just consider 3 things it gets hard. Suppose we're doing the sun, the earth, and Jupiter. Jupiter is so heavy that the center of mass of the Jupiter-sun system is outside the sun. The sun wobbles viewed from an inertial frame. If we ask what the earth's orbit is, given the wobbling sun, the answer is complex. I would guess that there is no closed-form solution -- certainly not an ellipse. It would not be an ellipse when viewed from any inertial frame, nor from one anchored to the sun.

The calculation is difficult, but I believe I have answered the question.

  • $\begingroup$ Even the famous "3-body problem" has no general closed-form solution. How much bodies the solar system consist of is a matter of debate and neither of these bodies is a point mass. The physics is all about the acceptable approximations. $\endgroup$ – fraxinus Jul 2 '20 at 5:15
  • $\begingroup$ And that's even before considering the motion of the sun through the Milky Way, or that of the Milky Way wrt. other (clusters of) galaxies... $\endgroup$ – cmaster - reinstate monica Jul 3 '20 at 9:07
  • $\begingroup$ Yes, but Cmaster, the wobble is quick on galactic time-scales. The motion of the sun through the galaxy is very low acceleration, almost inertial, so only adds a vector to the orbit. The wobble is a (relatively) large deviation from an elliptical orbit, viewed from any inertial frame. $\endgroup$ – Andrew Jul 4 '20 at 16:15

Yes, what you're talking about are reference frames.

For an outside observer looking at our solar system the whole thing, sun and planets including, would swoosh by in space and obviously not trace circular orbits. For an observer stationary with regards to the sun however they would indeed trace elliptical sources.


Well, which point in space? And how does the Sun move with regard to that point? If you have the answer to that question, then simply take the Sun's motion relative to your chosen point, and add it to the planets' motion relative to the Sun, to obtain the motion of the planets relative to your chosen point (principle of superposition).


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