Closed Kepler orbits are ellipses with the Sun at one focus.

The force felt by the planet points in the direction of the Sun. As such, it is not a central force, since the focus is not the center.

I am confused. The force should be central. What is my misunderstanding?

  • $\begingroup$ The appellation 'central force' makes a bit more sense after performing the canonical transformation for two body forces to center of mass coordinates. $\endgroup$ Sep 19, 2017 at 3:23
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    $\begingroup$ 'Central force' means that the force is a function of the scalar distance between the particle in question and a single point in space - it doesn't mean that the resulting orbit has that point as it's geometrical center. The two-body problem is, in the strictest sense, not a central force problem, but it can be easily reduced to one by performing the transformation @dmckee mentions above. $\endgroup$
    – J. Murray
    Sep 19, 2017 at 3:43
  • $\begingroup$ @J.Murray That looks like an answer, not a comment. $\endgroup$ Sep 20, 2017 at 12:14
  • $\begingroup$ Related: physics.stackexchange.com/q/323183/2451 , physics.stackexchange.com/q/331366/2451 and links therein. $\endgroup$
    – Qmechanic
    Sep 20, 2017 at 12:22
  • $\begingroup$ Pretty important, the force at any instant is at the center of the body in question. It is a force field so it doesn't care about the time parameterized path. $\endgroup$ Jul 5, 2018 at 15:43

3 Answers 3


We generally choose to use the location of the sun as our origin. You're correct in identifying this is at the focus of the ellipse. We then choose to take quantities relative to the Sun as origin. e.g. the radial distance for an ellipse from a focus is given by $$ r=\frac{a(1-e^2)}{1\pm e\cos\theta}, $$ whereas to give the distance from the center of the ellipse, the expression is $$ r=\frac{ab}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}}. $$

Further we define things like $\vec{r}$ from the Sun, and angular momentum $\vec{r}\times\vec{p}.$ Now I'll try to explain why we choose to do this.

Consider if we were to use the center of the ellipse instead. This position of the center depends upon the orbital elements of the orbit we're looking at. So the center of our origin for the Earth-Sun, would be different to that of Sun-Jupiter. That sounds incredibly fiddly to work with. E.g. the angular momentum for Jupiter would be meaningless to compare to the angular momentum of Earth, since it's taken about a different point. If instead we use the Sun, then the origin stays put, and the two angular momenta become meaningful to compare.

Further it has good physical meaning, since as you also correctly identify, it's where gravity is pulling us towards. This means we get to easily utilize the spherical symmetry, if we choose that origin.

So to conclude, the focus as origin is physically meaningful. The location of the center of the ellipse is something that depends on the particular orbit you're looking at, and not a very useful origin to do the mathematics with. Then when we say something is central, we mean around our chosen origin, which here is the focus.


The point in space you define as being the origin cannot change the nature of the force since nature does not care about coordinate systems.

When we say that a central force satisfy $$\vec F=F(r)\hat r,$$ we are defining the origin as the center of force. However we can consider the center of force at position $\vec r_0$ and in this case the central force is given by $$\vec F=F(|\vec r-\vec r_0|)\frac{(\vec r-\vec r_0)}{|\vec r-\vec r_0|}.$$ As you can guess you can consider the Sun as off the origin but the force is still central.

  • $\begingroup$ Yes, but when you write things like $r(\theta)=\alpha/(1+\epsilon\cos(\theta))$ and $\vec{L}=\vec{r}\times\vec{p}$ the quantities $\vec{r}$ and $r$ are measured with respect to the center and not the Sun, right? $\endgroup$
    – thedude
    Sep 19, 2017 at 0:57

Both the Sun and the Earth are actually moving in an ellipse and each of their ellipses has a focus at the center of mass of the two, also known as the barycenter.

Our Sun is about 333k times more massive than the Earth. Therefore its ellipse is tiny in comparison to the ellipse of the Earth and so we can ignore it. I did read somewhere (and it would be easy enough to do the calculation) that our Sun is pulled about 4km by the mass of Earth, as earth travels around it. 4km from the Sun's center would still be inside the Sun.

If both objects have the same or similar mass such as in a binary system, it is more obvious that each travels along its own ellipse with their foci at the center of mass of the system.

The force acts toward this center of mass not exactly toward the center of the Sun.

The Processing code here might help and demo.

  • $\begingroup$ Even if you disregard the difference between the center of mass and the center of the Sun, they may still be far from the center of the orbit. That is the point. $\endgroup$
    – thedude
    Sep 19, 2017 at 0:58
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    $\begingroup$ When you say 'center of orbit' do you mean the focus of the ellipse or are you thinking of the center of the ellipse as in where the major and minor axes intersect? $\endgroup$
    – ofey
    Sep 19, 2017 at 8:51
  • $\begingroup$ The centre of mass of the system is at one of the foci of the ellipse and each object moves on a separate ellipse with a common focus. $\endgroup$
    – ofey
    May 5, 2018 at 17:31

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