Why are orbits elliptical instead of circular?

Why do planets rotate around a star in a specific elliptical orbit with the star at one of it's foci? Why isn't the orbit a circle?

• Eduardo's answer sums up the most of it. Although you can see my answer to a similar question on Physics SE. physics.stackexchange.com/questions/56657/… – Cheeku Dec 23 '13 at 17:39
• Circular orbits are a special case of elliptical orbits. – asawyer Dec 23 '13 at 20:49

Assume the planet has a negligible mass compared to the star, that both are spherically symmetric (so Newton's law of gravitation holds, but this normally happens to a very good approximation anyway), and that there aren't any forces besides the gravity between them. If the first condition does not hold, then the acceleration of each is going to be towards the barycenter of the system, as if barycenter was attracting them a gravitational force with a certain reduced mass, so the problem is mathematically equivalent.

Take the star to be at the origin. By Newton's law of gravitation, the force is $\mathbf{F} = -\frac{m\mu}{r^3}\mathbf{r}$, where $\mathbf{r}$ is the vector to the planet, $m$ is its mass, and $\mu = GM$ is the standard gravitational parameter of the star.

Conservation Laws

Because the force is purely radial $(\mathbf{F}\parallel\mathbf{r})$, angular momentum $\mathbf{L} = \mathbf{r}\times\mathbf{p}$ is conserved: $$\dot{\mathbf{L}} = \frac{\mathrm{d}}{\mathrm{d}t}\left(\mathbf{r}\times\mathbf{p}\right) = m(\dot{\mathbf{r}}\times \dot{\mathbf{r}}) + \mathbf{r}\times\mathbf{F} = \mathbf{0}\text{.}$$ If the initial velocity is nonzero and the star is at the origin, then in terms of the initial position and velocity, the orbit must be confined to the plane of all points with vectors $\mathbf{x}$ from the origin that satisify $\mathbf{L}\cdot\mathbf{x} = 0$. If the initial velocity is zero, then the motion is purely radial, and we can take any one of infinitely many planes that contain the barycenter and initial position.

The total orbital energy is given by $$\mathcal{E} = \frac{p^2}{2m} - \frac{m\mu}{r}\text{,}$$ where the first term part is the kinetic energy and the second term is the gravitational potential energy of the planet. Its conservation, as well as the fact that it invokes the correct potential energy, can be proven by the fundamental theorem of calculus for line integrals.

Define the Laplace-Runge-Lenz vector to be $$\mathbf{A} = \mathbf{p}\times\mathbf{L} - \frac{m^2\mu}{r}\mathbf{r}\text{.}$$ It is also conserved: $$\begin{eqnarray*} \dot{\mathbf{A}} &=& \mathbf{F}\times\mathbf{L} + \mathbf{p}\times\dot{\mathbf{L}} - \frac{m\mu}{r}\mathbf{p} + \frac{m\mu}{r^3}(\mathbf{p}\cdot\mathbf{r})\mathbf{r}\\ &=& -\frac{m\mu}{r^3}\underbrace{\left(\mathbf{r}\times(\mathbf{r}\times\mathbf{p})\right)}_{(\mathbf{r}\cdot\mathbf{p})\mathbf{r} - r^2\mathbf{p}} - \frac{m\mu}{r}\mathbf{p} + \frac{m\mu}{r^3}(\mathbf{p}\cdot\mathbf{r})\mathbf{r}\\ &=& \mathbf{0}\text{.} \end{eqnarray*}$$

Finally, let's also take $\mathbf{f} = \mathbf{A}/(m\mathcal{E})$, which has the same units as $\mathbf{r}$, and since $\mathbf{L}\cdot\mathbf{f} = 0$, it lies along the orbital plane. As it's a conserved vector scaled by a conserved scalar, it's easy to show that $\mathbf{f}$ is conserved as well, as long as $\mathcal{E}\neq 0$.

Simplifying

By employing the vector triple product, we can write $$\begin{eqnarray*} \frac{1}{m}\mathbf{A} &=& \frac{1}{m}\left[p^2\mathbf{r}-(\mathbf{p}\cdot\mathbf{r})\mathbf{p}\right] -\frac{m\mu}{r}\mathbf{r}\\ &=& \left(\mathcal{E}+\frac{p^2}{2m}\right)\mathbf{r} - \frac{1}{m}\left(\mathbf{p}\cdot\mathbf{r}\right)\mathbf{p}\\ \mathcal{E}(\mathbf{f}-\mathbf{r}) &=& \left(\frac{p^2}{2m}\right)\mathbf{r} - \frac{1}{m}\left(\mathbf{p}\cdot\mathbf{r}\right)\mathbf{p}\text{,} \end{eqnarray*}$$ the norm-squared of which is easy to crank out: $$\mathcal{E}^2|\mathbf{f}-\mathbf{r}|^2 = \left(\mathcal{E} + \frac{m\mu}{r}\right)^2r^2\text{,}$$ where $\mathcal{E}$ was used throughout to switch between kinetic and potential terms.

Why Ellipses?

Since $\mathcal{E}$ is energy relative to infinity, to have a bound orbit we need $\mathcal{E}<0$. Thus, from the previous section, $|\mathbf{f}-\mathbf{r}| = -\mathcal{E}^{-1}\left(\mathcal{E}r + m\mu\right)$ and therefore $$|\mathbf{f}-\mathbf{r}| + |\mathbf{r}| = -\frac{m\mu}{\mathcal{E}}\text{,}$$ which defines an ellipse with foci $\mathbf{0},\,\mathbf{f}$ and major axis $2a=-m\mu/\mathcal{E}$.

Why Not Circles?

The circle is a special case where the foci are the same point, $\mathbf{f} = \mathbf{0}$, which can be restated as $$\mathcal{E} = -\frac{1}{2}\frac{m\mu}{r} = -\frac{p^2}{2m}\text{.}$$ In other words, circular orbits require the orbital energy to be the negative of the kinetic energy. This is possible, but almost certain not to hold exactly. Since any values of $\mathcal{E}<0$ are allowed for bound orbits, there are many more ways to have elliptic orbits. (Although some of them would actually crash because the star and planet have positive size.)

Note that hyperbolic orbits have $\mathcal{E}>0$, and we can still find the foci using the above method, though being careful with the signs. For $\mathcal{E}=0$, the second focus $\mathbf{f}$ is undefined because this is a parabolic orbit, and parabolas only have one focus within a finite distance from the center.

Additionally, the eccentricity vector $\mathbf{e} = \mathbf{A}/(m^2\mu)$ is an alternative choice for the LRL vector; as the name suggests, its magnitude is the orbital eccentricity.

It is posible for a planet to have a circular orbit, a circle, after all, is an ellipse where both foci are in the same place; this is known as having an eccentricity of 0. Eccentricity is defined in the following way: $$e = \frac{r_{a} - r_{p}}{r_{a} + r_{p}}$$ where $r_{a}$ is the apoapsis (farthest point in the orbit from the center of mass), and $r_{p}$ is the periapsis (the closest distance). Just to build some intuition here, if the apoapsis is twice the distance of the periapsis, the eccentricity will be $e=0.333$.

From all the planets of the solar system, Venus, with an eccentricity of 0.007 has the most circular orbit.

As to why all orbits aren't round, it comes down to kinetic energy. The kinetic energy is proportional to the square of the speed. In the orbital plane and in polar coordinates about the star, we can decompose this is into a combination of radial velocity $\dot{r}$ and angular velocity $\dot{\phi}$: $$v^2 = \dot{r}^2 + r^2\dot{\phi}^2\text{.}$$ Since circles have constant radii, for the orbit to be circular around the star, the planet's radial velocity must be exactly zero. Additionally, the angular speed must be such that the centrifugal force in the corotating frame exactly balances the gravitational force--a little more or a little less, the imbalance will change the radial velocity, spoiling the circle.

Given the fact that velocities vary for a large number of reasons, it's no wonder that only a few orbits end up being circular, and considering that actual orbits change with time, we know they can't stay this way for long.

If you're looking for a mathematical proof, this link shares some details about it.

Here's an image showing eccentricity of some bodies in the solar system extracted from here: • This is completely wrong: "For the orbit to be round, the planet's speed must be exactly the minimun needed to be in orbit;... a little less and it would crash into the planet it's orbiting." The paragraph is also pretty confused about what orbits what. Obviously, they minimize the radial speed, but that's different and doesn't connect with the kinetic energy discussion. Breaking up kinetic energy into radial and angular parts, circular orbits also minimize the effective potential if angular momentum is held fixed. – Stan Liou Dec 24 '13 at 8:50
• @Stan you can propose an edit or give your own answer. Could you go into detail about why that statement is wrong? If a satellite is describing a circular orbit and you slow it, it'll crash into the planet; if you speed it up, it'll form and elliptical orbit. – Eduardo Serra Dec 24 '13 at 13:23
• A circular orbit has $r_a = r_p$. A small change in satellite speed is going to produce a small change in these quantities. The satellite will only crash if its new $r'_p$ is less than or equal to the planetary radius, incl. atmosphere, but since the changes are small, that can only happen if the satellite orbit was already almost hugging the planet. ... I'll suggest an edit that keeps the tie-in to kinetic energy. – Stan Liou Dec 24 '13 at 22:49
• @EduardoSerra - Slow down an object in a circular orbit and it will be in an elliptical orbit with the former circular orbit radius now being the apofocal distance. – David Hammen Aug 31 '17 at 12:50

I always prefer answers which try to avoid any formula and reply on argumentation instead. Concerning the part of the question why not all orbits are circular, an argumentation would be like this:

Consider a stationary star and a moving planet. For each impulse the planet can have, a curve for its further movement can be predicted. If this impulse is directed exactly orthogonal to the line from the star to the planet, and if the velocity has the exact amount, then this curve of movement can be an exact circle.

But for every deviance of this one exact impulse, the resulting curve cannot be a circle:

• If the speed is too low, the planet will fall towards the star (in the extreme case of an impulse of zero, this fall will be in a straight line).
• If the speed is too high, the planet will gain distance from the star (similar to a slingshot).
• If the impulse is not directly orthogonal to the line to the star, the first movement will move the towards or from the star, so again the curve will not be a circle.

So, one can simply argue, a circle is a very special case for the curve a planet can take around a star.

• (1) The initial orthogonality argument is a good start. (2) But the "speed is too [low/high]" considerations are unjustified: how would one know that circular orbits at multiple speeds disallowed for the same distance? One can argue against the possibility of multiple speeds by balancing gravitational and centrifugal forces, but then both (1) and (2) turn into exactly what's outlined in Eduardo Serra's answer. – Stan Liou Jun 5 '14 at 9:01
• So you mean that one could be under the impression that the gravitational force could be like a tight rope in the sense that it will apply more force on the planet towards the star when more force is "needed" to keep the planet on a circular path? Hmm … yes, depending on the background of the layman this could be what one expects. Thank you for the notion; maybe I can improve my answer to address this issue as well! – Alfe Jun 5 '14 at 12:26