Q: Why are orbits elliptical? Why is the Sun in one focus, and what’s in the other?

Physicist: This question always bothered me too.  The short answer is: it falls out of the math.  Specifically, the math of first year physics and second year calculus.  The fact that the Sun is in one focus is just one of those things.  It’s nothing special.  Even less special is the other focus, which contains nothing at all.

Ellipses and their foci have a lot of useful properties. It so happens that an orbiting object traces out an ellipse, with the thing it orbits around at one of the focuses. Coincidence? Yes.

I can’t find a good intuitive reason why orbits are elliptical.  In fact, I can’t even find a mathematical derivation.  So, because it should be found somewhere, I’ll leave the derivation floating in the answer gravy.


Answer gravy: The force of gravity is usually written as ma=F=-\frac{GMm}{R^2}.  You can rewrite this using vector notation as m\ddot{\vec{x}}=-\frac{GMm\vec{x}}{|\vec{x}|^3}, where the dot on top is a time derivative.  To keep the notation both standard and confusing, \vec{x}=(x,y).

\begin{array}{ll}m\ddot{\vec{x}}=-\frac{GMm\vec{x}}{|\vec{x}|^3}\\\Rightarrow\ddot{\vec{x}}=-\frac{GM}{|\vec{x}|^3}\vec{x}\\\Rightarrow\dot{\vec{x}}\cdot\ddot{\vec{x}}=-\frac{GM}{|\vec{x}|^3}\vec{x}\cdot\dot{\vec{x}}\\\Rightarrow\dot{\vec{x}}\cdot\ddot{\vec{x}}=-\frac{GM}{(\vec{x}\cdot\vec{x})^{3/2}}\vec{x}\cdot\dot{\vec{x}}&\left\{\vec{x}\cdot\vec{x}=|\vec{x}|^2\right.\\\Rightarrow\frac{d}{dt}\left[\frac{1}{2} \dot{\vec{x}}\cdot\dot{\vec{x}}\right]=\frac{d}{dt} \left[\frac{GM}{(\vec{x}\cdot\vec{x})^{1/2}} \right]&\left\{\frac{d}{dt}\left(\vec{x}\cdot\vec{x}\right)=2\vec{x}\cdot\dot{\vec{x}}\right.\\\Rightarrow \frac{d}{dt}\left[|\dot{\vec{x}}|^2 \right]=\frac{d}{dt}\left[\frac{2GM}{|\vec{x}|}\right]\\\Rightarrow |\dot{\vec{x}}|^2=\frac{2GM}{|\vec{x}|}+c\end{array}

c is an “integration constant“, it can be any number.  Jumping over to polar coordinates \left(\begin{array}{l}x=R\cos{(\theta)}\\y=R\sin{(\theta)}\end{array}\right) you can rewrite the usual velocity in terms of how fast you’re moving toward or away from the Sun (\dot{R}) and how fast you’re going around (\dot{\theta}).

\begin{array}{ll}\Rightarrow\dot{R}^2+R^2\dot{\theta}^2=\frac{2GM}{R}+c&\left\{\begin{array}{ll}|\vec{x}|=R\\|\dot{\vec{x}}|^2=\dot{R}^2+R^2\dot{\theta}^2\end{array}\right.\\\Rightarrow\left(\frac{dR}{d\theta}\dot{\theta} \right)^2+R^2\dot{\theta}^2=\frac{2GM}{R}+c&\left\{\frac{dR}{dt}=\frac{dR}{d\theta}\frac{d\theta}{dt}\right.\\\Rightarrow\left(\left(\frac{dR}{d\theta}\right)^2+R^2\right)\dot{\theta}^2=\frac{2GM}{R}+c\\\Rightarrow\left(\left(\frac{dR}{d\theta}\right)^2+R^2\right)\frac{L^2}{R^4}=\frac{2GM}{R}+c&\left\{R^2\dot{\theta}=L\right.\\\Rightarrow\left(\frac{1}{R^2}\frac{dR}{d\theta}\right)^2+\frac{1}{R^2}=\frac{2GM}{L^2R}+\frac{c}{L^2}\\\Rightarrow\left(\frac{1}{R^2}\frac{dR}{d\theta}\right)^2+\frac{1}{R^2}=2\alpha\frac{1}{R} +C&\left\{\begin{array}{l}C=\frac{c}{L^2}\\\alpha=\frac{GM}{L^2}\end{array}\right.\end{array}

L is the angular momentum of the planet in question, and it’s constant.  It may seem silly but, with the advantage of foresight, it’s better to solve this problem in terms of 1/R instead of R.

\begin{array}{ll}\Rightarrow \left(-\frac{dS}{d\theta}\right)^2+S^2=2\alpha S +C&\left\{\begin{array}{l}S=\frac{1}{R}\\\frac{dS}{d\theta}=-\frac{1}{R^2}\frac{dR}{d\theta}\end{array}\right.\\\Rightarrow-\frac{dS}{d\theta}=\sqrt{-S^2+2\alpha S+C}\\\Rightarrow d\theta=\frac{-dS}{\sqrt{-S^2+2\alpha S+C}}\\\Rightarrow\int d\theta=-\int\frac{dS}{\sqrt{-S^2+2\alpha S+C}}\\\Rightarrow\theta+D=-\int\frac{dS}{\sqrt{C+\alpha^2-(S-\alpha)^2}}\\=\int\frac{\sqrt{C+\alpha^2}\sin{(u)}du}{\sqrt{C+\alpha^2-(C+\alpha^2)\cos^2{(u)}}}&\left\{\begin{array}{l}S-\alpha=\sqrt{C+\alpha^2}\cos{(u)}\\dS=-\sqrt{C+\alpha^2}\sin{(u)}du\end{array}\right.\\=\int\frac{\sin{(u)}du}{\sqrt{1-\cos^2{(u)}}}\\=\int\frac{\sin{(u)}du}{\sqrt{\sin^2{(u)}}}\\=\int du\\\Rightarrow\theta+D=u\\\Rightarrow\cos{(\theta+D)}=\cos{(u)}\\\Rightarrow\sqrt{C+\alpha^2}\cos{(\theta+D)}=\sqrt{C+\alpha^2}\cos{(u)}\\\Rightarrow\sqrt{C+\alpha^2}\cos{(\theta+D)}=S-\alpha\\\Rightarrow\frac{1}{R}=S=\alpha+\sqrt{C+\alpha^2}\cos{(\theta+D)}\\\Rightarrow R=\frac{1}{\alpha+\sqrt{C+\alpha^2}\cos{(\theta+D)}}\\\Rightarrow R=\frac{P}{1+\epsilon\cos{(\theta+D)}}&\left\{\begin{array}{l}P=\frac{1}{\alpha}=\frac{L^2}{GM}\\\epsilon=\sqrt{\frac{C}{\alpha^2}+1}\end{array}\right.\end{array}

The choice of P and ε may seem arbitrary (and it is), but it has some historical relevance.  P is called the “semi-latus recturn” and it basically describes the size of the orbit.  ε is called the “eccentricity”, and it describes how lopsided the orbit is.  ε=0 means the orbit is a circle, 0<ε<1 means the orbit is elliptical, and 1≤ε means that the orbit is open (not actually orbiting).  For reference, the Earth’s eccentricity is ε=0.01671123 and Halley’s comet’s is ε=0.967.

D just describes what direction the far side of the ellipse points in, so it’s not actually important to the overall shape.

It turns out that this last equation relating R and θ is all you need to define an ellipse, such that the center of the system, (0,0), is at one of the foci.  Here’s a proof:

An ellipse with a focus at (0,0) can be written \frac{(x+F)^2}{A^2}+\frac{y^2}{B^2}=1 where F is the distance from the center of the ellipse to the focus and F^2=A^2-B^2.

\begin{array}{ll}R=\frac{P}{1+\epsilon\cos{(\theta)}}\\\Rightarrow R+\epsilon R\cos{(\theta)}=P\\\Rightarrow \sqrt{x^2+y^2}+\epsilon x=P\quad\quad\quad\left\{\begin{array}{l}x=R\cos{(\theta)}\\y=R\sin{(\theta)}\end{array}\right.\\\Rightarrow \sqrt{x^2+y^2}=P-\epsilon x\\\Rightarrow x^2+y^2=P^2-2P\epsilon x+\epsilon^2x^2\\\Rightarrow (1-\epsilon^2)x^2+2P\epsilon x+y^2=P^2\\\Rightarrow x^2+2\frac{P\epsilon}{1-\epsilon^2} x+\frac{y^2}{1-\epsilon^2}=\frac{P^2}{1-\epsilon^2}\\\Rightarrow x^2+2\frac{P\epsilon}{1-\epsilon^2} x+\left(\frac{P\epsilon}{1-\epsilon^2}\right)^2+\frac{y^2}{1-\epsilon^2}=\frac{P^2}{1-\epsilon^2}+\left(\frac{P\epsilon}{1-\epsilon^2}\right)^2\\\Rightarrow \left(x+\frac{P\epsilon}{1-\epsilon^2}\right)^2+\frac{y^2}{1-\epsilon^2}=\frac{P^2(1-\epsilon^2)}{(1-\epsilon^2)^2}+\frac{P^2\epsilon^2}{(1-\epsilon^2)^2}\\\Rightarrow \left(x+\frac{P\epsilon}{1-\epsilon^2}\right)^2+\frac{y^2}{1-\epsilon^2}=\frac{P^2}{(1-\epsilon^2)^2}\\\Rightarrow \frac{\left(x+\frac{P\epsilon}{1-\epsilon^2}\right)^2}{\left(\frac{P^2}{(1-\epsilon^2)^2}\right)}+\frac{y^2}{\left(\frac{P^2}{1-\epsilon^2}\right)}=1\end{array}

Put it all together, and you’ll find that this is definitely an ellipse with a focus at the point (0,0), the location being orbited around (like the Sun for instance).

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22 Responses to Q: Why are orbits elliptical? Why is the Sun in one focus, and what’s in the other?

  1. Scott says:

    Standard and confusing is right. Someone must have been reading Jackson recently. My favorite David Jackson story: a friend of mine met him once at a conference and asked him why he dedicated the book to his father. His reply? “So everyone would know I had one.”

  2. The Physicist Physicist says:

    Jackson has a father?

  3. James says:

    Just a curious question. Is this one of those things that you explicated that is known or is it something that you can just figure out by looking at it HAVING NEVER SEEN IT. I’m just curious. I’m a math and physics double major and I can follow the math (I’m only in my first year >.>) but I always feel stupid when someone pulls out a huge derivation like this (don’t get me wrong, I love them) but I’m always like, “is this something they thought of or is this like, written down? Should I be able to do this?” I’m not sure what I’m trying to say I guess lol.

  4. The Physicist The Physicist says:

    I’d never seen a derivation, which is why mine is so clunky.
    My thinking was: write down “F=MA” and go from there. Knowing that the end result is an ellipse means that the solution can be in a form that’s independent of time (it’s just a shape). I chose to used R and theta instead of x and y because the conservation of momentum has a simpler form, and it gives you a chance to get rid of the time dependence.
    Also, this was my third or fourth attempt. The first ones were tumbling train wrecks. The derivations you see are almost never the first attempt, and generally it’s streamlined so much that you can barely see what the author was thinking.
    In your first year it would be unusual to be able to do this. Having experience seeing tricks other people use is in derivations is very important. When a professor is deriving stuff it helps (both of you) to ask “why did you do that?” or “what is the goal in this step?”. Harass them during office hours as often as you can, and keep a dialog open while they work through problems. It takes the mystery away from the whole process.

  5. Nainan says:

    Orbits are elliptical, because Kepler told so and Newton proved them to be so!!!
    However, simple mechanics tells us that no free body can orbit around another moving body in closed geometrical path. [Central body of a planetary system is also a moving body]. Therefore, elliptical orbits are imaginary constructs, which may show relative positions of concerned bodies in space. They do not depict paths of these bodies in space. In reality there are neither elliptical orbits or foci for the central bodies to be in. Real orbital path of a planet is wavy about the path of its central body.
    Details at http://vixra.org/abs/1008.0010

  6. The Physicist The Physicist says:

    The calculation in this post assumes that the central mass is stationary, however in reality each body orbits the center of mass of the system, and both of those orbits are elliptical (with the center of mass in a focus).
    For more than two bodies, and taking into account general relativity, you find that the orbits are only very, very, very nearly elliptical. The orbit of Mercury, due to general relativistic effects, shifts the farthest point in it’s orbit by about 1 degree every 8.5 thousand years (so it doesn’t quite close).
    So, for an imaginary construct, it’s pretty good.

  7. Pingback: Q: What is the three body problem? | Ask a Mathematician / Ask a Physicist

  8. Lindsay says:

    Elliptical orbits are perhaps due to the fact that the universe is expanding! Thus, the expanding movement imparts an excentricity to orbits that would have been circular if the universe were stationary!

  9. Verbion says:

    Good basic point in that it is the expansion of our universe that imparts movement on large heavenly bodies, thereby causing eccentric movements of lesser heavenly bodies around these larger ones… makes sense? The only difference I see would be if there is no expansion of the universe, in which case there would be no significant movements (circular or eccentric) of bodies around their larger counterparts. In the sense of a retraction(collapse) of our universe… I would think that everything would then begin to operate in a reverse, perhaps even eccentric, order.

  10. The Physicist The Physicist says:

    The expansion of the universe is important on really, really gargantuan scales. Even on the scale of a few million light years (the distance to the nearest galaxy) it’s negligible.
    The elliptical-ness of orbits is purely a product of regular, dull as dishwater, Newtonian gravity.
    There’s no mystery and no unknown causes behind elliptical orbits.

  11. Verbion says:

    Thank you so much for the reply. I’m assuming then that the difference in mass between the orbital bodies, is somewhat responsible for the elliptical-ness effect powered by Newtonian gravity.

  12. The Physicist The Physicist says:

    Nope!
    Even if you have two equal-mass bodies they’ll still follow elliptical orbits. The only change is that, instead of the Sun being in one focus it’s the center of mass of both bodies. Technically, this is always the case, it’s just that when the Sun is involved the center of mass and the center of the Sun are almost the same.

  13. Verbion says:

    Okay, so then an elliptical orbit is pretty much the norm.
    What I now gather is that the Sun is not at the center of an elliptical orbit, but rather.. it is a little off to one side; at a point called a “focus” of the ellipse (just as you stated earlier). And it is because of this offset, that causes a planet to move closer to(perihelion) and further away(aphelion) from the Sun every orbit. Thanks for your time.

  14. Guillermo says:

    Veribion. What he said is that the Sun is not exactly at the focus, but orbiting it. The thing is that the focus is inside the Sun and very close to its center so its really jut waving a bit. This also means that we can approximate the Sun as being in the focus.
    James, I feel the same a lot when I see derivations!

    For the OP, I have a question: from step four to five, where you go from velocity dot force to derivative of energy, where has the negative sign go?

  15. The Physicist The Physicist says:

    It got absorbed into the derivative of (\vec{x}\cdot\vec{x})^{-\frac{1}{2}}.

  16. umm says:

    um so I don’t mean to seem disrespectful but I’m trying to understand this. I’m in year 10 at college at the moment and i was wondering if you could maybe try to explain it simply?

  17. alcazar84 says:

    Back to the question about the sun being at one focus, but what is at the other focus? Years ago I was playing around with drawing ellipses with LOGO software (remember Turtle Graphics?), and I made an observation that I no longer have the mathematical skills to prove. As I watched the ellipse draw, with the center of the window at one focus, from 0 degrees to 90 degrees it drew the ellipse very slowly. Then from 90 degrees to 180 degrees, it traveled far to the left and got faster and faster. Then from 180 degrees to 270 degrees, it came back towards the center and drew slower and slower. Then finishing from 270 degrees to 360 degrees, it drew the last little bit of the ellipse slowly.

    I realized that as the angle changed with a constant rate of change, sitting at the focus on the right, it traced out the path of the ellipse with what seemed to be the appropriate speed a planet would travel at as if the sun was at the focus off to the left. There is no physical object at this focus on the right, but rather the center of a polar coordinate system tracing the planet’s path. For any given time, the exact location of the planet can be calculated relative to this focus.

    If this turns out to be true, and somebody out there can prove it to be true, and it wins that person some sort of physics/astronomy award, please reference this post and give me the credit as the inspiration for your computations. I can give my contact information if needed.

  18. nizar96 says:

    Thanks for the derivations! When you introduce the angular momentum L, do you use a unit mass (dimensional analysis of R²θ dot gives L²T⁻¹) ?

  19. Greg says:

    I was looking into this after my daughter asked why orbits are elliptical, and I couldn’t answer. I think that the simplest answer might be that they are elliptical because they can be. As you show above, an ellipse leads to a stable orbit, so it is possible. I think that the alternative that most people wonder about is “Why not a circle?” It seems that the only reason against a circle is that it is only one of an infinite number of possible ellipses (where f1=f2). So a circle could happen, but would be extremely unlikely since there are infinitely more other possible ellipses that are equally stable.

  20. Enos says:

    I believe orbits are elliptical because one of the two focuses is due to the resulting mass of the rest of the universe pulling the planet in that direction.
    That is, the center of mass of the universe is in the direction from the sun to the other focus of the ellipse.

  21. Buck Rogers says:

    Not all orbits are elliptical, as well: there are parabolic orbits, hyperbolic orbits, and circular orbits.

    One can see the shape of the four different orbital shapes by use of a conical dissect. Take a cone and slice it diagonally from one side to the other and you have the shape of an ellipse; slice it perpendicular to its longitudinal axis and you have a circular pattern; slice it at a sharper diagonal angle so the end of the cut exits the bottom of the base, but passing beyond the base center and you have the shape of a parabola; slice it at an even sharper angle to exit the base before crossing the center of the base and you have the shape of a hyperbola… these are know as the four shapes of a conic section.

    An elliptical orbit will accelerate as it approaches perigee and decelerates as it nears apogee. On the other hand, circular orbits have no periapsis or apoapsis and therefore the orbit velocity remains constant.

    To orbit Earth in a circular path the orbital path of satellites are positioned approximately 24,000 miles from the Earth’s center (dependent on the mass). Most communication satellites are positioned in a synchronous circular orbit around the equator in the same direction as the Earth’s rotation, but if the circular orbit is in reverse of the Earth’s rotation or in a polar orbit it would no longer be synchronous.

    Moreover, due to changing gravitational forces, an orbital path is slightly irregular, this concept holds true for all four types of orbits, but more pronounced in a circular orbit not synchronous to the Earth’s rotation and due to the irregular shape of the Earth.

    Furthermore, all elliptical and circular orbits are closed decaying loops and are eventually succumbed by gravity. Retro-rockets on satellites are used to occasionally reposition the satellite to sustain the orbital path for a longer period of time. Once the retro-rocket fuel is exhausted the satellite orbit can no longer be repositioned unless refueled while still in a sustainable orbital path; Skylab’s orbit for example decayed rapidly and crashed into Earth on July 11th, 1979.

    Hyperbolic and parabolic orbits are open orbital paths that never return. These type orbits are used to send probes into outer space far away from Earth. Trajectories through space are always in a orbital curve and never in a straight line. To accelerate the velocity of an orbit, a slingshot method is used by directing the orbital path of an object towards a celestial body and using retro rockets with predetermined burn times to exit the trajectory path at the right moment and to enter into another orbital path with increased velocity.

  22. Inquest says:

    What does our sun orbit?

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