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 
![\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}](http://s0.wp.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bll%7Dm%5Cddot%7B%5Cvec%7Bx%7D%7D%3D-%5Cfrac%7BGMm%5Cvec%7Bx%7D%7D%7B%7C%5Cvec%7Bx%7D%7C%5E3%7D%5C%5C%5CRightarrow%5Cddot%7B%5Cvec%7Bx%7D%7D%3D-%5Cfrac%7BGM%7D%7B%7C%5Cvec%7Bx%7D%7C%5E3%7D%5Cvec%7Bx%7D%5C%5C%5CRightarrow%5Cdot%7B%5Cvec%7Bx%7D%7D%5Ccdot%5Cddot%7B%5Cvec%7Bx%7D%7D%3D-%5Cfrac%7BGM%7D%7B%7C%5Cvec%7Bx%7D%7C%5E3%7D%5Cvec%7Bx%7D%5Ccdot%5Cdot%7B%5Cvec%7Bx%7D%7D%5C%5C%5CRightarrow%5Cdot%7B%5Cvec%7Bx%7D%7D%5Ccdot%5Cddot%7B%5Cvec%7Bx%7D%7D%3D-%5Cfrac%7BGM%7D%7B%28%5Cvec%7Bx%7D%5Ccdot%5Cvec%7Bx%7D%29%5E%7B3%2F2%7D%7D%5Cvec%7Bx%7D%5Ccdot%5Cdot%7B%5Cvec%7Bx%7D%7D%26%5Cleft%5C%7B%5Cvec%7Bx%7D%5Ccdot%5Cvec%7Bx%7D%3D%7C%5Cvec%7Bx%7D%7C%5E2%5Cright.%5C%5C%5CRightarrow%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%5B%5Cfrac%7B1%7D%7B2%7D+%5Cdot%7B%5Cvec%7Bx%7D%7D%5Ccdot%5Cdot%7B%5Cvec%7Bx%7D%7D%5Cright%5D%3D%5Cfrac%7Bd%7D%7Bdt%7D+%5Cleft%5B%5Cfrac%7BGM%7D%7B%28%5Cvec%7Bx%7D%5Ccdot%5Cvec%7Bx%7D%29%5E%7B1%2F2%7D%7D+%5Cright%5D%26%5Cleft%5C%7B%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%28%5Cvec%7Bx%7D%5Ccdot%5Cvec%7Bx%7D%5Cright%29%3D2%5Cvec%7Bx%7D%5Ccdot%5Cdot%7B%5Cvec%7Bx%7D%7D%5Cright.%5C%5C%5CRightarrow+%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%5B%7C%5Cdot%7B%5Cvec%7Bx%7D%7D%7C%5E2+%5Cright%5D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%5B%5Cfrac%7B2GM%7D%7B%7C%5Cvec%7Bx%7D%7C%7D%5Cright%5D%5C%5C%5CRightarrow+%7C%5Cdot%7B%5Cvec%7Bx%7D%7D%7C%5E2%3D%5Cfrac%7B2GM%7D%7B%7C%5Cvec%7Bx%7D%7C%7D%2Bc%5Cend%7Barray%7D&bg=ffffff&fg=000&s=0 "\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 
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.
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 
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).
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.”
Jackson has a father?
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.
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.
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
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.
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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!
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.
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.
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.
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.
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.