# Understanding Kepler's problem

I wanted to know a few things about elliptical motion, that I do not understand so far.

Let's imagine that we have an attractive gravitational two body problem where both bodies carry out elliptical motion around the center of mass. In this case, the center of mass is in one focus of the ellipse, right? Further, we are able to determine the closest distance between the center of mass and the 1st planet(perigee) and the largest distance(apogee). further we have the semi-major axis, that will enter our third kepler-law, but is different from apogee and perigee.

but is there also a physical meaning behind the semi-major axis? and is there any meaning behind the center of the ellipse, cause until now, we only discussed the one focus of the ellipse?

• – Qmechanic Jun 4 '13 at 0:26
• The semi-major axis is one-half of the distance between the apogee and perigee. Note that the previous sentence hasn't used the center of the ellipse at all but even if I had to use the center of the ellipse, you would have to have bizarre reasons not to count the center of an ellipse as a "physical meaning of a point". It doesn't matter whether something is located at the point. It's more important that we may physically determine where the point is and such a point may play role in various propositions and laws, too. – Luboš Motl Jun 4 '13 at 5:14

You could argue that everything that's used in the mathematical description of a physical problem indeed has physical meaning. For many things that meaning may be hard to find, or requires a completely unintuitive perspective of the situation (see for example, the gas mileage analysis in this XKCD what-if).

Because of this, not many people will find this physical meaning useful to think about during the advancement of their models or theory. I still think it is a very good habit, if only to practice your physical intuition and theoretical understanding (but of course, only up to a certain point :)

• For circular and elliptical orbits, the semi-major axis $a$ is simply half the distance between pericentre and apocentre. Half, because that's often more useful in symmetrical shapes and prevents factors of 2 popping up all over the place (in some senses it's also strange that we use $2\pi$ everywhere). The full distance $2a$ (the major axis), in combination with the mass of the central body $M$ and the "relative strength of gravity" expressed by Newton's gravitational constant $G$, gives you the total binding energy $\epsilon$ of the orbit: $$\epsilon = -\frac{GM}{2a}$$

and the orbital period $T$ again relates to only half the length,

$$T = 2\pi \sqrt{\frac{a^3}{GM}}$$

which is equivalent to but more convenient than

$$T = 2\pi \sqrt{\frac{(2a)^3}{8GM}}$$

where the factor of $2\pi$ of course stems from the traversal of a complete circle (or elliptical angle, in this case).

• There are however still an infinite amount of possible orbits that have equal binding energies and equal orbital periods; there's more to the orbit than just the semi-major axis. This is often expressed in the eccentricity $e$, which is simply the scale factor to use on the semi-major axis to get the distance between the ellipse center $C$ and the focus $F$:

$$CF = ae$$

This gives you the meaning of the centre $C$ as well; the distance between the centre $C$ and the focus $F$ is a measure for how non-circular the orbit is; it relates to the ratio of the distances at closest and farthest approach. If the distance $CF=0$, you have a circle. if the distance $CF$ is small, you have an elliptical orbit with the same orbital period and the same binding energy as the circular one. And, applying Kepler's laws blindly, if the distance is a billion lightyears, you have a orbit that is so elliptical you can no longer distinguish it from a straight line, but still with the same orbital period and binding energy as the original circular orbit (this is completely non-physical, because this would require faster-than-light travel at the close approach, and a black hole with mass $M$ at the centre, but near-zero Schwarzschild radius...).

These interpretations only go up to a certain "philosophical level", so to speak. You could advance your understanding of the problem (the general problem that is, not restricted to circles or ellipses) by asking yourself what the physical meaning of the equivalent cone is.

In the framework of classical mechanics, the problem is equivalent to the motion of balls thrown on the insides of a cone. So in principle, you can place a huge plastic cone with a specific opening angle and orientation in space, with the central body replaced by an equivalent everywhere-parallel gravitational field, and get the exact same orbit for your "balls".

• great, this was truely helpful. one further question: the distance r and the angle phi, that describe my motion in plane polar coordinates are taken between the focus and the trace of the ellipse, right?-cause one of my colleagues proposed that r and phi have to be taken from the center, but I guess this is wrong. – Xin Wang Jun 4 '13 at 8:41
• @Lipschitz: both are possible. It's more convenient to have $r$ and $\phi$ from the focus $F$ to get to the equation of time $\phi(t)$. It's often more convenient to have $r_C$ and $\phi_C$ from the center $C$ when doing pure geometrical computations, like plotting the orbit or computing closest-approach distances between orbits. – Rody Oldenhuis Jun 4 '13 at 8:44