In the previous post, I mentioned sharing some things I have learned whilst preparing notes for my novels. Today’s topic relates to the orbital motion of planets.

Consider a very simple model of a point moving along the circumference of a circle. We give no reason for the point to change its speed, so let us assume it stays constant. Now, a circle seems an unlikely thing, a very special type of ellipse. So let us alter the model so that the ellipse has two foci instead of one. Is the point still moving at constant speed? Why would it not? Intuition might say that if this were a physical system where the point is pulled to a focus, the point would be moving at a constant speed (unless some external force were altering the system). So this simple model, one might have expected, applies to the orbits of planets around a sun. But when Johannes Kepler analyzed Tycho Brahe’s big data set on planetary motion, he discovered that the planets travel faster near the Sun and slower when far from the Sun. This is called Kepler’s Second Law of Planetary Motion.

Isaac Newton was able to prove that this law is true, and it holds for all two-body systems bound together in gravitational orbits. But wait, how could Newton deductively prove an observational discovery which seems dependent on the contingent nature of a physical system? Here is an attempt to outline why Kepler’s Second Law is a matter of deductive reasoning, largely independent from the exceptional physical nature of gravity, or planets, or stars.

But first let us acknowledge that we shall be talking about a toy model of a planetary system; we shall not be considering relativistic effects, quantum effects (i.e., a system of objects with planet-like orbits cannot exist at subatomic scales), or more mysterious galactic-scale effects. Additionally, for simplicity we can assume that a low-mass planet is orbiting a high-mass star so that one focus of the elliptical orbit aligns with the star’s center of mass. (In reality, this is never the case. Observe the diagrams below.

The size of the white dots indicates the relative mass of the objects in orbit. The objects orbit the center of mass of the system, the barycenter, not the center of mass of the star. As the difference in mass between the two objects increases, the barycenter approaches the center of mass of the more massive object. In discussing Kepler’s law, the important thing is that we place the origin of our coordinate system at this barycenter, which is one focus of an ellipse.)

With the preliminary qualifications of the model out of the way, let us turn to the argument itself. (Please note I am keeping the tone informal and conversational; this is not a formal mathematical proof.) Every argument begins with a set of assumptions (not to be restricted to the colloquial usage of “assumption”, as argumentative assumptions may be empirically sourced facts). We shall assume that angular momentum is conserved, a fact which can be explained in the following way. First, note that the mass scalar () multiplied by the acceleration vector () equals the force vector () (Newton’s Second Law of Motion):

Second, recall Newton’s Law of Gravity, which states that the force equals the gravitational constant multiplied by the two objects’ masses, divided by the square of the distance, and then multiplied by the position vector:

So these two expressions of force are equal to each other:

Divide both sides of the equation by the mass of the planet to simplify the formula:

Now it is easy to see that you can take that quotient to the left of the position vector and collapse it into a constant. So the acceleration vector equals the position vector times a constant:

And vectors which are multiples of each other are parallel. The cross product of parallel vectors equals a zero vector:

which is

Now, take a moment to consider the derivative of the cross product of the position vector and the velocity vector:

This derivative equals two summands: velocity cross velocity, and position cross acceleration:

The cross product of identical vectors is a zero vector, so the augend is zero:

And we just concluded above that the addend is zero. So the sum is a zero vector:

And if a function’s derivative is zero, then it is a constant function. We shall call this constant vector : because it is their cross product, is perpendicular to the position vector and the velocity vector. So the orbiting objects move orthogonally to . Since never changes, the objects’ movements are restricted to a plane. The orbit is never warped in a third dimension; in other words, angular momentum is conserved.

With the establishment of the momentum conservation assumption, we can enter the heart of the argument. The vector function denoting the orbiting object’s position can be expressed in polar coordinates as cosine of the angle () times the unit vector plus sine of the angle times the unit vector all multiplied by the length of that position vector:

Next, we find velocity: . Then find the cross product of the position and velocity vectors. If you do the algebra, you should see that it equals the distance squared times the time derivative of the angle times unit vector :

Well, earlier we had already decided that this cross product equals , so this expression also equals :

Because is just a unit vector, the distance of is squared, times the derivative of the angle:

Now consider the variable angle at two specific angles, and . We know that the area bounded by these angles should be half the integral of squared times the differential of from to :

If we set our clock to zero when the angle equals , and one time unit later the angle reaches , we can re-write the integral like this: half the integral of squared times the time-derivative of the angle times the differential of time from to :

We already learned that the integrand is the length of :

So this integral equals half of multiplied by the time difference:

Now you can easily see that for any two time intervals of equal length, the area is the same, as swept out by the distance line from the orbiting object to its barycenter. Hence the famous refrain “equal areas in equal times.” But equal areas mean unequal arc lengths along the ellipse traveled in equal times, and hence the planets change their orbital speeds.

So how much of this argument relies on empirical observation, and how much on armchair reasoning? You can see that the main body of the argument relies on properties of vectors. So let us go back further, to the assumptions. Again, we used vector properties to obtain conservation of momentum, and the main argument’s deductions are a consequence of this conservation, but we really started with Newton’s Second Law of Motion and his Law of Gravity. Note that most of the empirical details, such as the masses of objects and the value of the gravitational constant (), disappear into the constant we labeled . It is not only that the values of the variables and constant are irrelevant; even much of the detail in the law itself is abstracted into one simple term. I did not expect that so much of Kepler’s Second Law relies not on the contingent properties of gravitation, but on the geometry of vectors, which is basically logic.

— Ander Nesser, the 29th of April, 2017

References:

https://plato.stanford.edu/entries/kepler/#CopRefThrPlaLaw

Newton’s original proof is in Book 1, Section 2 of his *Principia*: http://www.17centurymaths.com/contents/newtoncontents.html

The mathematics of the above argument is based on Lecture 14 of the course *Understanding Multivariable Calculus* by Prof. Bruce Edwards, University of Florida: http://www.thegreatcourses.com/courses/understanding-multivariable-calculus-problems-solutions-and-tips.html

The images of orbits are drawn with Celestia: https://celestiaproject.net/

The barycenter animations are provided by Wikipedia: https://en.wikipedia.org/wiki/Barycenter

The gif animating Kepler’s Second Law was made by Antonio González Fernández of the Engineering School of the University of Seville: https://en.wikipedia.org/wiki/File:Kepler-second-law.gif