Calculating Orbits: Kepler's Second Law

Kepler’s Second Law

Our aim is to derive Kepler’s Second Law:

The radius from the sun to the planet sweeps out equal areas in equal times.

To put this in language better suited to mathematical manipulation, let $A$ denote the area swept out from time zero to time $t$.  Then Kepler’s Second Law saws $DA$ is a constant.  We shall show that not only is it a constant but a constant with meaning to physics.

In this derivation, we shall have occasion to use a formula derived from Newton’s Law of Gravity:

The acceleration of a planet is inversely proportional to the square of the distance from the sun and is directed toward the sun.

Mathematically:

or

So

as the angle from $\vec{r}$ to itself is zero and so $\sin \theta =0$.  Thus,

Let us also observe that

So that $\vec{r}\times\vec{v}$ is constant.  We will denote this constant by $\vec{H}$ and note that

where $\vec{L}$ is angular momentum.  We can think of $\vec{H}$ as being angular momentum per unit mass.

Now consider the following diagram of our two-body system as time $\Delta t$ has elapsed.

For small $\Delta t$

so that the altitude of this triangle is $v\sin \theta$ where $\theta$ is the angle from $\vec{r}$ to $\vec{v}$,  If $\Delta A$ is the area of this triangle, then

so that

where $H$ is the magnitude of $\vec{H}$.

Therefore,

which is Kepler’s Second Law.

Calculating Orbits: Angular Momentum and the Cross Product

Angular Momentum and the Cross Product

The physical concept of angular momentum is a key piece of our approach to the derivation of Kepler’s Laws.  Let us begin our study of angular momentum with a thought experiment. This thought experiment is low-tech enough for you to carry out in your backyard, if you should so desire.

In an ill-fated attempt to teach me the basics of hitting a baseball, my father created the following device:  He put a tennis ball in a sock and tied the sock to the end of a cord.  He then swung it around his head.  Even when he stopped moving his arm, the ball persisted in a roughly circular path for a time.  This persistence was caused by angular momentum.  (We will ultimately replace the ball by a planet and the string by the sun’s gravity.)

There are three things that affect the tennis ball’s angular momentum in this system: its mass; its distance from the center of rotation; its velocity.  Mathematically this is captured in an equation that I will first give and then carefully explain

The variable $\vec{L}$ denotes the angular momentum itself.  Note that it is a vector.  Its direction is the axis of rotation of our tennis ball.  The variable $m$ stands for mass, of course; it will enter into our calculations surprisingly little.  The vector $\vec{r}$ goes from our center of rotation to the tennis ball, as it were.  It will vary in magnitude and direction as it gives us the position of the moving object.  The vector $\vec{v}$ denotes the velocity of our tennis ball.  It is the derivative of $\vec{r}$ with respect to time.

Within this formula, you will notice the symbol $\times$, the cross product.  It is very tempting to give a full treatment of it here, but I will resist that temptation.  Instead, I will give a simple definition; a list of the results we need; and refer the interested reader to any decent calculus text.

Given vectors $\vec{Y}$ and $\vec{Z}$, we define a third vector $\vec{W}$ whose magnitude is equal to

where $\theta$ is the angle from $\vec{Y}$ to $\vec{Z}$.  The vector $\vec{W}$ lies in a line that is mutually orthogonal to $\vec{Y}$ and $\vec{Z}$ and has a sense so that the triple $\vec{Y},\vec{Z},\vec{W}$ preserves the right hand rule.

Let us now consider vectors $\vec{A},\vec{B},\vec{C}$.  We will need the following pair of formulas in the sequel.  The first is called bac-cab:

where $\cdot$ is the familiar dot product.  The second is called the scalar triple product:

We will also be taking derivatives.  I will, for the most part, eschew both the Newtonian and Leibnizian notation.  Instead, $D$ will denote the taking of a derivative with respect to time.  So $\vec{v}=D\vec{r}$, for example.  The usual product rules for differentiation hold for both the dot and cross products:

We will, as far as we can, try to follow the convention of using the plain letter to denote the magnitude of a vector, i.e. $r$ is the magnitude of $\vec{r}$ and $v$ is the magnitude of $\vec{v}$.  However, $a$ will never denote the magnitude of $\vec{a}$: the former is the length of the semimajor axis of an ellipse; the latter is acceleration.  This will not cause any confusion.

Let us now return briefly to

As the magnitude of $\vec{L}$ is

we can think of $v\sin \theta$ as being the portion of the velocity that is perpendicular to $\vec{r}$.  This will be useful at a crucial moment.