Calculating Orbits: Kepler’s First Law

Kepler’s First Law

Kepler’s First Law tells us that planetary orbits are ellipses.  The following derivation, taken from Wiesel, is very clever: it solves a second order differential equation as if by magic.  We begin with our second order differential equation, which is an expression for the acceleration of a planet taken from Newton’s Law of Gravity:

We will now cross each side of this equation on the left with the vector $\vec{H}$:

Consider the left side of this equation and note the following

because $\vec{H}$ is constant with respect to $t$.

Now consider the vector portion of the right hand side

By the bac-cab formula

Clearly, $\vec{r}\cdot\vec{r}=r^2$.  It is also true that $\vec{r}\cdot\vec{r}=rDr$.  This is clear when we resolve $\vec{v}$ into a coordinate system in which $\vec{r}$ lies along one of the axes.  It follows that

Therefore,

It is at this point in the derivation that the god comes out of the machine.  Note

so

which can be written as

Therefore, the vector whose derivative we’ve taken is constant.  Choose the vector $\vec{e}$ so that

and as a consequence

There is yet another god in the machine, however.  Dot both sides of this equation with $\vec{r}$ so that

By the application of the triple scalar product, the left hand side may be transformed as follows

If we let $\nu$ be the angle between $\vec{r}$ and $\vec{e}$, the right hand side becomes

Therefore,

and consequently,

Calculating Orbits: The Ellipse

The Ellipse

The ellipse is defined in the following way.  Let $F_1$ and $F_2$ be two distinct points on the plane.  Each will be referred to as a focus and together they will be referred to as foci (fo-sai).  An ellipse is a set of points $P$ such that the sum of the distance from $P$ to $F_1$ and the distance from $P$ to $F_2$ is a constant particular to the ellipse.   The language become less cumbersome when we introduce a coordinate system, so we shall.

Let the $x$-axis run through the foci so that the point $x=0$ is the midpoint between the foci. Choose orientation so that $F_1=(c,0)$ and $F_2=(-c,0)$.  Let $H$ be a point on the positive $x$-axis that is on the ellipse, say $H=(a,0)$. Note that the distance from $H$ to $F_1$ is $a-c$ and from $H$ to $F_2$ is $a+c$.  The sum of these two distances is $2a$. Note that $a$ is the length of the segment from the origin $O$, which is the center of the ellipse, and $H$.  This is called the semimajor axis of the ellipse.

Now let $P=(x,y)$ be a point on the ellipse; let $D_1$ be the distance from $P$ to $F_1$; let $D_2$ be the distance from $P$ to $F_2$.  Then

Now $D_1+D_2=2a$, so $D_1=2a-D_2$.  We can square both sides and rearrange so as to determine

As this difference is also

we can see that

This can be rearranged to

and squaring both sides will yield

which can be reduced and rearranged to

As $a^2-c^2>0$, we can choose $b>0$ so that $b^2=a^2-c^2$.  We now can define that important quantity $e$, referred to by the name eccentricity, by

For convenience sake, let us also quickly note that

Our aim now to put this into an appropriate polar form. In our application of this, we will want the sun to be at the focus $F_1$.  Consequently, we will perform a translation of the plane along the $x$-axis so that it is.  The foci $F_1$ and $F_2$ will have coordinates $(0,0)$ and  and $(-2c,0)$, respectively, while the center of the ellipse will be at $(-c,0)$. Taking the equation we obtained for the ellipse and replacing $a^2-c^2$ with $b^2$  we have

We now multiply both sides by $a^2b^2$, expand (x+c)^2, and distribute to obtain

In the coefficient for $x^2$, replace $b^2$ with $a^2-c^2$, carry out the multiplication, and rearrange the equation to be

We are aiming to put this into polar coordinates, so we replace $x^2+y^2$ with $r^2$.  A little algebra will allow us to replace $a^2b^2-b^2c^2$ with $b^4$. Therefore our equation becomes

That is to say

By taking the square root of both sides, we get that either $ar=cx-b^2$ or that $ar=b^2-cx$; that is $ar-cx=-b^2$ or $ar+cx=b^2$.  Let us now substitute $rcos (\theta)$ for $x$, $ae$ for $c$, and $a^2(1-e^2)$ for $b^2$.  In that case, the first alternative will give us

where the LHS is always positive and the RHS is always negative.  However, the second alternative will yield

and so