Calculating Orbits: Relating E to Nu

Relating E to Nu

Relating $E$ to $\nu$ is a problem of trigonometry.  We will be referring to the diagram

which we have extracted (with slight modifications) from one of our previous diagrams wherein we juxtaposed an ellipse and a circle.  Recall that $P$ was a point on the ellipse, $P’$ a point on the circle, $S$ the sun (one focus of the ellipse), and $C$ the center of both the circle and the ellipse.  The point $Q$ is the foot of the perpendicular line through $P$ and $P’$ to the major axis.

In this section, we will use the trigonometric identity

In the diagram, $QF=-r\cos \nu$.  Recalling our formula for $r$ from a previous section, this becomes

We may also observe $CQ=a \cos E$ and $CF=ae$, so noting that $CF=CQ+QF$, we get

Solving this equation for $\cos E$ gives us

From basic trigonometry, we may calculate that $QP’=a\sin E$ and $QP=r \sin \nu$.  Let us be mindful that $P$ is on the ellipse and $P’$ is directly above it on the circle, and that the first can be obtained from the second by shrinking by a factor of

So

Solving for $\sin \nu$ and substituting for $r$ will give us

It is an easy series of calculations (he wrote with a malicious grin) to see

and

and

Using our formula for the tangent of the half-angle, we may combine our formulas for $1-\cos \nu$ and $\sin \nu$ to obtain

Some algebra and another application of the tangent of the half-angle formula will reveal

Calculating Orbits: M and E, Kepler's Equation

M and E, Kepler’s Equation

In this section, we will at last have occasion to refer to Kepler’s Second Law which states that the segment from the sun to the planet sweeps out equal areas in equal times. This is to say that the area swept out is proportional to time, exactly as if the ellipse were a circle and exactly as if the planet were moving at a constant rate.

Before we explore this too deeply, let us discuss the matter of calculating the area of an ellipse and regions of an ellipse by making a relation to the circle.

Let us imagine our ellipse as having a semimajor axis of length $a$ and eccentricity $e$, and of sharing a center with a circle of radius $a$.   We may impose a coordinate system wherein both have a center at the origin.  In this case the circle and the ellipse have the following equations respectively:

and

It is a simple calculation to show that if $(x_0,y_0)$ is a point on the circle then

is a point on the ellipse.  This multiplication of the $y$-coordinate corresponds to a vertical shrinking of the plane by a factor of

Suppose that $R_{CL}$ is some region of the circle and that $R_{EL}$ is the region of the ellipse obtained by this vertical shrinking.  We may approximate the area of $R_{CL}$ by covering it with rectangles whose sides are parallel to the $x$- and $y$-axes, each of which has some area

After the vertical shrinking, each area will have an area of

Given an appropriately developed theory of area, which is beyond the scope of this work, it is easy to conclude that the area of $R_{EL}$ is

times the area of $R_{CL}$.

From this it follows that the area of an ellipse is

Having gotten a handle on the area of an ellipse, let us now relate it to Kepler’s Second Law. The value of Kepler’s Second Law resides in its relating area to time.

For a circle, the area of a sector can be used found by the formula

By allowing $\theta$ to vary from $0$ to $2\pi$, we can cover the whole circle.  For our purposes, it will serve us well to think of $\theta$ as a linear function of $t$.

Then if we set

inserting the vertical shrinking factor of

we obtain a function which gives the area swept out by the segment from the sun to the planet in terms of time.  Equal areas will be swept out in equal times, as promised by Kepler. It all depends on the value of $n$.

If we let $T$ denote the period of the planet, we know two things

and

Therefore,

Solving for $n$ gives us

So

I do not simplify this because I am interested in $\frac{2\pi}{T}t$ as a separate expression.  

We will now define $M$, the mean anomaly, as

This $M$ is the key piece in determining $\nu$ in terms of time.

Let us now recall our circle-ellipse pair.  We will ignore the coordinate system in favor of the synthetic approach.

We will determine an expression for $A_{\nu}$, the area of the region of the ellipse bounded the angle clockwise from the ray from $S$ to $H$ to the ray from $S$ to $P$.  Let $A_E$ be the area of the circle bounded by the angle from the ray from $S$ to $H$ to the ray from $S$ to $P’$; this latter ray is not pictured in our diagram.  However, let us note that

as the region in the circle can be shrunk to the second by multiplication by our shrink factor.

We may find the value of $A_E$ by reducing the area of the sector of the circle bound by the ray from $C$ to $H$ on one side and the ray from $C$ to $P’$ on the other by the area of the triangle $SCP’$.  The area of the sector is

and the area of the triangle is

so

A comparison of this to the equations for $A(t)$ and $M$ will give us

Given a calculated value for $M$, the most practical way to solve for $E$ is to make use of Newton’s Method.

In the next section, we will relate $E$ to $\nu$.

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