## Orbits: the elliptical orbit

I want to show you the geometry of an elliptical orbit.

Most of this post is reference, but note that it includes a calculation of the Hohmann transfer orbit between two circular orbits. We measure the angle $\nu\$ from the horizontal, counterclockwise. All our training in trigonometry says we should label the horizontal line as the x-axis — and I will subsequently do so, but it’s not essential.

What is essential is that periapse (closest point) is always $\nu\ = 0$, apoapse (farthest point) is always $\nu\ = \pi\$ , a positive angle $\nu\$ is always measured counterclockwise from periapse (and, of course, clockwise for a negative value).

Oh, and it uses the occupied focus F rather than the empty focus (not shown here) or the center of the ellipse. That is, the origin is at the occupied focus — which is a change from our customary practice of putting the origin at the center of the ellipse.

Let’s look at the geometry of that ellipse: The major axis is of length 2a, so a is the semi-major axis.

The distance between the foci is 2c, so the center of this ellipse is at (-c, 0) and the empty focus F’ is at (-2c, 0).

Periapse rp is the distance of closest approach, and it occurs at $\nu = 0\$. Apoapse is the furthest distance on the orbit, and it occurs at $\nu = \pi\$. Incidentally, the two, periapse and apoapse, are collectively called the apsides.

The angle $\nu\$ is called the true anomaly.

(And, for an orbit about the earth, we speak of perigee and apogee; for an orbit about the sun, perihelion and apohelion.)

We see that

ra + rp = 2a.

“p” denotes the “semi-latus rectum”, also called “the parameter”. We see that it is the distance at $\nu = \pm \pi/2\$.

There is another appropriate drawing. From this drawing, we have $a^2 = b^2 + c^2\$.

Note that this is not the usual way we write the Pythagorean theorem.

“b” is the semi-minor axis of the ellipse.

The two lines marked “a” are equal by symmetry… and they are each of length a because:

• the sum of the distances from the two foci is constant;
• looking at the apsides, that constant is in fact 2a (2 a = rp + ra)

The base of the blue triangle is of length c, again by symmetry: it is half of 2c, the distance between the two foci.

Now, what are all the things we know about this ellipse? And how do we know them?

The orbit is… $r = \frac{p}{e \cos (\nu )+1}\$.

No matter how many auxiliary parameters we introduce, remember that the orbit is defined by just two parameters. We just have a lot of choices for which two parameters we use.

We know that $\nu = 0\$ and $\nu = \pi\$ are periapse and apoapse because they are the maximum and minimum values of $cos(\nu)\$ respectively: $r_p = \frac{p}{1+e}$ $r_a = \frac{p}{1-e}\$.

(Yes, we could have used calculus to derive those without using our intuition.)

We know geometrically that…

ra + rp = 2 a

and this will give us a relationship between p, e, and a: That is, $a=\frac{p}{1-e^2}\$.

One way to remember what symbols are where is to consider the limit as e -> 1, i.e. as our ellipse tends to a parabola. p is finite, but the ratio tends to infinity, as does a.

Can we find the geometric meaning of the eccentricy? (Would I ask if we couldn’t?) Let’s see. The distance between the foci is 2c, and we see that

2c = 2a – 2 rp.

Substitute for periapse… So now we have a simple geometric interpretation of the eccentricity e, as the ratio of c to a… or, take the distance between the foci (2c) and divide by the major axis (2a):

e = 2c / 2a = c / a.

Let me show you something cute that may or may not be useful. (Well, two of the following three are certainly useful.) We can rephrase

2 a = rp + ra

as $a=\frac{\text{ra}+\text{rp}}{2}$

and then we can say that a, the semi-major axis, is the arithmetic mean of the apsides.

Wow. Not.

But let’s compute the harmonic mean of the apsides: $\frac{2}{\frac{1}{\text{ra}}+\frac{1}{\text{rp}}}\$. Well, well. The harmonic mean of the apsides is p.

At this point I have no choice; I simply have got to compute the geometric mean of the apsides. Hey, that’s b.

So, given the apsides ra and rp:

• their arithmetic mean is a;
• their harmonic mean is p;
• their geometric mean is b.

“b” is the one that may or may not be useful.

What is particularly useful is this: if we knew ra and rp, we could get a and p as the arithmetic and harmonic means… then we could get e from… $p=a \left(1-e^2\right)$

and we would have the orbit.

## The Hohmann (minimum energy) transfer orbit

This in fact is a standard computation. Suppose we have a spacecraft in a circular orbit at one altitude, and we want to find the (minimum energy = Hohmann) transfer orbit to another altitude. (I’m not going to prove that it’s the minimum energy transfer, but when you see it, I think you’ll find it plausible. Actually, seeing this the next day, I think the proof is clear from the picture. Since the energy of an orbit goes as -1/a, any shorter transfer orbit won’t get there, and any longer transfer orbit has more energy.)

We’re going to find the “nicest” elliptical orbit connecting two given circular orbits. Let’s keep it easy: one orbit is at 1 DU altitude, the other is at 2 DU altitude, and the primary is shown as a disk of radius 1 DU. (I told you canonical units let us use nice numbers.) This means that the orbits have radii 2 and 3 (altitude + radius of primary).

Let me remark that I had not realized when I wrote this that it is, in fact, an example from pp. 165-166 of BMW. How do we find an elliptical orbit tangent to each of those circles? Well, it has periapse equal to the radius of the smaller orbit, and apoapse equal to the radius of the larger orbit.

So we are given ra = 3 and rp = 2 and we want p and e.

a is the arithmetic mean: a = 5/2.

p (which comes from h) is the harmonic mean: then from p=a (1-e^2)… There were two solutions, but I took the positive one, so our orbit is… $r = \frac{12}{5 \left(\frac{\cos (\nu )}{5}+1\right)}\$.

Add it to the drawing of our circular orbits: Beautiful. (OK, just my opinion.)

Oh, while we’re here, let’s compute the speeds at the transititions.

The key to the computation is that the r and v vectors are perpendicular at the apsides. (For this purpose, every point on a circular orbit is both periapse and apoapse.) This means that the magnitude of h is the product of the magnitudes of r and v; the sine of the angle between them is 1.

I’m going to use the following equations. In general, in canonical coordinates,

h = rp vp = ra va;

p = h^2;

and for a circular orbit

r = p.

In fact, we will be using $h = \sqrt{p}$

and $v = h/r\$.

We have three values of p: 2, 3, and 12/5. We get three corresponding values of h by taking square roots (each of our three orbits has a different specific angular momentum): $h = \{1.41421, 1.73205, 1.54919\}\$.

(The last number is for the elliptical orbit.)

And then we get four values of v by dividing each h by an appropriate distance r; at each of the two transition points, there are two speeds — one for the circular orbit, one for the elliptical orbit. There are only two distances, rp and ra for the ellipse are also the radii of the two circles.

For the smaller circular orbit, r = rp; we have

h = 1.41421,

rp = 2,

and the speed is the ratio: v = 0.707107 .

For the elliptical transfer orbit at its periapse, r = rp; we have

h = 1.54919,

rp = 2,

and the speed is the ratio: v = 0.774597 .

For the elliptical orbit at its apoapse, r = ra;

we have h = 1.54919,

ra = 3,

and the speed is the ratio: v = 0.516398 .

Finally, for the larger circular orbit, r = ra;

we have h = 1.73205,

ra = 3,

and the speed is the ratio: v = 0.57735 .

That is, we were initially moving in a smaller circular orbit with speed .707… we switch to the elliptical transfer orbit by (instantaneously) increasing the speed to .775… at apoapse on the transfer orbit, our speed has fallen to .516 . Then we increase our speed (again, instantaneously) to .577 — and that puts us in the larger circular orbit.

So, we accelerate at both transitions, but the net effect is a lower speed. We have both increased our energy, but also traded kinetic enegy for potential energy.

Yes, our three orbits have three different values of a: 2, 2.5, and 3, and therefore three different values of energy. We could confirm that the changes in kinetic energy at the transitions coorespond to the changes in total energy… but I’ll leave it to you.

While I’m collecting equations and pictures, let me remind you of the definition of the flight path angle.  $\gamma\$ is the angle between r and v; $\phi\$ is the flight path angle.

Because $\phi + \gamma = 90{}^{\circ}\$, we are replacing the scalar equation (for the magnitude of the cross product) $h = r\ v\ sin(\gamma)$

by $h = r\ v\ cos(\phi)\$.

(To be more specific, $\gamma\$ is the angle between r and v; by using its complement $\phi\$, we get to write the magnitude of h as though it were some kind of dot product — but the fact remains that $\phi\$ is not the angle between r and v, and h is not the dot product of r and v.)

They say, plausibly, that the sign of $\phi\$ will be the sign of r.v, which is also the sign of $cos(\phi)\$. I infer that if $\gamma > 90{}^{\circ}\$, then $\phi\$ will be chosen negative, so that we still have $\phi + \gamma = 90{}^{\circ}\$.

## Summary:

We have the following dynamical equations:

The specific (unit mass) angular momentum is $\vec{h} = \vec{r} \times \vec{v}\$, and it is constant.

The magnitude of specific angular momentum can be written $h = r\ v\ sin(\gamma) = r\ v\ cos(\phi)\$,

where $\gamma\$ is the angle between the h and v vectors, and $\phi\$ is the flight-path angle, defined by $\phi + \gamma = 90{}^{\circ}\$ ( $\gamma\ \text{ and }\phi\$ are complements).

It is very valuable to know that $\gamma = 90{}^{\circ}\$ at the apsides.

The specific mechanical energy is $\mathcal{E} = \frac{v^2}{2} - \frac{\mu}{r} \$, and it is constant.

The vector equation of the orbit is $\dot{\vec{r}} \times \vec{h} - \frac{\mu\ \vec{r}}{r} = \vec{B}\$.

The scalar equation of the orbit is $r = \frac{h^2/\mu}{1 - B/\mu\ cos(\nu)}\$.

By setting $e = B/\mu\$

and $p = h^2 / \mu \$,

we get what I might call a geometric equation of the orbit. If the orbit is an ellipse, then e < 1 (if a circle, e = 0).

We might say that those two equations relate the dynamics to the geometry. There is another such equation: $\mathcal{E} = -\frac{\mu}{2\ a}$.

We have the following geometric equations for an elliptical orbit.

The eccentricity is the distance between the foci divided by the major axis:

e = 2 c / 2 a = c / a.

We have an equation relating p, a, and e: $p = a\ (1-e^2)$

The semi-major axis a is the arithmetic mean of the apsides:

ra + rp = 2 a.

a, b, c form a right triangle: $a^2 = b^2 + c^2$

We can find ra and rp from the scalar orbit:

rp, ra = $\frac{p}{1 \pm e}$

p is the harmonic mean of the apsides;

b is the geometric mean of the apsides.

Whew! I hope I didn’t forget anything we’ll need down the road.

Many of these equations are true for a hyperbolic orbit; in fact, I think only one needs to be altered (the one for b). We’ll review this when we get there.

### 2 Responses to “Orbits: the elliptical orbit”

1. sunadu Says:

in thus cases are considered many aspects observed prof dr mircea orasanu and prof drd horia orasanu so that followed for LAGRANGIAN OPERATOR AND CONSTRAINTS OPTIMIZATIONS important points views Johannes Kepler, working with data painstakingly collected by Tycho Brahe without the aid of a telescope, developed three laws which described the motion of the planets across the sky.
1. The Law of Orbits: All planets move in elliptical orbits, with the sun at one focus.
2. The Law of Areas: A line that connects a planet to the sun sweeps out equal areas in equal times.
3. The Law of Periods: The square of the period of any planet is proportional to the cube of the semimajor axis of its orbit.
Kepler’s laws were derived for orbits around the sun, but they apply to satellite orbits as well.
he quantity T2/a3 depends upon the sum of the masses of the Sun and the planet, but since the mass of the Sun is so great, adding the mass of the planet makes very little difference.
The distance from one focus to any point on the ellipse and then back to the second focus is always the same. A planet’s orbital speed changes, depending on how far it is from the Sun. The closer a planet is to the Sun, the stronger the Sun’s gravitational pull on it, and the faster the planet moves. The farther it is from the Sun, the weaker the Sun’s gravitational pull, and the slower it moves in its orbit.

2. marthalindeman Says:

A note of thanks! Mathematics does not get old, and this page pulled together several pieces of info into a very useful whole.