So, I've been doing endless research, and I've found a bit of information, all in different places, that is. Now, I'm getting really confused about it; I can't keep up with it all anymore, so, now, it's time I release it here. What are the Keplerian Elements, and how do you calculate them?

I'm particularly interested in getting an explanation of angular orbital elements, such as the longitude of the ascending node (as well as the ascending node itself), the angular momentum, the argument of perihelion, the mean and true anomaly, and some other variables like the Orbital Momentum Vector.

In an answer, I want a feasibly understandable explanation of what the orbital elements above are, and equations describing how to calculate them. If there are any variables that require further research to grasp, please do the same for them. Afterwards, at the end of a response, please link the source where you got it from (if there is any), but if you got it from in-person courses, which have no link, please ignore this part, and do not try to do treacherous research to try and find one.

Hoping to find an answer soon, thanks in advance!

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    $\begingroup$ I've tried that link, even before I posted this question. $\endgroup$
    – Questioner
    Apr 13 '21 at 17:52
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    $\begingroup$ Answering this question well is probably the equivalent of an introductory course on orbital mechanics. The calculations aren't widely posted, because they aren't very easy. There are many ways to calculate them, including starting with {az, el, time} observations, then estimating state vectors, and then orbital elements. Here is a paper with a "simple" method that also includes an example at the end: articles.adsabs.harvard.edu/pdf/1981A%26A...102...59N . $\endgroup$
    – Connor Garcia
    Apr 13 '21 at 19:38
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    $\begingroup$ @ConnorGarcia that's a pretty cool paper! It's the complete recipe for taking 3 or more Earth-based observations of a heliocentric orbiting body and constructing the 3D position vectors $\mathbf{E}$ and Earth observer unit vectors $\mathbf{e}$ via analytical propagation (Eq. 16, see also 1, 2). That's where the fitting is done. Then ex post facto the cartesian positions are finally converted to Keplerian elements via Eqs. 1-5. They even give an example at the end. Sweet! $\endgroup$
    – uhoh
    Apr 13 '21 at 23:34
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    $\begingroup$ I agree. That PDF contributed in answering my question greatly. $\endgroup$
    – Questioner
    Apr 14 '21 at 17:13
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    $\begingroup$ I have made a follow-up question about Launch Elements. Here's the link: Launch Elements $\endgroup$
    – Questioner
    Apr 27 '21 at 12:45

What Kepler (and others before and after him) wanted to do is predict where a planet would be. To do this we need some set up:

First we want a coordinate system. This is a system of axes: x, y and z, at right angles to each other. And it should be an inertial coordinate system, so Newton's laws work. This means that the axes should not be rotating. And we will be able to choose a coordinate system that is not moving relative to the centre of mass of the objects. Along with the coordinate system, we need a system of time, with some particular moment designated t=0 (and because this isn't relativity we shan't worry about clocks running at different speeds or curved space-time etc...).

So now we meaningfully ask "where is the planet at t=3 years" and understand the reply: "It is at coordinates (0.6, 1.2, 0.1) AU".

Now it turns out that to completely describe the future motion of the planet, as it obeys Newton's laws of motion and gravity, you need 6 numbers. You could, for example, give the position and velocity at t=0. Three coordinates for position and three for velocity would give sufficient information.

But this ignores facts that Kepler discovered about orbits (and Newton was able to prove from his laws of motion). The planet moves in an ellipse. So we can describe the orbit in terms of the ellipse. Ellipses are squashed circles and there are two special points called foci that are off center along the long axis of the ellipse. The centre of mass will be at one focus.

First there is the size of the ellipse. Now it turns out that the length of the long axis is important, since the period of the orbit (the year) is related to the length of the long axis. Conventionally we speak of the "semimajor axis" which is the half the length of long axis, or the distance from the centre of the ellipse to the furthest point. We label this $a$, and this is the first of our six numbers.

Next there is the shape of the ellipse. Ellipses can be fat and circular or thin and skinny. The nearly circular ones have foci that are close to the center. So we measure the distance from the centre to either focus: $c$. And then define the eccentricity to be $e=c/a$ All ellipses with the same $e$ have the same shape, in the sense that they are all mathematically similar. $e$ is the second number.

Now we need to describe the orientation of the ellipse in space. To do this we come back to our coordinate system. The x-y plane is a flat plane with the origin at the focus of the ellipse (remember that we chose coordinates with the origin at the the centre of mass). The ellipse won't be in the x-y plane but sometimes be above (in the sense of positive z-coordinate) and sometimes below this plane.

The two planes: the plane of the ellipse and the x-y plane, will be at some angle to each other. We call this angle the inclination. This is $i$ and is the third number. An inclination of 0 means the planet orbits in the x-y plane. An inclination of 90 means the orbital plane is perpendicular to the x-y plane. And an inclination of 180 means it orbits in the x-y plane on a retrograde orbit (to define retrograde we need some definition of positive and negative angles. Conventionally negative angles are taken to be clockwise).

To describe directions in the x-y plane, we will use angles. The x-axis will be at 0 degrees. And then we measure the angle from the x-axis, from 0 to 360 degrees. There will be a point in the x-y plane where the orbit crosses it, and where the planet's motion causes it to move from below the plane to above the plane. This is called the ascending node. It is a fixed point and the direction of this fixed point is called the longitude of the ascending node, $\Omega$. The inclination and $\Omega$ together define the plane that the planet orbits in.

We next look at the location where the planet is closest to the focus. This is called the periapse. This is an angle in the orbital plane. It is the angle between the ascending node and the direction of periapse. This is the argument of periapse and is labelled $\omega$.

These five numbers define the shape and orientation of the ellipse. They have been chosen partly for convenience, and partly for historical reasons. But the principle reason is that each describes a different aspect of the orbit. Unlike the state vectors where the three position coordinates are the same kind of value, each orbital element is distinct and meaningful.

The sixth element is simply the location of the planet at time $t=0$. This is an angle in the orbital plane, and there are different ways to describe this angle. The simplest is the angle between the periapse direction and the planet. This is the True Anomaly $\nu$ (nothing to do with anomalous data!) But sometimes you might see the Mean Anomaly. This looks like an angle, but is actually a measure of time. 0 means "at the periapse" and 360 means "one orbital period (year) later", so 90 means "one quarter of a year through the orbit". Because the planet doesn't move at a constant speed, a mean anomaly of 90 doesn't correspond to a true anomaly of 90 degrees.

Those are the six orbital elements:

$a$ size of ellipse
$e$ shape of ellipse
$i$, $\Omega$ position of orbital plane relative to x-y plane
$\omega$ orientation of ellipse within orbital plane
$\nu$ position of planet on ellipse

It is a standard problem in celestial mechanics to convert from 6 orbital elements to the state vectors of position and velocity, and convert back from state vectors to orbital elements. It is also a standard problem to take the six orbital elements, which give the true (or mean) anomaly at time 0, and calculate the true anomaly at an other time. But the maths involved is rather involved: See https://web.archive.org/web/20170810015111/http://ccar.colorado.edu/asen5070/handouts/kep2cart_2002.doc or https://web.archive.org/web/20160418175843/https://ccar.colorado.edu/asen5070/handouts/cart2kep2002.pdf to go the other way

For the solar system, it is conventional to take the x-direction to be in the direction where the plane of the equator crosses the plane of the Earth's orbit around the sun. And the y-direction to be perpendicular to it, in the plane of the Earth's orbit (so by definition the Earth has zero inclination). This points towards the constellation of "Pisces" but it does slowly move and 2000 years ago it pointed towards "Aries" and so is called "the first point of Aries", or (since the sun is at the position in spring) the vernal equinox, $\gamma$.

No discussion of orbital elements is complete without the Wikipedia image which describes them.

enter image description here

By Lasunncty at the English Wikipedia, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=8971052

  • $\begingroup$ Actually, I can get the calculations of orbital elements from Wikipedia, right? $\endgroup$
    – Questioner
    Apr 14 '21 at 13:03
  • $\begingroup$ Orbital elements are like a giant celestial coordinate system. If I got it wrong, please point out my mistakes. $\endgroup$
    – Questioner
    Apr 14 '21 at 13:05
  • $\begingroup$ They are not like a giant celestical coordinate system. They are like what I described in my answer. $\endgroup$
    – James K
    Apr 14 '21 at 14:01
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    $\begingroup$ Orbital elements are way of describing the size, shape, position and orientation of an ellipse in space, and the position of a body on that ellipse. The state vectors are based off the coordinate system. $\endgroup$
    – James K
    Apr 14 '21 at 14:08
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    $\begingroup$ @Questioner do you need it typed, single-sided, double-spaced? $\endgroup$
    – uhoh
    Apr 14 '21 at 21:53

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