Kepler’s Impossible Equation

[jmr]

To clarify: There is no *analytical* solution for this equation with a finite number of terms. "Analytical" means "Can be manipulated via algebra to place a single 'E' term on one side of the equation."

The equation can be solved for E via an infinite series. Additionally, there are numerous iterative methods that produce a value of E given a value of M (via "root finding").

Nevertheless, the lack of an analytical solution makes rendezvous calculations far, far more difficult than is desirable. This is because M ("Mean Anomaly") provides the *only* method to generate "time since periapsis, " and if you are attempting to rendezvous two objects it is kinda important to know that the two objects reach the same point at the same time.

[charlespeterson]

When I was in college in the early 80's, I had found some equations in an astronomy magazine that I was able to bash up into an orbit simulator for the Apple II. I tried to take the next step and create some code that would let me jump orbits ahead by some arbitrary amount of time... I could never figure it out. For 40 years I've just figured I was too dumb; now I see that maybe it was because I actually picked a hard problem.

[simcityman]

Hmmm, ok…..I know some of these words.

[youtube]

if only motion were so simple we could solve the 3+ body problems

[iteerrex]

What it actually took to go from circular to elliptical orbits, as I learned from that two part documentary, is just nutz.

[revolutionofmedicine]

I definitely read the title of the paper as, "On the Bestest Solution of Kepler's Equation."

[nattananchunbunluesook]

“Oh okay…”
“Umm….”
“Sorry what are we talking about again?”
“I…. think I’m not supposed to be in this FYP”

[jeremydening]

Iterative methods can be used to solve this. You can use a Newton Raphson method, a fixed-point iteration, or a zeroth order method like line search to get the value of E. The Newton Raphson method can get you a good value in just a few iterations.

[JivanPal]

I thought, "uhh, that's easy, M = E (1 - e sin), so E = M / (1 - e sin)..." massive facepalm.

I swear, I have a Master's in mathematics.

[pootca]

Stop reminding me of the headaches I got from trigonometry

[samorgan]

I can't believe how incredible your videos are!

[kolbyking]

I actually made my own approximation in a spreadsheet, so I could calculate orbital behaviors in Kerbal space program. It's only accurate within 2%, but that's good enough for my use case.

[tinytroll]

The issue with his equation is it requires algebra manipulation otherwise the time constant is broken

[phospenguillite]

That is some cool visuals! I wonder what you're talking about though?

[helicalactual]

wow!!!! THANK YOU you inspired me with an idea!

[davidgillies]

There's no analytic solution. The problem isn't numerically calculating E(M) as such, but stability of the solutions.

[stewartbugler]

Something that caught my attention... is that E is simply the middle point between the furthest and nearest orbit.

So ultimately the distance of the nearest approach is porportional to its furthest reach. The close the approach the faster but more elliptical the orbit will be. That ellipse only occurs because the larger body is off centre.

In the grand scales everything is off centre because of each other... the combinational small objects influence the giant object thats influencing them.

Kinda like the chicken or the egg... but the answer is the Chicken not because it laid the eggs... but as its simply the more complex of the two. It may have came from something like the eggs it lays but it certainly was a chicken before the question was relevant.

[Robzem-mj]

It will be where it will be. There, i solved it.
Where do i pick up my doctorate?

[Khashayarissi-obyj]

With luck and more power to you.
hoping for more videos.

[AmanSingh-nwlw]

Thanks, helpful in my satellite communication course