Orbital Circumference

[Old Duck]

You guys are smart, so tell me, how do I calculate the circumference of a planet's orbit based on the various orbital data provided us? I want to do this using the data recorded in our journal, so any formula will have to take into account whatever units the journal data is recorded in. I'd like my results to me in kilometers. This would allow me to generate more "meaningful" stats for systems of interest - things like how fast a planet is orbiting its parent body. For example, a moon that orbits a large gas giant in a day is more interesting to me than a moon that orbits its very close twin in half a day.

Someday I'll take the time to learn all of Kepler's equations for myself (I'd like to eventually make my own Orrery generator), but for now I'm looking for an easy answer. Thanks!

 

[Vetinari]

Probably missing the whole thing and simplifying but.

you have the distance of the planet from the star, its radius, so 2PixR is the C of the planet orbit?

Moons not a clue, do we get distance from planet/parent as a separate stat? You get the orbital period though so you can find a moon that orbits its GG once day from that?

 

[Old Duck]

Probably missing the whole thing and simplifying but.

you have the distance of the planet from the star, its radius, so 2PixR is the C of the planet orbit?

Moons not a clue, do we get distance from planet/parent as a separate stat? You get the orbital period though so you can find a moon that orbits its GG once day from that?

Orbits are elliptical, so the calculation must be more complex than this.

 

[Orvidius]

The easiest approximation is to just use the semi-major axis. If the orbit is anywhere close to circular, that's a good estimate. This number is basically half of the diameter on the long axis of the ellipse (like a radius). It functions well as an "average" distance from the parent object, sort-of, but it's not an average over time.

EDIT: The reason it isn't a time-based average is due to the varying orbital velocities in an eccentric orbit. The smaller body flies by very rapidly at periapsis (the nearest point in the orbit), but spends a huge amount of time moving slowly near apoapsis (the furthest point in the orbit).

EDIT 2: I misread this as asking for the radius, rather than circumference. Doh!

 

[Para Handy]

... (content deleted by me) ...

 

[iain666]

Orbits are elliptical, so the calculation must be more complex than this.

We have the semi-major axis (a) and the eccentricity (e) so we can calculate the semi-minor axis (b):

b = ((ea)^2 - a^2)^0.5

From that you can calculate h:

h = (a-b)^2 / (a+b)^2

The perimeter can then be approximated with the first few terms of an infinite series:

p = π(a+b)(1 + h/4 + h^2/64 + h^3/256).

The third term is from the sum for n=0 to infinity of [0.5 n]^2 x h^n (that square bracket thing is supposed to be the binomial coefficient with half integer factorials)

Just treating it like a circle with a radius of the semi-major axis unless the eccentricity is extreme seems like a lot less effort though

(edited to correct my silly mistake, with thanks to @Iron Body , add maths to flying on the list of things not to do drunk)

 

[Orvidius]

Yeah I misread the question as asking for a radius rather than circumference. So yeah, there are those equations.

 

[schlowi123]

So good you've asked that question old duck. Here I go into detail about that particular problem and present a rather simple, yet extremely good approximation for what you've asked for (the equation discovered by Srinivasa Ramanujan). I totally recommend reading that short older entry

 

[schlowi123]

And by the way, here are the equations regarding the orbital velocity … if you are interested in that.

 

[Old Duck]

Yeah I misread the question as asking for a radius rather than circumference. So yeah, there are those equations.

FWIW, I would be happy with a radius, as I'm currently using this more for comparative purposes. I also don't need exact numbers, so ballpark figures work. So, using simple grade school math, how can I find the approximate radius using all the different variables ED gives us?

 

[Iron Body]

We have the semi-major axis (a) and the eccentricity (e) so we can calculate the semi-minor axis (b):

b = ((ea)^2 - a^2)^0.5

From that you can calculate h:

h = (a-b)^2 / (a+b)^2

The perimeter can then be approximated with the first few terms of an infinite series:

p = π(a+b)(1 + h/4 + h^2/64 + h^3/256).

The third term is from the sum for n=0 to infinity of [0.5 n]^2 x h^n (that square bracket thing is supposed to be a 1 column, 2 row matrix)

Just treating it like a circle with a radius of the semi-major axis unless the eccentricity is extreme seems like a lot less effort though

The [0.5 n] is not a matrix, it is a "binomial coefficient with half-integer factorials".
Don't be misled by the 4,64,256 denominators, the next two terms are 49h^4/16384 and 441h^5/1048576.

The best Ramanujan formula is p = π(a+b)(1 + 3h / (10+sqr(4-3h) ) )

I looked at some numbers in the log.
For "semi-major axis" = 1409838438034.0576 is equated to 9.42 AU.
So I the s-m-a is measured in metres. Divide the answer by 1000, to get km.

ref: https://www.mathsisfun.com/geometry/ellipse-perimeter.html

 

[Orvidius]

FWIW, I would be happy with a radius, as I'm currently using this more for comparative purposes. I also don't need exact numbers, so ballpark figures work. So, using simple grade school math, how can I find the approximate radius using all the different variables ED gives us?

Yeah, like I said, for radius you can use the semi-major axis as an estimate. This will be exactly the radius for zero eccentricity, and will be less accurate the more eccentric the orbit is. But for "ballpark" it's OK. The perimeter is what's more messy.

Probably one of the simpler estimates for the perimeter (circumference) is this one (which is usually within 5% of correct, as long as "a" is no more than 3x the length of "b"):

perimeter = 2 * pi * sqrt((a^2 + b^2) / 2)

or expressed another way:

perimeter = 2 * pi * ((a^2 + b^2) / 2)^0.5

...Where "a" is the semi-major axis, and "b" is the semi-minor axis. That is, half the long axis, and half the short axis (two separate radii). The SMA is one of the parameters that ED provides. While ED doesn't provide the semi-minor axis (b), it does give you the eccentricity. Using these you can calculate "b" using "a" (SMA) and "e" (eccentricity):

b = a*(1 - e^2)^0.5

... and then that can be plugged into the equation above.

EDIT: Check out this page for a simple explanation of some methods for estimates. Also, there are online calculators, like this one.

 

[Old Duck]

Just so I understand what you are all saying, the semi-major axis IS the radius, measured in meters. I'm guessing it's the radius at the orbit's aphelion, being "major". This is also when the planet will be moving slowest, if I remember my grade school science from all those many years ago. So if I'm looking for "zippy" planets, then I might actually be more interested in the semi-minor axis, which IIRC is the radius at perihelion.

For right now, I'm basically looking to parse my logs for interesting planets to revisit, which include those that fly through the sky quickly.

Anywho, thanks everyone for your help - I will bookmark this thread and refer back to it as I try some more advanced stuff later on!

 

[Orvidius]

Getting close.

The semi-major axis is half of the long axis of the ellipse. But remember that the larger body is at one of the foci of the ellipse. For an eccentric ellipse, that means that the SMA is quite a bit longer that than the distance between the two bodies at closest approach (periapsis, perihelion, perigee, depending on what you're orbiting), and also quite a bit shorter than the distance between the bodies at apoapsis/aphelion/apogee).

But it will be exactly the radius when it's a true circle.

Here's a diagram to help. Note that the sun is pretty close to one end of the ellipse, and so the orbital "radius" varies quite a bit. The SMA is a good approximation of an "average" distance, but that exact distance will not apply at either apoapsis or periapsis.

https://www.astronomynotes.com/history/ellipse.gif

(As an aside, apoapsis/periapsis are generic terms, whereas apogee/perigee refer to orbiting the Earth, and aphelion/perihelion refer to orbiting the Sun).