Minmus Eccentric Synchronous Orbit

[0something0]

Note: this is a math-heavy question.

I am attempting to put an object in a synchronous orbit around Minmus with a periapses of 5 kilometers from surface (65 km from the focus) and an orbital period of one Minmus day (1 Kerbin day 5 hours 13 minutes  = 40380 seconds). Using a modified version of t Kepler's Third Law, we have

t² / a³  =  4π² / GM 

where  t is the orbital period, a is the semi-major axis, G is the universal gravitational constant and M is the mass of Minmus. This can be isolated for a (the semi-major axis) to get

a = ∛(GMt² / 4π²)

We want M = 2.6457580 * 10¹⁹ kg,and t = 40380 seconds,  which gives us 

a = ∛(G(2.6457580 * 10¹⁹ kilograms)(40380 seconds)² / 4π²) = 417800 meters

for the semi-major axis. From there, we can calculate the linear eccentricity c, since the periapses is the difference between the semi-major axis and the linear eccentricity.

417800 - c = 65000
417800 - 65000 = c = 352800

gives us a linear eccentricity of 352800 meters. From here, we can calculate rest of the needed orbital parameters.

Orbital Period (given) = 40380 seconds

Semi-major axis = 417800 meters

Periapsis (given) = 65000 meters

Apoapsis =  a + c = 417800 +  352800 = 770600 meters

Eccentricity = a / c = 0.84423169 

However, when all of these parameters are applied, the orbital period is actually shorter than the Minmus day, causing the periapsis to drift across the surface. Help!

Edited June 28, 2019 by 0something0

[OhioBob]

It's my understanding that Minmus' rotation period is 40400 seconds.  How much are you drifting?  Can the 20 second difference account for it?

[bewing]

3 minutes ago, OhioBob said:

It's my understanding that Minmus' rotation period is 40400 seconds.  How much are you drifting?  Can the 20 second difference account for it?

According to the wiki, that's the siderial period. Don't we want synodic, since this is supposed to be referenced to a point on the ground?

[OhioBob]

For geosynchronous we want sidereal.  Sidereal is the time it takes to make a full 360 degree rotation relative to the stars.  We want the orbital period to match that to stay above the same point on the surface.  Synodic period is relative to the sun.

 

[0something0]

14 minutes ago, OhioBob said:

It's my understanding that Minmus' rotation period is 40400 seconds.  How much are you drifting?  Can the 20 second difference account for it?

My initial maths were done using the 40400 second figure, which had somewhat less of a drift, but were still significant. The wiki lists the 40400 seconds as the sidereal day while the in-game info tab lists 40380.

[OhioBob]

10 minutes ago, 0something0 said:

… while the in-game info tab lists 40380.

Everything I've seen says the period is 40400 s.  The in game info tab is just not showing the seconds.

According to my calculations, the semimajor axis of a synchronous orbit should be 417,941 metes.

[0something0]

2 hours ago, OhioBob said:

Everything I've seen says the period is 40400 s.  The in game info tab is just not showing the seconds.

According to my calculations, the semimajor axis of a synchronous orbit should be 417,941 metes.

My calculations show (G(2.6457580 * 10^19 kilograms)(40400 seconds)^2 / (4π^2))^(1/3) = 417940.868 meters. I also learned that for this application, merely putting it in the Google search bar is more accurate since Wolfram appears to round it to significant figures. This is confirmed using the Wiki's figure for a synchronous orbit gives me 60000 + 357940 = 417940 meters for the semi-major axis. Using the first figure to calculate the mass of Minmus of sanity-checking, I get ((417940.868 meters)^3 * 4 pi^2 )/((40400 seconds)^2 * G) = 2.64575801 × 10^19 kilograms, consistent with the Wiki.

Calculating using the precision figures gives me orbital parameters that have no noticeable drift.
Parameters:

Eccentricity: 0.8444756065

Linear eccentricity: 352940 meters

Semi-major axis: 417940.868 meters

Periapsis: 65000 meters (5000 meters from "sea level")

Apoapsis: 770881.736 meters

 

Edited June 29, 2019 by 0something0

[Zhetaan]

Of course a cool question like this appears when I'm away from my computer.

I only have to add this:

56 minutes ago, 0something0 said:

I also learned that for this application, merely putting it in the Google search bar is more accurate since Wolfram appears to round it to significant figures.

Wolfram is nice but it's made to be consumer-friendly.  It is possibly also made to be professor-friendly; I can envision a few physics professors cackling with glee when they spot the mark of Wolfram on a student's paper.  Either way, if you want good answers, then you should invest in a good calculator.

Edited June 29, 2019 by Zhetaan

[OhioBob]

@0something0, for future reference, if you need information about the celestial bodies, like mass, rotation period, etc., the best place to look are the kittopia-dumps.

[0something0]

4 hours ago, Zhetaan said:

you should invest in a good calculator.

So actually learn the Mathmatica language? 

[Zhetaan]

10 hours ago, 0something0 said:

So actually learn the Mathmatica language? 

Oh, heavens, no.  We're not trying to decide whether a particular N-dimensional manifold compactifies about the origin or play with surreal infinitesimals, or whatever else Mathematica can do (which, as I understand it, is most anything).  Kepler predated Newton:  a lot of the important equations in orbital mechanics were derived with algebra, geometry, and trigonometry, not calculus.  A scientific calculator that can raise values to rational exponents and take logarithms is enough for most anything you might want to do in KSP.