Keosynchronous Help

[Nicklas]

So I wonna place 3 satellites in Keosynchronous orbit around Kerbin, if thats what its called when a satellite is orbiting a fixed spot on Kerbin Sorry for my english.

I found a video on youtube, a guy has packed 3 satellites, and use Infernal Robotics to make it more compact, he explaines the orbital period, so that every time he gets to apoapsis, he release a satellite and its timed so they get spread even. I can't find the video again. Its not in history.

So do you guys have a link to a video that explaines how to get this timing. Or maybe give me some data and can work with. I know what my apoapsis should be. But what about the periapsis, what should that be, to get that near perfect spread ?

ofc he raises the satellites periapsis after release.

Edited March 10, 2015 by Nicklas

[Taki117]

Alright, I suggest getting a mod like Kerbal Engineer Redux or something else that shows you your orbital period. The Altitude for Kerbosynch orbit is found on the Wiki page. Put your AP around that altitude and make sure your orbital period is 2 Hrs (1/3 the 6 Hrs for a Synchronous orbit) Then you can release a Sat at each Apoapsis and circularize. The important part is that the final orbit is 6 Hrs, AP and PE are not as important.

[Kryxal]

You're looking for an orbit with a period of 2 hours and apoapsis of 2868.75 km

T = 2*pi*sqrt(a^3/u) by Kepler's Third Law, where T = orbital period in seconds, a = semi-major axis, u = GM = 3.5316000×10^12 m^3/s^2 for Kerbin

sqrt(a^3/u) = T/2*pi

a^3 = u*(T/2*pi)^2 = u*(3600/pi)^2 = 4.637 * 10^18 m^3

a = 1667.55 km = (ap + pe) / 2 + 600 km ... which doesn't quite work.

Redo for 4 hours:

a^3 = u*(T/2*pi)^2 = u*(7200/pi)^2 = 1.855 * 10^19 m^3

a = 2647.15 km = (ap + pe) / 2 + 600 km

ap + pe = 2047.15 * 2 = 4094.3, ap = 2868.75, pe = 1225.55 km

[Nicklas]

Thank you both that helped even though you lost me there Kryxal

[superm18]

I think you might have been watching Bob Fitch

[Eric S]

You're looking for an orbit with a period of 2 hours and apoapsis of 2868.75 km

A period of 4 hours will also work, and is what I usually use so that the satellites don't have to expend as much delta-v circularizing.

[Kryxal]

The first set of equations ended up being a proof by contradiction that a 2 hour orbit wouldn't work, actually. In a practical sense, I'd just set apoapsis to what's needed, drop a node there and get the periapsis to be 2 hours later, then begin releases on the next orbit (maybe 5 minutes beforehand using a delivery vehicle with trilateral symmetry).

[purpletarget]

A quick in-game way to do the same thing, is push the Ap up to the KEO altitude, and let the package travel up to it. At the Ap, raise the Pe, and keep an eye on your Time to Pe in the Map View. If you want a 4 hour period for the orbit, raise the Pe until it's 2 hours from Ap to Pe.

At that point, you can release your first satellite for circulation, and the remainder can sync themselves into positions with 2 hour offsets from the original.

[footforhire]

I think you might have been watching Bob Fitch

Bob Fitch does this at the beginning of his Odyssey series, but it's not from one craft. He does three separate launches and aligns them so that the two on the other side of Kerbin can communicate with the one in Keosynchronous orbit above Mission Control.

He also uses MechJeb to determine his orbital period to match Kerbin's, which is far more accurate that using the Keosynchronous orbit altitude mentioned in the Wiki here.

+Rep to previous mentioners

[superm18]

i know this is a bit old, but it's a great guide by Bob Fitch:

[GoSlash27]

*edit*

After re-reading your OP, according to my math he would have made the periapsis 1,225,554 m.

This would make an eccentric orbit with a period of exactly 4 hours.

period(secs)=sqrt(4pi^2r(meters)^3/std gravity)

simplifying,

r= cube root(p^2*K)

Since our keosynchronous radius is 3,468,750m and the period is 6 hrs, we know that our constant (K) must be 1.159355E+18.

so a transfer orbit with a period of 4 hours would have a semimajor axis of 2,647,152 m.

Since the apoapsis is 821,598m higher than the semimajor axis, the periapsis must necessarily be the same amount less than the semimajor axis, or 1,825,554m. Subtracting Kerbin's sea level gives an altitude of 1,225,554m.

Best,

-Slashy

Edited March 16, 2015 by GoSlash27