How to get into keostationary orbit?

[Ncog Nito 19]

Hello fellow kerbonauts.  I have accepted a contract to put a satellite in a keostationary orbit above a specified location over Kerbin.  The AP and PE are both 2,863,334 M at 0 degrees inclination.  What is the best way to approach this and complete the contract?  I have yet to ever achieve an eccentric orbit.  I believe that my definition of eccentric orbit is an equal distance from the planet throughout an objects entire orbit, is this correct?

Edited May 1 by Ncog Nito
clarifying definition of eccentric orbit

[Kryxal 142]

This should be fairly easy:

1: Get into a circular orbit at 871 km.

2: When over the required spot, burn to increase apoapsis to your final target.  This orbit should have a period of 216 minutes.

3: Wait till you've made 2 full orbits and are approaching apoapsis for the 3rd ... you should be right on top of your target point.  If you're trailing a bit, leave periapsis a bit low and wait an orbit, if you're leading burn a bit extra and wait an orbit.

[swjr-swis 1454]

A few tips:

• It helps a lot to pack a very weak engine of some sort in your final payload (the satellite), like a linear RCS port (monoprop), ion engine (xenon gas and EC), or Spider/Ant (LFO). At KEO altitudes, most engines have way too much thrust and even tiny taps of acceleration can overshoot the target Ap/Pe by a lot.
• Additionally, it helps to fine tune the final orbit if once you have Ap/Pe roughly right, you rightclick your engine and lower the thrust to the very minimum of 0.5. This should allow to get Ap/Pe within even 1m of accuracy, no matter how light your payload is.

[Abastro 150]

Satellite on specific orbit contract is usually generous in my experience. You could have 1 degree of inclination and still complete equatorial orbit contract. So you don't have to try to do it precisely.

Besides, I think what you mean by 'eccentric orbit' was circular orbit. It's an orbit with circular trajectory, and the Ap and Pe has the same altitude. It's really hard to get into one, even with 10m of margin on LKO!

Though as mentioned above, contracts are generous enough to allow over 1~10km of margin(for keostationary). Just make an orbit similar to the contract orbit with burns on Pe/Ap, and it will be marked as complete.

[FullMetalMachinist 694]

13 hours ago, Ncog Nito said:

I believe that my definition of eccentric orbit is an equal distance from the planet throughout an objects entire orbit, is this correct?

Nope, that's called a circular orbit. Or if you prefer it's an orbit with 0 eccentricity. An orbit that's elliptical will have an eccentricity somewhere between 0 and 1. And an orbit with eccentricity of 1 or greater is an escape trajectory. 

[Ncog Nito 19]

On ‎5‎/‎1‎/‎2017 at 2:23 PM, Kryxal said:

This should be fairly easy:

1: Get into a circular orbit at 871 km.

2: When over the required spot, burn to increase apoapsis to your final target.  This orbit should have a period of 216 minutes.

3: Wait till you've made 2 full orbits and are approaching apoapsis for the 3rd ... you should be right on top of your target point.  If you're trailing a bit, leave periapsis a bit low and wait an orbit, if you're leading burn a bit extra and wait an orbit.

 

On ‎5‎/‎1‎/‎2017 at 10:30 AM, swjr-swis said:

A few tips:

• It helps a lot to pack a very weak engine of some sort in your final payload (the satellite), like a linear RCS port (monoprop), ion engine (xenon gas and EC), or Spider/Ant (LFO). At KEO altitudes, most engines have way too much thrust and even tiny taps of acceleration can overshoot the target Ap/Pe by a lot.
• Additionally, it helps to fine tune the final orbit if once you have Ap/Pe roughly right, you rightclick your engine and lower the thrust to the very minimum of 0.5. This should allow to get Ap/Pe within even 1m of accuracy, no matter how light your payload is.

Excellent advice @Kryxal!  Followed your steps to the letter, with the side advice from @swjr-swis about the thrust limiter, and it worked perfectly on the first attempt.  Thanks guys, you are the kind of Kerbals that should be answering questions regularly.  Great work!

[Spricigo 437]

A 'simplified' version of what @Kryxal said:

From a lower orbit, perform a maneuver that make you reach apoapsis at the required altitude, over the required area and then circularize. The relative position(angle) between the target area and the maneuver node need to so both orbits 'synch up'.

He kindly provide a particular case that works, but you may also calculate from the astrodynamics equations if you are facing a similar, but not quite equal, transfer.

[Stoney3K 178]

I would like to see the calculations that @Kryxal used for this. An orbital period of 216 minutes is not resonant to KEO, which has an orbital period of exactly 6 hours (one Kerbin day). You'd expect a 1/3 resonant transfer orbit into KEO to have an orbital period of 120 minutes, so something else is going on here.

[Spricigo 437]

4 minutes ago, Stoney3K said:

I would like to see the calculations that @Kryxal used for this. An orbital period of 216 minutes is not resonant to KEO, which has an orbital period of exactly 6 hours (one Kerbin day). You'd expect a 1/3 resonant transfer orbit into KEO to have an orbital period of 120 minutes, so something else is going on here.

With 216min orbital period you complete 5 orbits while kerbin completes 3. Or 2,5 to 1,5 as described by Kryxal. 

[Stoney3K 178]

Just now, Spricigo said:

With 216min orbital period you complete 5 orbits while kerbin completes 3. Or 2,5 to 1,5 as described by Kryxal. 

Which raises my question: Why choose that ratio instead of, for example, 2:1, 3:1 or 4:1?

I understand the need to get at Apoapsis while Kerbin completes a full revolution, which would be a 'half' orbit on your own.

[Spricigo 437]

1 minute ago, Stoney3K said:

Which raises my question: Why choose that ratio instead of, for example, 2:1, 3:1 or 4:1?

Personal taste,  I suppose.   Choosing any particular ratio is an arbitrary decision. 

[Kryxal 142]

Honestly, it was a fairly simple set of conditions.  I wanted a transfer orbit where you started at a clear point (over the destination, at periapsis) ended at a clear point and altitude (over the destination, at apoapsis, at the right altitude for a synchronous orbit), and didn't have to do too much fiddling.  Clearly you're not getting there in a half-rotation, so set up a resonant transfer for rotation-and-a-half.  That's 540 minutes, and you're not getting the transfer orbit below 165.6 minutes (70 km periapsis) so the numbers just worked out that way.  You'd need 154.3 minutes for your orbital period for a lower transfer orbit, at which point you're lithobraking first.

[FullMetalMachinist 694]

4 hours ago, Stoney3K said:

An orbital period of 216 minutes is not resonant to KEO, which has an orbital period of exactly 6 hours (one Kerbin day).

Minor nitpick. A stationary orbit's period needs to match the sidereal rotation period of the planet. On Kerbin's surface a day is exactly 6 hours. However it's sidereal rotation is slightly less, 5h 59m 9.4s.

[Blaarkies 234]

On 5/1/2017 at 5:30 PM, swjr-swis said:

A few tips:

• It helps a lot to pack a very weak engine of some sort in your final payload (the satellite), like a linear RCS port (monoprop), ion engine (xenon gas and EC), or Spider/Ant (LFO). At KEO altitudes, most engines have way too much thrust and even tiny taps of acceleration can overshoot the target Ap/Pe by a lot.
• Additionally, it helps to fine tune the final orbit if once you have Ap/Pe roughly right, you rightclick your engine and lower the thrust to the very minimum of 0.5. This should allow to get Ap/Pe within even 1m of accuracy, no matter how light your payload is.

That is certainly the best approach.

Also, when using RCS, press "caps lock" for fine control. For RCS translation, this gives very small changes in speed(press "h" to thrust forward). An added benefit is the slight balancing of RCS thrust power(when the RCS ports are not aligned to center of mass)

With main engines, this stupid trick can help get tiny bursts of thrust. hold down "control" key, and just tap the "shift" key (it sounds really stupid i know, but very useful and many player don't know about it)

[Spricigo 437]

50 minutes ago, Blaarkies said:

With main engines, this stupid trick can help get tiny bursts of thrust. hold down "control" key, and just tap the "shift" key (it sounds really stupid i know, but very useful and many player don't know about it)

Its a smart trick tht i would use if not because muscular memory  to tap shift & X.

another point about main engines(but also usefull for RCS) no one mentioned yet: thrust limit is very usefull to tame overpowered engines, and with already precise engines at low throttle allow for cirurgical adjustments.

[Dr.LoveJoy 20]

On 5/3/2017 at 9:03 AM, FullMetalMachinist said:

Minor nitpick. A stationary orbit's period needs to match the sidereal rotation period of the planet. On Kerbin's surface a day is exactly 6 hours. However it's sidereal rotation is slightly less, 5h 59m 9.4s.

Kerbin precesses? I don't think it does.. or do you mean its "stellar day" is that amount? I suppose, this is a nitpick.. OF the nitpick... ;-)

[Spricigo 437]

4 minutes ago, Dr.LoveJoy said:

Kerbin precesses? I don't think it does.. or do you mean its "stellar day" is that amount? I suppose, this is a nitpick.. OF the nitpick... ;-)

Question about the nitpick: since kerbin don't precesses, it don't mean that sideral day = stellar day?

[swjr-swis 1454]

1 hour ago, Dr.LoveJoy said:

Kerbin precesses? I don't think it does..

Nothing to do with precession (*) - the slight difference between solar and sidereal day is due to the planet moving a slight bit ahead in its own orbit around its sun while it makes one full rotation, which causes the sun's angle to the surface to 'fall behind' slightly, which means the planet has to rotate just a tiny bit more than an exact 360 degrees for the sun to be at the exact same location in the sky. Hence the solar day (measured by the position of the sun in the sky) is always (#) a slightly bit longer than the sidereal day (measured by the 'fixed' position of the distant stars, which in KSP is effectively fixed).

(*) - precession on the other hand is the shifting of the rotation axis of the planet, which does not exist at all in KSP, and in our familiar case of Earth is a very slow mostly circular movement noticeable by the stellar background shifting in the sky over the course of ~26000 years, give or take a dynasty.

(#) in the most common case of a prograde rotation. If the planet's rotation is counter to its orbital direction, then it would be the other way around.

Edited May 16 by swjr-swis
#

[Dr.LoveJoy 20]

On 5/16/2017 at 11:30 AM, swjr-swis said:

Nothing to do with precession (*) - the slight difference between solar and sidereal day is due to the planet moving a slight bit ahead in its own orbit around its sun while it makes one full rotation, which causes the sun's angle to the surface to 'fall behind' slightly, which means the planet has to rotate just a tiny bit more than an exact 360 degrees for the sun to be at the exact same location in the sky. Hence the solar day (measured by the position of the sun in the sky) is always (#) a slightly bit longer than the sidereal day (measured by the 'fixed' position of the distant stars, which in KSP is effectively fixed).

(*) - precession on the other hand is the shifting of the rotation axis of the planet, which does not exist at all in KSP, and in our familiar case of Earth is a very slow mostly circular movement noticeable by the stellar background shifting in the sky over the course of ~26000 years, give or take a dynasty.

(#) in the most common case of a prograde rotation. If the planet's rotation is counter to its orbital direction, then it would be the other way around.

Oh I see. Well, chalk one up for Astronomy not being quite clear when you look up references online. :-)

Thanks very much for the clarification! Makes perfect sense.