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How would I calculate how often an orbital plane intersects a given location on the surface of a body?


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I'm trying to determine the maximum length a lander would need to stay on the surface of a body (i.e. the Mun) prior to being able to efficiently return to an orbiting station

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It would be a function of the orbital period and the body rotational period.  If the orbiting station is in a polar orbit of Mun at about 30km then I can tell you that you can land, do science, plant a flag, and return to it before the orbiting station's LAN has changed too much for a rendezvous.  You can even catch it on its next orbit, iirc.  But after a few orbits you will need to make some inclination corrections and if you stay for awhile you will need to wait for nearly half the Mun rotational period, which I vaguely recall is about 19 hours, for the orbiting station orbit to be near overhead.

You can use the wiki to look at the "sidereal rotation period" for a body and KER or other can give you the orbital period

(on left in sidebar)

https://wiki.kerbalspaceprogram.com/wiki/Mun

Edited by darthgently
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20 minutes ago, dlrk said:

Ok, thanks. I'm interested in this in the context of a long-stay base.

What do I do with the orbital period and sidereal rotation?

I'm not sure exactly what you mean then.  I guess I need to know what you mean by "long term".  Do you mean to stay long enough to mine ore and convert it to fuel?  Or do you mean how long would you need to wait for the orbiting station's orbit to cross over your lander location again?  My previous answer assumed the second case.

You probably don't need to worry too much about orbital period really.  The orbit will "stay" in place while the body rotates underneath it.  So it is all about how fast the body rotates under the orbit.  In the case of Mun, if you are staying on the surface longer than an hour or so your location on the surface will have rotated away from the orbit quite a bit and so you will need to wait an hour or so less than half the body's rotation period (about 38.5/2 or ~19 hours for Mun) for your location to cross under the other "half" of the orbit.  And to be clear, the orbit will be overhead every ~19.25 hours so you can always catch the next one if you miss the window

Of course if the orbiting station is in polar orbit and you land near the pole, the orbit will always be near overhead.  And it also simplifies quite a bit if the orbiting station is in an equatorial orbit and your land on the equator, as then the orbit will always be near overhead also.

 

Edited by darthgently
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So, let me clarify a bit:

If I have a station in an inclined or polar orbit, and I detach a lander from it and land on the Mun (or some other body), and I leave the lander there long enough that it goes out-of-plane to the maximum extent possible, how would I determine how long it would take to get back in-plane? Or, if I have a surface base needing resupply, how often will it have the opportunity to send a spacecraft to an orbiting inclined station

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3 minutes ago, dlrk said:

So, let me clarify a bit:

If I have a station in an inclined or polar orbit, and I detach a lander from it and land on the Mun (or some other body), and I leave the lander there long enough that it goes out-of-plane to the maximum extent possible, how would I determine how long it would take to get back in-plane? Or, if I have a surface base needing resupply, how often will it have the opportunity to send a spacecraft to an orbiting inclined station

Ok, I literally just did the math for you.  lol  read last message more carefully  :)

An inclined orbit would be no different than a polar orbit in this case.  I could have mentioned that

Edited by darthgently
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I'm assuming you merely want to launch into the same orbital plane as the orbiting station, not launch directly to rendezvous.  That is a bit more work and very craft dependant

1 minute ago, dlrk said:

Ok. Am I understanding correctly that the answer is half the sidereal period?

Sidereal rotation period.  But in Mun's case the sidereal rotation period is the same as the sidereal orbital period.   Because it is tidally locked to Kerbin

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Any multiple of half the sidereal rotation period will do.  It just keeps going round and round.  Like wheels on a bus

Once you launch into near the same orbital plane, you can adjust to match planes with the target where the orbits cross.    If you time it just right you will launch very close into the same plane, but you can always adjust your relative inclination where the orbits cross by burning either normal or antinormal

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Knowing which direction to launch can be hard when orbiting station is in inclined orbit.  That is why I usually do polar or equatorial.  That way one either launches east, north, or south.  If you are shooting for an inclined rendezvous, wait until the orbiting station has crossed overhead, set it as "target", then aim in its heading direction on the navball but adjust your pitch as you normally would for a launch.  Kind of hard to explain, but if you think about it and try it a few times it will make sense.  I usually use my kOS script which does things a bit differently so I have to consider how I'd do it just using the navball

Edited by darthgently
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  • 2 weeks later...

https://sourceforge.net/p/obtlib/code/HEAD/tree/head/Ships/Script/obtlib/launchwait.ks

... it requires a hell of alot of ugly maths if you want to do it accurately (that is by no means accurately), but a "dumb" human can do it equally as well by eye...

It just requires an understanding of where the target orbit and the "assumed" orbit the ground object is under cross. Thus the mathematical name for it: "cross product"? :)

 

 

Edited by surge
extra rambling
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