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rocketBob

Resonant synchronous orbit for repeat Mun landings

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Hello

I'm attempting to build a base on the Mun. Each section is parked together in orbit with a lander craft to drop them on the surface. What I would like to achieve is an orbit to facilitate landing the pieces in the same spot. Ideally this orbit would have a periapsis around 15km. How to I calculate the apoapsis in which the craft returns to the same longitude at periapsis every N orbits? The landing site for this base is above the equator, is such an orbit possible with inclination?

Cheers

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@rocketBob:

There is an equation, but it is not a pretty one.

Also, inclination is not a problem provided that the orbital period is resonant with the Mun's rotational period.  For example, a circular, Kerbin-synchronous, one-day, polar orbit will still pass over the same longitude once every orbit (that is, once per day).  An equatorial version of the same orbit holds position over that longitude indefinitely.  Intermediate inclinations and eccentricities show greater and greater apparent motion with respect to a surface observer, but so long as the semi-major axis of the orbit remains the same, it also remains synchronous (and the motion will appear periodic ... because it is).

The equation for orbital period is this:

T = 2 * pi * (a3 / GM)1/2

where:

T = period
a = semi-major axis
GM = gravitational parameter, which for the Mun is 6.5138398×1010 m3/s2

In your case, the easy solution is to have your T equal to 138984 / n seconds, where n is some integer and 138984 is the Mun's rotational period in seconds.

However, you don't want the period; you want the apoapsis, which means we need to rearrange the equation:

a = (GM * T2 / 4 * pi2)1/3

To simplify this for the Mun, we can pre-calculate a number of the parameters and combine constants:

a = (6.5138398×1010 * T2 / 4 * pi2)1/3

= (1.649974896×109 * (138984 / n)2)1/3

= (3.18718×1019 / n2)1/3

For n = 1 (a synchronous orbit), the value of a is 3.170556×106 m, which, less the 200,000 m of the Mun's radius, gives a circular orbit at an altitude of 2.9706×106 m.  This cannot be used because the Mun's sphere of influence ends at 2.4296×106 m.

For n = 64, the value of a is 198159.8 m, which puts the orbit under the Mun's surface.  The reason this is undesirable is left as an exercise for Jeb's piloting.

For n = 58, the semi-major axis is 211600.6 m, and that means a circular orbit at 11,600.6 m because such a low orbit is only barely enough to clear the Mun's mountaintops. 

It also means that you will have to wait 58 orbits to get a resonance.  Keep in mind that for all of these values of n, the number of orbits is largely irrelevant; excepting cases of crash or escape, the orbits will add up to one overhead pass of your chosen landing site for every Mun rotation.  The spacecraft will pass over the correct longitude many times--58, to use the maximum safe value--but those passes will also but for one be at the wrong latitude unless the landing site is on the equator.  If the orbit could precess, then it would be possible to achieve your goal a little more often (at the cost of some truly frightening calculations), but precession requires n-body gravity.

To take these values with a fixed Pe and obtain a useful Ap requires one more bit of mathematical wizardry.  The semi-major axis is to an ellipse something like a radius is to a circle, but because it is an ellipse, you can have varying Pe and Ap values that still give the same semi-major axis.  For the Mun, the semi-major axis is:

a = (Ap + Pe + 400000) / 2

where the 400,000 is the Mun's surface diameter and the Ap and Pe are taken as altitudes from the surface (as KSP displays them).  If we hold the periapsis at 15,000 m then the equation simplifies to:

a = (Ap + 415000) / 2

and combining this with the earlier equation gives:

(Ap + 415000) / 2 = (3.18718×1019 / n2)1/3

and once rearranged, this gives:

Ap 2 * (3.18718×1019 / n2)1/3 - 415000

This constrains n by quite a bit:  for example, you can no longer go above n = 56 because that lowers the 'apoapsis' to below the periapsis.  You can also no longer go below n = 4 because smaller values of n require either your Ap to be outside the Mun's sphere of influence or your Pe to be above 15 km.

I will assume that your mission profile is such that you will prefer either a long, languorous time at a high Ap or a close-in, near-circular orbit.  For this reason, my suggestions to you are to use either n = 4 or n = 56.

For n = 4 and a Pe of 15000 m, the Ap is 2.101473×106 m.  The orbital period is 34,746 seconds.

For n = 56 and a Pe of 15000 m,  the Ap is 18218 m ... approximately.  The important thing is to make certain that the orbital period is 2481.9 seconds.

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The Pe of your orbit stays at a fixed position forever while the Mun spins beneath it, once every 6 days. If you want them to line up occasionally, you need to set your orbit to some precise fraction of 138984 seconds.

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Posted (edited)

Smarter guys than me can tell you about calculations and whatnot (I fly mostly by feel and experience; I know what I wanna do and how to do it, but not how to explain why it works). Airless worlds are pretty easy, though. If you're in an equatorial orbit, just land your first module where you want. Then set it as your target and drop everything else right next to it. After a while, you'll be able to land within 100 meters consistently. I really enjoy seeing how close I can get without busting a solar panel.

 

Oops. Just realized you said your LZ is above the equator. Still don't think it matters. Once you land that first module and target it, it'll give you the inclination just like any other target. You can adjust at the node like normal. You'll have to be aware of the rotation, though. A little adjustment on the way in. Nothing major.

 

Oh, yeah, and welcome to the forums. Neglecting to say that is frowned upon. :)

Edited by Cpt Kerbalkrunch

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It certainly is possible, but I'm not really sure if it is practical. How many degrees off equator is you base? If it is just a few degrees, I would not bother with synchronous orbits, but rather park everything in equatorial orbit and just pack a bit of extra fuel for the lander.

If the equatorial orbit is not an option, because the base is too far north, I'd suggest another solution - Send everything to low, inclined orbit (doesn't really matter how inclined, as far as it crosses your base location). Since the Mun's rotation is really slow (a bit over six days), you can wait until the base "drifts" under your orbit, burn retrograde on the other side, and you'll be set to land near your base. It's not really hard to correct your approach to land reasonably near, I suggest using target mode, and keeping the "retrograde" and "anti target" markers as close as possible.

I hope it helps,

Michal.don

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As Bewing said, you want the period of your orbit to be an exact fraction of the mun's rotation. That way every n orbits your orbiting craft will pass over exactly the same track on the ground. As long as the orbit is inclined enough to pass over your desired landing site and passes it over it once then you return to it later.

Unfortunately I don't think stock KSP shows you the orbital period. You need to either install a mod that will show you (Kerbal engineer etc) or calculate it by hand.

https://en.wikipedia.org/wiki/Orbital_period

One example - choosing a resonance of 1:20 and a PE of exactly 15km you get a required period of 6949.219 seconds and an AP of 445.623km

With that you will pass over your chosen spot once every 20 orbits

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31 minutes ago, tomf said:

Unfortunately I don't think stock KSP shows you the orbital period. You need to either install a mod that will show you (Kerbal engineer etc) or calculate it by hand.

It does show you, just indirectly. Right click on the AP and PE markers to pin them. Then subtract the the' time to PE' from the 'time to AP', then double it. 

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nit-picking: the ratio between orbital period and Mun's rotation need to be a rational number. That is:  you need it to be a/b where a and b are integer numbers. Just notice:  for the smallest a and b that makes a/b  a rational number the synchronization occurs every 6a hours and every orbits .

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10 hours ago, Cpt Kerbalkrunch said:

Smarter guys than me can tell you about calculations and whatnot (I fly mostly by feel and experience; I know what I wanna do and how to do it, but not how to explain why it works). Airless worlds are pretty easy, though. If you're in an equatorial orbit, just land your first module where you want. Then set it as your target and drop everything else right next to it. After a while, you'll be able to land within 100 meters consistently. I really enjoy seeing how close I can get without busting a solar panel.

 

Oops. Just realized you said your LZ is above the equator. Still don't think it matters. Once you land that first module and target it, it'll give you the inclination just like any other target. You can adjust at the node like normal. You'll have to be aware of the rotation, though. A little adjustment on the way in. Nothing major.

 

Oh, yeah, and welcome to the forums. Neglecting to say that is frowned upon. :)

Thank you for your welcome Captain. Apologies for not having said I am new here. Perhaps I should also mention that I have been Hooked on Kerbal for the past 3 years! :D It amazes me that this game continually offers new ways to learn.

This is solid advice, and what I have previously done for landings in the past. When landing somewhere above/below the equator, I would put myself in an inclined (usually ~45deg) orbit and wait for the LZ on the body to drift slightly before being under my craft. Once the target was set, it was fairly easy to correct course during the descent using the target markers on the navball. This time however, I was looking to expand my knowledge of the game and learn a more precise/predictable way to get the job done.

Thank you to you and everyone else for the input!

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4 hours ago, Zhetaan said:

@rocketBob:

There is an equation, but it is not a pretty one.

Also, inclination is not a problem provided that the orbital period is resonant with the Mun's rotational period.  For example, a circular, Kerbin-synchronous, one-day, polar orbit will still pass over the same longitude once every orbit (that is, once per day).  An equatorial version of the same orbit holds position over that longitude indefinitely.  Intermediate inclinations and eccentricities show greater and greater apparent motion with respect to a surface observer, but so long as the semi-major axis of the orbit remains the same, it also remains synchronous (and the motion will appear periodic ... because it is).

The equation for orbital period is this:

T = 2 * pi * (a3 / GM)1/2

where:

T = period
a = semi-major axis
GM = gravitational parameter, which for the Mun is 6.5138398×1010 m3/s2

In your case, the easy solution is to have your T equal to 138984 / n seconds, where n is some integer and 138984 is the Mun's rotational period in seconds.

However, you don't want the period; you want the apoapsis, which means we need to rearrange the equation:

a = (GM * T2 / 4 * pi2)1/3

To simplify this for the Mun, we can pre-calculate a number of the parameters and combine constants:

a = (6.5138398×1010 * T2 / 4 * pi2)1/3

= (1.649974896×109 * (138984 / n)2)1/3

= (3.18718×1019 / n2)1/3

For n = 1 (a synchronous orbit), the value of a is 3.170556×106 m, which, less the 200,000 m of the Mun's radius, gives a circular orbit at an altitude of 2.9706×106 m.  This cannot be used because the Mun's sphere of influence ends at 2.4296×106 m.

For n = 64, the value of a is 198159.8 m, which puts the orbit under the Mun's surface.  The reason this is undesirable is left as an exercise for Jeb's piloting.

For n = 58, the semi-major axis is 211600.6 m, and that means a circular orbit at 11,600.6 m because such a low orbit is only barely enough to clear the Mun's mountaintops. 

It also means that you will have to wait 58 orbits to get a resonance.  Keep in mind that for all of these values of n, the number of orbits is largely irrelevant; excepting cases of crash or escape, the orbits will add up to one overhead pass of your chosen landing site for every Mun rotation.  The spacecraft will pass over the correct longitude many times--58, to use the maximum safe value--but those passes will also but for one be at the wrong latitude unless the landing site is on the equator.  If the orbit could precess, then it would be possible to achieve your goal a little more often (at the cost of some truly frightening calculations), but precession requires n-body gravity.

To take these values with a fixed Pe and obtain a useful Ap requires one more bit of mathematical wizardry.  The semi-major axis is to an ellipse something like a radius is to a circle, but because it is an ellipse, you can have varying Pe and Ap values that still give the same semi-major axis.  For the Mun, the semi-major axis is:

a = (Ap + Pe + 400000) / 2

where the 400,000 is the Mun's surface diameter and the Ap and Pe are taken as altitudes from the surface (as KSP displays them).  If we hold the periapsis at 15,000 m then the equation simplifies to:

a = (Ap + 415000) / 2

and combining this with the earlier equation gives:

(Ap + 415000) / 2 = (3.18718×1019 / n2)1/3

and once rearranged, this gives:

Ap 2 * (3.18718×1019 / n2)1/3 - 415000

This constrains n by quite a bit:  for example, you can no longer go above n = 56 because that lowers the 'apoapsis' to below the periapsis.  You can also no longer go below n = 4 because smaller values of n require either your Ap to be outside the Mun's sphere of influence or your Pe to be above 15 km.

I will assume that your mission profile is such that you will prefer either a long, languorous time at a high Ap or a close-in, near-circular orbit.  For this reason, my suggestions to you are to use either n = 4 or n = 56.

For n = 4 and a Pe of 15000 m, the Ap is 2.101473×106 m.  The orbital period is 34,746 seconds.

For n = 56 and a Pe of 15000 m,  the Ap is 18218 m ... approximately.  The important thing is to make certain that the orbital period is 2481.9 seconds.

Zhetaan, this is exactly what I was looking for. Thank you for taking the time to guide me in this math, and I appreciate the detail that you have provided here. This makes sense to me now, and I am looking forward to being able to apply this knowledge towards other missions! The max/min constraints is especially interesting. Next up is a Minmus refueling station/base, and this immediately strikes me as being super useful for transporting fuel/ore back up to orbit.

I've played around a little with the n value trying to find a good balance between fuel efficiency of a lower apoapsis and time efficiency of a higher one. Ultimately I'm going with your advice of n = 4, which will require about 200m/s to circularize down to 15km. n = 8 for example only saved me about 25m/s, and in the end I'd rather plan for spending more fuel than spending more time up in orbit. Not to mention that at n = 56 time warp would be limited to 10x (yikes).

Thanks again!

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Disregard it if you mentioned the deltaV to circularize at 15km just for reference but personally I think part of the reason to such orbit is missed if before landing you first circuralze. Ideally you want a single burn at periapsis to change from a orbital trajectory to a landing trajectory. 

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