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Moon exit trajectory question


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The best way to think about it is not in terms of "vertical velocity" or "horizontal velocity", because neither of those is a conserved quantity. Think about what's conserved, then the math will flow from that.

For an orbiting body, two things are conserved:

1. Energy (kinetic + potential)

2. Angular momentum

Whenever you're flying a rocket, and you want your orbit to be something different than it is now, really what you want is two things: "I want to have a different amount of orbital energy, and I want to have a different amount of orbital angular momentum." Every ounce of rocket fuel you expend is aiming to change those two things. Therefore, when deciding where (and in what direction) to burn, the optimal choice will be the one that maximizes your contribution to those two things. So it comes down to four questions:

1. In what direction should I burn to maximize my impact on orbital energy?

2. In what location should I burn to maximize my impact on orbital energy?

3. In what direction should I burn to maximize my impact on orbital angular momentum?

4. In what location should I burn to maximize my impact on orbital angular momentum?

Taking these in order:

1. Burn direction for energy change: This one's easy. Always burn prograde (to increase energy) or retrograde (to decrease it). Any time you're not thrusting directly prograde or retrograde, you're "wasting" fuel to do something other than change your energy, since only the prograde/retrograde component affects your speed. Your efficiency here is basically the cosine of the angle between prograde (or retrograde) and what you're thrusting, so keep that as close to zero as possible.

2. Burn location for energy change: Here's where Oberth effect comes in. You want to do it where you're moving the fastest, e.g. at periapsis.

3. Burn direction for angular momentum change: The optimum direction is when your thrust is at right angles to your "actual radial" direction (i.e. the vector from your ship to the body that you're orbiting around). I put that in quotes and called it "actual radial" because it's not what KSP nav-ball "radial" and "anti-radial" indicators mean. Those indicators are always at right angles to prograde/retrograde, but the direction I refer to is perpendicular to the radius, not perpendicular to your prograde vector. Those two aren't the same as each other unless you're at periapsis or apoapsis. Again, the efficiency factor here is the cosine between your thrust direction and the optimal direction.

4. Burn location for angular momentum change: Angular momentum is the product of your "sideways" velocity and your orbital radius. Therefore, you get the most "bang for your buck" when you're at a really high altitude, i.e. at apoapsis.

This is why gravity turns are so important. If you're sitting on the planet and want to be in orbit, you need to increase both your energy and your angular momentum, by a lot. A well-designed gravity turn maximizes both energy and angular-momentum efficiency: you're always thrusting prograde (good for #1 above), you're doing all your burn at low elevation (good for #2), you're doing as much of it as possible in a "sideways" direction (good for #3). The only place it falls short is #4, but you can't do any burning way up high without getting there in the first place. :)

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This is why gravity turns are so important. If you're sitting on the planet and want to be in orbit, you need to increase both your energy and your angular momentum, by a lot. A well-designed gravity turn maximizes both energy and angular-momentum efficiency: you're always thrusting prograde (good for #1 above), you're doing all your burn at low elevation (good for #2), you're doing as much of it as possible in a "sideways" direction (good for #3). The only place it falls short is #4, but you can't do any burning way up high without getting there in the first place. :)

Great reply Snark! Something I've actually been wondering is this (I was gonna make the calculation but I might as well ask now):

If you've accidentally put your apoapsis too high (ascent from body with atmosphere) whilst gravity turning, is is better to thrust sideways (to help #2 & #3) or should you coast to apoapsis and then thrust (to help #1, #3, & #4)?

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Yes, heading east lets you gain the rotational speed of the body you are leaving for free. Going straight up is actually probably optimal at the exact right time with perfect piloting. Its just that in practice even the tiniest errors in either timing or piloting will likely cost you more than circularizing and making an optimal maneuver node before your ejection burn.

Note that with the Mun and Minmus (and most other moons in the game) their rotational angular velocity is so slow -- because they're tidally locked --, that it doesn't really matter which direction you launch. On both of Kerbin's moons, the difference between a prograde and a retrograde launch is a whole 18 m/s. (Laythe is the only moon where the difference is more than 50 m/s.) And returning to Kerbin from a retrograde equatorial orbit around the Mun takes exactly as much energy as returning from a prograde orbit, provided your escape vector is parallel to and opposite the Mun's orbit.

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Note that with the Mun and Minmus (and most other moons in the game) their rotational angular velocity is so slow -- because they're tidally locked --, that it doesn't really matter which direction you launch. On both of Kerbin's moons, the difference between a prograde and a retrograde launch is a whole 18 m/s. (Laythe is the only moon where the difference is more than 50 m/s.) And returning to Kerbin from a retrograde equatorial orbit around the Mun takes exactly as much energy as returning from a prograde orbit, provided your escape vector is parallel to and opposite the Mun's orbit.

Actually, Minmus isn't tidally locked. I think it rotates ~ 25 times for each orbit. Surprisingly, Gilly is the only other moon that isn't tidally locked to it's parent.

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"I want to have a different amount of orbital energy, and I want to have a different amount of orbital angular momentum."

I can't understand what you mean by orbital angular momentum. My understanding of angular momentum is purely based on rotational motion in a given frame of reference.

I did a couple of quick searches on the term and only turned up references to quantum mechanics.

Happy landings!

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Answering a few different questions from different posts:

If you've accidentally put your apoapsis too high (ascent from body with atmosphere) whilst gravity turning, is is better to thrust sideways (to help #2 & #3) or should you coast to apoapsis and then thrust (to help #1, #3, & #4)?

Depends a lot on your specific trajectory, and also what your desired end state is. For example, is your only goal "be in orbit and I don't care what it looks like"? Or is it "be in a circular orbit at a particular radius"?

Assuming that all you care about is "lift my periapsis above a certain minimum height" (e.g. "out of the atmosphere"), then the most efficient way to adjust periapsis is to thrust prograde (if you're raising) or retrograde (if you're lowering) when at apoapsis. However, if (for example) you've already raised apoapsis of 200km, want to be at a circular orbit of 100km, and are currently rising past 80km, it may be a different story (e.g. may be worth burning a bit radially inward from prograde). Working out exact numbers would require orbital parameters, though.

Note that with the Mun and Minmus (and most other moons in the game) their rotational angular velocity is so slow -- because they're tidally locked --, that it doesn't really matter which direction you launch.

Absolutely correct about the "doesn't matter" part, the numbers are tiny... just want to note that Minmus isn't actually tidally locked; it rotates fairly rapidly. However, that's irrelevant to the current discussion-- you're right that it doesn't matter, since you can practically sneeze your way off Minmus. :)

I can't understand what you mean by orbital angular momentum. My understanding of angular momentum is purely based on rotational motion in a given frame of reference.

What I mean is "the angular momentum of the object as it orbits around the parent body." Sorry, I guess I should be more careful with my terminology. :blush: I mean this:

http://en.wikipedia.org/wiki/Specific_relative_angular_momentum

This is the vector cross product of its momentum with the radius vector from the primary to the orbiting object. Easiest way to think of it in KSP terms is the product of orbital radius times the "sideways component" of your velocity.

It's conserved, meaning that unless you apply thrust, it stays constant. That's what Kepler's 2nd law is about:

http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion

If you're in an orbit with an apoapsis of 1000km and periapsis at 500km, you'll be moving twice as fast at periapsis as apoapsis. Note that for purposes of doing this math, I refer to periapsis/apoapsis as measured from the center of the primary (which is what physics cares about), not from its surface (which is what KSP reports to you).

Conservation of angular momentum is really important in calculating orbital mechanics. Pretty much every math problem (e.g. "I'm currently at this location moving at this speed in this direction, what will my periapsis and apoapsis be" or "how much of a burn will I need to make my orbit like thus-and-so") involves writing down the equations for energy (call it E) and angular momentum (call it L), and then solving for the case where E1=E2 and L1=L2.

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  • 4 years later...
On 5/24/2015 at 9:36 PM, MathmoRichard said:

The best way to work out the angle McKermack is with manoeuvre nodes. It is actually the best way to understand all this. Try setting one up in a circular Munar orbit with a set delta V of 300 m/s say, and then slide the node about to see which gives you the smallest Kerbin periapsis. I use this technique every single time, it's the only way to be most efficient without mods.

I know this is an old thread, but in high Munar orbit, it seems every escape puts me in a higher orbit than the Mun. I'm talking about orbits which are near escape like 2.2Mm. At this altitude the orbital period of the Mun is over 4 days while the Mun orbits Kerbin in about 6.5 days. It seems to me like every time I set a node on the front side of the Mun, in prograde direction, by time my ship actually gets there the Mun has orbited around Kerbin sufficiently such that I am still behind the Mun and my ejection is prograde throwing me into a higher Kerbin orbit. Its almost like I'm trapped in Munar orbit without slowing down and doing a slingshot, which seems to be wasting fuel. It' kind of frustrating. In theory I should just have to wait a month or so until me ship is actually on the prograde side of the Mun and do a burn relatively soon thereafter without giving Mun the chance to move with respect to Kerbin. 

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I'm having trouble picturing your question. If you set a node, though, you can see which direction your ship will be going after the burn and you can slide the node around the orbit until the ship leaves Mun going in a way that results in what you want. Something like this, maybe? 

77VyDex.png

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On 3/19/2020 at 8:35 AM, r3_141592654 said:

I know this is an old thread, but in high Munar orbit, it seems every escape puts me in a higher orbit than the Mun. I'm talking about orbits which are near escape like 2.2Mm. At this altitude the orbital period of the Mun is over 4 days while the Mun orbits Kerbin in about 6.5 days. It seems to me like every time I set a node on the front side of the Mun, in prograde direction, by time my ship actually gets there the Mun has orbited around Kerbin sufficiently such that I am still behind the Mun and my ejection is prograde throwing me into a higher Kerbin orbit. Its almost like I'm trapped in Munar orbit without slowing down and doing a slingshot, which seems to be wasting fuel. It' kind of frustrating. In theory I should just have to wait a month or so until me ship is actually on the prograde side of the Mun and do a burn relatively soon thereafter without giving Mun the chance to move with respect to Kerbin. 

Congratulations; you've discovered an edge case!

One thing to remember is that your orbit about the Mun is fixed with reference to the stars, not the Mun or Kerbin.  The Mun still rotates (albeit slowly) under you, and it still orbits Kerbin, so yes, this absolutely will happen.  You're at a point where you need to consider your departure almost as a transfer window, and set your node in such a way that will have you leaving the Mun in a retrograde direction at the point that you arrive there.  This is somewhat more involved than typical interplanetary transfers because in the grand scheme, while Kerbin doesn't move much over the several days it takes to escape to interplanetary space, the Mun's six-day year-equivalent is fast.  Therefore, you need to visualise your position and path, not only with respect to your position and motion about the Mun, but also with respect to your motion about Kerbin.

The best way I can think to illustrate your problem is this:  you have to remember that if your ship is orbiting the Mun and the Mun is orbiting Kerbin, then you're orbiting Kerbin, too, albeit in an extended-family sort of way.  If you were to graph your position with respect to Kerbin rather than the Mun, then you'd see that the orbit always appears to be a variation of the Mun's own orbit.  It depends a bit on whether your orbit is polar or equatorial, but the variation will be some sort of helix or spiral (or a combination of the two).  Eccentricity also plays a role and changes the character of the spiral, but the point is that if you were to look at your vessel from a top-down perspective focused on Kerbin's north pole, then your vessel's orbital track would usually appear to be some attempt at Spirograph in space.

As your semimajor axis extends farther and farther out and away from the Mun, the orbit changes.  Let's assume for a moment that the Mun is intangible; that is to say, you won't crash into its surface and may set your vessel exactly at the Mun's centre of mass or otherwise orbit arbitrarily close.  At the exact centre, your vessel's orbit is equivalent to the Mun's orbit.  Very close to the centre, the orbit is still mostly circular with a tiny bit of periodicity--or, should you be unfamiliar with that term, 'wobble'.  As you get closer and closer to the boundary of the sphere of influence, your orbital track 'expands' to fill the volume of the Mun's sphere of influence as it traces a toroid (that is a shape like to an inner tube or a doughnut) about Kerbin.  See the attached image for a visual example:

Om0kAkt.jpg

The red track is the closest orbit, the blue track is in the middle, and the green track is the outermost.  Note that the path 'expands' in both directions; so does the sphere of influence extend to points inside the orbit as well as outside.  All three tracks are nonetheless centred on the same orbit circle, and in KSP, this circle would be traced by the Mun's centre of mass.

Eventually, you simply leave the Mun's sphere of influence, but because you're so close to escape velocity at that point anyway, your resulting orbit is essentially (not completely!) co-orbital, which is to say that your orbit is equivalent to the Mun's, but not in orbit of the Mun--which is exactly what you would expect for a vessel that orbits at the Mun's altitude with enough energy to remain there, but is not gravitationally bound to the Mun itself.  The resulting orbital track, then, is eccentric with a periapsis somewhere on the circle traced by the innermost point of the Mun's sphere of influence, and the apoapsis, in like fashion, at the outermost point.

This is the difference between orbiting freely and getting captured into Mun orbit:  when you execute a Hohmann or other transfer to get to the Mun's altitude and enter its sphere of influence, you initially have enough energy that, without a second manoeuvre, you will eject from the Mun's sphere just as quickly as you entered--albeit in a different direction, which is the essence of a gravity assist.  When you make your capture burn, you lose energy with respect to the Mun--that's how you get captured--but you gain energy with respect to Kerbin, because from Kerbin's perspective, you are remaining in a higher orbit.

By increasing your orbital altitude to the limits of the Mun's sphere of influence, you gain energy with respect to the Mun, but because you remain gravitationally bound to it, your energy with respect to Kerbin doesn't change appreciably; it only varies between increasing extremes.  In that respect, you've already wasted the fuel.

To efficiently return to Kerbin, it would be best to dive towards the Mun's surface and set up the burn so that you can make use of the Oberth effect.  Again, this is for reasons that are more commonly encountered when making interplanetary transfers, but the point is that no one makes an effecient interplanetary burn from the edge of Kerbin's sphere of influence.  Sometimes, people will burn for a high apoapsis and then escape at the next periapsis, but that's it.

Edited by Zhetaan
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