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# Higher vs. Lower Orbits

## Question

I'm sure there's a mathematical solution to this: if you're planning to send a ship out of Kerbin orbit, is it better (or is there a trade-off) to launch it into a lower or higher parking orbit prior to the transfer burn to the destination?

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47 minutes ago, camacju said:

Fly it yourself and see.

What can I say? Done.

Despite the lack of any maneuver after the initial burn  the orbital energy changed with each subsequent encounter with the same celestial body.  No encounter with a different celestial body also.

Granted, the resulting trajectory is not as useful as what I can get a few course correction along the way (I'd rather reach Jool in the same century) but that is besides the point.

Perhaps what is throwing me off the track is that you are talking about things that don't change in regard the assisting body when I'm talking about what happens in regard to the parent body?

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Yes, that's it. Your velocity relative to the assisting body will not change, which means that at some point you'll saturate the gravity assists and not be able to get any more energy relative to the parent body. See the comment just above yours for more detail.

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@OHara @camacju if i get it right this time:

Is a matter of practicality. At some point the cost of correction burns is more than what you get from the gravity assist.

Oh well...

That is why I go for a high energy transfer if I can afford the deltaV cost to pull it off. Burn like a madman and get done with it.

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You're close enough. At some point you don't get significant extra benefit from repeating gravity assists, so you need to swing off a different body. That's why in a K-E-K-K-J route some people stick an extra Eve and Kerbin assist between the two Kerbin assists, to "reset" the gravity assists and get more Kerbin-relative velocity.

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2 minutes ago, camacju said:

At some point you don't get significant extra benefit from repeating gravity assists, so you need to swing off a different body.

Well,  we are not in agreement then.

It may surprise you how far, far away that point is.

K-E-K-K-J is a think not because K-K-K-K-J is not possible(or practical). It is a thing because Eve is more massive than Kerbin and as such can give an assist bigger enough to compensate the need to lower the orbit to make it happen. Also, you can wait for Eve to get in a more convenient position (while you need to settle for resonant orbits for Kerbin) and see the closer approach marking while you depart from Kerbin. Eve is a bit mroe practical, that is it.

Anyways, at this point I really don't see how to settle the divergence other than careful testing. This can be quite laborious, my game time is limited and I rather not tell how would you expend yours. Thus, I rest my case.

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I've done missions that quite easily prove the point I'm making. And you're not understanding the reason for the Eve assist.

First, I hope you can agree that the most efficient use of a gravity assist is to eject perfectly prograde relative to the assisting body's orbital velocity. I also hope you agree that no matter what you do, if you only use one assisting body, you cannot get more energy than a perfectly prograde ejection. If you're still confused, the Wikipedia article on gravity assists should clear some things up.

Let's do some example orbits to demonstrate.

First, I do a 1200 m/s prograde burn from low Kerbin orbit, such that I eject perfectly prograde.

My speed relative to Kerbin upon SOI exit is 1350 m/s. My apoapsis is 25708 Mm. This is the highest altitude that I could possibly achieve with a 1350 m/s exit velocity from Kerbin, for what should hopefully be obvious reasons. (Kerbin's orbital velocity plus my relative velocity are both directly prograde).

Then, I do the same burn, but adjusted so I eject radially  and I encounter Kerbin one year later.

Then, to get the maximal effect from the gravity assist, I make sure that my exit vector is perfectly prograde. Note that my Kerbin exit velocity is 1350 m/s and my Kerbin entry velocity is the same. In other words, it's exactly the same as the previous relative velocity.

Note also that my final apoapsis is the same as if I did the initial exit burn properly in the first place! No matter how many Kerbin flybys you attempt, you cannot get any extra energy from them! You need a second body. This is the true reason for the Eve assist, which I will show later.

Next, I do the same magnitude of burn, but adjusted so I eject retrograde and get an Eve gravity assist. I purposely don't use the full effect of the Eve gravity assist, to simulate the effect of a less massive body. Hopefully this will satisfactorily disprove your assertion that Eve's greater mass plays any essential role in this.

Because of my normal and radial component to the burn, my Kerbin exit velocity is lower - 1005 m/s. However, this is still instructive.

Now, after my Eve gravity assist, I encounter Kerbin again. Note that my Kerbin entry velocity is higher than 1005 m/s! This is the real reason for the Eve gravity assist - to increase your relative velocity to Kerbin. An encounter with a body of any size is sufficient for this, which I've hopefully proven with my low powered gravity assist. Eve's mass just makes this require fewer assists. Mass doesn't really matter with gravity assists as much as the relative energy differences between orbits.

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2 hours ago, camacju said:

I also hope you agree that no matter what you do, if you only use one assisting body, you cannot get more energy than a perfectly prograde ejection.

How much energy "a perfectly prograde ejection" is supposed to be?!!

Also  "I hope you agree about the thing you are asking about" is not convincing.

I asked because I don't see the argument/evidence that made you get from [I will use a gravity assist] to [Kerbin will not help enough]*. Conversely, it seems I'm not able to present you what makes me think it may not be the case. As I said in the previous post, that is an impasse I don't how to resolve and at this point I'd rather let it be.

* I get that is not exactly what you said, just pointing where is the missing link.

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Ok this'll probably be my last reply here. If what I've explained already hasn't been clear enough, nothing will ever be.

The measure of energy that I use is to get apoapsis as high as possible. When I do a gravity assist, that corresponds to leaving the assisting body in a direction that is exactly prograde. There's no need to say why this is the case; it should be enough that it is the case. This is why I say that a prograde ejection maximizes your energy.

The best way to realize this is, again, to try it yourself. Without doing that, it's a bit like explaining primary colors to someone who's been blind since birth.

Eject prograde from Kerbin. Without using any subsequent maneuvers, try to reach a significantly higher altitude over the Sun than you originally had. You won't.

Edit: I'm racking my brain trying to figure out what's being missed here. It's probably something that I think is obvious enough not to need explanation, but is actually quite unintuitive at first and only becomes obvious through practice. If that's the case, then this communications mishap is probably my fault. (For the record, the members of the more technical KSP Discord servers are all of the same opinion as I am).

Edited by camacju
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16 minutes ago, camacju said:

The measure of energy that I use is to get apoapsis as high as possible.

Problem is: "as high as possible" is not a quantity.

If I understood correct your hypothesis can be rephrased as:

With an initial ejection burn of no more than [valueA]m/s , using only one assisting body, you cannot get enough energy to raise you apoapsis above [valueB]Gm

If you provide valueA and ValueB (along some other restriction I could have missed), then we can look for counterexamples of that. Otherwise we simple have no way to test it.

Inability to test don't prove the hypothesis correct,  only make impossible to demonstrate if it is wrong.

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I hope the original poster has been happy to tolerate this deviation from his original question: "are lower or higher starting orbits better for interplanetary trips?"

The answer: lower is better . . .  unless you have the option to refuel in orbit, in which case Zhetaan points out that the 'gate orbit' is the ideal altitude for your asteroid or other refueling station.

18 hours ago, Spricigo said:

How much energy "a perfectly prograde ejection" is supposed to be? [in the context of gravity assists]

To be quantitative, the specific energy (energy-per-mass) of a craft, in orbit at distance r from central body with mass M, is  v²/2 − GM/r.  This energy is negative for bound orbits.

The trick with gravity assists is to bounce off an assisting body that has orbital velocity vassister.  Suppose we approach the assisting body with relative speed vrel, measured when the assister's gravity has not affected the crafts velocity, so in KSP when the craft is barely inside the assister's SOI.  It happens to be true that the interaction with the assister only changes the direction of vrel, so after the assist the craft has velocity around the central body v = vassister + vrel   using bold-face to indicate the sum of vectors.

So the idea is that the full potential of the assist is had when the relative velocity of the craft vrel is in the same direction as the orbital velocity vasssister, and that full potential is to reach a specific energy (vassister + vrel)²/r − GM/r in the case where the magnitude of the sum of the parallel vectors is the sum of the speeds.

For the case of Kerbin assists as in Kerbin-Eve-Kerbin-Kerbin-to-Jool, the gravitational parameter GM of the Sun can be found in the wiki, but maybe a more intuitive way to understand the size of GM is to use another fact from Newtonian gravity:
for circular orbits v² happens to equal GM/r, so for the sun GM = (vassister)² rassister= (9285 m/s)² 13600 Mm, figured from parameters of Kerbin's orbit.

This way it is easy to see how gravity assists, if we can approach the assisting body with relative speed comparable to its 9285m/s orbital speed, have the potential to give us a v² big enough that v²/2 can overcome −GM/r --- enough to escape the system.

For an example, the craft might exit Kerbin's SOI at 2500m/s relative to Kerbin.  (A burn of 1800m/s in low-kerbin orbit would give this speed at SOI-exit.)  Depending on the direction of that SOI exit, the specific energy in orbit of the Sun is between
( vassister − vrel)²/2 − GM/r   = (9285−2500)²/2 − ((9285 m/s)² 13600 Mm)/13600 Mm   = − 63'190'000 m²/s²  for a retrograde exit, and
( vassister + vrel)²/2 − GM/r   = (9285+2500)²/2 − ((9285 m/s)² 13600 Mm)/13600 Mm   = − 16'768'000 m²/s²  for a prograde exit

An orbit with semi-major axis a has specific energy −GM/(2a),  so the energies above correspond to orbits with
2a = 18550 Mm,  one end at Kerbin 13600Mm the other at  4950Mm just below Moho, and
2a = 69920 Mm,  one end at Kerbin 13600Mm the other at 56300Mm not quite at Jool.

The idea discussed above is that repeated encounters with Kerbin alone will have the same 2500m/s speed  on each entry into Kerbin's SOI, so  the potential for final orbits around the sun is between the bounds above.

Edited by OHara
undo typos
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Thanks OHara, that's a good summary of how to get a grip on the concept of gravity assists, why they work and a good start to "building" your own mission plan using them. Its a bit beyond me, I've always gone direct or let the randomness of the maneouver node thing suggest something by chance; but one day I may get to grips with it all!

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5 hours ago, OHara said:

I hope the original poster has been happy to tolerate this deviation from his original question: "are lower or better starting orbits better for interplanetary trips?"

Everyone left me in the dust LONG ago!

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10 hours ago, maddog59 said:

Everyone left me in the dust LONG ago!

You and me, both.

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17 hours ago, maddog59 said:

Everyone left me in the dust LONG ago!

Even with all of the simplifying assumptions that KSP makes with respect to physics, there's still a lot going on.  KSP attracts a lot of intelligent people who want to understand some of the technical depths (including you:  you did ask the question, after all) so perhaps it was inevitable.

For the purposes of directly (and briefly!) answering your question, lower is better, except in certain very specific circumstances.

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