Higher vs. Lower Orbits

[maddog59]

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?

[VoidSquid]

Lower: Oberth effect.

[king of nowhere]

lower, always lower. the reason is that you get more oberth effect from being closer to the planet.

a couple of  not-really-exceptions:

- if you are using ion engines with really low thrust, you may prefer a higher orbit because it is slower, and it will give you more time for your burn. in some cases, what you lose in oberth effect you may gain in lesser cosine losses

- if you have ISRU, then it's convenient to go to minmus to restock in fuel before leaving. but that's another matter entirely

[maddog59]

4 minutes ago, king of nowhere said:

if you are using ion engines with really low thrust, you may prefer a higher orbit because it is slower, and it will give you more time for your burn.

Early on in the sandbox I tried to do a long duration burn with just the nuke engines from something like 75 or 80km, and ended up driving my ship right back into the atmosphere. My spider sense telling me this just wasn't right finally kicked in at around 35km!  

[VoidSquid]

Yeah, for such longer burns an orbit a bit higher is advisable, you're right.

[king of nowhere]

actually, there are other ways to work around those low thrust limitation too, but i could write treatises on the argument without exhausting it

long story short, oberth effect make it so that it is always more efficient to burn close to the planet. but sometimes other considerations may force you to revise priorities

[Guest]

I usually eject from a 150 km orbit. Lower than that and I have to split them in two or get very tricky about them as they typically take about 10 minutes.

[paul_c]

Lower, for 2 reasons: Oberth effect, and also a shorter orbital period means you can choose the time to burn/which orbit to burn (which is handy if you timewarp and oops go past...) without the target having moved along as much.

46 minutes ago, VoidSquid said:

Yeah, for such longer burns an orbit a bit higher is advisable, you're right.

There is a technique to do a series of multiple burns. Of course, the last burn is the one you "shoot for the target" from, so you need to have made an allowance and the burn before, needs to take into account both 1) the extra time for this previous orbit and 2) that the previous orbit is itself, a bit elliptical, being 'half way there' or whatever, takes a longer orbital period in itself.

The most I've done is 6 burns......and no, I couldn't work out where the eventual ejection occurred, I relied on random chance and/or just waiting an orbit or two for the moon to come round again!

[VoidSquid]

14 minutes ago, paul_c said:

There is a technique to do a series of multiple burns

 

[paul_c]

Indeed, they are entirely calculable, and intuitive to do. I may look at the mod at a later date, thanks for the suggestion.

[Zhetaan]

On 2/22/2021 at 3:42 PM, maddog59 said:

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?

It's a little more complex than a simple higher/lower choice.  Yes, as others have said, the Oberth effect is important and yields great returns for expended propellant, but there is more going on than a straightforward relationship of 'go lower = go faster = go better'.

One thing to keep in mind is that there are different requirements for different destinations.  In order to make an interplanetary transfer, you need a specific hyperbolic excess velocity that is different for each planetary destination (and different for each transfer window, too, owing to inclination and eccentricity variations).  This excess velocity is called that because it is the velocity that your escaping craft still has at the point of escape from its original sphere of influence (i.e., Kerbin's), and thus is 'extra' velocity not needed to escape.  However, because you need to add that velocity in with the initial ejection burn (in essence, 'storing' it for later), it interacts mathematically with the other energy of the burn and thus affects the outcome.

Also, in order to effect a transfer from any closed orbit, you need to reach escape velocity first--otherwise, you can't leave!  Therefore, any manoeuvre to escape a closed orbit about a planet or other gravity source has to account for both the escape velocity (to begin the transfer and break free from the parent body) and also the hyperbolic excess velocity (to finish the transfer and carry the vessel the rest of the way after leaving the parent).   The equation that combines these yields a value called the burnout velocity, which is the velocity your vessel needs at engine shutdown.  Burnout velocity, in turn, is the sum of your orbital speed and your manoeuvre delta-V.  Rearranging it a bit, your manoeuvre delta-V is thus the required burnout velocity minus whatever assistance your orbital speed can give you.

This implies that either decreasing the required burnout velocity or increasing orbital speed saves delta-V.  This is true, but the Oberth effect only affects the second part.

Orbital speed increases as altitude decreases, but so does escape velocity.  To illustrate the concept better, consider this:  at infinite distance, your orbital speed is zero, because you're not in orbit.  At infinite distance, your escape velocity is also zero, because you've escaped.  However, when in a closed orbit, your orbital speed must be less than escape velocity, because it's a closed orbit.  Therefore, as your orbital altitude increases to infinity, both your orbital speed and escape velocity go to zero, but escape velocity goes to zero slightly faster because it starts from a higher value.

Your needed ejection delta-V is a combination of three values.  First, there is the hyperbolic excess velocity, which is a constant for a given transfer.  Second, there is the escape velocity, which is inversely proportional to altitude.  Third, there is the orbital speed, which is also inversely proportional to altitude.  At low altitudes, orbital speed is high and so is escape velocity.  Specifically, the escape velocity dominates the burnout velocity term.  As the altitude rises, the escape velocity decreases faster than the orbital speed does, so the orbital speed gains a slight edge and that reduces the manoeuvre delta-V costs.  However, soon after that, the escape velocity reduces to a point that it no longer dominates the burnout velocity term--eventually, the constant hyperbolic excess velocity portion takes over.  Since hyperbolic excess velocity is not dependent on altitude, increasing altitude after this point only reduces the orbital speed and thereby increases the manoeuvre cost, which is the Oberth effect in action (albeit in reverse).  The concept is called the gate orbit, and is built entirely on the notion of striking a balance between the Oberth effect and the ejection requirements at different altitudes.

As an example, I took the first Kerbin-Duna transfer window and a fairly typical hyperbolic excess velocity of 920 m/s.  With that value in mind, the optimal transfer altitude to conduct the ejection burn is 7,745 km (as seen in the altimeter; the distance-from-centre is 8,345 km), and a transfer from this altitude costs 650.5 m/s.  The best transfer I could find in that window has a hyperbolic excess velocity of 855 m/s, and its optimal altitude is 9,062 km with a manoeuvre cost of 604.6 m/s.  By contrast, transferring from a 100 km low Kerbin orbit on that same transfer costs 1,060.9 m/s.

The problem, of course, is that it costs 1,200 m/s to get to 9,062 km from 100 km.  That trade-off actually trades away all of the savings from choosing the optimum altitude and costs extra besides, but the information is still useful.

For example, if you are assembling a large interplanetary vessel in orbit, there's something to be said for assembling it in the gate orbit.  The cost of putting the vessel there is borne by the assembly craft and propellant tankers, not the vessel itself.  Also note that the gate orbit for Duna is reasonably close to the Mun, which is a reason for having a fuel infrastructure there--and if you want to transfer from the Mun directly to Duna, then you can multiply the effect (that is to say, choose the gate orbit for a Mun hyperbolic excess velocity equal to an ejection burn to Duna from Kerbin at the Mun's altitude) and see even greater savings.

Duna and Eve gate orbits are fairly high because they don't have large hyperbolic excess velocity requirements for the transfer.  A gate orbit to Jool, on the other hand, is at about 270 km (but that makes it a good place to put a Kerbin-Jool propellant depot or embarkation station if you want one).  This is because, as the hyperbolic excess velocity increases, you need to get more from the Oberth effect to supply that velocity.  Alternatively, to use my illustration from above, a higher hyperbolic excess velocity comes to dominate the burnout velocity term that much sooner as altitude increases.

The other reason that gate orbits are useful is because they assist with interplanetary operations at the destination.  Suppose that you want to send a vessel to some distant place, but you also want to return it to Kerbin.  Sending it to the gate orbit at the destination allows you to 'start' at that higher orbit for the return trip and thus obtain all of the delta-V savings that you can.

Edited February 24, 2021 by Zhetaan

[paul_c]

Sounds good - I nearly followed it!

[camacju]

As other people have wonderfully explained in this thread, a lower orbit is better. Ideally you'd want to be as close to the atmosphere as possible - and if your craft is properly drag optimized, you'd even want to be inside the atmosphere. I've seen missions that circularized at a 10 km orbit to maximize the Oberth effect as much as possible, since the crafts had so little drag. I prefer a 72-75 km orbit, except for when I'm feeling particularly lazy, in which case I go into a higher orbit.

Low TWR actually isn't an issue for low orbits, as long as you plan properly. Even if you don't have the requisite thrust to weight ratio, you can split the burn into multiple smaller burns so you don't accidentally do an atmospheric dive. This also does more of the burn at periapsis when you're moving faster, taking extra advantage of the Oberth effect. This also has been addressed in previous comments.

The main problem with periapsis kicks is that if your destination is outside Kerbin's sphere of influence, you need to perform an ejection burn, which might take a long time. If your TWR is low enough, then something like a 10 minute ejection burn just isn't going to work. However, if you've needed to optimize cost and/or mass enough to require that low of a TWR, then you are probably going to benefit from gravity assists.

Using gravity assists, you can actually get anywhere in the Kerbol system for only the cost of the initial Mun transfer. This means that all you need to do is wind up your orbit to the Mun's orbit, which doesn't suffer from any TWR limitation. This is actually the cheapest possible way to get anywhere in the system, so it's a very useful trick even when you don't strictly need it. I've done a bunch of missions using this flight path.

The first step is to get an orbit that is just below the Mun's orbit, and get a gravity assist that passes behind the Mun, throwing you into a higher orbit of Kerbin. Then, do it a couple more times to eject from Kerbin with as much energy as possible.

There's a problem with this, however: the energy from Mun assists isn't actually enough to get anywhere by itself. It's just short of an Eve or Duna transfer. This means that instead of ejecting with maximum energy, you need to eject into an orbital resonance with Kerbin. What this means is that the period of your orbit, divided by the period of Kerbin's orbit, forms a nice fraction, like 6/5 or 5/6 or 4/5. Then, after a few orbits, you'll encounter Kerbin again, letting you do another gravity assist.

The gravity assist won't be off of Kerbin, however, Since you left Kerbin with a certain velocity, you'll encounter Kerbin with that same velocity. You won't gain any extra energy from swinging around Kerbin. Instead, you use the Mun another time, passing behind the Mun while you're on a hyperbolic trajectory. (To encounter the Mun in the right place in its orbit, do a burn one or two orbits before, in order to adjust your orbital period). This gravity assist gives you less energy than the initial Mun assists, but it'll probably be enough to get you to Eve. If not, just eject into another orbital resonance, and do the same thing one more time.

Once you've gotten an Eve assist, you're practically home free. Pass in front of Eve to get an orbit crossing Moho's orbit; pass behind Eve to reach Kerbin with more speed than you left it and continue to Duna, Dres, or Jool.

[Spricigo]

52 minutes ago, camacju said:

The gravity assist won't be off of Kerbin, however, Since you left Kerbin with a certain velocity, you'll encounter Kerbin with that same velocity.

I wonder why you assume this. Multiple gravity assist from the same celestial body were used extensively both in KSP and in real life.

[camacju]

9 minutes ago, Spricigo said:

I wonder why you assume this. Multiple gravity assist from the same celestial body were used extensively both in KSP and in real life.

You may be thinking of something else. Normally, when you do multiple assists from one body, it's because you have a deep space maneuver half an orbit away, in order to increase your relative velocity. While this is more efficient than a naive direct burn, it's still less efficient than a completely ballistic trajectory. This was used in the MESSENGER mission, Parker Solar Probe, and the Stardust mission (although I'm not 100% sure on the last one).

The one exception to this is when your relative velocity is high enough that you can't bend your trajectory all the way around with just one assist. However, if you look at the numbers, your relative velocity before and after the flyby will be exactly the same. There's a pretty simple argument involving potential and kinetic energy as to why this is the case. To actually gain energy, you need to do gravity assists off 2 or more bodies, hence the Mun and Kerbin flybys before getting to Eve.

You can see this in the Kerbin-Eve-Kerbin-Kerbin-Jool route that many people use to get to Jool for a reasonably low delta-v cost. The Kerbin->Kerbin assist does not change your Kerbin relative velocity at all. Instead, it redirects that velocity closer to prograde, raising the orbit. Meanwhile, Kerbin->Eve->Kerbin does arrive back at Kerbin with greater relative velocity than you left, since you passed by a second body during that time - and that's the important part.

Edited February 24, 2021 by camacju

[Spricigo]

 

1 hour ago, camacju said:

You may be thinking of something else.

No. I'm just considering the physics of the situation and wondering why you assume that it will be different when it happens a 2nd time. Why you think one cannot do this?

Spoiler

 

 

[camacju]

2 hours ago, camacju said:

it's because you have a deep space maneuver half an orbit away, in order to increase your relative velocity

 

27 minutes ago, Spricigo said:

Why you think one cannot do this?

I think there may be some misunderstanding here. Note the deep space maneuver in your screenshot. This means you expend extra delta-v where that's not necessary if you're smarter with gravity assists.

Edit:

It seems you haven't actually read my post.

2 hours ago, camacju said:

The one exception to this is when your relative velocity is high enough that you can't bend your trajectory all the way around with just one assist.

This means that you can use two gravity assists in a row off the same body. However, neither gravity assist will change your relative velocity. It simply redirects your pre-existing relative velocity into a different direction.

28 minutes ago, Spricigo said:

wondering why you assume that it will be different when it happens a 2nd time

You seem to be putting words in my mouth here. My point is that it isn't different. Each gravity assist will have a similar effect. No single gravity assist will change your relative velocity to the assisting body. I'm not sure where the block is here.

Edited February 24, 2021 by camacju

[Spricigo]

6 minutes ago, camacju said:

Note the deep space maneuver in your screenshot.

Let's try again.

Spoiler

I get that Kerbin is not the Sun and Mun is not Kerbin. But the physics is the same.

 

[Zhetaan]

@camacju:

Quote

Low TWR actually isn't an issue for low orbits, as long as you plan properly. Even if you don't have the requisite thrust to weight ratio, you can split the burn into multiple smaller burns so you don't accidentally do an atmospheric dive. This also does more of the burn at periapsis when you're moving faster, taking extra advantage of the Oberth effect. This also has been addressed in previous comments.

While avoiding an atmospheric dive is certainly helpful, the main benefit of periapsis kicking is that it reduces cosine losses:  for this reason, periapsis kicks make sense for low-thrust craft even when they are not at risk of falling into an atmosphere, and they also make sense for high-thrust craft that need to make very long burns (for whatever reason that they may need to do so--escaping the star, perhaps).  This comes of the fact that velocity is both a magnitude and a direction, and when thrust is low, the burn takes so long that the direction can change substantially over the duration of the burn.

Also, the benefit of the Oberth effect does not increase with periapsis kicks.  The Oberth effect only operates on one's speed; whether you acquire this speed in one burn or in several is of no consequence.  For example, let's say that you need to increase your velocity from 2,300 m/s to 2,600 m/s for some manoeuvre.  You'll get the same benefit from the Oberth effect when going from 2,599 to 2,600 m/s no matter how many burns it took to get there.  I will concede that with enough precision, one can have an arbitrary number of infinitesimal burns at the exact periapsis, and that would maximise the orbital speed at which one makes the burn.  However, that is no different from an instantaneous burn, which is how we generally calculate these things anyway.

10 hours ago, camacju said:

The one exception to this is when your relative velocity is high enough that you can't bend your trajectory all the way around with just one assist. However, if you look at the numbers, your relative velocity before and after the flyby will be exactly the same. There's a pretty simple argument involving potential and kinetic energy as to why this is the case. To actually gain energy, you need to do gravity assists off 2 or more bodies, hence the Mun and Kerbin flybys before getting to Eve.

I would like to respectfully qualify that.  The energy will be the same, but energy is a scalar, not a vector--one can gain energy from a purely ballistic repeat flyby provided that the direction changes on leaving the sphere of influence, which is, of course, how gravity assists work in the first place.  The purely ballistic K-E-K-K-J transfer from Kerbin (eventually) to Jool is fairly well-established, and it includes a Kerbin repeat-flyby.

I will add the caveat that you are correct in that you cannot start from Kerbin and use it again for the next flyby, but that is because in that case, Kerbin would be at one of the apsides of the orbit.

Maybe if you did a bad transfer with a lot of radial, you could use Kerbin to fix the trajectory, so perhaps even the caveat has caveats.

Edited February 24, 2021 by Zhetaan

[OHara]

10 hours ago, camacju said:

Since you left Kerbin with a certain velocity, you'll encounter Kerbin with that same velocity.

That seems like a very safe assumption.  When the craft's elliptical orbit returns to the same point, it has the same velocity, because orbits under central gravity are periodic.  The same holds for Kerbin, so I expect the same relative velocity.  (There are two intersections, and if both orbits were significantly elliptical you could leave one intersection and then re-encounter the other, but that opportunity does not come up often.)

Substituting Kerbin->Mun, I would expect if your craft leaves Mun with some speed, and does no burns, the next encounter will have the same relative velocity.

Given that, I need help to understand the intended message in the image:

6 hours ago, Spricigo said:

Let's try again. [Screenshot of two encounters with Mun]

It looks like the elliptical purple orbit leaves the Mun and re-encounters it at the same point on the Mun's orbit.   

The Mun is in a slightly different position on the second encounter, so the craft goes much deeper into the Mun's SOI, has a higher speed at Mun-periapsis, and gets a much larger assist on the second encounter.  Is that the point?

Usually, you can plan how close to fly-by to choose the strength of the first assist.   The exception is what you mentioned above, "when you can't bend your trajectory all the way around with just one assist."  The image could be an illustration of that exception, simply choosing to stay far from the Mun on the first assist.

For its use in planning gravity assists, the expectation "you come back with the speed you left" still seems to hold.

[camacju]

2 hours ago, OHara said:

(There are two intersections, and if both orbits were significantly elliptical you could leave one intersection and then re-encounter the other, but that opportunity does not come up often.)

Yes - I should have clarified that I was talking about a body in a mostly circular orbit.

8 hours ago, Spricigo said:

Let's try again.

You again miss my point. You have two Mun assists, but your relative velocity to Mun will not change at all. Again, I've been talking about relative velocity this whole time, which is the most important factor in gravity assists. If you actually fly this path then you'll realize that while the direction of your exit velocity will change, because that's the entire point of a gravity assist, your Mun relative velocity won't.

2 hours ago, Zhetaan said:

Also, the benefit of the Oberth effect does not increase with periapsis kicks

Not to be pedantic but you say this and then immediately come up with a case where you do benefit from periapsis kicks. You want to do as much of the burn as close to periapsis as possible for this exact reason. Cosine losses aren't actually a problem because gravity is redirecting your velocity. If cosine losses were a problem, then doing two prograde burns on opposite sides of the planet would cancel each other out. Instead, they raise you into a higher orbit (Hohmann transfer).

2 hours ago, Zhetaan said:

I would like to respectfully qualify that.  The energy will be the same, but energy is a scalar, not a vector

You're right. I meant to say relative velocity.

[Spricigo]

1 hour ago, camacju said:

You again miss my point.

Yes, I do. That is why I'm asking.

I follow what you are telling will happen if you meet the body a second time with the same velocity of the first encounter. But what I'm wondering is why you apparently assume you will get the same velocity in both encounters to begin with.

As @Zhetaan points out: velocity is a vector, has both a magnitude and a direction. All it takes is to meet the celestial body slightly ahead/behind in it's orbit and the velocity will be in a different direction, it will be a different velocity. It really appear to me that you are not taking change in the direction in account.

 

4 hours ago, OHara said:

The Mun is in a slightly different position on the second encounter, so the craft goes much deeper into the Mun's SOI, has a higher speed at Mun-periapsis, and gets a much larger assist on the second encounter. 

The point is that it took me a small, but noticeable, lesser amount of deltaV in the initial maneuver to make it in 2 consecutive flybys instead of a single one. Shouldn't that be preferred if the idea is to use the lesser amount of deltaV?

4 hours ago, OHara said:

the expectation "you come back with the speed you left" still seems to hold.

Oh well, it will hold forever if you don't get an encounter with the celestial body.  It's the encounter with the celestial body that breaks the cycle.

The encounter with the celestial body will push you from the blue orbit to purple orbit and then again to green orbit. Every subsequent encounter will change your orbit(the velocity, the energy) in some manner.

So, I don't get why we are talking about what happens in between the encounters when we are interested in what happen during the encounter. If nothing happens in between the encounter is inconsequential.

 

[camacju]

29 minutes ago, Spricigo said:

But what I'm wondering is why you apparently assume you will get the same velocity in both encounters to begin with.

Not necessarily same speed and direction - only same speed relative to the body. Fly it yourself and see.

29 minutes ago, Spricigo said:

All it takes is to meet the celestial body slightly ahead/behind in it's orbit

How do you propose to do this without a deep space maneuver? Sure, ellipses can intersect multiple times, but you don't get any benefit you wouldn't get from a simple resonant orbit.

30 minutes ago, Spricigo said:

change your orbit(the velocity, the energy) in some manner.

Changes relative speed to Kerbin, yes. Changes relative speed to Mun, no.

31 minutes ago, Spricigo said:

Oh well, it will hold forever if you don't get an encounter with the celestial body.  It's the encounter with the celestial body that breaks the cycle.

Again, fly it yourself and see what your relative velocity to Mun is at each encounter.

[Zhetaan]

4 hours ago, camacju said:

Not to be pedantic but you say this and then immediately come up with a case where you do benefit from periapsis kicks. You want to do as much of the burn as close to periapsis as possible for this exact reason. Cosine losses aren't actually a problem because gravity is redirecting your velocity. If cosine losses were a problem, then doing two prograde burns on opposite sides of the planet would cancel each other out. Instead, they raise you into a higher orbit (Hohmann transfer).

Not to compound the pedantry (sorry; I couldn't resist), but my example was to illustrate that the benefit of periapsis kicks over a single burn does not come from the Oberth effect.  Yes, as the apoapsis rises higher, the speed at each periapsis encounter for subsequent kicks increases, and therefore the Oberth effect also increases; however, these increases occur whether you do it in one burn or in ten.

The principal benefit of doing periapsis kicks is that it allows one to concentrate the manoeuvre in space (at the cost of time) at the periapsis, thereby coming closer to a practical realisation of an instantaneous burn than would be otherwise possible.  In that respect, since the proximity to the periapsis means that the velocity at the time of the burn is closer to the maximum possible, it is also true that there are tiny increases to the Oberth effect from kicking over not kicking.  However, I contend that these increases are irrelevant because failing to kick results in far greater cosine losses, and the point of kicking is to avoid those losses.  Furthermore (for the sake of completeness), a more powerful engine that does not need to kick also concentrates the manoeuvre at the periapsis, which also means that the velocity is closer to the maximum possible, which gives the same increase to the Oberth effect--in other words, the principal benefit of kicking is still in that it allows a weaker engine to pretend that it is a stronger one by burning multiple times in (nearly) the same place and with reduced cosine losses, not that kicking in itself somehow grants more benefit from the Oberth effect than one may otherwise realise.

Put in a different light, consider the vis-viva equation.  The orbital speed at a specific radial distance on an orbit, given that the primary is constant (for simplicity's sake, let's assume that we're just working in Kerbin orbit) is dependent only on the semi-major axis and that radial distance.  Unless you are at an apsis or on a circular orbit, any valid radial distance corresponds to only two points on a given orbit.  Let's further assume that the radial distance we're interested in is that of the periapsis, since we're talking about periapsis kicks.  That reduces the number of corresponding points to one.  Since the vis-viva equation is in three variables, and one of them (the radial distance at periapsis) is held constant for the purpose of periapsis kicking, the other two variables can only vary with one another.  What this means is that for a given orbital speed at a given periapsis, there is only one possible semi-major axis, but it also means that for a given semi-major axis, there is only one corresponding orbital speed at that periapsis.  As you kick at the periapsis and increase speed, the semi-major axis increases with that speed (and so does the Oberth effect), but that will happen anyway whether it is done in one burn or a pulsed series of kicks.

The reason that this is important is because it means that the interval between kicks is irrelevant.  By necessity and the requirement of burning at the periapsis, the interval is limited to a discrete integer number of orbits, whatever the orbital period may be, but that interval can be zero and it still works because a single (instantaneous) burn can be considered as the limiting case of an infinite number of periapsis kicks with an infinitesimal interval between them.  The Oberth effect operates on that orbital speed regardless of whether the speed is reached in one burn, which is to say that the Oberth Effect does not increase because of periapsis kicking.  It increases because the periapsis speed increases, however one may choose to accomplish that and no matter how long it takes to do so.

Furthermore, I contend that cosine losses are a problem precisely because gravity is redirecting your velocity.  Prograde burns at the apsides are efficient because the prograde direction at those points is exactly perpendicular to the force of gravity.  Outside of those points, either the prograde direction is not perpendicular, in which case some of the burn is lost fighting gravity, or else the perpendicular to gravity is not prograde, in which case some of the burn is lost fighting the orbit.  This is mathematically inescapable because the perpendicular to a tangent of an ellipse (this tangent corresponds to prograde) does not go through the focus except when it is the major axis (which is the line of apsides for an orbit).

Lastly, the reason that burning prograde at opposite sides of an orbit does not cancel out is not because of the direction, but because velocities in orbit are not strictly additive.  There are certainly exceptions at various times and places, to be sure, but it is the energies that add, not the velocities.  A prograde burn at periapsis and an equal prograde burn at apoapsis are burns in opposite directions, but they both correspond to an energy increase in despite of their cancelling directions.  Total specific orbital energy is time-invariant (meaning that it is constant for a given orbit), but that total is the sum of a perpetually-shifting exchange between specific potential and specific kinetic energy.  Specific potential energy at a given point relates only to your radial distance from the primary, specific kinetic energy relates only to your speed, and total specific orbital energy relates only to the semi-major axis of the orbit.  Burning prograde at the periapsis raises the apoapsis because it increases the specific kinetic energy, and since the specific potential energy does not change (since you are still at the periapsis), it must also increase the specific orbital energy, which is reflected by an increased semi-major axis.  Burning prograde at the apoapsis also increases the specific kinetic energy, and since the specific potential energy does not change (since you are still at the apoapsis), it also increases the specific orbital energy by increasing the semi-major axis.

[camacju]

I haven't actually put that much thought into this. Thanks for explaining it quite well!

I tend to sidestep this issue by putting enough thrust into my transfer stages that I don't need any periapsis kicks in the first place. So I haven't had much experience with it. I'm better with chaining gravity assists together (: