Khaur

I think a lot of the confusion comes from that the term "effect" is a bit of a misnomer, at least in how it's used. It's not like there's a basic price in m/s to pay for a given amount of orbital energy, and the Oberth effect comes on top of that to apply a discount or a surcharge. Discarding the Oberth "effect" is akin to discarding Newton's second law of motion: one gives the rate to convert dv into orbital energy, the other to convert thrust into acceleration. Back to the Mun escape case: what we want is a given escape velocity. Up to patched conics approximation errors, that's only tied to the orbital energy around the Mun. How much dv is required depends exclusively on the average speed when performing the burn(s) (weighted to compensate for variable thrust over time). In terms of energy, as long as the burns are along the (orbital) prograde vector, it doesn't matter which way it is pointing. To maximise this we want to stay as close to the centre as we can. Enters constraint number 1, the lithosphere: we really don't want to go through it (atmosphere isn't as hard a constraint and makes things messy, but there's none in our case). That means we have to stay above a certain altitude, so we have to settle for as low a gravity turn as ground clearance will allow. Constraint 2 is on the escape vector. If we want to return to Kerbin we want it close to retrograde, but the reasoning applies to all cases. Unless we're really lucky, that means we have to give up on our grazing single-burn trajectory. If we're on the right hemisphere (actually a little more), we can still try a direct ascent, but it will get less and less efficient the closer to the zenith our desired escape vector is. The other (and only if we're on the wrong hemisphere) option is to transition through an orbit (or a quasi-orbit). This means going a little bit higher to steer clear of additional obstacles, and possibly coasting to apoapsis to circularise, both of which lower efficiency slightly. The upside is that we'll get to perform the rest of the escape burn from a rather low altitude and we have fine control over the escape vector. Which one is more efficient depends on your craft, and how quickly it can reach the desired orbital energy. If it can get up to speed before going higher than the required clearing altitude for the orbit, go for a direct ascent. Otherwise, use an orbit. Additionally, you may want to consider the "free" velocity from the surface rotation, and higher piloting error in the case of direct ascent. If your desired escape vector is straight up, it would requires insane amounts of thrust for option 1 to be better. TL;DR: Save yourself the trouble and just get into a low orbit first if you care about dv.

It doesn't disappear into thin air. You're getting altitude (potential energy) in exchange for that. And that altitude will get converted back to velocity (kinetic energy) when you're on the way down. Horizontal velocity isn't yours forever: look at your velocities at periapsis an apoapsis: unless you're on a circular orbit, they'll be different, despite both being completely horizontal. What is yours forever is orbital energy (kinetic+potential). Well, as long as you don't collide with anything, but at high altitude, even a tiny nudge sideways is enough to avoid collision with the orbited body.

This is just wrong. Thrust is not wasted because it is against the direction of gravity, it is wasted because you could have netted a better Oberth effect by going sideways. Take this simple case: moving radially away from Kerbin at 1km/s, with a TWR of 1 (adjust throttle to stay at no acceleration as you move away and gravity gets weaker). If as you say "the portion of thrust fighting gravity (i.e. along the direction of gravity's pull) is completely wasted", then you'd never escape Kerbin, even though you're moving steadily at 1km/s away from it. That's an obvious contradiction. How much energy you get per dv is dictated by the average speed at which you spend dv. At equal thrust (along the prograde), the lower you stay the faster you'll go on average, therefore the less dv you need to spend to reach escape energy. Now just to be clear, I'm not saying burning radially is an efficient way to escape (it's not in most conditions), I'm saying that your justification of why is based on a misconception.