K^2

If the gravitational force is applied to individual modules, tidal forces will rip a 5km long ship apart. Though, that should happen long before you manage to land.

I was actually assuming a constant mass, but since m'(t) is a constant under constant thrust, I doubt it'd make that much difference. And yeah, after giving up trying to solve it by hand, I ran it through Mathematica.

Heh. I wanted to see if I can at least get an analytic solution to constant altitude takeoff/landing. No dice. The solution can be written in terms of elliptic integrals, which means an infinite series is the closest you can get to a closed form.

You can't use standard variation techniques to optimize the trajectory. Any trajectory where full thrust is applied is locally optimal. (Locally, I can go to coordinate system that's free-falling and x axis is aligned with direction of thrust. Then T = m x'', and Euler-Lagrange equation is the trivial 0 = 0.) Global variation is very difficult, because it's not clear what the boundary condition should be, and how to optimize the overall trajectory. The only thing that even comes to mind is just doing numerical optimization. I was having same difficulties trying to find optimal gravity turn for the ascent through atmosphere. Speaking of which, I was thinking of take-off instead of landing. Neglecting mass changes, they are time-reversals of each other. If we simply assume that we should be diverting as much thrust as possible to horizontal acceleration, we get constant-altitude right until the point that the circular orbit at zero altitude is achieved. Thereafter, it's just a matter of going to a transfer orbit. So it really does sound like optimal landing will include a switch to constant altitude only after you circularize. Of course, this only opens up the question of whether a direct pro/retrograde burn is the optimal way to transfer/circularize.

Yes, I also expect it to be in between. And maybe you're right. Maybe it is all about keeping gravity perpendicular to the velocity vector. As far as starting the burn a bit before periapsis, it might make sense to go from transfer orbit to a circular with a pure retrograde burn, and then switch to constant altitude. One more thing to try, I guess. But damn, these thrust equations are starting to get complicated.

There is an actual force acting on the craft. It has nothing to do with the r'Æ' term in the equation. Consider a particle moving in a straight line in polar coordinates. The straight line x=r0 is r(Æ) = r0 sec(Æ). For a particle moving at constant velocity, Æ(t) = arctan(vt/r0). Therefore, we get r(t) = (r0² + t²v²)1/2. Acceleration in polar coordinates is given by a = (r'' - r Æ'², Æ'' + 2 r' Æ' / r). Substituting the above r(t) and Æ(t), you get a = (0, 0). The r'Æ' term is still there, but it cancels perfectly with the Æ'' term. This is the difference between proper acceleration and coordinate acceleration, as this article discusses.

Yes, but the r-dot phi-dot term is just coordinate acceleration. It's not true acceleration. The fact that it goes away in Cartesian coordinates is the simple test for that. More importantly, it will never alter the speed of the craft. So I don't have to worry about it with the landing. But at any rate, the best way to resolve this is just to run the test. I just need to finish programming the thrust function to make sure it maintains proper orbit. (For constant altitude I did use polar coordinates to simply counter the effective weight of gravitational + centrifugal acceleration, but I really don't want to mess with all of the relevant terms for elliptic orbit. I'll just do constrained motion with undetermined multiplier.) Edit: Oh, and yes, I kind of forgot about moon's rotation, so I'm also doing a non-rotating case. For the Mun, though, I doubt 9m/s of sidereal velocity will make that much difference.

Well, the constant-altitude approach is effectively flying a circular-orbit-only-slower. So I don't really see the difference in that regard. And I only have to consider Coriolis acceleration in the rotating frame. And yes, for the landing I was looking at, the burn towards the ground is only 2s out of 40, but still, if you are right about why it works, coming down along elliptic instead of circular should still improve on that, even if by very small amount.

I know my intuition on the matter has been compromised, but I don't think it's that simple. Yes, you don't let gravity speed you up, but you are burning at an angle as well. In fact, early on you have to burn towards the ground. So you have to assist gravity in keeping you at zero vertical velocity. I would test one more procedure. Instead of starting the burn at the periapsis and maintaining constant altitude, how about starting the burn before periapsis in such a way that you always stay on the same elliptical trajectory and come to halt exactly at the periapsis. Now you are always burning angled down, preventing gravity from altering your trajectory, but your altitude and vertical velocity are changing. If the main reason why constant altitude woks has to do with not allowing gravity to speed you up, this should work even better, since you never burn towards the ground. If the reason constant altitude works well is because all of the burn is done at minimal altitude, then this new approach should perform worse.

You can do high power burst transmissions with very simple/light equipment. Continuous communication is out of the question, of course, so the thing would have to be very autonomous. But it can send back some data, including pictures.

Actually, it is the same thing. Gravity variations are going to be minimal. A burn that optimizes local d||u|| will therefore also minimize time spent under influence of gravity, and that's the only way you inflate required delta-V. In other words, if you minimize time required for braking and find locally optimal d||u||/dv, you have an optimal solution. It's all nicely monotonous, so there is just no room for it to be anything else. Edit: Turns out I'm wrong. At low TWR, maintaining 0 vertical velocity does reduce fuel requirement by a few percent. I'll upload the computations in a bit. Now I just need to figure out why it works out that way. Could be the fact that this allows for lowest periapsis. This would be consistent with the fact that increasing TWR seems to reduce the advantage almost to nothing. Naturally, with higher TWR, suicide burn can be initiated at a lower altitude.

It's pretty straight forward. You have a craft traveling at velocity u and you use dv of your delta-V reserve to try and slow down. The total change in velocity is du = dv + g dt. However, what you are interested in is optimizing d||u||. If you break it down to components and take appropriate limits, you get maximum d||u|| when dv runs along -u. But yeah, I can do numerical integration of these trajectories. I've been doing a lot of these for trying to optimize ascent with air resistance. In vacuum, it's almost trivial. I'll do a full write-up.

That is not true. Burning retrograde optimizes fuel consumption. It's simple calculus. Something wasn't done right in these experiments.

Correct. The actual altitude will read as 14km, of course, since the game will subtract the radius of the Mun. Because the ship will slow down as it races from periapsis to apoapsis. (Or vice versa, speed up on the way down.) This is just consequence of conservation of energy.

551.71 m/s is the orbital velocity at 14km altitude. You can use the formula I posted earlier with r1 = r2. As for the difference, I suspect rounding errors. They are more than sufficient to introduce a discrepancy of 1 part per 1,000.

Very much so. Yet 4gH/vt formula goes very nicely with all of the empirical results available on "best" ascent scenarios. While a loose approximation, it's a pretty good tool for estimating how much you are going to save by change of elevation, for example, which will change vt value.

Not once you started the deceleration burn. That drops apoapsis, causing your trajectory to cut through the planet/moon. So if it's just a matter of, "Oh, crap. Missed my window." Then yes, a small correction burn will get you to miss the surface and you can go for another try. But if you only realized that you are not going to make it half way through your burn, you're done.

Yes, these are the same equations, but for actual velocity, rather than change. Works out the same once you take everything into account. Without atmosphere, perfect landing is the exact reversal of perfect takeoff. So ideally, delta-V is going to be the same. However, "perfect landing" requires a suicide burn. So you perform a de-orbiting burn to put yourself on transfer orbit that just touches the surface, and then you burn to halt right before hitting the surface. Of course, if you mess up the suicide burn, splat. Hence the name. So in practice, you burn earlier which means you'll need more fuel to fight gravity as you do final adjustments. How much extra fuel you use will depend on how much of margin you want for errors. I would recommend about 20% in reserve. Then, of course, there are alternative landing techniques if you want to land at a precise location. These tend to require a lot more fuel, but it all depends on exactly what you are planning to do.

What's wrong with equations I gave you? They are absolutely precise.

You need to burn retrograde with respect to target velocity. Make sure that the little word before the orbital velocity indicator says "Target," rather than "Orbit" or "Surface". You can click that indicator to cycle through these.

You are sitting on the surface. At equator that puts you at sidereal rotational speed. On the Mun, that's 9.04 m/s. Periapsis velocity for Hohmann is sqrt((r2/r1)GM/a) = sqrt((214,000m/200,000m)6.5138398x1010/(207,000m)) = 580.26. This means your first burn is going to be 571.22m/s. (I just subtract off the velocity you already have, the 9.04m/s. Assuming Eastward launch, of course.) The apoapsis velocity, using the same formula, will be 542.30m/s. You need 551.71m/s to circularize, so that's another 9.41m/s burn at apoapsis. The total delta-V required to reach orbit, assuming Eastward launch from equator is, therefore, 580.63m/s. Realistically, it will be slightly higher, depending on how high your TWR is and how well you perform the gravity turn. For ascent in vacuum, the higher the TWR the better. For a rough estimate you can also take escape velocity, which is 807.08m/s and divide that by 1.41 to get 572.40m/s. It's always going to be an underestimate, but it gives you the absolute minimum delta-V to break away from the surface. Finally, the atmosphere. The simple formula to estimate additional delta-V to leave atmosphere is 4gH/vt. Here, g is surface gravity, H is scale height of the atmosphere, and vt is the surface terminal velocity. For Kerbin that works out to 1,832m/s. This assumes TWR of roughly 2 of your rocket, which optimizes fuel consumption on ascent.

Center of lift needs to be bellow center of mass. That's just something assumed by game's logic when it figures out thrust vectoring. The thrust vector itself should pass through the center of mass, or close to it. That will allow you to keep the craft stable. Finally, it will make takeoffs and landings much easier if the thrust vector is near-vertical.

Transfer orbit follows Kepler's 3rd same as any other. Compute the period of transfer orbit. Half of that is how long it will take your sat to reach synchronous orbit. Satellites in sync will shift forward 60° for every hour of transfer.

Yes. The "trick" is called the Kepler's 3rd Law. T² = ka³ for some constant k, where T is the period of time required to complete the orbit and a is semi-major axis. You can compute the later by taking planet's radius + (periapsis + apoapsis)/2. For the synchronous orbit, T = sidereal day, which is 6 hours for Kerbin. So say you have second satellite too close behind the first one. They both complete a full rotation in 6 hours. That's 360° around Kerbin. In 6 hours and 1 minute the first satellite will move 361°. So if you make the second satellite's period 6 hours and 1 minute, you will increase the separation between them by 1° for one revolution. If you keep your periapsis at synchronous altitude of 2,868.7 km and raise apoapsis to 2,881.6 km, the period will be exactly 1 minute longer. Feel free to check that with the Kepler's 3rd above. Now, if you only need to move by 1°, you raise apoapsis to that altitude, wait for 1 full revolution, and bring apoapsis back down to circularize back at synchronous. If you need to move 5°, you can wait for five revolutions. Or you can re-compute the new altitude to go 5° in one turn. Naturally, you can also catch up by lowering the periapsis. In either case, a few maneuvers like that, and you should have all your sats aligned. Shouldn't take much delta-V, either.

That works. Flux will go through the hole, along with a good chunk of the plasma. It wouldn't be 100% effective, of course, but it should be pretty good. Nice idea.