Negative ejection burns at Laythe?

[skaterzero807]

I have been doing some delta-v calculations for where to put a Joolian system station, but I ran into an issue when calculating dV requirements from Laythe orbit to Vall or Tylo.

Even the calculator at ksp.olex.biz seems to agree with the issue: some ejection burns are negative.

For example, a 3000 km circular orbit around Laythe has orbital velocity = sqrt(mu/3e6) ~ 800 m/s.

For a transfer to Vall at Laythe's altitude, you need to exit the SOI at about 350 m/s moving prograde so Laythe's velocity (~3200 m/s) adds and you end up at Vall's orbit at apoapsis.

To be going 350ish m/s exiting the SOI you need to slow down by about 400 m/s, but we all know slowing down your circular orbit won't get you closer to the SOI.

So it is obvious we would want to increase our speed, a positive ejection burn to escape, but this seems to indicate we will be moving too fast when leaving.

So my question is, why are there negative ejection burns? Is it just a poor assumption of scalar addition of velocity upon leaving the SOI? Or do we just have to burn positive to escape Laythe then slow down to bring the apoapsis to Vall's orbit?

When I have more time, I'll have to launch a test probe and play with maneuver nodes to see what the deal is but maybe someone knows already. Thanks in advance.

Edited August 5, 2014 by skaterzero807

[NecroBones]

So my question is, why are there negative ejection burns? Is it just a poor assumption of scalar addition of velocity upon leaving the SOI? Or do we just have to burn positive to escape Laythe then slow down to bring the apoapsis to Vall's orbit?

Based on your description, I think this is the case. I think I see three details that lead to this scenario:

1. Escape velocities are almost like an "all or nothing" sort of deal. Either you're over that threshold, or you aren't.

2. This is exacerbated by one of the shortcuts that KSP takes, namely the on-rails, one-body gravity model, so you end up with a harsh SOI transition, rather than a more realistic, gradual boundary.

3. Jool's moons have orbits closer together than would probably be sustainable in a true N-body physics model, so you're more likely to encounter something like this than would probably happen in the real world.

Edited August 5, 2014 by NecroBones

[Mr Shifty]

It's because 350 m/s is less than escape velocity at the edge of Laythe's SOI. You can't leave Laythe's SOI going less than about 1025 m/s. I'm not sure if KSP uses the standard calculation: r_SOI=a(m/M)^(2/5), but it's similar. Jool dominates its own system; its gravitational field sort of squishes the SOIs of the moons, especially the close ones, which is why even though Laythe is 55% the mass of Kerbin, its SOI is about 22 times smaller. If you want to calculate a Laythe->Vall transfer, use alexmun's calculator. Ejection dV should be about 12 m/s if you start at 3000 km.

Edited August 5, 2014 by Mr Shifty

[EdFred]

I think MrShifty has it right It appears that the OP confused some numbers.

[Mr Shifty]

I think MrShifty has it right It appears that the OP confused some numbers.

No, the OP's numbers are correct. I have my own Hohmann calculator that gives out intermediate results for all the steps and it showed the same thing: 350 m/s at Laythe SOI, which meant a 400 m/s slowdown at 3000 km.

EDIT: I, however, did garble it a bit up above. The problem is that under certain conditions in the patched conics paradigm, it's possible to exit the SOI of celestial bodies going less than escape velocity, i.e. with an elliptical rather than hyperbolic orbit. For circular orbits at higher altitudes, this creates a situation where there are SOI-edge-velocities that simply can't be achieved. For a 3000 km around Laythe, your circular velocity is 749 m/s, and escape velocity is 1059 m/s, but you only have to raise your periapsis velocity at that altitude to 783 m/s to push your apoapsis out past Laythe's SOI edge. And when you reach the SOI edge, you'll be going 648 m/s. This lower limit exists for all orbits in every SOI, but in most cases, the SOI is big enough and/or periapsis low enough that the lower limit is negligible compared to any transfer velocities you might need. (Example: lower limit for an escape trajectory from 100 km LKO is about 26 m/s.)

A Hohmann transfer from Laythe to Vall requires a velocity at Layth's SOI-edge of about 350 m/s. But you can't get to that velocity starting at a 3000 km circular orbit. So you either have to do a non-Hohmann transfer using alexmun's calculator, or start from a lower circular orbit (~100 km or below for Laythe->Vall.)

Edited August 5, 2014 by Mr Shifty

[skaterzero807]

Ok, thanks Mr Shifty. I think that explains it pretty well. I guess to do a transfer to Vall would require you to leave the SOI moving at ~648 m/s but in a direction such that when added as a vector to Laythe's motion would result in an orbit (no longer Hohmann) that brings you to apoapsis at Vall. I'll have to see if I can formulate this as a solvable equation to fix my code so I can simulate realistic dV's from these high orbits.

[Mr Shifty]

I guess to do a transfer to Vall would require you to leave the SOI moving at ~648 m/s but in a direction such that when added as a vector to Laythe's motion would result in an orbit (no longer Hohmann) that brings you to apoapsis at Vall.

Or one that departs prograde, but arrives at Vall before or after apoasis. Or something in between:)

[James_Eh]

Glad to hear that my standard method of moving from Laythe to Vall (which invariably involves a stand-on-the-brakes burn) is only partially a result of my bad piloting...

[skaterzero807]

Just for anyone who is interested, I figured out how to get around this problem.

Here is a picture of the orbits to hopefully explain better:

The main problem was coming from assuming we would transfer to Vall using the dashed cyan orbit. The speed leaving the SOI would have to be smaller than physically possible so calculators might show a negative burn.

Anyway, the way to get around it is:

1. Solve for the slowest speed you could leave the SOI from your orbit

2. If the speed you would need to go is slower than this minimum speed, then solve for the flight-path angle (or zenith angle, whichever you prefer, see http://www.braeunig.us/space/orbmech.htm) required so that leaving at the minimum speed and that angle, you will have your desired apoapsis (or periapsis if going toward a target lower). This involves solving a quadratic function for cos(theta) (or sin(theta) using zenith angle).

3. From the known flight-path angle and speed just before leaving the SOI, you can calculate the speed and flight-path angle after leaving by adding the velocity vectors.

4. Given the altitude, speed and flight-path angle just outside the SOI, you know your new trajectory (blue solid orbit)

5. The delta-V in this case is simply the delta-V required to lift your apoapsis to the SOI. From there, you are moving at the minimum allowable speed and if the point of exit is correct, you will exit with the right speed and direction to make it to your target at apoapsis (or periapsis if going lower)

I won't go into the mathy details but this would at least give you an idea of how to do it. I may end up doing a small article on this once I have a definitive answer of where to put a space station in Jool orbit to minimize future trips' delta-V.

[skaterzero807]

And here is the post on my results for those interested: http://forum.kerbalspaceprogram.com/threads/90322-Optimal-Jool-Space-Station-Orbit