Hohmann transfer screwiness around the moons of Jool

[WafflesToo]

Having some trouble doing Hohmann transfer calculations. I was working out transfer maneuvers between Vall and Laythe and I'm getting an ejection velocity of 2451.91 m/s, and exit velocity of 347.21 m/s and an error for the ejection angle.

The problem is, escape velocity from Laythe is 2635.36 m/s

I ran this through that rather excellent Hohmann calculator by olex and came up with identical results...

http//ksp.olex.biz/

Am I to take this to mean that there is no Hohmann transfer window between Laythe and Vall, making transfers more of a brute force affair? I do find that hard to beleive... but it's not outside the relm of possiblity I suppose. :/

Anyway, here's my data and my math for a Laythe -> Vall transfer. Maybe somebody else can spot my mistake.

r1 (Parking orbit) = 565,000 meters (Laythe's radius + 65km altitude orbit)

r2 (SOI radius) = 3,723,646 meters

v2 (SOI exit velocity) = 347.21 m/s

u (Laythe gravitational parameter) ~ 1.96E+12 m^3/s^2

v1 (Ejection velocity) = SQRT((r1*(r2*v2^2-2u)+2*r2*u)/(r1*r2)) = 2451.91 m/s

ve (Escape velocity) = SQRT(2*u/r1) = 2635.36 m/s

Edited February 1, 2014 by WafflesToo
answered

[Taki117]

Try This It's a tad more accurate for such things. I got the following:

Departure

Year 1, day 198 at 23:55:12

Arrival

Year 1, day 199 at 11:33:36

Time of flight

11:38:24

Phase angle

71.56°

Ejection angle

103° to prograde

Ejection inclination

0.00°

Ejection ÃŽâ€v

827 m/s Transfer periapsis

20,131 km

Transfer apoapsis

37,166 km

Transfer inclination

0.00°

Transfer angle

214°

Insertion inclination

0.00°

Insertion ÃŽâ€v

387 m/s

Total ÃŽâ€v

1,215 m/s

[alexmun]

Your problem is that Laythe's sphere of influence is relatively small for its mass because it is so close to Jool. As a result, the ejection orbit that gets you into Jool orbit with the correct velocity is not actually a hyperbolic escape orbit, but just an elliptical orbit with an apoapsis outside of Laythe's sphere of influence. This means you can't use the true anomaly at infinity to calculate the ejection angle, because an elliptical orbit does not go to infinity. Specifically, instead of:

theta = acos(-1/e)

use:

e (eccentricity) = r1 * v1^2 / u - 1

a (semi-major axis) = r1 / (1 - e)

theta (ejection angle) = acos((a * (1 - e^2) - r2) / (e * r2))

This equation also works for hyperbolic ejections and will be marginally more accurate than the simpler version at the expense of a lot more terms.

[WafflesToo]

Your problem is that Laythe's sphere of influence is relatively small for its mass because it is so close to Jool. As a result, the ejection orbit that gets you into Jool orbit with the correct velocity is not actually a hyperbolic escape orbit, but just an elliptical orbit with an apoapsis outside of Laythe's sphere of influence...

This equation also works for hyperbolic ejections and will be marginally more accurate than the simpler version at the expense of a lot more terms.

AAAAAAAHHHHHHHHHHHHH!!! PERFECT! EXACTLY WHAT I NEEDED! Um... thanks :$

And I have no idea how this got double posted.

[alexmun]

I've done some more work and it turns out there's another wrinkle to this case. There is no orbit with a periapsis above the surface of Laythe that will leave you with 347 m/s when leaving the sphere of influence. The reason for this is that an orbit with a periapsis altitude of 0 (on the surface), and a apoapsis equal to the edge of the sphere of influence, has a speed at apoapsis of 353 m/s. The higher your periapsis above the surface, the faster your speed at apoapsis will be, so 353 m/s is your minimum excess velocity for a single burn ejection from Laythe.

As a result, it is not possible to execute a pure Hohmann transfer from Laythe to Vall (which is too bad, since they're the only two bodies in the game with zero inclination and eccentricity around a common parent, i.e. the only two bodies for which a Hohmann transfer is actually correct). You will either need to depart or arrive at a different time, or you will need to make midcourse corrections at some point during the transfer.

Thanks for bringing this question up. I've updated my calculator to give more accurate results in situations like this.

[WafflesToo]

Interesting... I was wondering if something like that were afoot but I lacked the math skills to pursue it in more than a haphazard way.

I actually have never been to Jool before (despite having more flight hours in-game than a few IRL astronauts XD). This only came up because I was in the early planning stages for my first Jool mission and stumbled into it. I actually didn't think situations like this could actually happen to be honest and I'm surprised it hasn't come up before now.

[Mobjack]

As a result, it is not possible to execute a pure Hohmann transfer from Laythe to Vall

Is a pure Hohmann transfer really the best way of traveling to the two moons?

I don't know the math behind it all, but the SOI are so big compared to each other that I didn't find the calculators useful at all. For me I just try to play with the maneuver nodes so that I can use a gravity assists from the moon to slow me down. I found that slight variations of how you approach the Valle can save you hundreds of dV.

[ComradeGoat]

Jool is fun because the three inner moons let you do all sorts of funky stuff with gravity assists. You can often get places cheaply just by letting encounters shift your orbit around Jool about.

For bonus points, use a gravity assist from Tylo to get back to Kerbin almost for free. Good luck getting the angles right!

[Kegereneku]

If we where in the real world with multi-body physics I know there's the "Interplanetary Transport Network", a set of orbital parameter making use of Lagrange points for extremely economical transfer.

There's just a catch, using it going from Earth to Moon could take something like 10 month.

But in KSP where there's only patched conic I don't think there's better than Hohmann transfer