Mun ejection angle

[RizzoTheRat]

I'm trying to do as much as I can with KOS in this career, which means actually working out what's going on rather than just eyeballing it with the manoeuvre nodes.

Calculating the dV and phase angle to get to the Mun is easy enough, but I can't get my head around coming back. I know from previous experience that 250ish m/s at an angle somewhere between pointing at Kerbin and Mun retrograde will get me the most efficient return to Kerbin atmosphere, but can't see how to work it out. 

For a given Ap altitude above Kerbin I can work out the velocity I need to achieve my required Kerbin Pe, and for a given ejection burn I should be able to calculate the velocity and altitude relative to Kerbin at the point it leaves the SOI (not done this yet but I think I know how), but as the required Kerbin velocity will change with Ap altitude I can't see how to link the two together.

Any suggestions? 

[Mark Kerbin]

I find that firing retrograde in reference to the mun at about 5 degrees past when your ship, the mun and kerbin line up is the best. Not sure about dv tho.

if you want, install mechjeb and have it create a return from moon node. That was how I figured it out back in the day. 

Edited May 8, 2018 by Mark Kerbin

[Not Sure]

Hmm, well you want to be exiting towards the muns retrograde vector. I do not know how to translate that to kOS right now. Maybe you can look around in the script library for the Mun mission script, it should be in there.

[OhioBob]

After existing Mun's SOI you want to be in an orbit with an Ap near Mun (~12,000 km) and a Pe near Kerbin (~625 km).  So the semimajor axis is,

a = (12000000 + 625000) / 2 = 6312500 m

Using the vis-viva equation, the velocity at Ap is,

v = SQRT(3.5316E+12 * (2/12000000 - 1/6312500) = 170.7 m/s

Mun's orbital velocity is,

v = SQRT(3.5316E+12 / 12000000) = 542.5 m/s

Therefore the spacecraft's velocity relative to Mun must be,

v_rel = 542.5 - 170.7 = 371.8 m/s 

For the next part we use the following equation,

v_ꚙ^2 = v_bo^2 - v_esc^2

where v_ꚙ is hyperbolic excess velocity, v_bo is burnout velocity, and v_esc is escape velocity.  What this says is if we take the square of the velocity we're traveling after completing the ejection burn, and subtract the square of the escape velocity, what we get is the square of the velocity we'll be traveling after escaping Mun's gravity.  Hyperbolic excess velocity is the result of the extra energy we give the vessel at ejection over and above that needed just to escape.

v_ꚙ is the velocity we'll be traveling relative to Mun after escaping, so we set v_ꚙ = v_rel.  So we first calculate v_esc and then v_bo.  Let's say our orbit around Mun is at an altitude of 20 km.

v_esc = SQRT(2 *  6.51384E+10 / 220000) = 769.5 m/s

v_bo = SQRT(371.8^2 + 769.5^2) = 854.6 m/s

So the Δv that we require for the ejection burn is simply v_bo less the original orbital velocity.

Δv = 854.6 - SQRT(6.51384E+10 / 220000) = 310.5 m/s

I find that in practice I generally need about 270 m/s.  The reason this is less than 310.5 m/s is because of the way KSP uses patched conics.  We don't need to accelerate all the way to escape velocity in order to leave Mun's SOI.  Therefore it takes less Δv to acquire the same v_ꚙ as it would in real life.

As far as ejection angle goes, it's easier to show than explain.  Take a look at this,

http://www.braeunig.us/space/orbmech.htm#hyperbolic

You want one of the asymptotes of the hyperbola to be pointed in the retrograde direction.  This means that the ejection angle from retrograde would be the angle η shown in figures 4.14 and 4.15.

 

Edited May 8, 2018 by OhioBob

[RizzoTheRat]

Awesome stuff @OhioBob, thanks.

I more or less understood the dV requirement before but it's crystal clear now thanks to that explanation.  As we know the SOI radius for each body presumably I could also calculate the patched conics version with the Vis Viva equation to find the velocity at my 20km periapsis needed to reach the SOI limit at Apoapsis?

I think I need to work through the asymptote bit with a pencil one evening though. 

So to make sure I've understood this, I need to

• Calculate the dV for escape velocity, which effectively puts me in a kerbin orbit at the same radius as the Mun
• Add the amount dv needed to decelerate it enough to get my required Kerbin Pe.
• Calculate the asymptote angle I'll get that from that burn from my current Mu orbit
• Centre my burn that angle before I reach retrograde to the Mun orbit.

Is that right?

[OhioBob]

7 hours ago, RizzoTheRat said:

So to make sure I've understood this, I need to

• Calculate the dV for escape velocity, which effectively puts me in a kerbin orbit at the same radius as the Mun
• Add the amount dv needed to decelerate it enough to get my required Kerbin Pe.
• Calculate the asymptote angle I'll get that from that burn from my current Mu orbit
• Centre my burn that angle before I reach retrograde to the Mun orbit.

Is that right?

That's correct, but the first two items are a single maneuver, not two separate ones.  Maybe you understand this, but that's not the way it sounds from reading your description.  That is, you don't (1) perform a burn with just enough velocity to escape, and then (2) perform a second burn after escaping to lower the Pe.  What you do is perform a single burn near Mun that provides the velocity to both escape and lower the Pe at the same time.  You provide more velocity during this burn than what is needed just to escape.  The energy that this extra velocity provides is leftover after escaping in the form of hyperbolic excess velocity.  The hyperbolic excess velocity is what lowers the Pe.  This method is far more efficient than performing two separate burns because we take advantage of the Oberth effect by performing the burn near Mun.

[RizzoTheRat]

Thanks for the help, it works, my KOS scrip can now get me back from the Mun, which means I should have the needed maths for transfers to other planets too.