From Kerbin to solar polar orbit.

[Aethon]

I've got the dv, but I'm wondering whether it's more efficient to burn Kerbin retrograde and Normal, or wait for the Mun to line up Kerbin retrograde, and go over Muns' pole. Doing some sanbox tests now. I know Mun gravity assists don't help much if any, but the plane change is going to be a killer. What say ye?

Thanks.

[Laie]

The Mun is only good for 200m/s or something. At best. If you really want to save dV, set your sights higher. You need to change the plane relative to the sun, right? In that case, an Eve or Jool gravity assist seems like the way to go. Jool would probably be better, being both farther out and even heavier.

EDIT: previous statement was silly.

Edited February 27, 2015 by Laie

[SRV Ron]

Jool would be the best bet for that much of a plane change. Being further out from the sun, you could probably get close to the full 90* change needed with a slingshot around Jool. This slingshot around Ike shows the possibility of achieving that much plane change using little delta v.

[Aethon]

Yeah. I'd hafta redesign (more Photovoltaics or RTGs) plus the Jool window is 40 E days away, not to mention the added travel time out to Jool and back. Thanks though

[Kryxal]

There's still Eve, that would get you a significant part of the way there ... what sort of apoapsis and periapsis are you looking for?

Also, how well will the slingshot line up with the plane of the final orbit, it won't be a lot of use if you're 90 degrees off after.

[Aethon]

KER says the craft I'm using has 25,000 dv but it's not accounting for my funkily staged xenon drop tanks, so there is significantly more than that. (Not to bash KER. It has a tough job and does it amazingly well).

I'm only on day 5 so the Eve window is over a hundred days away iirc.

I want to get low science, so I'm shooting for a periapse under 1000 km. A circular SAR ScanSat altitude of 750 km or even lower would be nice. I know there will be no map, but the ScanSat settings screen will give some % completion.

I guess I'll just burn Normal and Kerbin retrograde.

Thanks all.

[Taki117]

Why not launch into a polar kerbal orbit, and then burn to solar orbit?

[SRV Ron]

Why not launch into a polar kerbal orbit, and then burn to solar orbit?

The plane change by doing so will be minimal.

[Kryxal]

Little enough point, your Kerbin-polar orbit still has all the velocity of Kerbin around their sun.

[eddiew]

I had to do this recently; needed a highly inclined solar orbit, with high eccentricity. Best option was to kick the apoapsis out to beyond Duna and do the plane change there. Think it cost about 6k delta-v.

[Sirrobert]

Why not launch into a polar kerbal orbit, and then burn to solar orbit?

When you escape Kerbin SOI, you still have Kerbin's orbit around the sun, plusminus the velocity you used to escape. So you are still rhougly equatorial

[KerikBalm]

For pretty much any 90 degree plane change, the most efficient way is to kick your PE way out, and do the plane change there.

If you can kick it out to Jool, and use Jool, even better.

[Aethon]

I of course realize that I'm not being very efficient here. I could do what I normally do, extend my Ap, or wait for Jool or Eve, but I just want to get there quickly (from LKO), but as efficiently as possible within those constraints. I asked the cool kids in Kerbal Space Program Forum chat (shameless plug) and no one seemed to know, which is unusual.

I appreciate the well meaning and helpful responses here but I think we've not quite gotten to the heart of the matter. I guess I'll run some sandbox tests this weekend and report back here with the results.

Cheers.

Edited February 27, 2015 by Aethon

[Yasmy]

Well, since you are not overly concerned with efficiency, just be direct about it.

Kerbin orbits Kerbol at 9284.5 m/s prograde, and you want to go, say, 9284.5 m/s north from Kerbin instead.

Add in about 1000 m/s of Kerbin escape velocity relative to LKO. Using good old Mr. Pythagoras,

you want to burn approximately sqrt( 9284.5^2 + (9284.5 + 1000)^2 ) = 13855 m/s. Call it 14 km/s.

The angle you want to burn is atan2( 9284.5 + 1000, -9284.5 ) = 132 degrees counter-clockwise from prograde, ie, pretty close to north-west.

So launch from the surface Kerbin into a NW orbit and burn 14 km/s delta-v.

[Aethon]

Are you talking about my ejection angle from Kerbin, or launching into an inclined LKO of 310 degrees?

[Yasmy]

More or less both. Kerbin surface velocity changes the angle slightly, adding a bit of prograde velocity to your orbital velocity relative to surface velocity. So to compensate,

launch a bit more westerly than your intended orbit. Probably a couple degrees. Just use Pythagoras's theorem if you care to do the math.

Launch into an inclined orbit which is already aligned with your ejection burn. More or less northwest instead of the usual easterly launch.

Upon crossing Kerbin's SOI, your trajectory should switch from inclined at 135 degrees to inclined at 90 degrees.

Edit: This is for a Kerbol orbit of the same size as Kerbin's Kerbol orbit. Not what you were looking for. More below in a bit...

Edited February 27, 2015 by Yasmy

[MarvinKitFox]

Without fiddling with gravity assists, etc... The most efficient way to do a 90 degree plane change from a circular orbit to another circular orbit at right angles would be:

1) accelerate prograde to escape velocity - 0.000000000001

2) wait until apoapsis is achieved. Enact your plane change, the cost should be ZERO (or a very good facsimile of it). Adjust desired periapsis height and angle.

3) When reaching the desired periapsis, circularise.

This maneuver will allways have a cost of exactly:

For initial circular orbit velocity A, and final orbital velocity B (for any starting orbit, any ending orbit, circular. Plane irrelevant. Orbital radii irrelevant!)

cost = A * (1-sqrt(2)) + B * (1-sqrt(2)) = 0.41421 (A +

Whereas brute-forcing the plane change, for A=B but 90 degree plane change:

Brute force cost = sqrt(A+^2 = 1.414 * A in this case

== 1.707 times as much.

[Yasmy]

Ok. After reading MartinKitFox's (absolutely correct) post, I went back and reread what you were trying to do: Get to low Kerbol polar orbit. I missed that.

Though correct, I didn't think MartinKitFox's post was relevant because you said you didn't want to burn way out of the way (ie, to Jool) to reduce delta-v costs.

Turns out his post is actually on the money because for a Kerbin to low Kerbol orbit, you are already on an extremely elliptical orbit at apoapsis.

With no fancy slingshots, just a single ejection burn:

1) LKO to LKerbolO. Use the vis-viva equation.

AP = radius of Kerbin's orbit

PE = radius of Kerbol + 1000 km

v = sqrt(mu (2/AP + 2/(AP + PE)))

v = 1807 m/s

So when you exit Kerbin's SOI, you want to be going 1.8 km/s on a 90 degree inclined orbit. (Might want to check my math.)

2) To get to a Kerbol polar orbit going 1.8 km/s at Kerbin, you need to shed Kerbin's 9.3 km/s (9284.5 m/s) orbital velocity.

So before crossing Kerbin's SOI, you need to be going 9.3 km/s west, 1.8 km/s north = 9.5 km/s inclined by 169 degrees.

9.5 = sqrt(9.3^2 + 1.8^2), 169 = atan2(1.8,-9.3)

Just trig.

So the plane change is cheap (9.5 - 9.3 = 0.2 km/s) since we combined it with a huge change in periapsis.

3) To get to the target velocity at SOI, you need to be going the target SOI velocity plus Kerbin escape velocity from LKO of approximately 1km/s.

So launch into a Kerbin orbit inclined at 169 degrees. Burn approximately 9.5 km/s + 1.0 km/s = 10.5 km/s.

4) If you want to circularize, ouch. At Kerbol periapsis you'll be going:

v = sqrt(mu (2/PE - 2/(AP+PE))) = 93.6 km/s

and to circularize you need to be going

v = 66.8 km/s

dv = 26.8 km/s

Yup. 26.8 km/s circularization + 10.5 km/s LKO ejection = 37.3 km/s. Ouch.

-------------

Instead of 1000 km PE, let's change it to 1,000,000 km PE, the limit of low Kerbol science:

PE = radius of Kerbol + 1e9 m.

v = sqrt(mu (2/AP + 2/(AP + PE))) = 3.8 km/s outside Kerbin's SOI.

Inside Kerbin's SOI: v = sqrt(9.3^2 + 3.8^2) km/s = 10 km/s, inclined at 157 degrees.

Adding LKO escape velocity, 11 km/s delta-v along a 157 degree inclined orbit.

At PE, delta-v = sqrt(mu (2/PE - 2/(AP+PE))) - sqrt(mu (1/PE)) = 41.2 km/s - 30.5 km/s = 10.7 km/s

Total delta-v from LKO: 21.7 km/s. Save you about 16 km/s over 1000 km PE.

-------------

That's the cost of going direct. Not cheap. I'd be inclined (heh) to use another method.

- - - Updated - - -

Third option, still with the inefficient direct method.

Don't completely circularize. Shrink your orbit to 1000km x 1000000km.

This would only cost 10.5 km/s ejection from LKO + 7.6 km/s apoapsis shrinking.

Edited February 27, 2015 by Yasmy
Numbers

[Aethon]

First test images.

Seems legit from out here.

Oh.

I think this image shows I need to launch into an inclined orbit going SE ( to preserve some orbital v ), coming over the top of Kerbin and initiating my burn there- and we're going to need a bigger boat (LV)/ higher LKO.

Edited February 27, 2015 by Aethon