best way to go from equatorial to polar orbit

[seyss]

Hello,

I got a satellite in a equatorial orbit (0º inclination) and I need to change it to a polar orbit (90º inclination) to use that Survey Scanner. I was expecting this so I put a Ion Propulsion engine on the sattelite.

The best way to achieve this orbit is to burn normal? I tired that but often I have to switch between normal and anti-normal.. Mechjeb won't work.

Any advices?

Thanks.

[Redstone]

Ooh! Ooh! I know this one! The best way (i.e., using least delta-v) is to boost your orbit up as high as you can, then make the plane change at apoapsis. Burn again at periapsis, and voila! You should now be in a polar orbit. This method may take longer, but as the plane change occurs when you are travelling more slowly, it uses less delta-v.

-Redstone.

[Capt. Hunt]

I've found the easiest way to get into a polar orbit is to launch into one to begin with. The easiest way to do it from an established equatorial orbit without raising your apoapsis is to make sure you're burning toward the normal at the ascending node.

[seyss]

Just realized my solar panels generate much less energy around Laythe when compared to Kerbin due to distance from the Sun... might not have been a good idea to use Ion propulsion.. this will take forever

[Redstone]

Sorry, you didn't say Laythe in your first post. Just use partial thrust, you will have lots of time with a high AP. Alternately, if you still have a save before you entered Laythe's Sphere of influence, focus on Laythe in map mode, and burn normal until your orbit passes over the pole and then circularise.

[FancyMouse]

Laythe? Then you should plan your polar orbit before you have the encounter. It should be just a few dV (10 or 20 or something) to accomplish that, by performing a mid-course correction maneuver before encountering Laythe that points either normal or anti-normal. Enter equatorial and then adjust to polar would cost a LOT.

But of course, if you already have one and you really need it to get polar, then you probably need to get a highly eccentric orbit, correct at Ap, then circularize back.

Edited June 7, 2015 by FancyMouse

[AlextheBodacious]

Using no deltaV:

Hyperedit.

[seyss]

Using no deltaV:

Hyperedit.

but no fun........

[Yasmy]

For planets or moons with atmospheres, the method of raising Ap to just below the SOI boundary, correcting the inclination, and returning to low orbit can be made even cheaper by the ability to aerobrake, if you don't have fixed solar panels or other breakables.

But if you don't want to do a multi-step process, burning normal is not the best you can do. Burning normal is fine for very small inclination changes, but for larger inclination changes, burn along the hypotenuse from your old velocity vector to your intended vector.

For example, to go from a circular equatorial orbit to a circular polar orbit at the same radius, burn in the direction half way between retrograde and normal. The cost is dv = sqrt(2) v = 1.41 v.

If instead you burn normally, following normal as it rotates from polar to equatorial, the cost is dv = pi/2 v = 1.57 v.

[seyss]

What I'll do is keep the satellite docked to main ship and just go always in a polar orbit, correcting the inclination right after SOI insertion. My main ship will stay in polar orbit too which is a change from the default equatorial orbit I always used. The good point is I'll be able to land anywhere in the planet with less fuel I guess.

[goldenpsp]

What I'll do is keep the satellite docked to main ship and just go always in a polar orbit, correcting the inclination right after SOI insertion. My main ship will stay in polar orbit too which is a change from the default equatorial orbit I always used. The good point is I'll be able to land anywhere in the planet with less fuel I guess.

Of course since it doesn't cost much dV you could always do that, undock the probe, and then set your main ship right back into an equatorial orbit if you wanted.

[Sharpy]

Entering LOW equatorial would cost a lot. The farther you are, the cheaper the change, so apoaxis of long elliptic is quite cheap.

Now, what I'd like to know...

Ooh! Ooh! I know this one! The best way (i.e., using least delta-v) is to boost your orbit up as high as you can, then make the plane change at apoapsis. Burn again at periapsis, and voila! You should now be in a polar orbit. This method may take longer, but as the plane change occurs when you are travelling more slowly, it uses less delta-v.

Nope. Boosting the orbit to the edge of sphere of influence and then reducing it back will consume a lot of dv. Less than turning polar in the low orbit, but still way more than needed. There is some golden middle, with the apoaxis pretty high but still far from edges of SOI, where performing the rise + turn + descent takes least dV. Now where this position is?

- - - Updated - - -

If the planet has atmosphere and you're not on time budget:

- enter the planet's SOI at any angle with a low periapsis, preferably skirting the upper atmosphere for added aerobraking.

- Burn retrograde at periapsis to reach an orbit. Just barely enough for the ends to meet, at the edge of the sphere of influence.

- Go back to your apoaxis.

- Detach your probe. Use its RCS to put it in polar orbit. (seriously, at the edge of SOI you need like 15m/s of dV, and with a lightweight probe that's nothing. The nice thing is there's an infinite number of polar orbits, so unless you want to go fancy (with sun-synchronous or the likes) you'll get any of them from any place, far Apoaxis is just extremely cheap. Burn retrograde for aerobraking (the normal/antinormal burn has raised your periapsis, again, this will be like 0.5m/s.

- Wait for your ship to enter the equator latitude - the farther of the ascending/descending nodes of the equatorial orbit. Burn to put it in equatorial orbit; it should still cost very little if you use the farther node. If both nodes are very close, burn at apoaxis to move one of them far (e.g. enter near-polar orbit as well). Note, these are all radial-anti-radial burns. Burn retrograde for aerobraking (the normal/antinormal burn has raised your periapsis)

- Allow the atmosphere to do it work. Occasionally burn very little prograde at apoaxis to return periaxis back into safe upper regions of the atmosphere.

- When your periaxis is at desired altitude (after multiple runs of aerobraking) burn prograde at the apoaxis to circularize.

[seyss]

thanks

[NASAHireMe]

For example, to go from a circular equatorial orbit to a circular polar orbit at the same radius, burn in the direction half way between retrograde and normal. The cost is dv = sqrt(2) v = 1.41 v.

If instead you burn normally, following normal as it rotates from polar to equatorial, the cost is dv = pi/2 v = 1.57 v.

Can you point me to the proof of this? I believe you, but is there a longer explanation for why this is?

- - - Updated - - -

Nope. Boosting the orbit to the edge of sphere of influence and then reducing it back will consume a lot of dv. Less than turning polar in the low orbit, but still way more than needed. There is some golden middle, with the apoaxis pretty high but still far from edges of SOI, where performing the rise + turn + descent takes least dV. Now where this position is?

The answer you are looking for is in this thread: http://forum.kerbalspaceprogram.com/threads/69659-Apoapsis-plane-change-manoeuvre-%CE%94v?p=971064&viewfull=1#post971064

"In summary, for less than 38.9 degrees, don't raise apoapsis. For 60 degrees or more, raise it as much as possible. In between those, raise it as given by the second equation. If aerobraking, these angles become 19.2 degrees and 28.96 degrees respectively, and in between use the third equation."

[Arkalius]

Nope. Boosting the orbit to the edge of sphere of influence and then reducing it back will consume a lot of dv. Less than turning polar in the low orbit, but still way more than needed. There is some golden middle, with the apoaxis pretty high but still far from edges of SOI, where performing the rise + turn + descent takes least dV. Now where this position is?

Actually, for any significant plane change, the larger you're willing to go with the orbit (without getting to escape velocity, obviously) the more delta-v you save on the plane change. Here is a graph:

This shows the ratio of delta-v of the plane change by expanding the orbit first to the delta-v of doing it without expanding the orbit. The x-axis is the amount of prograde delta-v applied to expand the orbit, and this takes into account having to apply that same delta-v again to shrink the orbit back. This situation is based on starting at a 100km orbit of Kerbin. Since at that orbit, 930 m/s is enough for escape velocity, I ended there.

As you can see, for plane changes of 60, 75, and 90, the more speed you add to your orbit initially, the better the savings in the end. For a 45 degree plane change, you can save a little delta-v by putting about 300 in to expand the orbit, but it will start to cost more again after that. For changes less than 45, expanding the orbit no longer has any benefit.

So really, it's up to how long you want to wait to complete the plane change, as obviously, expanding your orbit a lot will extend your orbital period.

I've also checked; these curves don't change at different starting altitudes, though obviously the range of extra velocity you can add shrinks as your altitude gets higher. I'm also pretty sure these curves stay the same regardless of the central body being discussed.

And I find it funny how you managed to spell 'periapsis' correctly, but not 'apoapsis'.

Edited June 10, 2015 by Arkalius

[Yasmy]

Can you point me to the proof of this? I believe you, but is there a longer explanation for why this is?

Suppose your initial speed is v. Let's make an inclination change of angle θ only by burning locally. For a pure inclination change, your final speed equals your initial speed, but the velocity vectors differ in direction by angle θ.

1) First, a truly horrible method: Burn dv = v retrograde, so that your speed is zero. Then burn dv = v in the inclined plane. The total cost is 2v, independent of θ. Horrible. Consider this the worst case scenario.

2) The best local, single-burn method is a burn along the hypotenuse of the triangle formed by the initial prograde velocity vector and the final inclined plane velocity vector. The triangle has two sides of length v at angle θ to each other. Use the law of cosines to find the length of the hypotenuse: c² = a² + b² - 2 a b cos(θ).

dv² = 2 v² - 2 v² cos(θ) = 4 v² ((1-cos(θ))/2) = 4 v² sin²(θ/2).

Thus dv = 2 v sin(θ/2). For θ = 90, sin(45) = sqrt(2)/2, and thus dv = sqrt(2) v.

3) Finally, the method of following the normal: Instead of burning along the hypotenuse of a triangle, you are sweeping out the arc of angle θ of a circle of radius v. Thus the arc length dv is equal to v θ. Remember to use radians for arc lengths instead of degrees. 90 degrees = pi/2 radians. So dv = pi/2 v. This method is always costlier than method 2. If this is following the arc of a circle, method 2 is the straight line between two points on a circle.

Derivation of this one is a bit more involved. Solve 2 coupled differential equations:

a_x = d/dt v_x = - c v_y

a_y = d/dt v_y = c v_x

The vector (-y,x) is normal to the vector (x,y). So the differential equations together describe thrust normal to the current velocity.

The constant c (assuming constant thrust -- not a great assumption if the burn reduces your mass significantly and you don't adjust your throttle) is your angular velocity. A solution is: v_x = v cos(c t), v_y = v sin(c t). The total burn delta-v is obtained by integrating the acceleration a = sqrt(a_x² + a_y²) = c*v from time t = 0 until time T, the time to get to angle θ. The integral of the constant c*v from 0 to T is simply dv = c*v*T. Now looking back at the solution, at what time T is the velocity inclined by angle θ? Ans: When c*T = θ, because the velocity points along angle θ when v_x = v cos(θ) and v_y = v sin(θ). Thus dv = c*v*T = v θ.

(What is the angular velocity c anyway? F = m a = thrust, so c = a/v = TWR*g₀/v.)

[Sharpy]

Yep, I asked the professionals. In a pure two-body system (planet+craft) anything above 60 degrees should be performed as a weak stability boundary maneuver - in KSP terms, at the edge of the sphere of influence. (and then aerobrake to bring the apoaxis back.)

But they gave a significantly cheaper way: perform a lunar fly-by, using a gravity turn to land in a polar orbit. (and aerobrake back again).

And a third way perfectly viable in KSP: Use wings and bank north in the upper atmosphere.

Edited June 10, 2015 by Sharpy