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At T+26:31 into the SpaceX ANASIS-II Mission Livestream, the host mentions that the mission trajectory calls for an orbital inclination change which is performed when the orbit of the second stage and payload cross the equator to minimize the maneuver's energy cost. (he says to maximise the maneuver's efficency, so I'm paraphrasing a little)

Why is this? Why is it the most energy efficient to change orbit inclination while crossing the equator?

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    $\begingroup$ Extreme example: if you are in polar orbit and perform the maneuver while over a pole - you are still in polar orbit $\endgroup$ – Hagen von Eitzen Jul 21 at 20:59
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Why is it the most energy efficient to change orbit inclination while crossing the equator?

Specifically, it's most efficient to do a plane change at one of the two "nodes" where the origin orbital plane intersects the destination plane. ANASIS-II is destined for geostationary orbit, so its destination plane is the plane of the equator.

Any orbit around a single massive body lies in a single plane. It should be clear that you can't enter an equatorial-plane orbit at any point except a point directly over the equator. Coming from any non-equatorial orbit, there are two points on the orbit where the planes intersect. If you try and do a burn to enter a particular destination orbit from anywhere else, you just push the intersection point a little further around the orbit.

(It's super easy to demonstrate this in Kerbal Space Program, but kind of hard to put into words!)

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    $\begingroup$ The first part of the answer says it's most efficient to perform the maneuver at the ascending/descending nodes, but the second part says it's not even possible to perform it anywhere else. I think it's the second one, isn't it? If you want to maneuver from one orbit to another, the burn must occur where the orbits intersect. $\endgroup$ – Nuclear Hoagie Jul 21 at 12:56
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    $\begingroup$ Doing a burn somewhere other than at an intersection will change your orbit, but not to an equatorial-plane orbit. To get to the equatorial-plane orbit after that non-intersection burn, you'll need a new burn -- at the new intersection point. So in that sense, burning elsewhere makes the process less efficient -- "never time to do it right, always time to do it over." $\endgroup$ – Ross Presser Jul 21 at 13:44
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    $\begingroup$ +1 for mentioning Kerbal Space Program. Just running through the tutorial of that game will teach you a ton about how orbits work. (OK, the +1 is also for the great answer) $\endgroup$ – MikeH Jul 21 at 17:56
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    $\begingroup$ @RossPresser You could burn elsewhere to raise the ascending/descending node, then do the inclination-change burn at the higher AN/DN, and do a final burn to return to the original orbital altitude. For large inclination changes when the AN/DN is low, this can be more efficient than just going straight for the inclination change in the original orbit. Burning elsewhere can actually make the process more efficient in some cases, although you always have to burn at the AN/DN to get the inclination right. For a small inclination change like this one, though, a single burn is almost surely best. $\endgroup$ – Nuclear Hoagie Jul 21 at 18:47
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    $\begingroup$ @NuclearWang, if you do the burn somewhere else, you'll need to do a second burn to finish the plane change later, once you're passing through the new node. The total thrust of these two burns is greater than that of a single burn at the original node. It's occasionally worth doing, though, if you're already doing a burn at the first location (eg. to change your orbital height), because combining two maneuvers into a single burn is more efficient than doing them as separate burns. $\endgroup$ – Mark Jul 22 at 23:17
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A great aid to intuition is to remember one principle about orbit changes: if the engine is off, the orbiter always returns to same point one orbit later.

So for any orbit change, if you want to do only a short burn, it has to be at a point that is common for both the current orbit and the destination orbit. This applies to inclination changes, altitude changes and basically any orbit change. If the orbits do not have a common point, the change requires two burns and an intermediate orbit, such as Hohmann transfer orbit.

Like Russell's answer details, for a geostationary target orbit those common points are always above equator. For e.g. polar target orbit, the points would be somewhere else.

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    $\begingroup$ This should be the accepted answer. The keypoint being that orbit changes are done in short burst(s) of thrust. If you had continuous thrust, then you could change from whatever inclination to an equatorial orbit, travelling on any desired path. $\endgroup$ – ysap Jul 21 at 18:34
  • $\begingroup$ @ysap I don't see how the continuous/discrete nature of thrust makes much of a difference here. With continuous thrust, you'll fire the engines for longer, but you still must burn when crossing the equator in order to achieve equatorial orbit. If your burn does not end when you're above the equator just entering equatorial orbit, you will not have achieved equatorial orbit. Regardless of thrust, you can only move between orbits where they cross, although I suppose continuous thrust gives you a continuum of orbits to move through. But you will always have to burn above the equator. $\endgroup$ – Nuclear Hoagie Jul 22 at 12:45
  • $\begingroup$ @NuclearWang - consider ideally continuous vs. ideally impulse trusts. With the former, you could choose whatever path you want in order to get to the equatorial trajectory. You could continue burning while in the trajectory to change its eccentricity. With the latter, you have "one shot". Then, as you correctly noted, you have to be on the target trajectory when changing the direction. $\endgroup$ – ysap Jul 22 at 16:18
  • $\begingroup$ @ysap Right, but those other paths will generally be less efficient than an impulse burn at the ascending/descending node, and will ultimately still require a burn above the equator. I don't see the nature of the thrust being a key point here, since it doesn't change anything about what's theoretically most efficient, or the fact that the burn must at some point occur above the equator. The continuous burn just lets you get there less efficiently, moving through a continuum of connected orbits, and ending in a burn above the equator just like the impulse thrust. $\endgroup$ – Nuclear Hoagie Jul 22 at 16:49
  • $\begingroup$ Of course they will be less efficient! That's why they do the one-shot impulse maneuver. $\endgroup$ – ysap Jul 22 at 16:49
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It's not just the most efficient way, it's the only way to achieve this particular target orbit.

As the other answers have pointed out, an orbital inclination change must occur at the so-called ascending/descending nodes, which are the two points in the orbit at which the current and target orbital planes intersect. Anytime a spacecraft moves from one orbit to another, the original and target orbits always share at least one point in common - it's where the burn occurred. If you want to move from an inclined orbit to an equatorial orbit, the burn must occur at one of two loci where the planes intersect, both of which are above the equator. If you don't perform the burn where the orbits intersect, you will never magically jump the distance between the two, and you will never reach the target orbit.

Adjusting inclination to reach an equatorial orbit must occur when above the equator - it's not just the most efficient way to do it, it's the only way to do it. You can potentially increase the efficiency of your maneuver by minimizing the spacecraft's speed and the resulting delta-v change, which requires moving into a larger orbit, performing the inclination change, and returning to the original orbit. But even if you move into a wider orbit, the inclination change still occurs when over the equator.

You can adjust inclination at either the ascending or descending node, and it'll be more efficient to do so at whichever one has a higher orbital altitude, since the spacecraft's velocity will be lower. So, it's possible that the mission in question chose to perform the burn at the ascending or descending node specifically to maximize efficiency. But the fact that the maneuver occurred over the equator has nothing to do with efficiency at all - it is in fact required if you're moving to an equatorial orbit with zero inclination.

To sum up, one cannot enter equatorial orbit from anywhere except above the equator.

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  • $\begingroup$ Your last sentence is absolutely correct. $\endgroup$ – ysap Jul 25 at 17:22
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In the case of ANASIS-II, the situation is far more complicated than can be explained in ten seconds of livestream. Some general rules regarding plane changes:

  1. You can only make a plane change at the point where your current orbit's plane intersects the target orbit's plane.
  2. The faster you're going, the more fuel it takes to perform a plane change.
  3. Because of how vector addition works, performing a plane change at the same time as you're doing an altitude-change burn is more efficient than performing the two separately.

The target orbit for ANASIS-II is a geostationary orbit. To get there from the Kennedy Space Center launchpad requires three maneuvers:

  1. A plane change into an equatorial orbit, taking place over the equator.
  2. An insertion burn into geostationary transfer orbit, taking place over the equator.
  3. A circularization burn from geostationary transfer orbit into geostationary orbit, taking place over the equator.

Notice something? All three maneuvers need to take place over the equator, so maneuver 1 can be combined with either maneuver 2 or maneuver 3, and by rule 2 above, it's most efficient to combine it with 3, the circularization burn (a satellite at the high end of a transfer orbit is moving far slower than at the low end).

So why did the ANASIS-II launch combine the plane change with the insertion burn? Because it could. The Falcon 9 upper stage can carry more fuel than is needed to put ANASIS-II into a geostationary transfer orbit, so it used that extra fuel for the plane change. This reduces how much fuel ANASIS-II needs to spend to get into the target orbit, increasing how much it has left for station-keeping.

(If you're paying close attention, you'll notice that the "after" trajectory line of the maneuver isn't following the equator. A 28.5-degree plane change in low Earth orbit is expensive, and Falcon 9's upper stage can't carry enough fuel to do it. Still, even a partial change means a reduction in the amount of change needed during circularization.)

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  • $\begingroup$ Wow! This is the most complete and understandable answer I've received! Makes complete sense now and I think explains why I don't recall hearing about the plane change maneuver on any other GTO-bound F9 launches... $\endgroup$ – ifconfig Jul 23 at 0:03
  • $\begingroup$ So, you're saying the Falcon took the more fuel-expensive option so the satellite would not need to burn as much of its own fuel? $\endgroup$ – jpaugh Jul 23 at 21:25
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    $\begingroup$ @jpaugh, yes. Fuel is not the expensive part of a Falcon 9 rocket, so if you've got a use for the extra capacity, there's no reason not to launch it with full tanks. $\endgroup$ – Mark Jul 23 at 21:35

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