Some satellites are injected into higher apogee than standard GTO apogee of 35 786 km. Next the satellite then has to increase its perigee ( usually 250 km ) to GEO altitude. Then the apogee has to be decreased to GEO altitude.

What is the benefits of this kind of orbital injection? Doesn't this require extra delta-v budget than standard GTO launches?

  • $\begingroup$ I don’t know but I’d suppose that spending time each day at synchronous altitude without being synchronous carries some risk. Related: Why put SunRISE in the graveyard? Why will it "fly slightly above geosynchronous orbit"? $\endgroup$
    – uhoh
    Commented Jan 10, 2022 at 10:56
  • $\begingroup$ Related question space.stackexchange.com/q/22231? $\endgroup$
    – Jack
    Commented Jan 10, 2022 at 10:57
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    $\begingroup$ GTO injection inclination is usually not 0degrees, but something like 27degrees. It it cheaper in delta-v to adjust the inclination towards 0 at higher-than-GSO, then/while raising the perigee to GSO and again adjust the last bit of inclination when/while cicularizing at GSO, than to have to force the plane change while at perigee in LEO or even at GSO. Plane changes are very expensive, but get cheaper very fast as apogee increases. In realworld one does a bit of plane change at each burn, that works out the mot efficient. posted as comment not answer, because will need a lot of mathjax. $\endgroup$ Commented Jan 10, 2022 at 11:35
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    $\begingroup$ @Cute Kitty ---- You made a good point that it is cheaper to change inclination in high orbits than low ones. But doesn’t that mean it would be even cheaper to adjust inclination after circularizing at GSO altitude? Is that because raising perigee is a necessary “big burn” and the inclination correction is relatively “free”? The math is above my pay grade. $\endgroup$
    – Woody
    Commented Jan 10, 2022 at 17:52
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    $\begingroup$ @Woody look on the NAsaSpaceflight forum (for example forum.nasaspaceflight.com/…), there are people that often discuss a specific mission's burns, when and where and how much and why. $\endgroup$ Commented Jan 10, 2022 at 18:39

2 Answers 2


This is a partially copied answer from this closely-related question:

The other answerer focuses on the straight-up dV savings which occur when you're launching from a very inclined launched site. I'm going to focus on a second reason you might want to do a supersynchronous (new-to-me term) transfer, but first, let me detail how a traditional GTO is used from an inclined launch site.

  1. The launcher will throw the satellite into a transfer orbit with an apogee of geosynchronous altitude (35,786 km).

  2. At apogee, the satellite will perform a burn which simultaneously reduces the inclination to 0° and raises the perigee to geosynchronous altitude.

  3. After this combined burn, the satellite is in geostationary orbit.

The reason why the inclination change and perigee raise maneuvers are combined is simple trigonometry--burning diagonally requires less total energy to achieve the same final velocity (not just speed, the direction matters a lot!) than burning one way and then burning perpendicular to that direction. Additionally, this is done at geosynchronous altitude because the satellite is traveling slower there than it was at low earth altitude. Thus, changing direction does not require as much change in velocity (delta-V).

As an example of a supersynchronous transfer orbit, let's consider Arabsat 6a. Falcon Heavy launched Arabsat 6a into a transfer orbit with an apogee of 90,000 km, well above the geosynchronous altitude of 35,786 km. At that altitude, it was traveling very, very slowly, so the combined inclination change and perigee raise burn required even less dV from Arabsat than if it had happened at GEO. But Arabsat would not yet have been in geostationary orbit. It needed to perform one more circularization burn at perigee to slow itself back down and lower its apogee from 90 Mm to geosynchronous altitude.

This kind of supersynchronous, "overshooting" transfer into a higher orbit is not a Hohmann transfer orbit, it is a bi-elliptic transfer. Despite requiring three burns (1: [over]raise apogee, 2: raise perigee, 3: lower apogee) rather than just the two of the Hohmann (1: raise apogee, 2: raise perigee), bi-elliptic transfers can require less dV in some cases. Raising from moderately-inclined LEO to GEO is not (*usually) one of these cases. The total dV required is greater than a traditional GTO, except in very inclined cases. (I was not aware of this when I wrote the original answer; thanks to @BrendanLuke15 for this surprising answer!)

But focusing on the other reason--the reason that Arabsat (which is only launching from the modestly-inclined Cape) used a supersynchronous orbit: the lower-energy burns required of Arabsat to enter GEO from its transfer orbit expended much less dV than the single burn to enter GEO from a traditional GTO.

The difference in energy was made up by Falcon Heavy, which placed Arabsat in the high energy geostationary transfer orbit which threw it out to 90 Mm instead of 35.7ish Mm.

In short, it's asking more of the launcher, but less of the payload. For various market reasons (namely, you're buying a pre-existing vehicle from a launch service provider, instead of building your own), payload owners/builders might find it cheaper to make a much smaller, lighter, and simpler payload which does not have to do the expensive combined GEO-altitude circularization/deinclination burn. This is doubly-so if they're going to be putting it on a large vehicle anyways. While the dV savings may be minimal or negative, there's a relatively large capability gap between mid-to-heavy launchers & smallsat launchers. If you're already buying slightly more rocket than you'll need, why not take advantage of it? It's the same price to launcher either way, but a much cheaper & simpler payload.

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    $\begingroup$ +1. It is important to remember that we are talking about commercial space. So, this is not just about orbital mechanics (how do we minimize total Δ-v) but also about cost (how do I minimize my Δ-v). $\endgroup$ Commented Jan 11, 2022 at 6:54
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    $\begingroup$ What is the second reason? The first reason seems to be $\Delta V$, then the second reason also seems to be ... $\Delta V$. How are they different? $\endgroup$ Commented Jan 11, 2022 at 12:27
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    $\begingroup$ @BrendanLuke15 The first reason is "minimizing total delta-v / cost of plane changes", the second reason is "trading cheap delta-v on the launch vehicle for expensive delta-v on the payload (even without plane changes)". $\endgroup$
    – TooTea
    Commented Jan 11, 2022 at 13:03
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    $\begingroup$ @BrendanLuke15: The "new" thing about "commercial space" compared to "military/government space" is that different entities pay for the ΔV of different stages; in particular, launch vehicle vs. spacecraft. So, it makes sense to use an overall less energy-efficient trajectory if it allows one entity to push off ΔV costs towards another entity. In this example, satellite ΔV is more expensive than launch vehicle ΔV, so it makes sense to move ΔV costs from the satellite towards the launch vehicle. $\endgroup$ Commented Jan 12, 2022 at 9:12
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    $\begingroup$ @BrendanLuke15: These optimizations are getting more interesting with reusable launchers. In the past, propellant cost was negligible compared to throwing away a brand-new rocket every time, so it made sense to use all available ΔV of the launch vehicle. But for reusable launchers, propellant cost as a fraction of launch cost is much higher. And, higher ΔV means landing further away, which has its own costs. For example, SpaceX's latest brand-new booster was heavily damaged in rough seas, which wouldn't have happened with an RTLS landing. $\endgroup$ Commented Jan 12, 2022 at 9:16

As the comments discuss, inclination is the missing variable in the equation.

The standard geostationary launch that you are thinking of looks something like this:

  • geostationary transfer orbit (GTO) injection burn:
    • perigee @ parking orbit height (e.g. 250 km)
    • apogee @ geosynchronous height (35786 km)
    • inclination is ~launch site latitude (i.e., ~28.5° for Florida launches)
    • this is typically done by the launch vehicle
  • combined plane change + circularization burn @ geosynchronous height:
    • change inclination to 0°
    • this is typically done by the spacecraft (although direct injection is becoming more and more common)

Assuming instantaneous burns, the second burn can be analyzed like this (excuse the crude MS Paint drawing):

standard GEO insertion

To which the cosine law can be applied to find the $\Delta V$.

If $|V1|=|V2|$, the equation reduces to:

$$\Delta V = 2 \cdot |V| \cdot \sin{\frac{i}{2}}$$

This presents the possibility of lowering the $\Delta V$ cost of the plane change maneuver if $|V|$ is lower (i.e., higher altitude). This is what the supersynchronous transfer orbit does.

These supersynchronous maneuvers break down like so:

  • geostationary transfer orbit (GTO) injection burn:
    • perigee @ parking orbit height (e.g. 250 km)
    • apogee > geosynchronous height (> 35786 km, supersynchronous)
    • no change in inclination
  • combined plane change + perigee raising burn @ transfer orbit apogee:
    • raise perigee to synchronous height (35786 km)
    • change inclination to 0°
  • circularize @ perigee

The total $\Delta V$ costs for the standard method (inclined 250 km parking orbit to geostationary orbit) are:

standard method

While the total $\Delta V$ costs for the supersynchronous method are:

super method

Interestingly, there is an iso-propic (if you will) line at an inclination of ~40° where the total $\Delta V$ cost of the maneuvers is independent of transfer orbit apoapsis (neat!).

The $\Delta V$ is lower for the supersynchronous method in the coloured regions of this plot:

super better than standard

However, when you only consider the maneuvers that the spacecraft is (typically) responsible for (i.e., post injection), it is always equal OR more efficient to use a supersynchronous transfer, though the $\Delta V$ savings are relatively meager for the most common inclinations/launch sites:

delta-V savings

The red dot represents the Thaicom 6 satellite that SpaceX launched in 2014 on a supersynchronous transfer orbit (90,000 km, 20.75°). The $\Delta V$ savings was about 10%.

A higher transfer orbit of course takes more time to reach GEO:

transfer time

(normalized for the sake of electric propulsion satellites where the maneuvers can't all occur in one orbit)

But, as pointedly put in this answer:

Lifetime is money.

and the $\Delta V$ savings are worth more life than they cost.


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