# When is the supersynchronous orbit more efficient than a typical Hohmann transfer where inclination change and circularization are simultaneous?

The title essentially explains it all. Some GEO launch vehicles like Proton, which launches from Baikonur at a latitude of 46 degrees, launch into a supersynchronous orbit. So at what point is the supersynchronous, bi-elliptic transfer more efficient than the Hohmann transfer when launching into geostationary orbit?

• Jan 11 at 17:13
• I would add one comment to the Op's question; normally the $\Delta_v$ in the previous answers on this forum usually give the $\Delta_v$ from LEO to GEO via either GTO or via super synchronous. But the $\Delta_v$ that matters is the $\Delta_v$ from GTO to GEO as compared with the $\Delta_v$ super synchronous orbit to GEO. That's because the launch vehicle may have that extra capability built in for the extra $\Delta_v$ to get all the way to super synchronous orbit instead of GTO. And if so, that saves on the additional $\Delta_v$ rockets in the payload required. Jan 11 at 17:43
• Also, the 2nd stage of the launch vehicle wants to stay in the GTO orbit (not GEO), so it can burn up as the perigee slowly degrades instead of becoming permanent GEO space junk. So the launch vehicle can't go all the way to GEO, but the launch vehicle can go all the way to either GTO or super synchronous without creating space junk. And I'm guessing it always requires less additional $\Delta_v$ from the payload's rockets if the launch vehicle goes higher to super synchronous instead of just GTO. Jan 11 at 17:56
• The effective latitude of Baikonur for space-launch purposes isn't 46°; it's over 54°, because they can't launch directly east (they have to go northeast to keep from dropping rocket stages on China, never mind that China does that to itself all the time). Jan 12 at 4:45

(mostly recycled from What are the benefits of supersynchronous transfer orbits?)

Excuse the wall of plots but I really do think they describe it better than my words ever could :)

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

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

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

The $$\Delta V$$ savings for the supersynchronous method is shown here (less than zero is less efficient):

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. The $$\Delta V$$ cost of the final burn for the standard method is:

For the supersynchronous method, the $$\Delta V$$ cost of the final two burns is:

The $$\Delta V$$ (final burn(s)) savings for the supersynchronous method are shown here:

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:

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

• Lets say the target orbit is GEO with zero inclination. Can you compare the GTO transfer orbit method launching from Cape Canaveral $28^{\circ}N$, vs the super synchronous orbit method launching from Cape Canaveral? Maybe same question from the Baikonur Cosmodrome launching from $46^{\circ}N$? For these two launch centers to GEO with zero inclination, how much lower is the $\Delta_v$ for the payload rocket's part of the $\Delta_v$? Also, can the 2nd stage do any of the orbital inclination adjustment at Perigee? To save additional fuel in the payload? Jan 11 at 18:09
• @Sheldon re: " 2nd stage do any of the orbital inclination adjustment at Perigee" I don't know, that is a much bigger analysis though. Jan 11 at 18:15
• I can eyeball the chart for $28^oN$ via GTO as $\Delta_V\approx$1.8, but having trouble with the corresponding number for super synchronous to 90km from $28^oN$. For the $46^oN$, case the GTO chart gives $\Delta_V\approx$2.25, again having trouble with the comparable super synchronous number. I think a concrete example of the payload rocket's $\Delta_V$ savings would be helpful .... :) Jan 11 at 18:21
• Looks like the other inclination of interest for the Baikonur Cosmodrome launching from $46^oN$ is actually $51.6^oN$ because the Russian don't want space debris falling into China when a Russian rocket fails, so the Russians typically launch into a slightly higher orbital inclination, just to be safe. The only other orbital inclination the might be interesting would be the Guiana Space Centre at $5^oN$. I can see no value whatsoever to any of the numbers in the chart .... Jan 11 at 18:46
• @CuteKItty_pleaseStopBArking I'm really sorry if I sound offensive. I just felt like the answer provided ... wasn't really an answer, and I was trying to explain why the colored charts didn't seem helpful to me. I think the chart says the savings from Cape Canaveral at $28^oN$ super synchronous to 90km is about 8%, and maybe about 20% from the Baikonur Cosmodrome launching at $51.6^oN$ inclination. But my guess is that I'm reading the charts wrong and maybe the savings is higher. Then again, maybe that's all the super synchronous launch saves, I don't know. Jan 11 at 19:06