In my answer to What are the benefits of supersynchronous transfer orbits? I stumbled upon an interesting outcome when considering the total $\Delta V$ cost from an inclined low altitude parking orbit to geostationary orbit using a supersynchronous transfer orbit. The maneuver profile is detailed below, requiring 3 burns:

  • supersynchronous transfer orbit injection burn:
    • perigee @ low parking orbit height
    • apogee > geosynchronous height (> 35786 km, supersynchronous)
    • no change in inclination
    • $\Delta V_1 = \sqrt{\mu (\frac{2}{r_{park}}-\frac{1}{\frac{r_{park}+r_{Apo}}{2}})} - \sqrt{\frac{\mu}{r_{park}}}$
  • combined plane change + perigee raising burn @ transfer orbit apogee:
    • raise perigee to synchronous height (35786 km)
    • change inclination to 0°
    • $V_{Apo1} = \sqrt{\mu (\frac{2}{r_{Apo}}-\frac{1}{\frac{r_{park}+r_{Apo}}{2}})}, V_{Apo2} = \sqrt{\mu (\frac{2}{r_{Apo}}-\frac{1}{\frac{r_{sync}+r_{Apo}}{2}})}$
    • $\Delta V_2 = \sqrt{V_{Apo1}^2 + V_{Apo2}^2 - 2V_{Apo1} \cdot V_{Apo2}\cos{i}}$ (cosine law)
  • circularize @ perigee:
    • $\Delta V_3 = \sqrt{\mu (\frac{2}{r_{sync}}-\frac{1}{\frac{r_{sync}+r_{Apo}}{2}})} - \sqrt{\frac{\mu}{r_{sync}}}$

The total $\Delta V$ is the sum of all 3 burns (though typically the satellite is on its own for the final 2). Plotting the total $\Delta V$ over a wide range of inclinations and transfer orbit apogees yields this (250 km parking orbit assumed):

delta-V to GEO

Around ~40° it looks like the contours go vertical, indicating a sort of iso-propic line (if you will) where the total $\Delta V$ cost is independent of transfer orbit apogee (!wow). But, note that the $\Delta V$ required of the satellite (final 2 burns) decreases with increased transfer orbit apogee.

Upon closer inspection there is no proper vertical line/invariant inclination, but these 5 m/s contours show that there is a region of quasi iso-propicness around 39° inclination (5 m/s in the scheme of 4+ km/s seems pretty invariant to me):

5 m/s contours

There is no benefit for a traditional upper stage + payload combination because while the total $\Delta V$ stays the same, the distribution between launcher and payload changes by as much as ~20% for the apogee scale shown in the plots.

Consider though, a single spacecraft responsible for all of the (admittedly high) $\Delta V$. Only one transfer orbit apogee can be used per mission, of course. However, is there any reason why sitting in a ~39° inclined low Earth orbit with the ability to target geostationary orbit from any transfer orbit apogee would make sense? Even if it does make sense, are there more practical solutions?

Re: Comments, here are the 3 burns individually. The first and last don't deal with inclination and thus only depend on transfer orbit apogee (click to enlarge):

3 burns individual

  • 2
    $\begingroup$ You made those beautiful plots - would you mind making them for the three burns individually as well? That would be great! $\endgroup$
    – asdfex
    Commented Jan 12, 2022 at 17:32
  • 1
    $\begingroup$ @asdfex good idea! $\endgroup$ Commented Jan 12, 2022 at 17:34

1 Answer 1


What you are seeing here is the result of the inefficiencies in the orbital transfer strategy employed, in the transition region between direct transfers and bi-elliptic transfers.

When transferring between two circular orbits with some inclination, eventually the "degenerate" bi-elliptic transfer (with infinite apoapsis) will become optimal, provided the target orbital radius or the inclination is large enough.

At that point, it's a flat transfer cost, independent of inclination change:

$$\Delta v = \left(\sqrt{2} - 1\right)\left(\sqrt{\frac{2\mu}{r_1}} + \sqrt{\frac{2\mu}{r_2}}\right)$$

The other strategy, or course, is to do a direct arc, splitting the inclination change between the two burns (not evenly!).

At ~40°, targeting synchronous orbits, you are in the transition region between these two, costs are about the same.

But they have the opposite reaction to increased altitude!

  • High-apopasis bi-elliptic transfers become cheaper when the target orbit becomes higher (after some inclination-dependent threshold)
  • Direct arc transfers become more expensive when the target orbit becomes higher.

You're using a hybrid approach, which has two theoretical inefficiencies:

  1. It has a limited transfer apoapsis. (bi-elliptic inefficiency)
  2. It does not put any of the inclination change in the first burn. (direct arc inefficiency)

With a hybrid approach like that, the altitude-dependent delta-v cost trends will cancel eachother out for some inclination-altitude region, and that's what you have.

Are there uses for these 'quasi iso-propic' supersynchronous transfer orbits?

I do not think so. The only reason to do large inclination changes, like ~40°, is to reach synchronous orbits from launch sites not on the equator. But that's only one useful altitude, so the rest of the quasi iso-propic line isn't used.


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