To move a satellite in geostationary orbit, 166°55′E, to the antipode, 13° 4' 3.2" W, what delta-V would be needed for this to be accomplished?

As for time constraints, I do not know what are the possibilities with contemporary technology. What would be the fastest? What would be the delta-V if the time required was four times slower than the fastest option already discussed?

  • 12
    $\begingroup$ What are your time constraints? An arbitrarily small manoeuvre will change the phasing such that the craft will gradually move to the desired orbit, but it will take an arbitrarily long time (ignoring perturbations) $\endgroup$
    – Jack
    Commented Dec 29, 2018 at 16:21
  • 4
    $\begingroup$ This is mathematically interesting as it depends on the initial and final longitudes. $\endgroup$
    – uhoh
    Commented Dec 29, 2018 at 16:25
  • 1
    $\begingroup$ @Jack have a look at the abstract at the bottom of page ii here: ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19660027977.pdf You need more than an arbitrarily small kick to get from one equilibrium point to another, otherwise, you can't get there from here. (also, audio). So the trick is to find out how deep the two stable points are. $\endgroup$
    – uhoh
    Commented Dec 29, 2018 at 16:31
  • 2
    $\begingroup$ I think an answer that explains that you can't even stay at some longitudes without expending delta-v, whereas you can't get away from other longitudes without expending delta-v, is sort-of the minimum answer, and what makes this question so cool! The time it takes is going to be weeks or months, but there are some hard minimums (in delta-v or its rate of use) that are the reality of GEO, and they can be estimated. $\endgroup$
    – uhoh
    Commented Dec 29, 2018 at 16:43
  • 2
    $\begingroup$ @Bob516 They're only important if you're looking for an absolute minimum ∆v solution -- if you're trying to do it in a reasonable amount of time instead the instabilities don't matter. $\endgroup$ Commented Dec 29, 2018 at 18:17

1 Answer 1


Theoretically, you can go anywhere in GEO for an arbitrarily small ∆v - you raise your apogee a little bit, which slows you down, wait until you've phased to your destination latitude, then re-circularize back into GEO.

In practice, though, as @uhoh mentions in comments, there are stable longitudes in GEO that require more than an infinitesimal maneuver to escape from. The maximum instability according to that paper requires only about 2 m/s per year to correct, however, so I would guess that any maneuver of more than a few m/s can escape the stable nodes.

So, the decisive figure is how long you want to take to move your satellite.

If you want to go to the opposite side of Earth in a month, you need to raise your orbit to the height where you're going 29.5/30.0 times as fast as GEO, so you lose half an orbit after 30 days.

The semi-major axis of an orbit with period $t$ is:

$$a = \sqrt[3]{\frac{\mu t^2}{4\pi^2}}$$

Where $\mu$ is Earth's standard gravitational parameter. For this orbit the SMA is about 42639 km (radius, not altitude). Holding perigee fixed, you get a 35736 km by 36750 km orbit. That apogee raise maneuver is equivalent to the first impulse of an ideal Hohmann transfer, the cost of which is given by:

$$\Delta v_1 = \sqrt{\frac{\mu}{r_1}} \left( \sqrt{\frac{2 r_2}{r_1+r_2}} - 1 \right)$$

Which works out to about 17.3 m/s to raise the orbit, and the same to recircularize a month later, for a total of about 35 m/s.

To do it in a week, you need to go to an orbit that's 6.5/7.0 as fast - an apogee of 40071 km. The cost here is about proportional to the speedup - 73.5 m/s to get into or out of the phasing orbit, for a total of 147 m/s.

If you can wait 6 months, you phase a degree per day and the cost drops to about 6.2 m/s.

The really fast way would be to drop perigee to 4595km, which is an orbit that takes only 12 hours to complete, re-circularize after one orbit when you're back at geosynchronous altitude, outrunning those slowpokes in GEO -- this takes 1099 m/s on each end for a total of 2198 m/s.

  • $\begingroup$ @RussellBorgove For the really fast option, how long would that take? $\endgroup$
    – Bob516
    Commented Dec 29, 2018 at 19:21
  • 1
    $\begingroup$ It takes half a day, one shortcut orbit -- rewrote to clarify. $\endgroup$ Commented Dec 29, 2018 at 19:22
  • $\begingroup$ Follow-up question: Quantitatively, how deep are the stable equilibrium points in GEO? $\endgroup$
    – uhoh
    Commented Dec 29, 2018 at 23:38
  • 7
    $\begingroup$ @RussellBorgove - at the daft end also have the 6 hour option of doing two 180 degree inclination changes and flying GEO retrograde, which I think totals in around 12000m/s and some sort of massive lawsuit from every other GEO operator in existence. $\endgroup$ Commented Dec 29, 2018 at 23:45
  • 1
    $\begingroup$ @GremlingWranger Another upper limmit would be to start by the speed of light in a straight line and then add constraints such as not hitting that pesky planet etc $\endgroup$
    – lijat
    Commented Dec 30, 2018 at 16:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.