I've been told that for low thrust ion spirals, delta V would be the difference between speeds of departure and destination orbits. For example the delta V between a 7.7 km/s LEO and a 3.1 km/s GEO would be 4.6 km/s. Is this correct? How is that derived?

For a ion drive accelerating a spacecraft 1 millimeter/sec^2, I get 11.6 days to accelerate 1 km/s.

To figure time it'd take to spiral from LEO to GEO I took 11.6 * 4.6 to get 54 days. Again, I don't know if this is correct.

Here is an attempt to draw a LEO to GEO spiral with 1 mm/s^2 acceleration:

enter image description here

This was my line of thought: end orbit is about 24 hours, beginning orbit is 1.5 hours so average orbit is about 12.75 hours. !2.75 hours goes into 54 days about a 100 times so I made a logarithmic spiral that turns 100 times from LEO to GEO. (hot colored areas are Van Allen Belts)

I strongly suspect this is wrong. It seems to me an ion spiral would be wound tighter at LEO and gradually relax as the spacecraft ascends. But at this point I have no idea how to model an ion spiral in an Excel spreadsheet.

In these questions I am using the specific example of LEO to GEO but I am hoping for guidelines for more general scenarios

  • 5
    $\begingroup$ If you want to play with this sort of thing, I highly recommend that you graduate from Excel. I like Mathematica, which currently has a Pi day sale for the home and student versions. (I have no connection with or stock in Wolfram.) $\endgroup$
    – Mark Adler
    Mar 12, 2015 at 5:02
  • 1
    $\begingroup$ @MarkAdler Rats! It is now June (and 2016 as well). It's too late. I'm wondering, do they have a Tau day sale as well? c.f. Tau explained by Vi Hart, Tau Day official site. $\endgroup$
    – uhoh
    Jun 10, 2016 at 4:01

2 Answers 2


The rule you have for the total $\Delta V$ of a low-thrust spiral is an upper limit arrived at as you let the thrust go to zero. However that takes an infinite amount of time. The total $\Delta V$ of a spiral with non-zero thrust is less, and the time is finite. But it is a good rule of thumb for quick calculations when trying to establish feasibility.

The derivation of the rule of thumb is quite simple. Look at an infinitesimally small Hohmann transfer. You will find that the $\Delta V$ total of the two infinitesimal burns at the initial orbit and at apoapsis of the transfer orbit is equal to the difference in the orbital velocities. Then if you add those up for a finite raise in orbit, you get the difference in $\Delta V$ of the initial and final orbit.

To find out the real total $\Delta V$ and to plot an actual trajectory that doesn't do an infinite number of orbits before it gets anywhere is best done using numerical integration.

Here is an example of a spiral from a circular orbit to escape ($C_3=0$):


This is normalized to the starting circular orbit, where the distances are in units of the initial orbit radius, and the acceleration is constant at $10^{-3}$ of the gravitational acceleration of the body at the initial orbit radius. The total $\Delta V$ to escape is 0.856 of the initial orbit velocity, as compared to 1.0 for the rule of thumb. The total time to escape is 136 initial orbit periods. It goes around the body about 40 times before escaping.

The first several orbits are close enough that you can't make them out at the resolution shown. This gets even worse for smaller accelerations. $10^{-3}$ is actually pretty high. I picked it so that you can see the spiral better. That time from a low Earth orbit is about 8.5 days. A typical spiral out might be more like months with accelerations of $10^{-4}$ of the initial gravitational acceleration, or less. Attempts at plotting that show a solid disk until near the end where you see the spiral escape.

Here is an example of a spiral from LEO (400 km) to GEO with the same normalization and a normalized constant acceleration of $10^{-4}$. It takes about two months over 945 orbits. In this case the total $\Delta V$ is very close to the rule of thumb. This is simplified, since the final flight path angle here is about half a degree. So there is some time and $\Delta V$ remaining to circularize the orbit.

spiral to geo

You could approximate this plot by advancing one orbit at a time, using the orbit period times the acceleration as the $\Delta V$ and raising the orbit the corresponding amount, connecting each with a linearly increasing spiral.


Just to add to your original question "is this correct?" - yes but only for spirals between two circular orbits.

Responding to your request for "guidelines for more general scenarios": If you wanted to find the ∆V to transfer from an elliptical orbit such as GTO and spiral to GEO then you could follow a method such as in Mark Adler's second paragraph above - with a significant adjustment. For such a transfer its best to stick to a short arc around the apogee and accept that the part of the arc away from the exact apogee will suffer a loss. Evaluating this loss needs numerical integration though if you really want just a ballpark figure you could assume a cosine dependency, i.e. efficiency 1 at the apogee, 0 at the ends of the semi-minor axis (1/4 of an orbit away).

  • $\begingroup$ I was thinking of circular to circular orbits. But now that you mention it I would like modeling elliptical to elliptical orbits. So I've upvoted your answer. though I have to confess that I'm not sure what you mean by a cosine dependency. $\endgroup$
    – HopDavid
    Oct 24, 2015 at 9:11
  • $\begingroup$ I confess for my part that the "cosine dependency" was something I thought of twenty years ago when I didn't have the time to look into it more. I meant to use the cosine shape to inspire what length of arc to apply full thrust. It will be trial and error in choosing a compromise between propellant efficiency and transfer time. You could also decide to modulate the thrust level, or direction (to remain orthogonal to the radius vector), away from apogee. I couldn't say off the cuff which would be better. $\endgroup$
    – Puffin
    Oct 25, 2015 at 18:10
  • $\begingroup$ You could get a rough idea of the effects of adding a small ∆V through the vis-viva equation, i.e. check the impact on the semi-major axis, at places progressively away from apogee. v^2 = GM (2/r - 1/a) $\endgroup$
    – Puffin
    Oct 25, 2015 at 18:13
  • $\begingroup$ One last thought that has just come back to me: the example I have desribed will not give you an exact circular orbit, you will need to tidy it up once you have got most of the way there. $\endgroup$
    – Puffin
    Oct 26, 2015 at 18:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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