What is the lowest altitude that an ion thruster can realistically be used for station keeping before its small amount of thrust is overcome by atmospheric drag? Please state your assumptions such as thrust, spacecraft drag area, mass etc.

The motivation behind my question is similar to this quote. From Wikipedia's article on orbital drag (even though the quote is about only one revolution):

Due to atmospheric drag, the lowest altitude above the Earth at which an object in a circular orbit can complete at least one full revolution without propulsion is approximately 150 km (90 mi).

I wonder if ion thrusters in continuous operation can meaningfully extend (or actually reduce) the limit of lowest non-degrading orbit.

Related question: What are the challenge to use atmosphere as a propellant for super low altitude (180-260 km) satellites?

  • $\begingroup$ The latter. The question is updated to clarify my intention. However, even satellites as high as geostationary ones require station keeping so I suppose any orbit is non-degrading only up to N revolutions unless there's a rigorous definition of non-degrading. But again, feel free to make any realistic assumptions. Since any normal satellite makes many revolutions you may average out the contribution of changing orientation. Thanks! $\endgroup$
    – Moobie
    Commented May 22, 2017 at 15:36
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    $\begingroup$ @ArthurDent I think the question makes perfect sense, and the one-revolution is just a sanity check or reference point to get a rough handle on the problem. The OP is interested in a quantitative analysis and doesn't pretend to be doing one here. $\endgroup$
    – uhoh
    Commented May 22, 2017 at 15:45
  • $\begingroup$ At geostationary, the drag is virtually non-existent. Instead, the SRP (solar radiation pressure) is the dominant perturbation. So even if your ion thruster was cancelling out drag in the velocity direction, the satellite would need to orient itself every once and a while to have the thruster cancel out SRP as well. $\endgroup$ Commented May 22, 2017 at 15:47
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    $\begingroup$ @ArthurDent what is the point of this comment, besides unabashed pedantry? Are you actually confused about what the OP is asking? If so, do you have specific suggestions for how to make his question clearer? For what it's worth, I agree with uhoh that it's perfectly clear as is. $\endgroup$
    – Chris
    Commented May 22, 2017 at 15:57
  • $\begingroup$ @Chris His question as currently stated is perfectly fine. Before, I was a little confused since he specifically mentioned a single orbit as his motivation. I was genuinely not 100% sure what his question was. His edits make it more clear imo. $\endgroup$ Commented May 22, 2017 at 16:20

1 Answer 1


For a hefty, substantial spacecraft like the 5000 kg, 5 square meter, drag coefficient = 1 simulation I did here, the final complete orbit was circularized at about 95 kilometers. A cube sat of the same average density would have been higher just because of the higher area to mass ratio.

For the atmosphere to match other tabulated values around 100km I used the very very rough approximation

$$\rho(h) \ = \ \rho_0 \ exp(-h/h_{scale})$$

With the sea-level density and scale height $\rho_0$ and $h_{scale}$ of 1.225 kg/m^3 and 7700 meters. This is just rough.

A simple equation for drag from Wikipedia might be

$$F_{drag} \ = \ -\frac{1}{2} \ \rho \ C_D \ A \ v^2$$

and $C_D$ for a randomly shaped satellite at LEO velocity might be 1, the area might be 5 meters, and the LEO velocity $v$ might be 7700 m/s.

$$F_{drag} \ = \ 0.5 \times 1.225 \times exp(-h/h_{scale}) \times 1.0 \times 5.0 \times 7700^2$$

If I assume the thruster has a force of 0.2 N as I did here, I can solve for the height $h$, which in this case turns out to be almost exactly your original number, 146 km.

Did I say roughly, and say it enough times?

Now the problem is that the upper atmosphere changes a lot in response to activity of the Sun, and so the density is not well approximated by a simple scale height, or is even constant or predictable. This is discussed in @DavidHammon's answer here and comment here. So if there is a solar event, you may suddenly need maybe even ten times the thrust to keep from burning up all of a sudden. So while estimates can give you a feel for it, your mileage may vary, a lot!

People talk about putting spy or Earth monitoring satellites at altitudes between 150 and 200 km because a smaller telescope (like a 10cm aperture inside a 3U cubesat) might get better resolution than one at 400km, or similar to a 50cm aperture at 750 km. Of course there are a lot of other factors that affect resolution, and there is a problem with atomic oxygen eating your satellite, but people still talk about it.

  • $\begingroup$ It is also worth mentioning that SRP would produce a force not in the velocity direction, so for pure ion station-keeping, there would need to be attitude shifts to counteract that force. $\endgroup$ Commented May 22, 2017 at 18:50
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    $\begingroup$ Nice! Kinda lame for me not to recall the drag equation. Is the density equation applied correctly and where does that 7700m (not escape velocity) come from? I got p = 6.6051294e-13 from this table: spacience.blogspot.com/2012/03/… When I solved for 0.5 * 8.77*10^-5 * e^(-h/6.549) *1.0*5.0*7700^2 = 0.2, I got h = 72.5766, which would indicate a useful change. (I've substituted a few values from the table but they will overshoot their intended altitude range.) But then I hadn't iteratively adjust the escape velocity. $\endgroup$
    – Moobie
    Commented May 22, 2017 at 19:22
  • $\begingroup$ OK this will be a good place to expand on the density, give me a few hours. Comments under this answer link to NASA's Standard Atmosphere 1976 tables Table 4, where 100 to 150 km falls on pages 68 and 69. It looks like I'd originally expected the ROI to be around 150 km and chose $h_{scale}$ to match density there, not 100 km, but a better approximation should be found because this going to keep coming up. Stay tuned! $\endgroup$
    – uhoh
    Commented May 23, 2017 at 1:23
  • $\begingroup$ @Moobie I've just asked Why does Earth's atmospheric density have a big “knee” around 100 km?. Take a look at the plot there. $\endgroup$
    – uhoh
    Commented May 23, 2017 at 6:39
  • $\begingroup$ @ArthurDent SRP produces a force but not necessarily a torque - if the centre of mass and centre of pressure are in similar places, attitude manoeuvres aren't needed to correct for SRP. The SRP forces balance quite nicely for a station over the full phase segment of an orbit. $\endgroup$
    – Diamond
    Commented Jan 16, 2019 at 16:44

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