The spate of K-questions got me thinking: At what point does atmospheric drag disappear1 with the pressure from sunlight, solar wind, or other forces becoming overwhelmingly dominant? I'm guessing it's somewhere in the outer reaches of Earth's magnetosphere so the answer might not be a fixed altitude. So, at what point does Earth's atmosphere cease having a dominant effect on an object's orbit2? Put another way, at what point is drag from Earth's atmosphere no longer able to cause an object's orbit to lower?

1 Properly-speaking, all gases/molecules/etc. that are above the Earth's surface and that are gravitationally bound to the Earth are within Earth's atmosphere. Atmospheric drag can thus occur anywhere within Earth's gravitational well. I suggest though that at some point the atmosphere should be swept away by e.g. solar wind; if it is no longer be present it won't be able to induce drag.

2 I reason that at some point other forces will either cancel out said drag's effect on an object or apply a force that has a net effect of boosting its orbit. I am unsure about this (especially the second assertion), hence the question. (This is clearly wrong since external forces on opposite sides of the orbit will cancel out while drag forces are additive.)

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    $\begingroup$ Looks like a good improvement overall. It subtly changed the question though so I may tweak it a bit. $\endgroup$ Commented Dec 28, 2018 at 2:21
  • $\begingroup$ Answer to the title "At what altitude do the effects of atmospheric drag completely disappear?" None, because it never completely disappears. $\endgroup$
    – uhoh
    Commented Dec 28, 2018 at 4:46
  • $\begingroup$ @uhoh I disagree. At one point one surely would get swept away by e.g. solar wind (albeit very slowly, maybe). At the very least, at some point the atmosphere itself should be swept away. At least that's my reasoning (which could well be flawed). $\endgroup$ Commented Dec 28, 2018 at 4:53
  • $\begingroup$ Drag does not disappear because density does not disappear. Even if there were other forces, they could be resisted by drag. I think you want to ask about the net effect of all the forces, but losses due to drag would still be there no matter what additional forces there are. It never "disappears" per se. $\endgroup$
    – uhoh
    Commented Dec 28, 2018 at 4:57
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    $\begingroup$ @uhoh I see what you're getting at. I'll update the question to clarify that. $\endgroup$ Commented Dec 28, 2018 at 5:00

1 Answer 1


Helpful factoids about orbits. To raise an orbit, you need a force that "pushes from behind", who's net effect accelerates you in the prograde direction. For circular orbits you can call that the $+\theta$ or tangential or $\mathbf{\hat{v}}$ direction. Counterintuitively this push from behind slows you down by an amount equal to the delta-v imparted, but it does raise your orbit.

Drag actually speeds you up by pushing against you, but lowers your orbit.

Aerodynamic drag always pushes in the exact opposite direction of your motion, because that's the definition. Aerodynamic effects on spacecraft in directions perpendicular to your velocity (radial or out-of-plane) are called lift.

The problem with comparing aerodynamic drag to effects from the solar wind or photon pressure is that those particles and photons are moving so much faster than the spacecraft's motion around the Earth, or the spacecraft plus Earth's motion around the Sun. These effects could be prograde at one point in the orbit and retrograde a half an orbit later, and ultimately substantially average out in some cases.

So it's a bit of an apples-to-oranges comparison.

However, in the plot below, the author has tried to show accelerations caused by various effects in an order-of-magnitude sort of way. For example, at an altitude of about 1300 km the acceleration due to solar radiation pressure is about the same size as that due to drag during maximum solar activity. Drag at high altitude is dramatically modulated by heating and expansion of the atmosphere by particles from the Sun, so actually both are caused by the Sun even though one also involves the atmosphere.

Below is borrowed from my answer to the question The sorting of perturbational effects by the power (used here as well).

I found the following plot in the book Satellite Orbits; Models, Methods, Applications by Oliver Montenbruck and Eberhard Gill, Springer, 2000. The figure and description can also be found in google books. It's a low quality snapshot but it's hard to capture a dozen different dependencies over 20 orders of magnitude without showing the whole thing.

Satellite Orbits; Models, Methods, Applications, Montenbruck & Eberhard Gill

Here is the bit of text that discusses the figure in more detail:

The effect of various perturbations a s a function of geocentric satellite distance is illustrated in Figure 3.1. For the calculation of the influence of atmospheric drag on circular low-Earth satellite orbits, exospheric temperatures between 500K and 2000K (cf. Sect. 3.5) have been assumed. The area-to-mass ratio used in the computation of non-gravitational forces is 0.01 m2/kg. For specially designed geodetic satellites like LAGEOS, the corresponding value may be smaller by one or two orders of magnitude. The perturbations due to various Geopotential coefficients Jn,m and the lunisolar attraction have been calculated from rule-of-thumb formulas by Milani et al. (1987). For the purposes of comparison it is mentioned that a constant radial acceleration of 10-11 km/s2 changes the semi-major axis of a geostationary satellite by approximately 1 m.

Aside from the aforementioned forces, various minor perturbations are considered in Fig. 3.1 which produce accelerations in the order of 10-15 to 10-12 km/s2. The most are due to the radiation pressure, resulting from the sunlight reflected by the Earth (albedo), as well as relativistic effects and the solid Earth tides.

  • $\begingroup$ Would you support the assertion that there is persistent atmosphere (albeit at vanishingly low density) throughout Earth's gravity well? $\endgroup$ Commented Dec 28, 2018 at 5:51
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    $\begingroup$ @AlexHajnal btw I switched $+\theta$ to MathJax since I'm already using it for $\mathbf{\hat{v}}$. Spelling out the word "delta-v" rather than using $\Delta v$ is really common, it's not for lack of access to a greek $\Delta$ character, it seems to be a real stand-alone term. en.wikipedia.org/wiki/Delta-v $\endgroup$
    – uhoh
    Commented Dec 28, 2018 at 6:02
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    $\begingroup$ @JoeJobs binary answer is yes, but "floating point" answer is so very very little that it would be very hard to notice, very hard to calculate exactly and so much smaller than other effects that nobody cares. For example when I take out a large library book and bring it home it changes the orbit of the international space station, but not by much! :-) $\endgroup$
    – uhoh
    Commented Dec 2, 2020 at 22:37
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    $\begingroup$ @OrganicMarble Disagree! An answer depends on A x B where A describes the satellite and B describes the atmosphere's density as a function of height and time. By far the hardest and most interesting part is B. An answer about A alone tells us nothing and one can look up or simply estimate ballpark numbers for A and use them with B if B is provided. So an answer providing B tells us almost everything. $\endgroup$
    – uhoh
    Commented Dec 3, 2020 at 23:21
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    $\begingroup$ @uhoh Agreed, I should have said what also matters here is the ballistic coefficient... $\endgroup$ Commented Dec 3, 2020 at 23:28

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