Astronomical seeing is the limiting factor for the resolution of all but the smallest Earthbound telescopes.

Astronomical Seeing Source

Stunning advances in adaptive optics (along with it's predecessor speckle interferometry and it's budget-minded cousin lucky imaging) get around this but only with substantial compromises in (some combination of) throughput, cost, complexity and wavelength range (see answers to Why aren't ground-based observatories using adaptive optics for visible wavelengths?)

If identical telescopes sat on the surface of Earth and Mars and looked at a distant and equi-distant body, how would the following differ between the Mars telescope and the Earth telescope?:

  • 2
    $\begingroup$ Lucky imaging is also mentioned in this recent answer and many others: astronomy.stackexchange.com/search?q=lucky+imaging $\endgroup$
    – uhoh
    Commented Oct 25, 2018 at 23:45
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    $\begingroup$ FWIW, the atmospheric pressure on top of Olympus Mons is about 12% of the mean Martian atmospheric pressure; OTOH, Olympus Mons creates orographic clouds. $\endgroup$
    – PM 2Ring
    Commented Oct 26, 2018 at 4:21
  • $\begingroup$ @PM2Ring here's some closer to home, for some of us at least ;-) earthscience.stackexchange.com/q/10478/6031 $\endgroup$
    – uhoh
    Commented Oct 26, 2018 at 5:15
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    $\begingroup$ Sure, the atmospheric pressure at the Mars datum level is already much lower than at Earth's sea level, and it's much more homogeneous, due to the low water content, but why not put your telescope as high as you can? The only problem is those pesky orographic clouds. $\endgroup$
    – PM 2Ring
    Commented Oct 26, 2018 at 14:09
  • 1
    $\begingroup$ The Moon is nearer and maybe better: no atmosphere and the far side is well shielded from Earth interference. $\endgroup$
    – badjohn
    Commented Jul 30, 2019 at 13:52

1 Answer 1


As far as I know, "seeing" (or rather the effects influencing optical wave propagation) is caused by turbulence in the atmosphere.

Using the Reynolds number Number $ Re = \dfrac{\rho L v}{\mu}$ as a measure for turbulence:

  • density $\rho$ drops due to the reduced pressure (about 1/100 earth pressure), additionally the gravity is smaller than on earth
  • characteristic length $L$ will stay similar
  • average wind speed $v$ is about 2 times higher than on earth
  • the dynamic viscosity $\mu$ of an (ideal) gas is independent on the pressure, and the temperature dependency can be approximated with $\sqrt{\dfrac{T_{mars}}{T_{earth}}} \approx 0.9$

So for an average martian day an atmospheric Reynolds number would be much smaller than on earth and I would expect a far better seeing due to less turbulence

Additonally, I would guess that the effects of a thinner and dryer atmosphere would help as well, since the index of refraction is dependent on the gas pressure as well.

  • $\begingroup$ Thank you for your answer! The first link in the question is Wikipedia's Astronomical seeing and the first sentence there says that it is "...due to turbulent mixing in the atmosphere of Earth, causing variations of the optical refractive index." and later mentions "different temperature layers and different wind speeds" so a detailed analysis would require more information about Mars' atmosphere than we probably have today. $\endgroup$
    – uhoh
    Commented Jan 29, 2020 at 23:37
  • $\begingroup$ But I think your argument is mostly that since at any optically relevant altitude the density of Mars' atmosphere is lower than Earth's by roughly a factor of 100, seeing is likely to be far better from Mars' surface than from Earth's simply because theres far less gas and so far less opportunity for index of refraction variation. $\endgroup$
    – uhoh
    Commented Jan 29, 2020 at 23:39

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