Would you be able to see the curvature of Mars from the top of Olympus Mons? And how much curvature would you see if it was visible, would it be only barely detectable, or would you see a lot of curvature?

Olympus Mons is 25km high and the tallest known mountain in the Solar System.

Mars is just over half the size of Earth.

Here on Earth, the curvature of the planet is hardly visible from the height of commerical aeroplane (roughly 10km). I was wondering, considering the height of Olympus Mons and the size difference of Mars, if the curvature from up there would be very pronounced.

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    $\begingroup$ Can you define what you mean by "curvature" ? You mean the human eye perceives the horizon as curved and not straight? Because that depends a bit on personal acuity and pattern recognition. But -- if you take your aircraft altitude as a "standard," it should be trivial to calculate what's on Mars via similar triangles. $\endgroup$ Apr 18, 2018 at 15:16

3 Answers 3


The impression I get from the Wikipedia page on Olympus Mons is that from the top of the volcano, you can't even see parts of Mars that aren't Olympus Mons. Quoted from above:

Because of the size of Olympus Mons and its shallow slopes, an observer standing on the Martian surface would be unable to view the entire profile of the volcano, even from a great distance. The curvature of the planet and the volcano itself would obscure such a synoptic view. Similarly, an observer near the summit would be unaware of standing on a very high mountain, as the slope of the volcano would extend far beyond the horizon, a mere 3 kilometers away.

Emphasis mine.

Wikipedia sources the bold line to this source in the Way Back Machine, which works out the math. Near the bottom of the page, there is a section starting with the phrase:

Years ago someone asked me if a person could see the horizon of Mars from the top of Olympus Mons. That is not as simple as one may think. If Mars was a prefect sphere and the slopes of Olympus Mons were smooth then the distance to the horizon could be calculated fairly easy. However, the surface of Mars has hills and valleys, and Tharsis bulges out from the mean spherical surface of Mars several miles, so one must consider that as well.

That's pretty much your question. The conclusion is:

Second, if this same [6 foot tall] person stood at the top of Olympus Mons (88704 ft) we would find that he or she could see a lot further to the horizon, some 265 miles away, if nothing was in the way:

$\cos A = (11,121,762.53 / (11,121,762.53 + 88704 + 6)$

$A = 7.2^\circ = 0.126\ \textrm{radians} \leftarrow (0.126 \times 11,121,762.53 ft) / 5280 = 265 \textrm{miles}$

At those angles a person 6 feet high would be looking down at an angle of 7.2 degrees in order to see the horizon of Mars, so they would surely look down into the hump positioned at 50 miles away or at the 2.5-degree grade of the slope instead of over it to the horizon.

So it would seem that the horizon, as viewed from the top of Olympus Mons, would still be part of Olympus Mons. Indeed, you wouldn't even be able to see most of Olympus Mons from the top.

  • $\begingroup$ It's possible that you couldn't see the base of Olympus Mons from the summit (because it would be hidden by the curvature of the slope), but that you could see the horizon beyond the base. If someone wants to do the math, a first order approximation could model Mars as a sphere and Olympus Mons as a wide cone molded to the surface of the sphere. $\endgroup$ Apr 22, 2018 at 20:43
  • $\begingroup$ You should never base an answer on the garbage that is sometimes spewn on wikipedia, and in this case, is garbage spewn on wikipedia. $\endgroup$ Dec 13, 2021 at 9:30

It would really depend on the local geometry of the volcano. At the altitude of Olympus Mons here on earth, there are several reports of being able to see the curvature (U2 and SR-71 pilots). But that relies on having a clear view of the horizon. Given the smaller radius of mars, if you were on a tower or in a vehicle at that altitude above a flat planet, the curvature would definitely be visible.

But Olympus Mons is a shield volcano and has a very gentle slope. I suspect that there's nowhere to stand on the ground that will give you the view you need. The shallow slope will get in the way of being able to see the horizon at a sufficient distance.

  • $\begingroup$ In an "ideal" situation, doing the math to determine if curvature is detectable, you would also need to input that the horizontal range of human vision is about 210 degrees. (I'm not so sure that detecting curvature would matter if the horizon is still part of Olympic Mons or not other than that you would have to take an altitude "difference" rather than the difference between the "height" of Olympus Mons and the "mean" radius of Mars (sea level is used on Earth). $\endgroup$ Apr 29, 2018 at 18:26
  • $\begingroup$ This is not an answer because it is based on info in a previous answer. Consider that your horizon is 50 miles away (the more realistic Olympic Mons situation). Then if you determine curvature in a 90 degree horizontal field of view (as opposed to the 210 deg. in previous comment), the "length" of your horizon in this field would be about PI x 50/2 miles. This length would be 25/4212 of a "great circle" of Mars. This means that a vertical pole at the far right (or left) horizon would appear to be tilted at about 1.07 degrees. With a horizon 265 miles away the tilt would be about 5.65 degrees. $\endgroup$ Apr 29, 2018 at 19:21

Found your old topic after seeing this tweet https://twitter.com/MAstronomers/status/1617139081779052545 While the structure is around 600 km wide, i would say i see around 1500 km on horizon, that would make around 175 km curvature for earth converting into 300+ curvature for mars?

enter image description here

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    $\begingroup$ Hello... I'm not sure if your answer is yes or no! Take a look at the existing answers - try not to duplicate what has already been said. If you can support your answer by actual maths, that would be great (you can use Latex notation for maths) $\endgroup$
    – James K
    Jan 23 at 17:39
  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Jan 27 at 17:26

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