Is the curvature of Mars visible from the top of Olympus Mons?

• 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.

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.

4 years ago

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.

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.