Last month Rhett Allain posted about the famous Bent Pyramid in Egypt. As part of his calculations he estimated how high a stack of stone blocks can be without collapsing. The value he came up with was 3200 m, which sounded very familiar to me for some reason. I realize that while back I had made an estimate as to how tall a mountain could be and the values were around 3 km. I don’t have those calculations any more but I thought it would be fun to start from scratch.

Assume the mountain is cone shaped (in the end the shape won’t matter) with height *h* and base radius of *r*. We can approximate the volume of the cone by:

Vol ≈ r^{2} h

Clearly we are leaving off a fractional factor that is roughly 1 (for a cone the factor π/3 ≈ 1) but since this is a back-of-the-envelope calculation we can ignore those factors.

The density of granite is ρ = 3 g/cm^{3} (actually, the densities of most liquids and solids are close to 1. Lead is only about 11 g/cm^{3} and gold is 19.3 g/cm^{3}). The total weight of the mountain is just the volume times density times *g* so:

Weight W≈ ρ*gr*^{2}*h*

To see when the rock will start to break, we’ll compare the stress of the weight of the mountain to the compressive strength of granite. (Most mountains aren’t made out of granite, but it should give us a good upper limit on mountain heights). The weight of the mountain is spread out over an area of roughly (ignoring constants such as π):

A ≈ *r*^{2}

so the stress σ the mountain exerts on the ground underneath it is:

σ ≈ W/A ≈ (ρ *g r*^{2}*h)*/*r*^{2} ≈ ρ*gh*

The compressive strength of a material is the maximum compressive stress a material can withstand before it starts to deform. For granite the compressive strength is σ_{C} = 200 megaPascals = 2 × 10^{8} N/m^{2} so the rock beneath the mountain will start to compress when:

σ = σ_{C}

or

ρ*gh _{max}* = σ

_{C}.

Rearrange this equation to solve for *h _{max}* yields:

*h _{max}* ≈ σ

_{C}/(ρ

*g)*

The max height for a mountain works out to be:

*h _{max}* ≈ 2×10

^{8}N/m

^{2}/(3×10

^{3}kg/m

^{3}˙ 10 m/s

^{2})≈ 10

^{4}m = 10 km

Since this is purely back-of-the-envelope I assumed that 2/3 ≈ 1. Given all the assumptions we end up making this helps keep us honest and makes the mental math a little easier.

So how does this compare to the real world? Everest comes in first at 8,848 m or a little over 8.8 km. I looked the mountain heights up on Wikipedia and the interesting thing to look at is not the mountain heights but the prominence. This is the height above the local landscape, not just the height above sea level. Prominence is more in line with what we want to look at. Other than Everest, with a prominence of 8.8 km, most of the highest mountains are only about 2-4 km above the surrounding landscape. Looking back at Rhett Allain’s height of 3 km for a stack of blocks we can see he ended up calculating the maximum height of a mountain.

Now this back-of-the-envelope calculation worked out pretty well, but there is a problem with our calculation. The height should depend more on the shear strength of the rock, so we should have used that value in our calculation. Unfortunately I’ve been told that calculating shear strengths is reserved for PhD dissertation-level work and there is a strong temperature and pressure dependence.

It would be nice to take our approximation out for a spin. We’ve got:

*h _{max}* ≈ σ

_{C}/(ρ

*g)*

If we had some idea what different mountains were made out of we could see how well this approximation holds up, but I’ll leave that to a geologist. As a physicist I’d much rather play with gravity, so lets look at mountains on other planets. The Wikipedia page also lists a few mountain heights on other planets so I can look up gravity on those planets and plug those numbers into the approximation. I’m going to stick with the density of compressive strength of granite for simplicity.

Planet |
Gravity (m/s^{2}) |
Approx Max Height |
Actual Height |

Venus | 8.9 | 7.5 km | 11 km |

Mars | 3.7 | 18 km | 21 km |

Io | 1.8 | 37 km | 17 km |

Looking at these numbers you can see we are still getting agreement to within an order of magnitude so maybe this isn’t too bad an approximation. The biggest drawbacks, besides the earlier concerns, is that this doesn’t take erosion or any other mountain shaping mechanism’s into account. Ignoring these factors explains why our estimate would come out high, but I’m not sure why some of the estimates are lower than the actual value.

Thanks to my friend, Matt Kuchta, over at Research at a Snail’s Pace for help with some of the geology.

Pingback: Glaciers Protect Alpine Peaks From Erosion | BirchIndigo

Pingback: Dağların yüksekliğinin bir sonu var mı? | DağDelisi

Pingback: How High Can Mountains Be? | Market Tamer

Hey, thanks for the calculation but after

http://www.azom.com/article.aspx?ArticleID=1114

the compressive strength is approx. 2000MPa and therefore 2 × 10^9 N/m^2.

Perhaps you get wrong approximates because you assume the mountains to be made entirely of granite. This is true for the highest montains on Earth, but on both Mars and Venus the highest peaks are volcanos, with the surface composed of lighther substances.

Pingback: antognini comments on "Alien Worlds Might Be Covered in Enormous Mountains"

I’m from Hawaii, and Mauna Kea mountain rises directly out of the ocean to 4.21km (13,802 feet), so in that case, Allain’s maximum estimate was a kilometer short of a mountain 30 miles from my house. A simple search of mountains by local prominence will show his estimate is way off and has no real world practicalities.

I assume you are referring to Rhett’s post here: https://www.wired.com/2011/07/does-the-slope-of-a-pyramid-really-matter/

He is looking at limestone, which is weaker than the granite I used for my calculation. Also, with back-of-the-envelope calculations, if you can get within a factor of 10 then you have a pretty good estimate. These are very simple estimates.