# How to calculate the highest theoretical artificial hill?

The biggest peak in the world is Mount Everest.

Imagine someone starting to make an artificial hill (like pyramide) from soil (earth).

So, when starting with an 200x200 Km base area, with 45degree slope, its mathematical height is 100km (Low orbital space height). Is it possible to make such an artificial peak? (Without taking into account financial issues and so on.)

If not, why not? What would happen? Is there a height limit?

-
This question is rather similar to this: physics.stackexchange.com/questions/7441/… –  Georg Jul 29 '11 at 17:48
I'm not actually really keen to close this as a duplicate because the accepted answer on 7441 describes the limits of building a bean stalk, and this questions is firmly limited to base supported objects. –  dmckee Jul 29 '11 at 18:44
The other question starts with a tower, but comes very soon to the core of bulding stability. But the gist is really the same. If You look in this question, its about hills made from soil! Is there much mention of soil in answers? But my opinion is a single one till now. –  Georg Jul 29 '11 at 19:02
I asked about the "soil" because i not want asking about the engeneering (buildings), but making a hill from "natural materials" like soils, sand, rocks and so on... –  jm666 Jul 29 '11 at 21:01

## 2 Answers

Your question is a classical college exercise. The limit is supposed to be the melting of the base of the artificial mountain under the pressure, which is linked with the energy of chemical bounds. You have an example of such a calculation here. You can also look it for an arbitrary planet size : the smaller the planet, the smaller the gravity is, and the bigger the biggest mountain can be. And when the mountain can be as big as the radius of the planet, you have roughly the dwarf planet / asteroid boundary.

-
Yesss! The link is exactly for what I'm looking. Thanx. ;) (Only don't understand the 4.9km result - because in the earth Everest is 8+Km - but will study the article more deeply.. (4.9x10**4m is 49km) –  jm666 Jul 29 '11 at 17:44
I'm sad to see that Georg deleted his answer because his point that on Earth the ductility and buoyancy of the crust becomes a major factor is important. –  dmckee Jul 29 '11 at 17:53
@dmckee - yes, becase it is importatnt for "setting" the base of the hill height. From the above link, the highest possible hill is 49km high, but the question is - where is its "base"? (it is not the sea level). –  jm666 Jul 29 '11 at 18:07
@dmckee Others did not appreciate my answer. BTW the limit a la Weiskopf is funny, but not the real limit. Crunching of basis due to "transverse tensile stresses" happens long before the liquefaction by chemical breakdown limit. The link I gave in my comment to the question above gives most of the relevant answers. –  Georg Jul 29 '11 at 18:17

The limiting factor will be that the shear strain inside the pyramid will eventually exceed the elastic limit for whatever material the pyramid is made of. For rock maximum shear strain is usually of order $10^{-3}$ if memory serves. I've seen people do order of magnitude estimates of this using dimensional analysis, and come up with something of order the height of Everest. To really get an upper limit though you'd need to do a more detailed modeling of the stresses.

-