Maximum height of (mostly) mountains and a flagpole:

Victor Weisskopf gave a derivation of the limit to the height of a mountain. This can be found in different sources (with minor numerical differences) [1].

The simplification is that he assumes a silicon dioxide (rock) mountain of area A and height h on a SiO₂ base. Its height is at its limit when its weight Mg bearing down on the base has a total potential energy Mgh that is greater than the liquefaction energy of SiO₂ E_l. That is, it would cause the base to turn into liquid, and it would sink into the liquid. The value of E_L is assumed to be 0.05X the binding energy of SiO₂ E_BE, and E_BE is O.2X the Rydberg energy (13.6 eV/atom). So, E_L = (0.05)(0.2)(13.6) = 0.136.

The mass of the mountain can be obtained from the volume of the mountain (V = Ah) multiplied by the number density of SiO₂ n_SiO2 (atoms/m³) and the mass of the protons & neutrons (assumed equal) m_p (1.67x10^-27 kg) and the mass number A_MS (the number of protons and neutrons in SiO₂):

M = Ahn_SiO2m_pA_MS

The mountain sinks by a distance z and displaces a volume Az of SiO₂ in the base, and the potential energy (on the l.h.s. equals the energy of liquefaction of the volume Az on the r.h.s.):

Mgz = n_SiO2 Az E_L

Or, cancelling z:

Mg = n_SiO2 A E_L

But M itself involves h, so:

Ahn_SiO2m_pA_MSg = n_SiO2 A E_L

The number density n_SiO2 and the area A both cancel, leaving:

h = E_L/(m_pA_MSg)

Numerically, this becomes:

h = [(0.136)(1.6x10^-19)]/[(1.67x10^-27)(60)(9.8)] = 21.6 kms

a)      For a conical mountain, the mass M will depend upon the volume, which is now Ah/3. So for this case, the height will go up by a factor of 3X.

b)      Weisskopf’s argument takes into account energy of liquefaction, whereas much before the base turn liquid, it would undergo plastic flow. This would lead to a lower and more realistic value of the maximum height.

A similar, though much simpler argument is given by Gnadig et al [2]:

The base of a mountain that does not melt under its load if:

h < L/g

where L = latent heat of melting of metals (200-300 kJ/kg), and the answer for Earth is 20-30 kms.

Gnadig  [2] also points out that on Mars g ~ 4 m/s², and in fact, the highest mountain on Mars is Mt.Olympus which is 26 kms high, according to Gnadig, but 21.9 kms high according to Caplan [3].

P.A.G.Scheuer [4] was more interested in the question of how high a mountain can be…on a neutron star (he gets a range of 0.04 to 0.4 mm for the maximum height of his mountain range!). But on the way, he deigns to look at Earth too. Scheuer gives two answers h₁ and h₂, where h₂ is much higher and refers to a broad-based pyramid:

h₁ =4 (1.5x10⁷)/[(2650)(9.8)] = 2.31 kms

Scheuer gives this number as 2.25 kms assuming a value of yield stress of Y = 1.5x10⁶ kg/m² which equals 15 MPa ( the latter is the kg force multiplied by g).

h₂ = (h₁b/4)^1/2 where b is the base of the mountain, such that b >> h.

Using an unconfined compressive strength value of 100-250 MPa for granite gives a much larger number for h₁ than the one Scheuer got:

h₁ = 4Y/(rg) where Y is the yield stress, Y = 250 MPa for granite and granite density r = 2650 kg/m³.

h₁ = 4(2.5x10⁸)/[(2650)(9.8)] = 38.5 kms

Miskinis [5] gives a somewhat different argument, using the Hooke’s law relation between the shear stress s_r on compressed rock and the shear angle q:

s_r = Gq

where G is the rock shear modulus.

Then he uses the weight of the mountain argument on the surface area S, to find the maximum height, giving:

s = mg/S = grh_maz

Solving:

h_maz = Gq/(rg)

Substituting values of G = 12.5x10⁹ Pa for basalt, r = 10⁴ kg/m3 for the increased density of compressed rock, and q = 0.1 rad (or 6°), he gets a maximum height of 12.5 kms.

The mountain argument can now be applied to an iron flagpole, assuming it to be an iron mountain on a silica base, adapting Weisskopf’s argument ( so why bring up Gnadig, Scheuer and Miskinis? Never mind!).

A similar argument can be used for an iron ‘mountain’ of area A and height h, on a SiO₂ base, giving:

Ahn_Fem_pA_MSg = n_SiO2 A E_L

In this case, the number densities are different on both sides of the equation: n_SiO2 = 2.66x10²² and n_Fe = 8.49x10²² cm^-3, and A_{M, Fe} = 55. The number densities are obtained from the density, the mass number A_M and the Avogadro number: n = rN_A/A_M. The areas still cancel, so:

h = [(2.66x10²²)(0.136)(1.6x10^-19)]/[( 8.49x10²²)(1.67x10^-27)(55)(9.8)] = 7.6 kms

So, if one goes by Weisskopf’s argument there is plenty of scope for building higher flagpoles, since even the Jeddah flagpole is just at 170 m!

Of course, a structural engineer could probably come up with a much lower – and more realistic -- limit by invoking other problems such as yielding or buckling… but that stuff is way too complicated… I found one reference on “Metal Flagpole Design” which is the American National Standard but it is much too iterative and demands too much knowledge of material properties.

And, why build a high flagpole when you can just as well stick your flag on top of some convenient hill or mountain? Tell that to the BSF!

References:

[1] http://www.hk-phy.org/articles/mount_high/mount_high_e.html

[2] Peter Gnadig et al ,”200 Puzzling physics problems” (Cambridge Univ.Press, 2001) p.197

[3] M.E.Caplan arXiv 1511.04297v1 “Calculating the potato radius of asteroids using the height of Mt.Everest”

[4] P.A.G.Scheuer “How high can a mountain be?” J.Astrophy.Astr.2 (1981) 165-69

[5] Paulius Miskinis “Mathematical modeling of mountain height distribution on Earth’s surface“ Geologija 53 (2011) 21-26