The size at which a body becomes round

A mountain stands on rock, and the rock under it is carrying the mountain’s weight spread over its base. The pressure there is $\rho g h$ — the same expression that gives the pressure at the bottom of a column of water, with the density of rock instead. Make the mountain tall enough and that pressure exceeds what rock can bear, at which point the base fails and the mountain sinks into its own foundation.

That gives a maximum height, $h_{\max} = \sigma/\rho g$, and it contains the body’s surface gravity. Which is the interesting part, because $g$ is not an independent quantity: for a body of uniform density, $g = \tfrac43 \pi G \rho R$. Bigger bodies have stronger surface gravity in direct proportion to their radius, so

$$h_{\max} = \frac{3\sigma}{4\pi G \rho^2 R}.$$

The tallest possible mountain falls as one over the radius of the world it is on.

The comparison, made carefully #

The argument is a competition between two stresses, and it is worth writing both down explicitly because the whole result is in their different scaling.

The disturbing stress is what a feature of height $h$ puts on the material beneath it: $\rho g h$. It grows with the feature and with the surface gravity.

The resisting stress is the material’s crushing strength $\sigma$ — the stress at which it stops behaving elastically and starts flowing or fracturing. For cold silicate rock it is of order $10^8$ pascals; for ice it is smaller by a factor of a few to ten.

The disturbing stress is hydrostatic in its ordinary form: pressure grows in proportion to depth and to density and to nothing else. The rock beneath a mountain carries exactly that, with the mountain’s height in place of the depth — so the whole argument reduces to a comparison between one line of hydrostatics and one material property, and neither of those has any astronomy in it.

Setting them equal gives $h_{\max}$, and substituting the self-gravity expression for $g$ makes it a statement about the body rather than about the mountain.

The crossing, and what it means #

Two curves — one falling as $1/R$, one rising as $R$ — cross once. Below the crossing radius, $h_{\max}$ exceeds $R$: the tallest structure the material can support is larger than the body, which means the material’s strength is deciding the shape and gravity is a passenger. Above it, gravity decides, and anything sticking up further than $h_{\max}$ collapses.

The crossing for rock comes out at 282 km in the figure. The number is worth comparing with the real inventory: in the solar system, objects below about 200 km in radius are irregular without exception, and objects above about 400 km are round with a few explicable exceptions. The transition zone matches.

For the Earth the figure gives about six kilometres. The tallest mountain measured from its own base is Mauna Kea at about ten, and Everest rises 8.8 km above sea level. The estimate is therefore low by a factor of under two, which for an argument with one material constant in it and no geology whatever is a good result — and the direction of the error is informative, since the strength used is conservative and real crust is stronger in compression than the value assumed.

The comparison also says something about what happens below the crossing, which is easy to miss because the interesting conclusion is at the other end. A small body’s shape is set by whatever last happened to it — a collision, an accretion history, the strength of the material holding it together — and gravity’s only role is to keep the pieces from drifting apart. Many small bodies are not even single pieces: at low enough strength a body can be a loose aggregate held together by nothing but its own weak gravity, in which case the relevant strength is not the rock’s but the pile’s, which is close to zero. Such an object cannot support a mountain at all in the sense used here, and its shape is decided by the angle of repose of a heap of rubble rather than by any crushing stress.

That is a useful reminder about what a scaling argument is doing. It compares two mechanisms and finds where each dominates; it says nothing about what a body in either regime actually looks like, and it does not claim that the losing mechanism is absent.

Why the answer is only an order of magnitude #

The result depends on $\sigma$, and $\sigma$ is the weakest number in the argument. Solving for the crossing radius,

$$R_{\text{round}} = \sqrt{\frac{3\sigma}{4\pi G \rho^2}},$$

which goes as the square root of the strength. That square root is a mercy — an uncertainty of a factor of ten in $\sigma$ becomes a factor of three in $R$ — and it is not enough to make the number precise.

There is a deeper reason for caution. Strength is not a single number even for one material: rock fails at one stress under a load applied for a second and flows at a much lower one under a load applied for a million years. What matters here is the long-term strength, which is smaller and much harder to measure, and which depends on temperature.

The density enters twice, which makes the composition matter more than the strength does. $R_{\text{round}}$ carries $1/\rho$ — once from the weight of the mountain and once from the body’s own gravity — so a body made of something light and weak rounds at a smaller radius than one made of something heavy and strong, and the two effects push the same way.

Where a real mountain’s weight actually goes #

The crushing argument is the right one for a body with no mobile interior, and the Earth is not one. What actually limits terrestrial mountains is a different mechanism with the same flavour, and the difference is worth stating because the crushing estimate happens to give a similar number for the wrong reason.

The Earth’s crust floats on a denser mantle that flows over geological time. A mountain therefore behaves like a floating body: it displaces mantle, and it sinks until the weight of what it displaces matches its own. That means a mountain has a root, and the root is several times deeper than the visible mountain is tall — the ratio being set by the density contrast, roughly $\rho_{\text{crust}}/(\rho_{\text{mantle}} - \rho_{\text{crust}})$, which is about five or six.

The consequence is that piling more rock on top does not make a mountain much taller: most of the addition goes into deepening the root. This is isostasy, inferred in the 1850s from a survey discrepancy — the Himalayas deflected a plumb line by far less than their visible mass should have, because the mass beneath them was deficient rather than excess.

So the honest account of Everest’s height involves both mechanisms, plus erosion, plus the rate at which the crust is being pushed up. The value of the crushing estimate is not that it is the operative limit on Earth but that it is the limit for a body with no fluid interior — which is most of the objects there are.

Round is what minimising energy gives #

The stress argument says what cannot stand up. There is a complementary argument that says what the resulting shape is, and it belongs to a family this collection meets repeatedly.

The shape of every stability argument is a system with a smooth energy function sitting where the function is least, with departures costing energy. For a self-gravitating body the energy is gravitational, and moving material from a bump into a hollow lowers it — so a body with any mobility at all runs downhill toward the shape with the lowest gravitational energy for its volume, which is a sphere. The question is never whether the sphere is preferred but whether the material can get there.

That the sphere is the minimum is a mathematical fact rather than a physical one, and it is the same statement in a different subject that makes a drop round.

The other route to a sphere is surface energy, which costs area — and among all shapes of a given volume the sphere has the least. That mechanism is completely different from gravity’s: it acts at an interface rather than throughout a volume, it scales as the square of the size rather than as the fifth power, and it dominates for a raindrop where gravity dominates for a moon. Two unrelated arguments, the same shape, which is a good example of why “it is round” says almost nothing about why.

The two mechanisms have wildly different reach. Surface tension wins for a drop of water below the capillary length, a couple of millimetres; self-gravity wins above a few hundred kilometres. Between those two scales, a factor of $10^8$, neither enforces anything, and every object in that range is whatever its history made it. Nearly all of the solid objects a person handles live in that gap, which is why irregularity feels like the normal state of matter and roundness like a special case.

The same comparison, applied to the inside #

The stress that decides a body’s outside also decides what happens at its centre, and running the same hydrostatic integral inward is a one-line extension with a large consequence.

For a uniform sphere the pressure at the centre is

$$p_c = \frac{3GM^2}{8\pi R^4} = \frac{2\pi}{3}G\rho^2R^2,$$

which grows as the square of the radius. Compare it with the material strength and a second crossing appears: above a certain size, the material at the centre of a body is at a pressure far beyond anything it can resist, so its behaviour there is decided by compressibility rather than by strength.

For rock at 3000 kg/m³ the central pressure reaches $10^8$ Pa — the crushing strength — at a radius of about 250 km, which is the same crossing as before and for the same reason: both comparisons are between $G\rho^2R^2$ and $\sigma$, once as a mountain’s weight and once as a central pressure. That the two agree is a check on the argument rather than a coincidence, and it is a good example of a scaling argument giving the same answer from two directions.

Above that size, a body’s interior is a fluid as far as the mechanics is concerned, whatever its temperature. That is what the phrase hydrostatic equilibrium is pointing at when it is used as a criterion for roundness — not that the body is molten, but that its material has stopped being able to hold a shape against its own weight.

The mountains that check the formula #

The Earth’s case was quoted as agreeing to within a factor of two, which is one data point. There are several more, and they are a better test than the Earth is, because the Earth’s surface is being remade by processes the argument knows nothing about.

Mars has no plate tectonics and a crust that has been rigid for billions of years, so its mountains are where they were put and are limited by whether the rock beneath them holds. Its surface gravity is 3.71 metres per second squared, so the crushing estimate for rock gives about eighteen kilometres.

Olympus Mons stands twenty-two kilometres above the Martian datum. That is the tallest mountain in the solar system, it is about twenty per cent above the estimate, and it is above it in the same direction and by about the same fraction as Everest is above the Earth’s.

Vesta is the more interesting case, because it sits almost exactly on the crossing. Its mean radius is 263 kilometres and its surface gravity is a quarter of a metre per second squared, so the tallest mountain the formula allows is about 270 kilometres — which is to say, larger than Vesta. The argument therefore predicts that Vesta is at the boundary: strong enough to hold a shape, and only just.

That is what is observed. Vesta is nearly round and is measurably not in hydrostatic equilibrium; the enormous impact basin at its south pole has left it with a central peak some twenty-two kilometres high, which is the same height as Olympus Mons on a body a twenty-fifth of the size. A feature like that cannot exist on a body above the crossing and is unremarkable on one below it.

Three bodies, three surface gravities spanning a factor of forty, and one expression with one material constant in it accounting for the tallest feature on each to within a few tens of per cent. For an argument that consists of setting $\rho g h$ equal to $\sigma$ that is as much as anyone should ask.

The shapes that are fossils #

The essay’s caveat that a shape records the conditions when the material was last mobile has a worked example, and it is one of the strangest objects in the solar system.

Iapetus has a mean radius of 735 kilometres, comfortably above any version of the crossing, and it is round — but it is too flattened. Its equatorial radius exceeds its polar radius by thirty-four kilometres, a flattening of four and a half per cent, and a body that size in hydrostatic equilibrium would take that shape only if it were spinning once every sixteen hours.

Iapetus rotates once every seventy-nine days. Its present shape corresponds to no rotation it has had for four thousand million years.

The reading is that Iapetus froze. It formed spinning fast, took the equilibrium shape for that spin while it was warm and mobile, and was then despun by Saturn’s tides — by which time its interior had cooled enough that its lithosphere could hold the old shape indefinitely. Its outline is a photograph of the first few tens of millions of years of the solar system.

It carries a second oddity of the same kind: a ridge thirteen kilometres high running most of the way round its equator, whose origin is still argued about. What is not in doubt is that a thirteen-kilometre ridge is easily within what the crushing argument permits on a body with a surface gravity of a fifth of a metre per second squared, so its persistence needs no explanation at all — only its formation does.

The awkwardness that produces is worth naming because it has been institutionalised. The definition of a planet adopted in 2006 requires a body to have enough mass for its own gravity to overcome rigid-body forces “so that it assumes a hydrostatic equilibrium (nearly round) shape”. Vesta is nearly round and is not in hydrostatic equilibrium. Iapetus is in a shape that is an equilibrium for a rotation it does not have. And for essentially every candidate beyond Pluto the shape has never been measured at all — only four bodies in the outer solar system have been resolved well enough to say — so the criterion is applied by assuming that anything large enough must be round, which is this essay’s argument used as a substitute for the observation it was supposed to explain.

The transition is genuinely not sharp, and two neighbours make the point. Mimas, at a radius of 198 kilometres, is round. Proteus, at 210, is not. The crossing for ice computed above falls at 189 kilometres, neatly between them — and the reason they differ is not their size but their history, since Mimas was warm early and Proteus was not.

What it costs, and where the model stops #

Uniform density is assumed twice. Once in $g = \tfrac43\pi G\rho R$ and once in treating the crushing strength as the same at the surface and at depth. Real bodies are differentiated, with a denser core, so the surface gravity is larger than the uniform estimate at a given mean density — pushing the crossing inward.

Rotation is ignored entirely. A spinning body is not a sphere but an oblate spheroid, flattened by an amount set by the ratio of centrifugal to gravitational acceleration at the equator. The Earth’s equatorial radius exceeds its polar radius by 21 km, which is more than three times the maximum mountain height computed above — so the dominant departure from a sphere is not topography at all.

The strength used is a static one. Over long times, materials creep: a stress far below the crushing strength will still deform rock if it is applied for millions of years. The relevant number for a body’s shape is therefore not the strength a laboratory press measures but something closer to a viscosity, and treating the problem as elastic-until-it-fails is a simplification that gets the scaling right and the number approximate.

Temperature does not appear. A warm body is weaker and rounds more easily; a body that was warm when it formed and has been cold since may be round for reasons that no longer apply. A shape is a record of the conditions when the material was last mobile, not of the conditions now.

The mountain is treated as a load and not as a structure. Real rock fails in shear rather than in pure compression, and a proper treatment uses a failure criterion with two stress components in it rather than a single crushing strength. That changes the coefficient and not the scaling, which is the general property of arguments of this kind: what they get right is the exponent, and what they get approximately is everything else.

“Round” has no sharp definition here. The stress argument gives a maximum departure from sphericity, not a yes-or-no answer, and any attempt to make it a criterion has to choose a tolerance. Definitions that lean on the phrase hydrostatic equilibrium are choosing one implicitly.

What the argument is worth #

The whole of this essay is two expressions and a comparison, with no differential equation anywhere, and it is worth saying what that buys and what it does not.

What it buys is an exponent and a threshold. The tallest mountain falls as $1/R$, the crossing goes as $\sqrt{\sigma}/\rho$, and both survive any refinement of the argument, because they follow from the way the quantities scale rather than from any detail. What it does not buy is a coefficient: every constant here is uncertain by a factor of a few, and any claim about a specific body has to be checked another way.

That is the characteristic shape of a scaling argument, and this collection meets it repeatedly — in when a fluid may be treated as a continuum, in whether radiation pressure or gravity wins for a grain, in when a magnetic field is frozen into a conductor. In each case two mechanisms are compared, their ratio is a dimensionless number, and the physics is in which side of one it falls.

The reason to reach for such an argument first is that it is nearly impossible to get wrong. A full calculation of a body’s shape requires a rheology, a thermal history and a formation model, any of which can be mistaken; the comparison of $\rho g h$ with $\sigma$ requires two numbers and cannot be.

The ladder from here #

Later rungs on this anchor: the flattening of a rotating body and Clairaut’s relation between it and the gravity field; the isostatic version of this argument, in which a mountain floats on a denser substrate rather than crushing it, and which is the correct treatment for the Earth; the maximum height of a mountain on a body of given strength derived with a proper failure criterion rather than a crushing stress; gravitational binding energy, and how much work it would take to disassemble a body; and the pressure at the centre, which is the same hydrostatic integral carried all the way down.

The neighbouring ladders are hydrostatics, which supplies the $\rho g h$ this whole argument stands on, and surface tension, which produces the same sphere by unrelated means at a scale $10^8$ times smaller.