# Why are some moons spherical while others are shaped like potatoes?

Short answer: Gravity likes to pull things together, which makes spheres. If the body is small enough gravity isn’t strong enough to deform it, which makes potatoes.

Long answer: Put a ball on top of a hill. What happens? It rolls down to the bottom. Why? Because gravity said so. This isn’t just how it works on the earth, but everywhere in the universe. Clearly, gravity is trying to make spheres. If you tried to dig a super deep hole stuff would fall in from the edges to fill it up. And what happen if we start to pile up rocks? Eventually, the pile of rocks reaches the point where it will all crumble down under its own weight. A sphere is the only shape that has no holes to fill or hills to crush. This is why every planet and star in the universe is round.

Of course, the earth and moon and planets aren’t perfect spheres. They’re lumpy. They’ve got hills and valleys and although none of them are that big compared to the planet, they’re still there. This is because gravity is strong enough to destroy (or prevent the formation) of a really big mountain, but not a small mountain. A small mountain’s own rigidness is enough to support its weight against gravity [1].

Two failure modes for mountains. The mountain on the left experiences shear failure, with the stress from the weight above the diagonal line exceeding the breaking point of the material. The mountain on the right fails due to compression of the base material.

Because materials have some intrinsic rigidity there must be bodies whose gravity isn’t strong enough to pull them into a sphere. Rather, the material is stiff enough to keep an oblong shape. After all, satellites and astronauts and cows don’t collapse into spheres in space.

The limit where gravity is strong enough to overcome the material properties of a body and pull it into a sphere is called the Potato Radius, and it effectively marks the transition from asteroid to dwarf planet [2]. It’s about 200-300 km, with rocky bodies having a slightly larger Potato Radius than icy bodies.

You can use some complicated math with material elasticity, density, and gravity to calculate the Potato Radius from scratch, or you could just look at Mt Everest. It turns out that the same physics determining the maximum height of mountains can be used to determine the Potato Radius – after all, they’re both just the behavior of rocks under gravity.

Check this out. The heights of the tallest mountains on Earth and Mars obey an interesting relation:

`    $\dpi{200} h_{earth} g_{earth} = h_{mars} g_{mars}$`

If you know the height of Everest and that Mars surface gravity is 2/5ths of Earth, then you know that Olympus Mons (tallest mountain on Mars) is about 5/2× taller than Everest! This relation also works with Maxwell Montes, the tallest mountain on Venus, but not for Mercury. Planetary science is a lot like medicine in this sense- there are always exceptions because everything is completely dependent on the body you’re looking at.

This is more than a curiosity. It tells us something important. The height of the tallest mountain a planet can support, multiplied by that planet’s surface gravity, is a constant. For this sake of this piece I’ll call it the Rock Constant because that sounds cool. So why am I spending so long on a tangent about mountains in a piece about potato moons? It’s because the Potato Radius and Rock Constant are determined by the same things – gravity and the elasticity of rock! We can use the Rock Constant to estimate the Potato Radius!

Consider an oblong asteroid. Let’s pretend this asteroid is actually a sphere with a large mountain whose height is equal to the radius of that sphere.

Potato asteroid with mountain.

As the radius of a body increases the maximum height of a mountain decreases. If the radius was any bigger the mountain would have to be shorter and our asteroid would be entering ‘sphere’ territory. Let’s check if the radius of this imaginary asteroid is close to the Potato Radius using our relation for the Rock Constant:

`$\dpi{200} h_{asteroid}g_{asteroid}=h_{earth}g_{earth}$`

We know the height of the mountain, so let’s find surface gravity:

`$\dpi{200} g = \frac{GM}{R^2}$`

And the mass can be expressed in terms of the density of rock (use earth’s density, ρ=5.5 g/cc) and the volume of the sphere:

```$\dpi{200} M = \rho V$
$\dpi{200} M = \rho (\frac{4}{3} \pi R^3)$```

And now we have everything we need:

```$\dpi{200} h_{asteroid}g_{asteroid}=h_{earth}g_{earth}$
$\dpi{200} R \frac{G \rho (\frac{4}{3} \pi R^3)}{R^2} = h_{Everest} g_{earth}$$\dpi{200} R = \sqrt{\frac{3 h_{Everest} g_{earth} } { 4 G \rho \pi}}$```

This works out to about 240 km [1], right in the middle of the 200-300 km range of the more rigorous calculation!

image credit: Wikimedia Commons

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