# Would rotating a large asteroid be a structurally viable way generate 0.3 Earth gravity?

This question is based off something I have seen in science fiction. The TV series The Expanse is known for having realistic physics when it comes to artificial gravity (i.e. no Star Trek gravity plates). Everything accelerates linearly or spins to simulate gravity (normally to about 0.3 g). However, two of the spinning bodies in the series are asteroids (433 Eros and Ceres).

If we assume we have the ability to generate enough thrust to sufficiently spin up large asteroids (and even dwarf planets), we are talking about spins generating outward forces greater than the surface gravity.

• According to Wikipedia, Eros has a surface gravity of 0.060 g. That means a rotation generating 0.3 g would be producing 5 times the asteroid's surface gravity (in the wrong direction). This would quickly quickly throw off all loss dust and rocks on the surface, but what about the rest of the asteroid? Are Eros-sized asteroids primarily held together by gravity (and therefore would spin apart)? Or has most of their mass been compacted into structure capable of withstanding such forces?
• What about dwarf planet sized asteroids like Ceres? Would something large enough to round itself before being spun up have a different fate (if Eros would not survive)?

Some asteroids are large snowballs. Some are large rocks. Some are nickel-iron.

A snowball would be pulled apart by the stress.

Up to a point, a rock would not, at least in the short term. Over a period of millions of years, rock does flow.

Also rock is strong in compression, but weak in tension. You have set the acceleration at $a = 0.3 g$. The centrifugal force, $F$, pulling it apart is proportional to the mass of the asteroid. $F \propto ma$, and $m \propto r^3$.

The asteroid would fail if the stress gets too big. $S = F/A$, and $A \propto r^2$

Putting it all together, $S \propto r$.

So a big enough asteroid will fail.

Metal is stronger in tension than rock. But even so, a large enough asteroid would come apart.

The math would change if the asteroid is hollow. Given a skin of constant thickness, $m \propto r^2$, and $A \propto r$. So the result is the same. $S \propto r$

A really large rock like the Earth has a thin rock skin over a liquid interior. It would definitely come apart.

There isn't much point to being more precise than this. The strength is not simply determine by the mechanical properties of a particular kind of rock. Cracks have a major effect. So you would have to know a lot of detail about the asteroid to really say.