# How fast does Haumea rotate?

I keep reading that Haumea spins the fastest of any planet (or dwarf planet) in the solar system. It's so fast it's been squashed into an elipsoid. But I can't find a figure on how fast it spins. (Its orbit round the Sun is commonly known, but not its rotation speed). I'm writing a kids' book about space and I'd really like to include this fact (and then compare it to other - slower - planets, like Earth). Does anyone know the answer? I'm aware that Haumea's day is only 4 hours long, but my maths isn't good enough to work out how fast that means it is spinning on its axis... Any help much appreciated. Adam Frost

• Do you mean it's rotation frequency or something else ? – StephenG Sep 13 '17 at 13:14
• It's rotating through 2$\pi$ radians in 4 hours, so $\omega=2\pi/(4×60×60)$ s $^{-1}$. Just multiply by the radius to get speed at the surface at the equator. – EL_DON Sep 13 '17 at 13:18

From Wikipedia, the sidereal rotation period is 3.91 hrs, so the angular frequency of its rotation is $2\pi/(3.91 \, \mathrm{hrs}) = 4.48 \times 10^{-4} \, \mathrm{rad/s}$, or $7.12 \times 10^{-5} \, \mathrm{Hz}$. Its "semimajor axis" is about 1000 km, so this corresponds to a surface speed of $2\pi (1000 \, \mathrm{km})/(3.9 \, \mathrm{hrs}) \approx 1600 \, \mathrm{km/hr}$ at the points furthest from its rotation axis. Note that this is almost identical to the speed of points on the Earth's equator, which is $2\pi (6370 \, \mathrm{km})/(24 \, \mathrm{hrs}) \approx 1670 \, \mathrm{km/hr}$.
So if the speed of points on Haumea's surface never even surpasses that of points on the Earth's equator, why is Haumea said to be rotating "fast"? The reason is that speed really doesn't mean much here; what is more meaningful is how close to its maximum speed it's rotating. In other words, there's a maximum rotational speed beyond which any self-gravitating object will come apart because gravity isn't strong enough to counteract the centrifugal force trying to pull it apart. For Earth, that angular velocity is roughly $\omega_\mathrm{max} = \sqrt{g/R} \approx 1.24 \times 10^{-3} \, \mathrm{rad/s}$, so the Earth's actual angular velocity of $2\pi/(24 \, \mathrm{hrs}) \approx 7.3 \times 10^{-5} \, \mathrm{rad/s}$ is only about 6% of this maximum (here $R$ is the radius of the Earth and $g$ is its surface gravity). For Haumea, on the other hand, the surface gravity is about $0.63 \, \mathrm{m/s}^2$ (also from Wikipedia, though I'm sure the number varies substantially since Haumea is so distorted), so its maximum angular velocity is roughly $\omega_\mathrm{max} = \sqrt{g/R} \approx 8 \times 10^{-4} \, \mathrm{rad/s}$. This means that its actual angular velocity is over 50% of the maximum it could have before being torn apart!
• Let's assume small objects in the solar system all have roughly the same density; then their mass should roughly scale like $R^3$, and thus their surface gravity should roughly go like $R$. That means that their maximum angular velocity (assuming they're in hydrostatic equilibrium) is roughly independent of their size, and thus the ratio of the actual angular velocity to the maximum just trivially scales like the actual angular velocity. Therefore (in this order-of-magnitude estimate, for small bodies) it doesn't matter whether you talk about actual angular velocity or fraction of maximum. – Sebastian Sep 15 '17 at 18:54