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

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    $\begingroup$ Do you mean it's rotation frequency or something else ? $\endgroup$ Commented Sep 13, 2017 at 13:14
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    $\begingroup$ 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. $\endgroup$
    – EL_DON
    Commented Sep 13, 2017 at 13:18

1 Answer 1


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!

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    $\begingroup$ I'm pretty sure people say it's rotating fastest not because of how close it is to breaking apart, but because it's just got the highest rotational frequency. Gravitational limits or absolute velocity at the surface don't come into it. $\endgroup$ Commented Sep 13, 2017 at 20:10
  • $\begingroup$ @user2357112 Wikipedia disagrees with you - en.wikipedia.org/wiki/Haumea#Orbit_and_rotation see paragraph 3. $\endgroup$
    – Tim
    Commented Sep 13, 2017 at 23:03
  • $\begingroup$ @Tim: I have indeed seen paragraph 3, and paragraph 3 supports my point. Paragraph 3 claims that Haumea's rotational period is faster than any other known equilibrium body in the solar system; the information about being close to breaking apart is supplemental to that point. $\endgroup$ Commented Sep 13, 2017 at 23:06
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    $\begingroup$ @Tim: That speck of dust is not in hydrostatic equilibrium. The statement is restricted to bodies in hydrostatic equilibrium. "Rotating faster" is a statement about rotational frequency, not surface speed or gravitational limits; while Haumea's rotation is indeed impressive relative to the absolute limit at which it could rotate, that isn't what makes its rotation the fastest. $\endgroup$ Commented Sep 13, 2017 at 23:15
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    $\begingroup$ 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. $\endgroup$
    – Sebastian
    Commented Sep 15, 2017 at 18:54

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