<p>The numbers are going to be approximate because <a href="https://en.wikipedia.org/wiki/Haumea" rel="nofollow noreferrer">Haumea</a>'s size is uncertain.  It may have a ring.  If it does, then it's smaller, if it doesn't, then it's larger.</p>
<p>The formula is simple enough.   For force:    <span class="math-container">$F= \frac {m v^2} {r} $</span> where v is the rotational velocity and r is the same r in the gravitational formula.   or <span class="math-container">$\frac {v^2} {r}$</span> if you want the reverse acceleration.</p>
<p>Haumea has a rotational period of 3.915 hours or 14,090 seconds (I'm rounding) and an average radius of 780,000 meters (assuming it has a ring, I'm going to go with that assumption), but an estimated equatorial radius of about 1,050,000 meters and that's the number we need.  Divide the larger radius by number of seconds, multiply by 2 Pi and the velocity is 468.2 m/s.   A tiny bit faster than Earth's equatorial velocity.</p>
<p>Some Maths, the lifting force on the equator is <span class="math-container">$ \frac {468.2^2} {1,050,000} $</span> = <span class="math-container">$ 0.208 \frac {m}{s^2} $</span>, slightly over half it's listed equatorial surface gravity of <span class="math-container">$ 0.401 \frac {m}{s^2} $</span>.</p>
<p>That's a pretty high ratio.   Possibly the highest in our solarsystem for objects that are massive enough to be spheroid shaped by their mass and gravity.   There are asteroids that are thought to <a href="https://www.iflscience.com/physics/bizarre-asteroid-has-%E2%80%9Cnegative-gravity%E2%80%9D/" rel="nofollow noreferrer">effectively have negative gravity at their equator</a>, held together by the stickiness of the material, but those are a lot smaller.</p>

CC BY-SA 4.0
2020-11-02T22:51:33.963Z
The given surface gravity is using the mean radius though. Using the equatorial radius of 1050 km and a mass of 4.006 * 10^21 kg (which seems like most people agree on) I get a surface gravity of ≈ 0.242 m/s^2. Subtract the rotational acceleration and the net acceleration felt is only ≈ 0.033 m/s^2. That just seems too low to be true.
user177107
CC BY-SA 4.0
2020-11-03T06:31:06.493Z
I used Wikipedia, which isn't recommended, but he asked for the formula, so I think I got the gist of the question.   Wikipedia's link says "Equatorial surface gravity" so . . . I don't know.   I think 0.033 is possible.   I think negative gravity with an object that size wouldn't be, but 0.033 could happen.   It's got a significant bulge.
userLTK
CC BY-SA 4.0
2020-11-02T17:13:51.557Z
2020-11-02T17:13:51.557Z
userLTK
<p>The equation for surface gravity is <span class="math-container">$\frac{GM}{r^2}$</span> but I'm not sure how to include the effects from its rotation.</p>

CC BY-SA 4.0
2020-11-02T15:02:33.860Z
Can you add some more details?
WarpPrime
CC BY-SA 4.0
2020-11-02T15:32:50.233Z
Related: [How fast would Earth need to rotate to fling someone off due to centrifugal force?](/q/16478)
Mike G
<gravity><rotation>
By how much does Haumea's fast rotation affect its surface gravity?
user177107
# By how much does Haumea's fast rotation affect its surface gravity?

(Asked by: username_0 on 2020-11-02)

The equation for surface gravity is $\frac{GM}{r^2}$ but I'm not sure how to include the effects from its rotation.

**Tags:** gravity, rotation

> **Comments:**
> **username_1**: Can you add some more details?  
> **username_2**: Related: [How fast would Earth need to rotate to fling someone off due to centrifugal force?](/q/16478)  

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## 1 answers

### Answer by username_3 on 2020-11-02. Score: 2

The numbers are going to be approximate because [Haumea](https://en.wikipedia.org/wiki/Haumea)'s size is uncertain. It may have a ring. If it does, then it's smaller, if it doesn't, then it's larger.

The formula is simple enough. For force: $F= \frac {m v^2} {r} $ where v is the rotational velocity and r is the same r in the gravitational formula. or $\frac {v^2} {r}$ if you want the reverse acceleration.

Haumea has a rotational period of 3.915 hours or 14,090 seconds (I'm rounding) and an average radius of 780,000 meters (assuming it has a ring, I'm going to go with that assumption), but an estimated equatorial radius of about 1,050,000 meters and that's the number we need. Divide the larger radius by number of seconds, multiply by 2 Pi and the velocity is 468.2 m/s. A tiny bit faster than Earth's equatorial velocity.

Some Maths, the lifting force on the equator is $ \frac {468.2^2} {1,050,000} $ = $ 0.208 \frac {m}{s^2} $, slightly over half it's listed equatorial surface gravity of $ 0.401 \frac {m}{s^2} $.

That's a pretty high ratio. Possibly the highest in our solarsystem for objects that are massive enough to be spheroid shaped by their mass and gravity. There are asteroids that are thought to [effectively have negative gravity at their equator](https://www.iflscience.com/physics/bizarre-asteroid-has-%E2%80%9Cnegative-gravity%E2%80%9D/), held together by the stickiness of the material, but those are a lot smaller.

> **Comments:**
> **username_0**: The given surface gravity is using the mean radius though. Using the equatorial radius of 1050 km and a mass of 4.006 * 10^21 kg (which seems like most people agree on) I get a surface gravity of ≈ 0.242 m/s^2. Subtract the rotational acceleration and the net acceleration felt is only ≈ 0.033 m/s^2. That just seems too low to be true.  
> **username_3**: I used Wikipedia, which isn't recommended, but he asked for the formula, so I think I got the gist of the question.   Wikipedia's link says "Equatorial surface gravity" so . . . I don't know.   I think 0.033 is possible.   I think negative gravity with an object that size wouldn't be, but 0.033 could happen.   It's got a significant bulge.  

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