<p>If it would rotate fast enough to throw a person off, it would throw everything off - dirt, houses, cars, etc., as the mass of an object is irrelevant - the centrifugal force (pulling out) and the gravitational force (pulling in) are both proportional to an object's mass.</p>

<p>Basically the whole thing would explode out at the same time. To calculate we equate the force of gravity $mg$ with the centrifugal force $mr\omega^2$, where $\omega$ is the angular velocity in radians per second </p>

<p>$$F = mg = mr\omega^2 $$ $$\Rightarrow g = r\omega^2 $$ $$\Rightarrow \omega = \sqrt{g/r} \approx 0.0012/s.$$</p>

<p>Insert the radius of the earth $r$ and the gravitational constant $g$, take the square root, and you have the necessary angular speed $\omega$.</p>

<p>with $\omega = 2 \pi/T$, you get $T = 2 \pi/\omega \approx 1.5 \mathrm{hours}$ [I did that in my head, so it might be wrong].</p>

CC BY-SA 3.0
2016-07-06T04:13:40.037Z
My math is badly out of practice but that looks right.  It's the same speed as orbital speed, which makes sense if you think about it.  But at such speeds, not only people would lift off the earth, or essentially be weighless, the atmosphere would fly off (at slightly lower speeds) and the ground would even begin to "lift" free of the earth due to expansion and rebound and heat below.   It would not be a fun day to be on Earth.
userLTK
CC BY-SA 3.0
2016-07-06T04:23:09.383Z
Curious sidebar, but if they every build that wacky Elon Musk super velocity train that rockets people around the earth in a vacuum tube at 4-6,000 kph, while riding in that thing, once acceleration stops people would weigh about 10%-15% lighter as a result of the same effect.  That also happens on airplanes that fly with the rotation of the Earth, but the weight loss there isn't noticeable.
userLTK
CC BY-SA 3.0
2016-07-06T05:40:02.203Z
I TeXified. Please check that I didn't break anything
James K
CC BY-SA 3.0
2016-07-06T11:40:19.487Z
+1, but note that these perils only apply to stuff at the equator. Polar bears and penguins would be fine.
pela
CC BY-SA 3.0
2016-07-06T02:43:58.410Z
2016-07-06T05:38:49.870Z
2016-07-06T05:38:49.870Z
Aganju
<p>Joe Blow commented about the orbital speed of satellites, correctly. As shown by Aganju, the earth would have to rotate 16 times faster in order to levitate equatorial residents. It's current rotational speed of 0.5 km per sec would then increase to 8 km per s (5 miles/s). We should all recognize this speed as that of orbiting satellites.</p>

CC BY-SA 3.0
2016-07-07T09:25:10.863Z
2016-07-07T09:25:10.863Z
yaron
<p>Aganju's answer is excellent, but I'd like to add one thing: The radius used is Earth's radius, which is applicable if you're at the equator. At higher latitudes, the effective radius is smaller by a factor $\cos\theta$, where $\theta$ is the latitude in radians.</p>

<p>For iMerchant, who seem to be located in Vancouver at
$$\theta = 49.3^{\circ}\times\pi/180 = 0.86\,\mathrm{rad},$$
the relevant radius is
$$
R_\mathrm{eff} = R_\oplus\cos\theta = 4150\,\mathrm{km},
$$
so at the $T=1.4\,\mathrm{h}$ of Aganju's answer (1.5 h is just a little too slow), the centrifugal (or <a href="https://xkcd.com/123/" rel="noreferrer">-petal</a>, if you prefer) acceleration would be only $g_\mathrm{spin} = 6.5\,\mathrm{m}\,\mathrm{s}^{-2}$.</p>

<p>Moreover, whereas Earth's acceleration is directed toward the center of Earth, in Vancouver the centrifugal acceleration would be directed at an angle $\theta$ toward South. Thus, only a component $g_\mathrm{spin} \cos\theta = 4.2\,\mathrm{m}\,\mathrm{s}^{-2}$ would point upward.</p>

<p>This is not enough to send iMerchant into space. However, a component $g_\mathrm{spin} \sin\theta = 4.9\,\mathrm{m}\,\mathrm{s}^{-2}$, or roughly half a G, would point horizontally toward South, enough to send him tumbling down the street, or at least walk awkwardly.</p>

<p><a href="https://i.sstatic.net/T5LKT.png" rel="noreferrer"><img src="https://i.sstatic.net/T5LKT.png" alt="Rotating Earth"></a></p>

CC BY-SA 3.0
2016-07-07T05:21:24.477Z
As the same forces act on the crust, the Earth would transform into a more oblate spheroid. Ultimately the resultant forces on things at the surface would be normal to the new surface.
Mike G
CC BY-SA 3.0
2016-07-10T05:46:13.197Z
+1 for the custom-made artwork (even though you drew me flinging to my death...twice...while the penguin laughs at the carnage)
iMerchant
CC BY-SA 3.0
2016-07-10T10:17:44.767Z
@iMerchant: Anytime :)
pela
CC BY-SA 3.0
2016-07-06T12:13:44.970Z
2016-07-06T18:44:33.337Z
2016-07-06T18:44:33.337Z
pela
<p>If Earth started spinning fast enough, would the centrifugal force eventually overcome Earth's gravity enough enough to fling someone (let's assume 75 kg) into orbit or into space? If so, how fast would it need to spin?</p>

<p>Being that I'm not a scientist, I'm hoping for a simple answer like, "one rotation per xx hours/mins/seconds needed to throw someone off Earth" or something to that effect. Complicated equations and/or jargon is fine for the sciencey types, of course. I just probably won't understand it too much. </p>

CC BY-SA 3.0
2016-07-06T02:36:59.953Z
Did you try to calculate it? What did you try?
Aganju
CC BY-SA 3.0
2016-07-06T02:53:59.543Z
@Aganju - I have no idea how to even try to calculate it. I majored in business administration.
iMerchant
CC BY-SA 3.0
2016-07-06T12:13:26.630Z
@iMerchant, an ingenious way to think about this is *it would be about the same speed as satellites move currently*...
Fattie
<gravity><earth>
How fast would Earth need to rotate to fling someone off due to centrifugal force?
iMerchant
# How fast would Earth need to rotate to fling someone off due to centrifugal force?

(Asked by: username_0 on 2016-07-06)

If Earth started spinning fast enough, would the centrifugal force eventually overcome Earth's gravity enough enough to fling someone (let's assume 75 kg) into orbit or into space? If so, how fast would it need to spin?

Being that I'm not a scientist, I'm hoping for a simple answer like, "one rotation per xx hours/mins/seconds needed to throw someone off Earth" or something to that effect. Complicated equations and/or jargon is fine for the sciencey types, of course. I just probably won't understand it too much.

**Tags:** gravity, earth

> **Comments:**
> **username_1**: Did you try to calculate it? What did you try?  
> **username_0**: @username_1 - I have no idea how to even try to calculate it. I majored in business administration.  
> **username_2**: @username_0, an ingenious way to think about this is *it would be about the same speed as satellites move currently*...  

---

## 3 answers

### Answer by username_5 on 2016-07-06. Score: 13 | Accepted answer

username_1's answer is excellent, but I'd like to add one thing: The radius used is Earth's radius, which is applicable if you're at the equator. At higher latitudes, the effective radius is smaller by a factor $\cos\theta$, where $\theta$ is the latitude in radians.

For username_0, who seem to be located in Vancouver at
$$\theta = 49.3^{\circ}\times\pi/180 = 0.86\,\mathrm{rad},$$
the relevant radius is
$$
R\_\mathrm{eff} = R\_\oplus\cos\theta = 4150\,\mathrm{km},
$$
so at the $T=1.4\,\mathrm{h}$ of username_1's answer (1.5 h is just a little too slow), the centrifugal (or [-petal](https://xkcd.com/123/), if you prefer) acceleration would be only $g\_\mathrm{spin} = 6.5\,\mathrm{m}\,\mathrm{s}^{-2}$.

Moreover, whereas Earth's acceleration is directed toward the center of Earth, in Vancouver the centrifugal acceleration would be directed at an angle $\theta$ toward South. Thus, only a component $g\_\mathrm{spin} \cos\theta = 4.2\,\mathrm{m}\,\mathrm{s}^{-2}$ would point upward.

This is not enough to send username_0 into space. However, a component $g\_\mathrm{spin} \sin\theta = 4.9\,\mathrm{m}\,\mathrm{s}^{-2}$, or roughly half a G, would point horizontally toward South, enough to send him tumbling down the street, or at least walk awkwardly.

[![Rotating Earth](https://i.sstatic.net/T5LKT.png)](https://i.sstatic.net/T5LKT.png)

> **Comments:**
> **username_7**: As the same forces act on the crust, the Earth would transform into a more oblate spheroid. Ultimately the resultant forces on things at the surface would be normal to the new surface.  
> **username_0**: +1 for the custom-made artwork (even though you drew me flinging to my death...twice...while the penguin laughs at the carnage)  
> **username_5**: @username_0: Anytime :)  

---

### Answer by username_1 on 2016-07-06. Score: 13

If it would rotate fast enough to throw a person off, it would throw everything off - dirt, houses, cars, etc., as the mass of an object is irrelevant - the centrifugal force (pulling out) and the gravitational force (pulling in) are both proportional to an object's mass.

Basically the whole thing would explode out at the same time. To calculate we equate the force of gravity $mg$ with the centrifugal force $mr\omega^2$, where $\omega$ is the angular velocity in radians per second

$$F = mg = mr\omega^2 $$ $$\Rightarrow g = r\omega^2 $$ $$\Rightarrow \omega = \sqrt{g/r} \approx 0.0012/s.$$

Insert the radius of the earth $r$ and the gravitational constant $g$, take the square root, and you have the necessary angular speed $\omega$.

with $\omega = 2 \pi/T$, you get $T = 2 \pi/\omega \approx 1.5 \mathrm{hours}$ [I did that in my head, so it might be wrong].

> **Comments:**
> **username_3**: My math is badly out of practice but that looks right.  It's the same speed as orbital speed, which makes sense if you think about it.  But at such speeds, not only people would lift off the earth, or essentially be weighless, the atmosphere would fly off (at slightly lower speeds) and the ground would even begin to "lift" free of the earth due to expansion and rebound and heat below.   It would not be a fun day to be on Earth.  
> **username_3**: Curious sidebar, but if they every build that wacky Elon Musk super velocity train that rockets people around the earth in a vacuum tube at 4-6,000 kph, while riding in that thing, once acceleration stops people would weigh about 10%-15% lighter as a result of the same effect.  That also happens on airplanes that fly with the rotation of the Earth, but the weight loss there isn't noticeable.  
> **username_4**: I TeXified. Please check that I didn't break anything  
> **username_5**: +1, but note that these perils only apply to stuff at the equator. Polar bears and penguins would be fine.  

---

### Answer by username_6 on 2016-07-07. Score: 2

Joe Blow commented about the orbital speed of satellites, correctly. As shown by username_1, the earth would have to rotate 16 times faster in order to levitate equatorial residents. It's current rotational speed of 0.5 km per sec would then increase to 8 km per s (5 miles/s). We should all recognize this speed as that of orbiting satellites.


---

