How fast would Earth need to rotate to fling someone off due to centrifugal force?

Question

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.

Answer 1

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.

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 -petal, 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 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.

Answer 2

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].

Answer 3

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.