It's a classic question with many answers all over the Internet, but none here so I figured I'd ask it:
How fast would the Earth need to spin for a person (or anything for that matter) to feel weightless while on the surface at the equator?
In this situation everything on the Earth's surface would essentially be in orbit around the Earth at the radius of the Earth's surface (let us assume the atmosphere was also spun up to this angular velocity so there would be no air drag slowing things down). Let us also say by "surface of the Earth" we mean mean sea level.
You can decide for yourself if/how to factor in the bulge of the Earth. You can assume that the Earth somehow is able to maintain its present shape while spinning up.
Any comments on whether an Earth spinning slightly faster than this speed will cause it to break apart or not will also be appreciated.
Answer
How fast would a sphere need to rotate for a dust speck at its equator to achieve balance between gravitational attraction and centrifugal force?
If you do the math (equating $G M m / R^2$ to $m \omega^2 R$ and using $M = \frac{4\pi}{3} \rho R^3$ as well as $\omega = 2\pi f$), it follows that the size of the sphere is entirely irrelevant and that only the density $\rho$ of the sphere enters into the equation for $f$, the number of revolutions per unit time: $$f^2 = \frac{1}{3\pi}G\rho$$ For $\rho = 5.5 \times 10^3$ kilogram per cubic meter (the density of planet earth) it follows that $f=0.197 \times 10^{-3}$ revolutions per second, corresponding to a revolution period of $5070$ seconds (1 hour and 24 minutes).
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