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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.

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    $\begingroup$ Would Earth have to be in roughly hydrostatic equilibrium, which causes the equator to bulge? $\endgroup$ – fibonatic Sep 20 '14 at 20:57
  • $\begingroup$ @fibonatic You can decide for yourself if and how to factor in the non-rigidity of Earth. $\endgroup$ – NeutronStar Sep 20 '14 at 21:38
  • $\begingroup$ Related: physics.stackexchange.com/q/10670/2451 and links therein. $\endgroup$ – Qmechanic Sep 20 '14 at 22:17
  • $\begingroup$ Joshua, I've updated my answer. Three hours and thirty eight minutes, assuming the volume of the Earth remains the same. $\endgroup$ – David Hammen Sep 21 '14 at 21:26
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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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  • $\begingroup$ That is of course assuming a spherical Earth. An Earth rotating at that rate would be far from spherical, and gravitational acceleration would be far from $GM/r^2$. But it's a close enough of a spherical cow answer as to be worthy of an up vote. $\endgroup$ – David Hammen Sep 20 '14 at 21:44
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Original answer

This question when applied to the Earth is purely academic. The easiest solution is the one posted by Johannes: One revolution per one hour and 24 minutes. Why go beyond that? The question is academic.

It's not academic when applied to asteroids. There's an interesting effect, the Yarkovsky–O'Keefe–Radzievskii–Paddack effect, aka the YORP effect, that can make asteroids spin faster and faster and faster. This theoretical spin-up was observed in the asteroid formerly named as 2000 PH5 (astronomers tend to have obscure naming conventions). After confirming that this asteroid was spinning faster than before, it was renamed as 54509 YORP.

An asteroid that spins faster than the gravitational attraction needed to hold it together is likely to split in two. This is now hypothesized to be the reason we see so many binary asteroids. See Walsh, Richardson, and Michel, "Rotational breakup as the origin of small binary asteroids," Nature 454.7201 (2008): 188-191 for details.

Updated answer: Going beyond Johannes's answer

Johannes's answer assumed a spherical Earth. That's not that a particularly good assumption; that the Earth is rotating creates an equatorial bulge. This bulge will grow ever larger as the Earth's rotation rate increases. This means the rotation rate needed to feel weightless at the equator will be substantially less than one revolution per 5070 seconds.

That the Earth has an equatorial bulge is a consequence of the second law of thermodynamics. The Earth's surface is an equipotential surface where potential energy is computed from the perspective an Earth-fixed frame (a frame rotating with the Earth). Anything but this would violate the principle of minimum energy.

Using a spherical harmonic expansion of the Earth's gravity field and ignoring higher moments, the rotating frame potential energy is given by (e.g., see http://ocw.mit.edu/courses/earth-atmospheric-and-planetary-sciences/12-201-essentials-of-geophysics-fall-2004/lecture-notes/ch2.pdf equation 2.41) $$U(r,\lambda) = -\left(\frac{\mu}{r} \left(1 - J_2 \frac {a^2} {r^2} \left(\frac 3 2 \sin^2 \lambda - \frac 1 2\right) \right) + \frac 1 2 r^2\omega^2 \cos^2 \lambda\right)$$ where $\mu \equiv GM_\text{earth} = 398600.4418\,\text{km}^3/\text{s}^2$ is the Earth's geopotential, $J_2$ is a widely-used measure of the Earth's oblateness, $\lambda$ is the geocentric latitude, $r$ is the radius of the Earth at that latitude, $a$ is the equatorial radius of the Earth, and $\omega$ is the Earth's rotation rate.

The perceived gravitational acceleration is the negated gradient of this potential: $$g(r,\lambda) = -\nabla U(r,\lambda) = -\left(\frac{\partial U}{\partial r} \hat r + \frac 1 r \frac{\partial U}{\partial \lambda} \hat \lambda\right)$$

At the equator, the non-radial component vanishes and the radial component becomes $$g_e = \frac{\mu}{a^2} \left(1 + \frac 3 2 \, J_2\right) - a \omega^2$$ We want to make this zero (this is the condition that makes things weightless at the equator). Thus $$\omega^2 = \frac{\mu}{a^3} \left(1 + \frac 3 2 \, J_2\right)$$

The potential needs to be the same at all points on the surface of the Earth per the principle of minimum energy. In particular, the potential at the poles and the equator need to be the same: $$ \frac \mu c\left (1-\frac {a^2}{c^2} J_2\right) = \frac \mu a \left(1+\frac 1 2 J_2\right) + \frac 1 2 a^2\omega ^2 $$ where $c$ is the polar radius of the Earth. Using the value of $\omega$ that results in weightless at the equation, this becomes$$ \frac \mu c\left (1-\frac {a^2}{c^2} J_2\right) = \frac \mu a \left(1+\frac 1 2 J_2\right) + \frac 1 2 a^2 \frac{\mu}{a^3} \left(1 + \frac 3 2 \, J_2\right) = \frac \mu a \left(\frac 3 2 + \frac 5 4 J_2\right) $$ or $$ \frac 1 c\left (1-\frac {a^2}{c^2} J_2\right) = \frac 1 a \left(\frac 3 2 + \frac 5 4 J_2\right)$$ Solving for $\kappa \equiv \frac c a$ yields $$\kappa^3 \left(\frac 3 2 + \frac 5 4 J_2\right) -\kappa^2 + J_2 = 0$$ So, apparently a cubic. It's worse than that. $J_2$ is a function of flattening and hence is not a constant. This document suggests using $J_2 = \frac 2 3 f - \frac{a^3 \omega^2}{3\mu}$. As a sanity check, this expression yields a value of 0.00108141, which agrees rather nicely with the established value of $J_2$, 0.00108263. The flattening $f$ is given by $f=\frac{a-c} a = 1-\frac c a = 1 -\kappa$, and from above, we have $\frac {a^3 \omega^2}{\mu} = 1+\frac 3 2 J_2$. Solving for $J_2$ yields $J_2 = \frac 2 9 (1-2\kappa)$. Substituting this in the expression for $\kappa$ yields the quartic $$5 \kappa^4 - 16 \kappa^3 +9\kappa^2 + 4 \kappa - 2 = 0$$ This quartic has four real zeros, three of which ($\kappa\approx-0.469$, $\kappa=1$, and $\kappa\approx 2.30$) can be rejected as non-physical. This leaves but one solution, $\kappa\approx 0.3709102193$. In other words, we have an Earth that is considerably flattened.

Applying $J_2 = \frac 2 9 (1-2\kappa)$ to the expression $\omega^2 = \frac{\mu}{a^3} \left(1 + \frac 3 2 \, J_2\right)$ for the Earth's angular velocity yields $$\omega^2 = \frac{\mu}{a^3} \left(1 + \frac 3 2 \, \frac 2 9 (1-2\kappa)\right) = \frac 2 3 \frac {\mu}{a^3} (2-\kappa)$$ We now know $\kappa$, but we don't yet know $a$, the Earth's equatorial radius. One way around this problem is to assume that the volume of the Earth is conserved, even in the face of this extreme flattening. This yields $a^3\kappa^2 = {a_0}^3$, where $a_0$ is the Earth's volumetric radius, 6371.0008 km. With this result, the angular velocity that results in objects at the surface of the Earth at the equator being weightless is given by $$\omega^2 = \frac 2 3 \frac {\mu}{{a_0}^3} \kappa^2(2-\kappa)$$ from whence the desired rotation period is about 3 hours and 38 minutes.

This is considerably greater than the value suggested by assuming a spherical Earth.

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This is an interesting question. Unfortunately it's impossible for everyone to be weightless at the same time because the spin effect would mostly effect people living on the equator and people on the poles would practically not notice it.

Also if it's at the sea level then seas would be weightless at the surface as well and eventually that water would fly off. Not to mention everything above the sea level, that is not strongly bound to earth, would fly off right away, meaning you would need to be tied to something in order to not become an astronaut if you jump (even a bit, even if you would just walk).

Oh yea, if you would spin earth that fast eventually it would stretch on the equators and become a disc and probably totally fall apart...

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It is not necessary to assume that the Earth has constant density, only that it retains the same shape when the rotation rate is increased. This is of course not realistic, because the Earth's crust rests on a thick fluid mantle which would deform if the forces acting on it changed.

According to wikipedia the effective strength of gravity at Earth's equator is $g_e=9.780m/s^2$, and where the radius is $r=6378km$. The rotation rate is $\omega_0=7.292\times 10^{-5}\text{rad/s}$.

If the effect of gravitational attraction alone is $g$ and the rotation rate of the Earth is $\omega$ then $g_e=g-r\omega^2$. In order to nullify effective gravity $g_e$ at the equator we would need a new rotation rate $\omega_1$ for which $g_e=0$ hence $g=r\omega_1^2$. Substituting the current value of $g_e=g-r\omega_0^2$ we get $$ (\frac{\omega_1}{\omega_0})^2=\frac{g_e}{r\omega_0^2}+1$$

Using the above figures we get $\frac{\omega_1}{\omega_0}\approx 17$. The Earth would have to spin 17 times faster before people at the Equator became weightless.

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  • $\begingroup$ Of course I agree with your answer. See my contribution. $\endgroup$ – ZeroTheHero Dec 30 '17 at 1:47
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In view of the difference in orders of magnitude of previous answers, let’s see what dimensional analysis can tell us.

We look for a dimensionless combination ${\varpi}$ of $v$, $M$, $G$ and $g$, where $v$ is the velocity, $M$ is the mass of the Earth, $G$ the gravitational constant and $g=9.8m/s^2$ the gravitational acceleration at the surface of the Earth. By elementary manipulations we get that $$ \varpi=v \left(GM g \right)^{-1/4} $$ is dimensionless. Thus, $$ v=R\omega = (GM g)^{1/4}\quad \Rightarrow \quad \omega =\frac{(GMg)^{1/4}}{R} $$ with $R$ the radius of the Earth. Plugging numbers we get $$ \omega\sim 1.2\times 10^{-3} \hbox{sec}^{-1}\, , $$ which would agree almost exactly with the answer of @sammy gerbil.

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