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On Earth we don't really notice many effects of the Earth spinning, apart from the day-night cycle. It took a long time for people of Earth to realise that it even was spinning, but there are measurable effects (differences in gravity, the Coriolis effect).

Would it be more obvious if the Earth spun 2 times faster? 10 times faster? 100 times faster? 1000 times faster?

The Sun would appear to move faster, but what else would we notice?

Could life survive on a planet spinning that fast or would we all be flung into space? I suspect at some point the Earth would break apart.

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    $\begingroup$ See "Dragon's Egg" by Robert L. Forward for exploration of this topic. smile.amazon.com/gp/aw/d/034543529X/… $\endgroup$ – SRM Dec 16 '16 at 18:03
  • $\begingroup$ Toilet flushes would rotate incredibly fast and it would be impossible to play pool. $\endgroup$ – jgadoury Dec 16 '16 at 21:03
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    $\begingroup$ Obligatory xkcd $\endgroup$ – Cody Dec 16 '16 at 21:51
  • $\begingroup$ @jgadoury: As the second and third answers indicate, at 17× as fast, the planet would probably self-destruct.  Life would probably become infeasible somewhere between 10× and 15×.  But why say, “it would be impossible to play pool”?  The Coriolis effect would affect all sports involving freely-moving projectiles, including baseball, basketball, football, lacrosse, and even bowling, hockey, and curling. (And, for that matter, even ice skating.) But athletes would learn to deal with it, just as baseball and football players deal with wind, and golfers deal with sloped greens. $\endgroup$ – Peregrine Rook Dec 16 '16 at 23:24
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    $\begingroup$ Read "Mission of Gravity" by Hal Clement, which does take place on such a world. $\endgroup$ – Thucydides Dec 17 '16 at 3:02
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There's an excellent answer on Quora about this.

  1. The Earth would become an even more imperfect sphere and this would reduce the acceleration due to gravity; it would decrease at the equator and increase at the poles. Also, the days would shorten.

  2. Since the effect of gravity has deceased near the equator, we would observe tides much higher than usual in these areas, much more land would go underwater during high tides. I assume Venice would cease to remain habitable.

  3. The precession of the Earth’s axis would change. As the Earth becomes an even more Oblate Spheroid, the gravitational differences (of the Sun on Earth) would be larger on different parts of the Earth, this would make the axis precess even faster. The current cycle is 26,000 years long for one complete precession, this would shorten depending on the increase in rotational velocity.

  4. North Star would change faster. Currently the North Star is Polaris which is set to be replaced by Deneb in 8,000 years. Due to the above reasons, it would occur faster and astronomical charts may cease to be relevant.

  5. It is also possible that the axial tilt of the Earth would increase. This would mean that winters would be colder and summers would be hotter.

  6. Since the Earth now rotates faster and the Coriolis effect depends on the rotation of the Earth, the impact of Coriolis effect would increase and we would experience faster wind speeds.

  7. This would be a boon for geostationary satellites as the operational altitude would reduce due to Earth’s increased rotational velocity.

  8. Since the Sidereal day (Sidereal time) would now be shorter, we would have to redefine our present units of time.

  9. Since the Coriolis Effect would alter the wind speeds, and the rotational velocity of Earth has increased, the travel time of aeroplanes would change. Watch: If Earth is spinning to the east, why isn't it faster to fly west?

Speeding up significantly could have effects on the Earth itself, to the point of nonexistence (See HDE's excellent answer for details!). My answer tells what would happen on a sped up earth, so keep in mind going certain speeds may amplify certain parts of this answer.

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    $\begingroup$ This seems to be answering the question of what would happen if our Earth sped up, not what would happen if it had just always been spinning fast. There is a lot of overlap, but you should edit out the "oh no, we'd have to change our units/satellites/GPS because we designed it for a different planet" stuff that doesn't apply at all. $\endgroup$ – Peter Cordes Dec 16 '16 at 20:54
  • $\begingroup$ (1) Why is the first paragraph not labeled #1?   (2) Paragraph #4 is a trivial restatement of #3, isn’t it?  I don’t see why they deserve separate paragraphs.  (Yes, I know you’re just quoting the entire list.) $\endgroup$ – Peregrine Rook Dec 16 '16 at 23:25
  • $\begingroup$ If the point in #2 about Venice is correct, then both coasts of the US should be flooded every high tide. Venice's latitude is almost the same as Montreal's. $\endgroup$ – SMS von der Tann Dec 16 '16 at 23:33
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    $\begingroup$ Can you tailor this answer a little bit more to this particular question? $\endgroup$ – HDE 226868 Dec 17 '16 at 1:07
  • $\begingroup$ 4 and 5 are essentially the same point. I don't think the question asked about the Earth speeding up, just the difference between an Earth that rotates "slowly" and an Earth that rotates "quickly", so 9 isn't really relevant. (No existing GPS to change, but GPS for the two Earths would be calibrated differently.) $\endgroup$ – chepner Dec 17 '16 at 13:35
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When will the planet break apart?

A solid body will break up if the centrifugal force at a point is equal to the gravitational force. At Earth's equator, the equation becomes $$\frac{GM_{\oplus}}{R_e^2}=\omega^2R_e$$ where $M_{\oplus}$ is the mass of Earth, $R_e$ is its equatorial radius, and $\omega$ is its angular speed. Rearranging and solving for $\omega$ gets $$\omega=\sqrt{\frac{GM_{\oplus}}{R_e^3}}$$ Assuming $R_e\approx\bar{R}_{\oplus}$, its mean radius, we have $$\omega=1.24\times10^{-3}\text{ rad s}^{-1}$$ Earth's current angular velocity is approximately $\omega=7.29\times10^{-5}\text{ rad s}^{-1}$. This is almost exactly 1/17th of Earth's breakup speed.

Now, this assumes that $R_e$ doesn't change with $\omega$ - which it does. In fact, the equatorial bulge is $$R_e=R_p+\frac{5}{4}\frac{\omega^2\bar{R}_{\oplus}^4}{GM_{\oplus}}$$ At the original breakup speed, we have $$R_e=6.356\times10^6+\frac{5}{4}\frac{(1.24\times10^{-3})^2(6.371\times10^6)^4}{6.673\times10^{-11}\times6\times10^{24}}=1.43\times10^{7}\text{ m}=2.24\bar{R}_{\oplus}$$ This decreases the gravity by a factor of roughly 5, so the breakup speed is actually a bit lower. So at 100 or 1000 times its current rotation speed, Earth wouldn't be here.

The Coriolis force

The maximum magnitude of the Coriolis acceleration is $$|\mathbf{a}_c|_{\text{max}}=2|\mathbf{\Omega}||\mathbf{v}|$$ where $\mathbf{\Omega}$ is the rotational velocity, and $\mathbf{v}$ is the velocity of a particular particle. This means that the force scales linearly with $|\mathbf{\Omega}|$, and thus can only be 17 times as large as it is right now, at a given point on Earth.

We'd see Rossby numbers (which characterize how much a system is affected by Coriolis forces) lower by no more than a factor of 17. As Anoplexian noted, this would lead to stronger winds and likely stronger hurricanes - although nothing catastrophic; an order of magnitude change here won't be devastating. I wouldn't be too concerned.

Daily life

At 17 times the Earth's current rotation speed, each day would last about 45 minutes, and each night would last about 45 minutes. Even at 12 hours per day-night cycle, you have to worry about humans getting the proper amount of deep sleep each cycle. 45 minutes is enough for a nice catnap, but not good sleep.

I'd assume that creatures would either adapt to such a short sleep cycle, or sleep through several days and nights, and then spend several days and nights asleep. Whether or not either of these is feasible is something I'll have to look into; I'm not overly optimistic.

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  • $\begingroup$ Glad to see we at least came to same rotation factor. $\endgroup$ – Feyre Dec 16 '16 at 18:35
  • $\begingroup$ I think your second equation needs an $R_e^3$ instead of $R_3^2$. $\endgroup$ – Devsman Dec 16 '16 at 21:02
  • $\begingroup$ @Devsman Yep, that was a typo on my part. I did the calculation with the $^3$ but wrote the equation with a $^2$. Thanks. $\endgroup$ – HDE 226868 Dec 16 '16 at 21:04
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    $\begingroup$ While this is certainly useful information, I don't think it really answers the question, which was about what it would be like to live there. $\endgroup$ – Ajedi32 Dec 16 '16 at 22:12
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    $\begingroup$ You seem to assume that the only force holding Earth together is gravity. Is this assumption well founded? $\endgroup$ – user58697 Dec 17 '16 at 8:00
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For us to "fly off the Earth", the centripetal force required to keep an orbit of the Earth's radius would have to be the same as the acceleration due to gravity.

This would be of the order $\frac{v^2(m^2/s^2)}{6.4\times10^{6}(m)}=10(m/s^2)$

Meaning the velocity at the surface would be $8000(m/s)=\frac{2\pi r}{t}=\frac{4\times10^6(m)}{t}$

Meaning one day would have to last ~5000 seconds, or one hour 12 minutes. For this to happen, Earth would have to rotate about 17 times as fast as it does in reality.

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  • $\begingroup$ If we were just under this speed, I wonder if we'd notice anything except for being nearly weightless at the equator. It would definitely be strange that our weight would increase as we moved towards the poles. At the poles I assume our weight would be near normal, but say you lived in Europe--would your weight still pull you straight down or would it pull you down and south? Would you still be able to stand? $\endgroup$ – Bill K Dec 16 '16 at 21:54
  • $\begingroup$ @BillK The pseudo-force vector would be 45 degrees south of the azimuth if you're 45 degrees north of the equator. Meaning you experience $\frac{1}{\sqrt{2}}$ times the force both away from the surface and to the south. However, the total velocity is also $\frac{1}{\sqrt{2}}$ there, so the total force in both directions would be $\frac{1}{2\sqrt{2}}$. That means you're pulled half as much to the south (0.3535:0.6465) as you are to the surface of the Earth. That's not pleasant. $\endgroup$ – Feyre Dec 16 '16 at 22:06
  • $\begingroup$ I guess it would feel like you were walking downhill if you were walking to the equator. So you'd probably have a good idea which way was north at all times. Cool stuff! $\endgroup$ – SpoonyBard Dec 19 '16 at 9:58
  • $\begingroup$ It won't feel so slopy after the land itself slides toward the equator. $\endgroup$ – Anton Sherwood Feb 16 '17 at 4:43
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For a fictional treatment of this, see Hal Clement's Mission of Gravity. In it, the planet Mesklin has a day approximately 18 minutes long, with a surface gravity of 3 G at the equator and 700 G at the poles. Much of the book involves the surface conditions and how the native species handles them. For example, they are instinctively terrified of even small heights, since at 700 G any fall is fatal.

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