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Let's take a planet identical to Earth, but with rotation speed multiplied by ten thousand. What would happen with the gravity if it was spinning madly around itself? Would the centrifugal force make objects seem lighter than on normal Earth? Would the people on that planet be able to jump higher?

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Possible duplicates: physics.stackexchange.com/q/44931/2451 , physics.stackexchange.com/q/10670/2451 , and links therein. –  Qmechanic May 7 '13 at 0:57
    
Also related: physics.stackexchange.com/q/12487 –  Brandon Enright May 7 '13 at 1:02
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I'm not as interested in the effects on the planet as on the objects on it. –  user1306322 May 7 '13 at 1:20

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The apparent radial accelleration due to moving in a circle is ω2R. For earth, ω = 2π rad/24 h = 73 x 10-6 rad/s. Earth's radius is about 6.37 Mm. Therefore the upwards accelleration at the equator is 34 mm/s2. That's pretty small compared to the 9.8 m/s2 downwards accelleration due to gravity, so we generally ignore it. Then there is also the issue that the planet is deformed due to the spin, which effects surface gravity, but let's ignore that too.

If the same earth were spinning 10,000 times faster, ω would be 0.727 rad/s, and upwards accelleration at the equator due to the spin would be 3.4 Mm/s2. Obviously this is much more than the downward accelleration due to gravity, so this system wouldn't be stable and the ridiculous factor of 10,000x higher spin speed makes no sense.

What if the earth was only spinning 10x faster? Then ω would be 727 x 10-6 rad/s, and the centrifugal accelleration at the equator 3.4 m/s2. That's a significant fraction of the accelleration due to gravity, which you'd definitely notice on a human scale. Yes, you'd be able to jump higher than on the real earth.

However this still doesn't make sense in the context of real earth. Such a extreme rotational accelleration would greatly flatten the planet, reduce apparent gravity holding the atmosphere at the equator, and lots of other effects that would make the result unlike the real earth in many ways.

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Okay, this question has assumed a lot. I think you're aware that Earth's equatorial rotational (spin) speed (465 m/s) is the left over speed after the Earth had formed from the dust clouds and matter, which means that it has already come to its equilibrium condition.

If on the contrary, you were to assume its spin to be multiplied by 10,000 all of a sudden, the outer layers start withering off. Finally, you'd be left with Earth made of barely some leftover mantle (maybe the outer core which would've cooled out). Depending on the remaining mass of the Earth $(M_{left}<<M_{before})$, the $g$ would be very less than $9.8$ and so, yes - people can jump higher!


It could also be argued this way. Earth completes one rotation every 24 hours. When it's increased to 10,000 times ($2\pi$ factor doesn't make much difference), it doesn't make sense! It's a mystical spinning horror! (10,000 rotations every day?). It's quite obvious that everyone would be thrown off.

Plugging in the numbers, $\omega=0.729069\ \mathrm{rad\ s^{-1}}$, which pushes the centrifugal acceleration to about $3\times 10^{6}\ \mathrm{m/s^2}$. I'm sure gravitational acceleration can't even come close to such a monstrous centrifugal acceleration!

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So, to be clear,you're saying that the effects of the hyperspin would cause a decrease in $g$, but the hyperspin itself wouldn't? –  WChargin May 7 '13 at 3:18
    
Hi @WChargin: Yes. If the Earth doesn't lose its mass, how come there would be any noticeable effect on the $g$ value? But, there wouldn't be existence of any other objects other than the matter that Earth was made of..! Because, that wizardo matter can only withstand the hyperspin. Others would be thrown away (assuming current size of Earth) ;-) –  Waffle's Crazy Peanut May 7 '13 at 11:47
    
"If the Earth doesn't lose its mass, how come there would be any noticeable effect on the g value?" If the earth is spun fast enough (and didn't fall apart) then objects on the surface might be in orbit, therefore constant freefall, therefor feel like the are in a 0 g environment. –  NPSF3000 Oct 13 at 0:51
    
@NPSF3000: Not exactly. That's what I mentioned above. The net acceleration might be zero, but not the "g" itself! When stuff is in orbit, the centripetal acceleration balances the "g", that doesn't mean "g" is zero... ;-) –  Waffle's Crazy Peanut Oct 13 at 2:05
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@NPSF3000: True. Obviously, the $g$ can't match the $mr\omega^2$ component, especially when $\omega$ is that large. Okay, I'll add it to my answer. Thanks for bringing that up. :) –  Waffle's Crazy Peanut Oct 13 at 12:34

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