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In my world a space civilization is colonizing a massive solid planet. The planet has an extraordinarily fast rotation. The gravity is severe at the poles, making life impossible, but life flourish within a wide belt at the equator. This “green belt” is possible thanks to the extreme rotation - the centrifugal force greatly reduces the high gravity.

Is this concept purely fiction or would such a planet (in theory) be possible?

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    $\begingroup$ This is the exact premise of Hal Clement’s 1953 novel Mission of Gravity. $\endgroup$
    – Mike Scott
    Jan 27, 2019 at 9:31
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    $\begingroup$ High gravity will make the poles uninhabitable for much Earth life but some Earth life forms like bacteria, as well as native life adapted to the gravity such as flat, pancake like animals, should do just fine at the poles. $\endgroup$
    – M. A. Golding
    Jan 27, 2019 at 20:45

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Yes

Example

Let's assume we have a planet with the gravitational acceleration of Jupiter and want to reduce it to earths gravitational acceleration.

Edit

This is meant as an in practice example of how fast a planet with the size and mass of Jupiter would need to spin to experience same forces as on earth.

Example continues

According to Wikipedia Jupiter has a Gravitational acceleration of $24.79 \;\frac{m}{s^{2}}$. And as we all know Earth has a gravitational acceleration of $9.81\;\frac{m}{s^{2}}$. So the rotation has to cancel out the difference between gravity on earth vs gravity on Jupiter ($24.79 - 9.81 = 14.98$). This means our centripetal acceleration has to be $14.98\;\frac{m}{s^{2}}$.

With this Formula we can determine the speed at which the Planet has to rotate: $a = \frac {v^2}{r}$, where $a$ = acceleration, $v$ = speed and $r$ = radius.

If we take Jupiter as an example again, $r$ would be $71 492 000\;m$. So when we plug in our values we get this: $v^2 = 14.98\;\frac{m}{s^{2}} \cdot 71492000\;m = 1 070 950 160\frac{m^{2}}{s^{2}}$. Because we have $v^2$, we have to take the root of it which gets us to $32 72\;\frac{m}{s}$ speed. Which is about 3 times faster than Jupiter is actually rotating, but possible.

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    $\begingroup$ @AtmosphericPrisonEscape Yes exactly you wouldn't feel it would simply reduce the gravitational pull of the planet and no one would feel the centripetal force. Don't get where your problem is. $\endgroup$
    – Soan
    Jan 27, 2019 at 13:49
  • $\begingroup$ @AtmosphericPrisonEscape I get different values? Which doesn't change my conclusion? Sorry I really don't get it. It is an EXAMPLE. $\endgroup$
    – Soan
    Jan 27, 2019 at 14:58
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    $\begingroup$ @AtmosphericPrisonEscape Even if you disagree, please do so respectfully. $\endgroup$
    – Tim B
    Jan 28, 2019 at 9:38
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    $\begingroup$ While the theory is nice, I don't think it would work out that nicely. For starters, the math only really works around equator. As you diverge from equator, the radius relative to the axis of rotation decrease, which changes your equation. Then secondly, as you move away from equator, the vectors don't line up anymore, and thus won't cancel each other. $\endgroup$
    – Spoki0 - Reinstate Monica
    Jan 28, 2019 at 15:06
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    $\begingroup$ The general principle works, but it's not quite that straightforward in reality, for the simple reason that the radius of the planet doesn't stay the same as you spin it up--it flattens out. Properly solving for the shape of the equipotential surface and the force of gravity at various latitudes for a given spin rate (let alone reversing that calculation) is rather hairy--enough so that even Hal Clement basically punted and assumed the shape of Mesklin was dominated by the gravity of a close-enough-to-spherical degenerate matter core. $\endgroup$
    – Logan R. Kearsley
    Jan 28, 2019 at 18:10
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If you put a high spin on something squishy like a planet it's going to bulge at the equator and the poles will descend (to keep the volume constant)

Both re-shaping effects themselves reduce surface gravity. At the poles you have less mass beneath you, and at the equator you're further separated from most of the mass

I don't think my calculus is strong enough to guess the shape, but this guy has some ideas.

https://www.science20.com/robert_inventor/so_you_thought_you_knew_what_spinning_planets_look_like_surprising_shapes_of_rapidly_spinning_planets-155538

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the centrifugal force greatly reduces the high gravity.

Yes, this is a physical concept, and in fact how orbits work. Any small mass $m$ orbiting a large mass $M$ has its centrifugal force balancing gravitational acceleration exactly, so that

$$g = \frac{v^2}{r} $$ and the gravitational acceleration $g$ is the result of the planetary mass $M$, gravitational constant $G$ and distance $r$ via

$$g = \frac{G M}{r^2}$$

so that any velocity that fulfills the force equality is $v^2= \frac{GM}{r}$, also called Keplerian or orbital velocity. Therefore this is the velocity at which centrifugal force can balance gravity. This velocity is only a function of the planet to be orbited and nothing else, particularly no other planet, like @Soans confused answer might suggest. It is $8 km/s$ for Earth's low orbit, and will be much higher for OP's rapidly spinning, high-mass planet.

This has now several important implications:

  1. A planet rotating with Keplerian speed at its equator, will not be a stable structure. Its surface would lift off into space.
  2. A planet rotating at less than Keplerian speed will feel a reduction in gravity. This reduction will become less and less as one goes from equator to the poles, so the concept in OP's question is also real, but the 'centrifugally assisted life' needs to develop at the equator, otherwise there is always an unstable region on the planet. The fine-tuning required for this effect to be viable (also we don't know at which gravity life can function) might make the setting unbelievable.
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  • $\begingroup$ Now I get what you meant but your interpretation of the question is simply another. You understood that the question is about how to completely cancel the gravitational force, while I simply thought about partial canceling. The same way it is on earth just scaled up (you weigh about 0.5% less at the equator than at the poles). $\endgroup$
    – Soan
    Jan 28, 2019 at 5:24
  • $\begingroup$ @Soan: Yes the idea is the same for all of us. But why you invoke Jupiter to cancel gravity on Earth centrifugally is completely beyond me. $\endgroup$
    – AtmosphericPrisonEscape
    Jan 28, 2019 at 11:23
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    $\begingroup$ No you totally misunderstood my answer. In the scenario that you want an earth like gravity but instead you have a planet with the gravitational force of Jupiter. For this example I calculated the speed at which Jupiter would need to turn in order to reduce its gravitational pull down to the level of earth $\endgroup$
    – Soan
    Jan 28, 2019 at 12:09
  • $\begingroup$ @Soan: In fact, now I understand your intent, but even after re-reading your answer I don't think you communicated it that well. I will however retract my comments below your answer, as with your additional explanation, they are not justified anymore. $\endgroup$
    – AtmosphericPrisonEscape
    Jan 28, 2019 at 14:45
  • $\begingroup$ I have edited my answer to better show what this example is for and what forces are achieved with this example. $\endgroup$
    – Soan
    Jan 28, 2019 at 17:58

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