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I have a planet which was created by a God, who is experimenting with it by making it spin at an incredible speed. At the equator, the gravity is reduced to half by the centrifugal force generated by spinning.

This God has been preventing collisions from meteors and the likes which would likely disrupt the experiment. The planet is made from a material which maintains its shape under these extreme conditions, so the planet remains spherical. The God places people on this planet. They can breathe, and have everything they need. This is an experiment, and the God doesn't care whether the people die or not.

My question is what the "gravity" will be like.

My question is not about atmosphere, whether the people survive, tectonics, the colour of the guy's T-Shirt, or any other details. This is a heartless God who is dropping humans onto this planet in an experiment, like how a child may drop insects into water to see what will happen.

The planet is earth-sized with 1G of gravity. The frequency of rotation is such that the centrifugal force reduces "gravity" to 1/2G at the equator. If you started at the equator, and moved towards the pole, what would the apparent "gravity" be like? I'm particularly interested in the strength of centrifugal force in proportion to gravity, their directions, and their resultant vector.

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Note: Yes. I'm aware that centrifugal force isn't a real force.

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  • $\begingroup$ "The frequency of rotation is enough to reduce this to 1/2G at the equator. I haven't worked out what this frequency is." You should have. $\endgroup$ – ksjohn Aug 25 '17 at 10:28
  • $\begingroup$ Such a planet, no matter how strongly build, cannot support Earthly life at all (considering all other things except the fast rotation speed, obey the laws of physics of our universe). Have you considered how terribly strong would be the coriolis effect and what monstrous storms would rage there? $\endgroup$ – Youstay Igo Aug 25 '17 at 10:29
  • $\begingroup$ @YoustayIgo I'm interested in gravity. Not habitability. $\endgroup$ – Aric Aug 25 '17 at 10:30
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    $\begingroup$ You might be interested in How would gravity change on a planet rotating around itself very fast?, and to allow for changes in the shape of the planet see Why is the Earth so fat? $\endgroup$ – John Rennie Aug 25 '17 at 10:45
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    $\begingroup$ You might want to have a look at Hal Clement's Mesklin Stories; have a look at Mission of Gravity. Not an earthlike planet, but one that has a very fast rotation with consequent impact on the net attraction (over a several hundred g range from equator to poles). $\endgroup$ – JerryTheC Aug 25 '17 at 12:35
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Centrifugal acceleration is $\omega^2 r$, so in order to have a centrifugal acceleration of $1/2 g$ at the equator you will need $\omega = \sqrt {g / 2r}$, giving $\omega = 0.000875 \text{ rad/s}$; the planet rotates about 12 times faster than Earth, completing a rotation in just a little under 2 hours.

($\omega$ is the angular speed in rad/s, $r$ is the radius of the planet, here assumed to be 6400 km.)

The apparent gravitational acceleration will increase from the equator towards the poles, with $1g$ at the poles and $1/2g$ at the equator. At the poles and at the equator the apparent gravitational acceleration will point towards the center of the planet; in between it will point off-center, with a component exerting a force towards the equator: any inhabitants of the planet will feel as if they were on an incline descending towards the equator; this effect will be maximum at middle latitudes where the force pointing towards the equator will be very large, reaching about one quarter of the apparent weight. (This effect is of course also present on Earth, but it's 144 times weaker...)

The Coriolis force is proportional with the angular speed, so would be only about 12 times stronger than on Earth; it will still be quite small.

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If planet is rigid (a God that can have something the size of a planet remain spherical under such extreme conditions can, obviously, do what He pleases) all of it will have the same angular speed.

In this condition centrifugal force will be proportional to radius (distance from axis, in this case): $F = M \omega^{2} r$ this means it will monotonically decrease from Equator to Poles. No way it can overcome gravity in any place.

Actual force, using latitude as free variable would be:

$$ F(\alpha) = M \omega^{2} r_{0} cos(\alpha)$$ where $r_{0}$ is radius of sphere.

Effect of being spherical, anyways, would be rather strange, because ground would seem to be on an incline pointing towards equator.

Your God must be working hard to keep all water in oceans from "going down" toward Equator.

Maximum distance between composed gravity and perpendicular to ground would be at $\pi/4 = 45°$.

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    $\begingroup$ He as Helium, the gas? $\endgroup$ – L.Dutch Aug 25 '17 at 10:48
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    $\begingroup$ Of course! That God is also Helium, God of the sun :p $\endgroup$ – ZioByte Aug 25 '17 at 10:50
  • $\begingroup$ So the gradient would appear to increase, as if the ground was curving upwards but it's actually convex so the ground is curving downwards.... interesting. $\endgroup$ – Aric Aug 25 '17 at 10:58
  • $\begingroup$ I , too, would be very worried about the water. $\endgroup$ – Lee Leon Aug 29 '17 at 18:45
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I've been thinking that if the structure is kept as spherical despite the fast rotation, then all liquid water in the oceans will flow towards the equator. If the planet is allowed to become oblate and balance the forces, this will not happen. People standing somewhere between the pole and the equator, would feel, under your set of conditions, that they stand on a slope and would incline themselves towards the pole. The reason :

EDIT: (visualizing the forces)

If the planet was held into a sphere no matter what, gravity would feel the same everywhere. Gravity will always point to the center of the planet. On the other hand, centrifugal force points away from the axis of rotation and is parallel to the equatorial plane, not the planet's surface. It is strongest at the equator and zero at the pole. Adding-up gravity and centrifuge will give inhabitants a weak gravity at the equator: gravity and centrifuge are opposed. Standing at the pole makes you feel heaviest as centrifugal force is zero. Between the two, the centrifugal force points away from the axis of rotation. This compares to standing on a carouselle, and you must lean towards its center to balance.

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  • $\begingroup$ I'm concerned about gravity, not water. See the bold text in the question: "My question is what the "gravity" will be like. My question is not about atmosphere, whether the people survive, tectonics, the colour of the guy's T-Shirt, or any other details." Your second point about the incline is what I'm after. Can you expand on this please? $\endgroup$ – Aric Aug 25 '17 at 12:25
  • $\begingroup$ A great deal of tensile strength is needed to hold the planet into a sphere. A tectonic plates system is likely to compromise the integrity of the sphere, because it requires cracks in the shell. The liquid mantle is likely to press hard on the equator due to centrifuge and this is where tectonics are not wanted. Regarding gravity, here is the explanation: $\endgroup$ – Christmas Snow Aug 26 '17 at 13:53
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I'm going to let Michael answer you, at least partially. I've linked the video to a precise moment but the whole video is worth watching, really.

To sum up what is said in the video, the gravity (as a force) would not be affected by a faster rotation. To be more precise :

• The centripetal force (generated by gravity) that pull you towards the Earth is constant.

• The rotation of the Earth gives you a velocity, and this force is always tangential to the surface of earth (if there was no gravity you'd be thrown in space because of that force).

Interesting numbers of the video :

• At the equator, if the earth wasn't spinning we'd weight 0.3% more.

• If Earth spun 17 times more faster, the sum of the two forces would be 0, meaning we would be floating around.

Edit : Apparently the link does not redirect where i wanted to in the video. It's @4"11 seconds

Other answers on this thread are far more mathematical - this one might be useful for someone looking up for approximations or basic principle.

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  • $\begingroup$ Please always try to summarize the important parts of linked resources. Otherwise your answer would be left basically useless for future readers if the source changes or is removed. This is currently basically a link-only answer, which is a reason to delete answers on this site. $\endgroup$ – Secespitus Aug 25 '17 at 11:26
  • $\begingroup$ New here, my mistake ! Correcting this right now. $\endgroup$ – CageyCat Aug 25 '17 at 11:30
  • $\begingroup$ No problem, we all started at some point and SE is a bit different from other sites. Have fun on the site! If you haven't done so already taking the tour is a pretty good start and the help center gives a lot of useful resources. And don't hesitate to ask if you have problems at some point. The chat is a great place to get quick feedback. $\endgroup$ – Secespitus Aug 25 '17 at 11:32
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Incredible speed isn't so incredible if it's not fast enough to modify the mass of your planet. As others have pointed out, if your planet rotates fast enough, it might get pulled apart by centrifugal forces. But, if God made your planet, he surely must have thought of this.

So, let it spin so fast, that it's equator moves close to the speed of light. Not only will your planet be seen by observers as a massive celestial body, but if its mass and speed were large enough, it could radiate detectable gravitational waves (I think).

I'm not a specialist in General Relativity, but to estimate the mass of the rotating planet, I'd solve this integral: $m=m_0\int_o^R\frac{1}{\sqrt{1-\omega^2r^2/c^2}}4\pi r^2dr$, where $\omega$ is the angular velocity of the planet, $m_0$ is its rest mass, $R$ its radius and $c$ is the speed of light.

Edit: for the low velocity case in which the objects at the equator experience half of the acceleration compared to the ones at the poles: $a=g-\omega^2R=g/2 $, meaning $\omega=\sqrt{\frac{Gm}{2R^3}}$, with $G$ the gravitational constant. There always will be a centrifugal acceleration $g\cos \theta/2$ perpendicular to the axis of the planet. $\theta$ is the latitude. The Coriolis acceleration would be $-2\omega \times \bf v$. Plugging in some numbers, for an Earth like planet, you would see a rotation period a little under 2 hours. The Coriolis forces should be one order of magnitude larger on your planet.

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  • $\begingroup$ I don't want it to spin so that the equator gets close to the speed of light. I want it to spin so that the equator has half gravity. This does not answer the question. $\endgroup$ – Aric Aug 25 '17 at 12:24

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