# Could a flat planet form by rotation?

Suppose a planet had an extremely fast rotation, so fast that the planet is flattened considerably and somewhat resembles fictional flat earth, Only at the north pile does life to exist, where the gravity is strong enough. This planet is tilted dramatically, like Uranus, so that this pole faces the sun almost directly during the summer. The planet orbits the sun very quickly, so that a year goes by in 24 hours, thus creating a “day/night” cycle despite the rotation being extremely fast.

How plausible is this?

• I feel like it would disintegrate under the centripetal force of spinning that fast, but I don't know where to start on calculating that. Nov 4, 2022 at 17:10
• ANSWERS, PEOPLE! Nov 4, 2022 at 17:16
• @user98816 PATIENCE, PERSON. Nov 4, 2022 at 17:24
• If the planet is made out of pizza dough, certainly. You have described near exactly the predicament in which sourdough microbes find themselves. Just imagine, spinning closer and closer to that brick oven. Needs tomato sauce and cheese though.
– Rab
Nov 5, 2022 at 5:13
• The gravity would be the same everywhere. See my comment on Nosajimiki's answer for more details. Nov 5, 2022 at 15:10

## A flat(ish) planet may be possible, but not quite as you've set it up

Haumea is a dwarf planet in our own solar system that is significantly flattened out by its own rotation. So, we know that stretched out, quickly rotating worlds can exist.

However, the gravity part of your idea does not work. No matter where you stand, gravity will always feel about the same. Planets are by definition hydrostatic; so, no matter how you change the properties of what a planet is made out of, its shape will always settle to rest in a form where the surface experiences nearly uniform gravity. The more massive the planet, the more true this is.

At your poles you may experience a bit more gravity because the solids in your world are not truly fluid so they would resist flattening to a degree, but if you were to stand on the North Pole, there would be significant amounts of gravity pulling you left and right in equilibrium, but much less pulling you down. As you walk towards the edge of the planet, the angle of gravity changes and becomes stronger... however, the strength of the planets centrifugal force also becomes stronger in proportion to the increase in gravity, because it is this equilibrium that defines the planet's flattened shape.

So while you may think that as you approach the "edge of the world", that you would feel like you are going up hill, this feeling would be negligible. In fact, this planet could possibly have a very large habitable zone and stable atmosphere everywhere you go. The Coriolis Effect may cause severe winds that might make it unsuitable for Earth Based life though.

Also, a 1 day Orbit around its parent star may lead to tidal locking making the fast rate of rotation difficult to explain. While its fast rate of rotation will resist tidal locking at first, it will also bleed off energy quickly (in geological time scales) in order to offer this resistance slowing the spin down.

• That seems so alien... I guess a realistic flat planet would be very fun to write about Nov 5, 2022 at 10:13
• Those comments about the gravity aren't correct. The ground will level out until the net effect of gravity and centrifugal force is the same everywhere - that's the reason the planet is flattened out in the first place! So even though the planet is flat-ish and spinning very fast, your weight at the equator will be more or less the same as at the poles, and walking from one to the other will feel like walking on level ground, unless there are mountains in the way. Nov 5, 2022 at 14:58
• It's also worth noting that Haumea is shaped like an egg or an American football, not like a pancake. Nov 5, 2022 at 15:00
• On the Earth the apparent gravity is actually a little stronger at the poles than at the equator. See physics.stackexchange.com/questions/141856/… -- so there is a difference in the net force holding you to the surface, but it is the opposite of what this answer says. Nov 5, 2022 at 18:53
• It will still feel like moving on level ground as you move from one to the other though. Nov 6, 2022 at 1:35

You can't get there from here with real physics.

The amount of oblateness is related to the amount of spin. If the centrifugal force is anywhere near the gravitational pull, then the atmosphere will all drain off of the edges.

This answer provides good calculations on how oblate your planet can get before it rips itself in half. Spoiler: it's roughly a 3:1 ratio.

On the rotation/orbit thing, I don't believe that you can have a 24 hour orbit at a distance that also corresponds to the goldilocks zone for habitable life. I'd have to do a bit of research to figure out if a planet could precess that fast without ripping itself apart, but I think that would significantly reduce the speed of spin.


So, sorry, in an environment where Niven couldn't get away with God's Easteregg, you couldn't get away with this. Maybe you can crank your suspension of disbelief up a couple notches.

Addendum: After a little research, I found the possibility of an actual moon-sized object orbiting a white dwarf roughly every 24 hours, within the habitable zone of that star. As such, your crazy-fast orbit is actually MORE likely than the pancake shaped planet.

• I don't think precession is an issue -- the question mentions the day-night cycle being due to the axis remaining fixed but "annually" pointing close to the sun as Uranus does. Nov 4, 2022 at 18:04
• You might be able to get a habitable zone with 24 hour orbit around a brown dwarf or deuterium dwarf (L class?) -- or maybe a white dwarf? -- but it probably wouldn't last long enough for life to evolve, surely not to sapience (at least at Earth's rate). Nov 4, 2022 at 19:19
• @ZeissIkon Are you saying the planet or the star would not last long enough for life to evolve because the smaller the star, the longer it lasts. Nov 4, 2022 at 19:50
• @ZeissIkon, precession was just presented as a way of getting the day/night cycle a little closer to 24 hrs without so close an orbit. He didn't request the thought in the original question. Nov 4, 2022 at 19:58
• @Nosajimiki, The problem isn't how long the star lasts, but how long before the planet falls into the star. Closer orbits suffer from orbital degradation due to constantly plowing through the solar wind. You MIGHT get this to happen if a non-binary white dwarf captured a planet after it collapsed, but not through the standard accretion process. Nov 4, 2022 at 20:02

From what I recall, Hal Clement's calculations for Mesklin were found to be wrong -- even with the mass of several Jupiters to work with, you couldn't actually have a planet with such a high rate of spin.

First, the accepted explanation (as far as I'm aware) for a planet's rotation rate is that it's due to the impacts of accreting bodies during planetary formation -- hence Venus wound up with a sidereal day longer than its year, Earth and Mars came out almost the same, all the gas giants rotate faster (fast enough, in the case of Jupiter, to be visibly oblate through a very modest telescope).

If you have a planet with high enough spin rate to be flattened to the level you want, it would have to have been struck many times on the same (receding) edge and this would have to have occurred after the one that upset the spin axis to be orthogonal to the spin of the rest of its system. And as the spin builds up beyond the (roughly) half kilometer per second Earth has, each subsequent impact adds less to the rotation rate than the last.

Then we run into the gravitational and mechanical issues that proved Mesklin was impossible. In short, Mesklin (or any other planet with a similar ratio of effective surface gravity from poles to equator) will lose its atmosphere, likely before life could evolve to make the atmosphere breathable. In your case, you'd have surface gravity of, say, 3G at the poles (same as Mesklin at the rim), but nearly nothing at the very fast-moving equator. Orbital velocity isn't enough above the equatorial rotation speed, and the atmosphere (far taller in the lower gravity) will reach up into orbit -- at which point it simply floats away.

You would not feel like you're walking uphill going to the edge -- the surface (of both land and sea) would feel level (because it rests at gravitational equipotential). It's just that you'd get to a point, if the spin rate is high enough, where you could jump high enough to suffocate...

• "because it rests at gravitational equipotential" <- I don't believe this can happen in practice. Your gravity still needs to be greater than your centrifugal force or the planet will fail to accrete. The rim may not feel like going straight up hill like it would on a slow spinning flat world, but it will still have to feel more-or-less like going up hill for the planet to exist at all. Nov 4, 2022 at 19:35
• Gravity would point towards the center of mass which would be in line with the center point of the disc. The "ground" surface won't be at equipotential because the center of the disc would be significantly closer to that center of gravity than the rim would be.
– bta
Nov 5, 2022 at 2:22
• @bta For a spherical object, gravity always points toward the center of mass. This fact does not necessarily apply to non-spherical objects. Nov 5, 2022 at 14:31
• @Nosajimiki If it feels like going uphill to you as you travel toward the equator, it "feels" like that to the rocks under your feet as well. It wouldn't just feel like a hill, it would be a hill. Something other than the planet's rotation would be required in order to create such an equatorial mountain range. Nov 5, 2022 at 14:40
• @DavidK Oh! yes, that is a very good point I did not think about. The planet would settle into an oval shape where down feels down no matter what. Thanks for the clarification Nov 6, 2022 at 18:52

It Needs to Rotate once Every Few Seconds.

The formula you want is:

$$\frac{a-b}{b} = \frac{5}{4}\frac{(2\pi/T)^2 R^3}{GM}$$

Where:

$$a$$ is the (larger) equatorial radius.

$$b$$ is the (smaller) polar radius.

$$T$$ is the time for one revolution.

$$R = (b+2a)/3$$ is the mean radius.

$$M$$ is the mass of the planet.

$$G$$ is Newton's constant.

The formula is derived in this thread where it seems to agree with the experimental numbers for Earth. This is good for you because it suggests we can indeed ignore how the planet is denser in the middle. See also Wikipedia and their source [8].

Since you want $$a$$ much bigger than $$b$$ we'll just use $$R \simeq 2a/3$$.

Since $$a/b$$ is very large we'll also write $$\frac{a-b}{b} = \frac{a }{b}-1 \simeq \frac{a }{b}$$. The formula becomes

$$\frac{a }{b} = \frac{5}{4}\frac{(2\pi/T)^2 }{GM}\frac{8 a^3}{27}$$

I'm going to ignore the constant factors because the goal is to show the thing has to spin ludicrously fast.

$$\frac{a }{b} = \frac{a^3 }{GM T^2}$$

Let's take $$a$$ as the 6000km radius of the Earth and $$M$$ the Earth's mass. In Si units:

$$G \simeq 6 \times 10^{-11}\simeq 10^{-11}$$

$$a \simeq 6000km = 6000 \times 1000 = 6 \times 10^3 \times 10^3 = 6 \times 10^6 \simeq 10^6$$

$$M \simeq 6 \times 10^{24} \simeq 10^{24}$$.

So we get:

$$\frac{a }{b} = \frac{10^{18} }{10^{-11} 10^{24} T^2}= \frac{10^{18} }{ 10^{13} T^2} = \frac{10^{5} }{ T^2}$$

If for example you want the ratio to be 1/100 we need

$$100 = \frac{10^{5} }{ T^2} \implies T^2 = 10^3 \implies T = 10^{1.5} \simeq 32$$

Meaning one revolution every 32 seconds.

How fast is that?

The equator of the pancake is 40,000 km long. So an ant standing on the rim moves more than 1000km each second. Much zippier than the ISS orbit speed of 10ish km/s but only a measly 0.3% the speed of light.

• The shape is correct, but as @Nosajimiki tried to show, there are severe problems with life at the north pole. The answer should at least discuss them, if only to make sure that the querent is aware of them. Nov 4, 2022 at 19:23
• @AlexP Nosajimiki has asserted many things, but I don't consider any of them to be shown. That answer is unsourced and patently incorrect. Nov 5, 2022 at 19:08