# What would a planet spinning fast enough to allow geostationary orbit near the surface look like?

From an answer to a previous question of mine:

A planet spinning fast enough to allow geostationary orbit near the surface would result in odd side effects. Any object at rest on the equator would be moving at speed near to orbital speed. It would have weight but much less than similar objects at the poles. A planet that formed spinning that fast would be flattened, with the equator at higher altitude. A planet spun up to that speed after solidifying as a sphere would result in any object just North or South of the equator experiencing a force towards the equator, resulting in a drift of loose rocks towards the equator. If the height of geostationary orbit was only just above the ground level, this could result in rocks finding their way into orbit simply by drifting towards the equator and then piling up. https://worldbuilding.stackexchange.com/a/303/90 by @githubphagocyte

1. How would such a planet look like, geography and climate wise?
2. Could earth-like biology evolve in such an intense context?
3. Would this planet be short lived? (Maybe so much so that actual evolution doesn't have time to take place...)
• Please define "near the surface". How near? 35786Km is Earth's geostationary distance. Is that near enough? Which distance is? Sep 20, 2014 at 6:37
• "Just above the ground level", I guess I mean less than a kilometer. I actually don't know: how flexible is such a thing? do a few meters matter? could rocks from the ground find their way into orbit with that distance? Sep 20, 2014 at 6:40
• I was looking for oblateness limits, and found this: en.wikipedia.org/wiki/Mission_of_Gravity#Setting Sep 20, 2014 at 14:25
• @NeilSlater Yes, I've mentioned that book somewhere else on worldbuilding, it's well worth a read. Sep 20, 2014 at 19:44
• Everyone would want to live on the equator, so they could use the geostationary orbit above their house to store their stuff. Sep 4, 2016 at 21:23

I'm taking "near ground" as meaning "the height over ground is negligible compared to the radius of the planet". That is, we can as good approximation we can assume that the radius of the geostationary orbit is the same as the radius of the equator.

# tl;dr

Such a planet would likely be a dead, airless rock, but with interesting physical effects. Nice to put a space station on, but not developing life on its own.

# Needed rotational speed

Let's first look at a perfectly spherical planet (ignoring for the moment that a planet under those extreme conditions won't be perfectly spherical). The relevant quantities for such a planet are its mass $$M$$, its radius $$R$$ and its angular velocity (rotational speed) $$\omega$$. We also assume the geostationary orbit at height $$h\ll R$$ above the equator (note that on earth, that condition would still be fulfilled at the top of Mount Everest, so it is not too limiting).

The condition of a circular orbit (which the geostationary orbit is) is that the centripetal acceleration equals the gravitational acceleration. The centripetal force is given by

$$a = \omega^2 (R+h)$$

and the gravitational acceleration is

$$g = \frac{G M}{(R+h)^2}$$

where $$G = 6.7\cdot 10^{-11}\,\rm m^3\, kg^{-1}\, s^{-2}$$ is the gravitational constant.

So the planet would have to spin at the angular velocity of

$$\omega = \sqrt{\frac{G M}{(R+h)^3}} \approx \sqrt{\frac{GM}{R^3}} \left(1 - \frac{3}{2} \frac{h}{R}\right)$$

To see what this means, let's insert a few numbers.

First, let's assume we've got a planet of earth mass, about $$6.0\cdot 10^{24}\,\rm kg$$ ($$GM = 4.0\cdot 10^{14}\,\rm m^3/s^2$$), and earth radius, about $$6.4\cdot 10^6\,\rm m$$. Then we have

$$\omega = 1.2\cdot 10^{-3}\,\rm s^{-1},$$

that is, you'd have one full rotation every 1.4 hours. Note that I neglected the height of the orbit here, since it would end up in the rounding error anyway.

Actually that's a lot less than I would have expected (but thinking again, the actual geostationary orbit's radius is just about 7 times the earth radius, so it should not have been that surprising).

Interestingly, if you look at the formula for omega, you see that the relevant quantity is the density of the planet. So we if we give the density of a planet as multiple of the earth density, omega scales with the square root of that number. So a planet of four times the earth's density would have twice the angular velocity, and thus have one rotation about every 42 minutes. On the other hand, a planet with only 1/4 of the earth's density would have 2.8 hours for each rotation. If you want to have an earth-length day (24 hours, neglecting the fact that you'd have to consider the sidereal instead of the solar day), the planet's density would have to be 0.34% of earth's density, or 19 kg/m^3. That's about 1/48 of the density of Styrofoam. A planet made completely of Styrofoam would therefore need to have a rotational period of 3.5 hours. (Note: It would be nice if someone cross-checked my numbers.)

# The effects of such a rotational speed

OK, so what would be the effects of such a rotation on the planet's surface? Well, the two forces to consider are the effective gravitational force (that is, gravitation + centrifugal), and the Coriolis force. As before, I'll use accelerations instead of forces; to get the force onto an object just multiply with its mass.

## Effective gravitation

The effects of course depend on the latitude, which I'll call $$\phi$$, in accordance to geographical conventions. It makes sense to split the acceleration into a vertical and a horizontal component, relative to the ground. The gravitational acceleration is, of course, always vertical and always the same (since we assume a spherical planet). The formula I've already given above (now we of course set $$h=0$$, since we are interested on the gravitation on the surface), but now we have to be careful about the direction: It points downwards, so we add a minus sign.

$$g = -G M/R^2$$

The absolute value of the centrifugal force depends on the distance from the rotation axis, which is

$$d = R \cos(\phi)$$

Otherwise, it's just the formula above, with $$R$$ replaced by $$d$$ (on the equator, of course we have $$d=R$$):

$$a = \omega^2 d = \omega^2 R \cos(\phi)$$

However, its direction is away from the axis, which means that we have to split it into a horizontal and a vertical component. The horizontal component is $$a\sin(\phi)$$, and the vertical component is $$a\cos(\phi)$$.

Putting everything together, we get the total vertical acceleration

$$g_{\text{eff}} = -\frac{G M}{R^2} + \omega^2 R \cos^2(\phi)$$

or, after inserting the "geostationary equator condition":

$$g_{\text{eff}} = \frac{G M}{R^2} \left(1 - \cos^2 \phi \left(1 - \frac{3}{2} \frac{h}{R}\right)^2\right) \approx \frac{G M}{R^2} \sin^2 \phi - 3 \cos^2 \phi = g \sin^2(\phi) - 3 \frac{h}{R} \cos(\phi)$$

This is exactly as expected, you're heaviest on the pole (where the rotation has no effect), and lightest at the equator (and for $$h=0$$, you'd be weightless at the equator).

And the force toward the equator is

$$a_{\text{eff}} = \omega^2 R \cos(\phi) \sin(\phi) \approx \frac{1}{2} \frac{G M}{R^2} \sin(2 \phi)) \left(1 - \frac{3}{2} \frac{h}{R}\right)^2 \approx \frac{g}{2} \sin(2 \phi) \left(1 - 3 \frac{h}{R}\right)$$

Note that this force is zero both at the equator and at the poles, and maximal at a latitude of 45°. At that latitude it would be half of the polar gravitation, so it would be a quite strong force. Indeed, at that latitude, the horizontal floor would have an apparent tilt of about 27°.

## Coriolis force

The Coriolis force is velocity-dependent. It is the force which is responsible for the rotation of air around high/low pressure regions (and thus also in part responsible for things like hurricanes).

The Coriolis force is always perpendicular both to the axis of rotation, and to the direction of movement. Therefore we now not only have to consider the position where we are, but also the direction we are running.

I'll define the cardinal directions as on the earth: The sun rises in the east and settles in the west. The poles are in the north and south. This means that the angular momentum vector points to the north. The formula for the Coriolis acceleration is

$$\vec a_C = 2\, \vec v \times \vec\omega.$$

The horizontal direction of the Coriolis force is to the right on the northern hemisphere, and to the left on the southern hemisphere. So for example if you are running towards the closest pole (to the north on the northern hemisphere, or to the south on the southern hemisphere), the force will push you in east direction.

The most interesting part is the vertical component, which will become relevant on the equator when you're running left or right. When running on the equator in east or west direction, all of the Coriolis force is vertical; you'll get

$$a_c = 2 v \omega = 2 \frac{v}{V} R \omega^2 \approx 2 \frac{v}{V} g \left(1-\frac{3}{2} \frac{h}{R}\right) = \frac{v}{V} g \left(2 - 3 \frac{h}{R}\right).$$

where I've introduced the equatorial speed $$V = R \omega$$. Note that when running to the east, the force will go downwards (make you more heavy), while when running to the west, it will go upwards (make you lighter).

Compare with the effective force on the equator (see above):

$$g_{\text{eff}} = -3 g \frac{h}{R}$$

So you get an effective upwards force if you run eastwards and $$a_c > g_{\text{eff}}$$, that is,

$$\frac{v}{V} > \frac{3 \frac{h}{R}}{2 - 3 \frac{h}{R}} \approx \frac{3}{2} \frac{h}{R}$$

Let's calculate that with earth mass/radius, and a geostationary orbit at 8000 meter height (about Mount Everest height):

$$V = R \omega \approx \sqrt{\frac{G M}{R}} = 7.9\,\rm km/s$$

$$\implies v > 0.015\,\mathrm{km/s} = 53\,\rm km/h.$$

That's slightly above the allowed maximum driving speed inside a settlement in Germany. It's definitely much below what cars are able to do.

# Would such a planet be able to develop life?

Given that above the equator there's a "gravitational leak" due to the centrifugal force, I'd not expect that planet to be able to hold an atmosphere. So if there were life on such a planet, it would certainly not be on the surface. Without much of an atmosphere, I guess also water would evaporate quite quickly, so I'd expect the planet to be mostly a dead rock. Without air, there would, of course, also not much of a climate.

# What advantage would such a planet have for colonization?

Despite the disadvantage of having an effectively space-like environment, such a planet could have the advantage that you have very low launch requirements, so it would be relatively cheap to get onto/off the planet. For a space station (and possibly mining), that would be ideal.

• The equations would benefit from using mathurl.com to improve readability (at least until we get MathJax on this site, if that is agreed). Sep 21, 2014 at 6:44
• I love this answer so if you don't have time to apply the equation formatting yourself I would be very happy to. Sep 21, 2014 at 6:45
• If I understand correctly, with mathurl, the equation source would no longer reside on this site, which I'd consider bad (if the site goes, so will the formulas). This is different to MathJax, which leaves the source on the page and only does the rendering. So I'd prefer to leave the equations as is until we (hopefully) get MathJax enabled. Also, I think there are already more than enough external deependencies. Sep 21, 2014 at 7:39
• While the quote was not my answer I suggested something along these lines. I figured the planet would have to be substantially more massive than the Earth in order to retain an atmosphere--atmosphere retention is based on the Δv from the top of the atmosphere to escape velocity, a sufficiently massive world will get this value high enough. Oct 14, 2014 at 20:02
• @LorenPechtel: YOu forget that the planet's atmosphere rotates with the planet. Thus the atmosphere above the geostationary orbit would be faster than the orbital speed, and thus rise and ultimately escape. Oct 15, 2014 at 7:40

With real-world physics, and not invoking any unknown or fantasy effects, a rocky planet would need to be very small before it could spin stably at such a speed. What would happen to a world that would be spherical when spinning slowly would be dramatic - within hours it would transform into a pancake of spinning debris. The high angular momentum would rebalance as a more extended set of objects in orbit around a common point.

I am not sure of the practical size limit/cutoff for this effect, but I think it will be quite small, lower than anything which could support liquid water or an atmosphere. Update: this link has some clear maths on the stability of rotating objects, and this page has run through the calculations including some nice graphics, but unfortunately without working though the details of possible orbits and where geostationary ones would be.

To make a large world practical you need to invoke some kind of fix to this problem. These things are pretty common in (non-"hard") science fiction and fantasy:

• Ignore or modify the problem physics, and focus on exploring consequences of a less extreme version.

• Posit some super-powerful materials or forces which keep the situation stable. You can then follow through with conjectures about experience of living in the environment, trusting to the magic you have invented to keep things stabe enough that life could evolve etc.

If we assume you work around the physical impossibility as above, then you can make some conjectures about life-forms on a planet which is somehow stabilised as a flattened spheroid with centrifugal (or centripetal if you prefer) force balancing.

The following things occur to me:

• There will be extreme forces operating in atmosphere towards the equator. Hyper-powerful vortexes hurricanes, tornadoes or some kinds of weird weather not experienced in "normal" planets. This would likely be too extreme for living creatures if you did the maths to figure out likely forces, but again you could reduce that to something that might work in a more fanciful description.

• The combination of powerful weather and low gravity should cause a major flow of material (basically masses of the atmosphere and whatever the world is made of) to make it into orbit at the equator, where it would either settle into an extended ring around the planet, or be recycled north or south to fall back onto the surface. You may need to posit this recycling scheme in order for the world to seem stable, and it could occur smoothly as a kind of weather, or in chaotic episodes, as catastrophic events, or likely both.

• A flattened world would experience more extremes of angles in sunlight. The effects of this will depend on angle of planet's spin to its orbit around the star. If you assume an Earth-like tilt, essentially a large proportion of the world's land surface would experience daylight similar to our Artic and Antartic circles. That doesn't necessarily mean "cold", that could be balanced by a closer position to the star . . .

• There would be a lot of atmosphere and debris in a band around the equator. Not a high atmospheric pressure though - in fact the opposite, the air would be highly rarified around the equator and creatures comfortable at the poles might not be able to breathe there. The extra air mass and debris should filter starlight that travels through it though, and the nature of the light would be very different in different parts of the planet. Sunsets might not even happen, so much as the "sun" would move towards the horizon becoming discoloured (more red in an Earth-like atmosphere), and more diffuse until it disappeared from view. At the equator it might be permanently shaded in dim red light with only a vague sense of where the sun is.

• Whether or not my conjectures above are on target, I would expect extreme banding of environments between pole and equator to influence the nature of any plants and creatures much more dramatically than on earth.

• Air pressure would equalize, so you would end up with very little atmosphere. Sep 20, 2014 at 14:50
• @TimB: Yes, that is why I suggest you need a fix to return material. A dynamic setup, where energy or material was flowing through the system, would not reach this equilibrium. Air pressure on Earth never equalises perfectly because of energy input from the sun, for example (much less extreme of course). Creating a "working" world with living creatures with this setup requires bending, breaking or ignoring physics in multiple places. Sep 20, 2014 at 17:09

The other answers are excellent but there is one important thing to add - the shape of the planet.

Planets normally form a sphere as they are condensing from a liquid form and gravity pulls it into a flat sphere as it does so. In this case though it would actually form a very squashed sphere, it may even form what looks a lot like a disk, as the level "effective gravity" point is modified by the spin.

At the extremes you are talking here the planet may only be (for example) 1 km tall and 6 million km wide. You would have very low experienced gravity no matter where you went on the surface.

This would be modified a little depending on whether the planet cooled before or after it acquired this extreme spin though.

• "The other answer" doesn't make much sense as a reference now that there are a total of three answers, including yours... Please, be more specific. :)
– user
Sep 20, 2014 at 19:22