# Life on a planet that spun just fast enough to negate gravity

Imagine a world that spun just fast enough (and had a strange enough composition to hold itself together as a spinning sphere) to create an outward, centrifugal force which was as strong as its gravity, causing the effects of gravity to cancel out. In other words, the planets surface would be located at the same position as geosynchronous orbit.

If the description seems to implausible for you: On this planet, the gravity on the surface is zero. As you move away from the surface (upward or downward) the effects of gravity would increase, pulling things towards the planet's core.

What would life on this planet be like?

• Umm , if its enough to negate gravity , would the planet tear apart? – user15036 May 23 '16 at 19:08
• a little perhaps. but as soon as it did, the bits would slowly fall back down. The entire surface would be in a state of flux. – Hoytman May 23 '16 at 19:10
• That's not how that works , when pieces separate , they'll be ejected from the planet , due to the decrease in the force of gravity the farther you get from the center of mass , along with the centrifugal force pushing mass away from the center – user15036 May 23 '16 at 19:21
• as objects move away from the surface, they slow, which lowers the effect of centrifugal force. – Hoytman May 23 '16 at 19:26
• How big did you say this planet was? – Schwern May 23 '16 at 20:54

First, some math to give you an idea about what's required for this to work:

The two biggest formulas to worry about are gravity $F=G\frac{m_1m_2}{r^2}$ and centripetal force $F=\frac{mv^2}{r}$. Another way to describe your planet is that the force of gravity provides exactly the centripetal force and no more. That means we can simplify the equation to get $Gm_2=rv^2$.

Now let's assume this planet is a rocky planet. Using the average density of the Earth (5510kg/m3) and the volume of a sphere ($\frac{4}{3}\pi r^3$) we get $m=23080r^3$. Using that in the above equation, we get $\frac{r}{v}\approx 800$. So for every 800 meters of radius, the required velocity increases by 1 meter per second. Because circumference is proportional to the radius, a planet of this density will have a day that is $2\pi*800=5026$ seconds long, regardless of how large the planet is.

For a planet the size of the Earth, we need a velocity of 7960 meters per second. While this is well below escape velocity (11.2km/s), there's some really bad news - the atmosphere can't spin at this same speed. The velocity required for a given orbit is proportional to the inverse of the square root of the radius of that orbit. In other words, the higher up you go the slower you need to go in order to have a circular orbit.

So here we have two mutually exclusive requirements. If the atmosphere is spinning as fast as the planet, the atmosphere would want to leave the surface of the planet and go to a higher orbit even without factoring air pressure in. If the atmosphere isn't spinning as fast as the planet, it will act as a brake on the planet and slow the planet's spin over time (a shorter timescale than what it takes for life to evolve). Also, when the atmosphere isn't moving as fast as the planet you've got a devastating global windstorm - Randall Munroe explored the idea in the first chapter of his "What If" book.

So what does life on such a planet look like? Either it is something that lives easily in the vacuum of space, or there's an unobtainum bubble around the planet keeping its atmosphere in, in which case life looks like whatever the advanced civilization that put the bubble there wants it to look like.

• At the required velocity part, don't you mean 1 radian per second? – Simply Beautiful Art May 24 '16 at 22:04
• @SimpleArt no, I do mean meters per second. But I did neglect to actually calculate how fast a Pluto-sized body would need to rotate and used an invalid shortcut. – Rob Watts May 24 '16 at 22:29
• Yeah, I just thought your Pluto day was wrong. :) – Simply Beautiful Art May 24 '16 at 22:31

Assuming this planet is made out of unobtanium that can hold itself together... it still cannot exist.

Rotational velocity on a rotating sphere is different depending where you are on the sphere. As you move towards the poles, you slow down. As you move towards the equator, you speed up. On the Earth, the rotational velocity at the equator is about 1675 km/h. At 45° latitude it's about 1185 km/h. At the poles it's 0. This is just geometry, the circle you're making as you rotate is smaller at higher latitudes, but you do it in the same time as everyone else, so you must move slower. If it wasn't this way, the planet would have to be liquid or something.

You can only have one latitude on your planet where the rotational velocity equals orbital velocity. If that's the equator, then the high latitudes will be a bit more normal. If that's not the equator, then everything below the equator is moving above orbital velocity and is constantly flung around and eventually shed into space.

Then there's the weather. Just as on Earth, the surface will drag the atmosphere with it. Just as on Earth, the differences in velocity between the various latitudes will produce cells of prevailing winds around the planet... except at this velocity they'll be crazy. Still assuming an Earth-sized planet, instead of going 1675 km/h at the equator, it's going 28500 km/h! That's about 24 times the speed of sound in air.

No material could accumulate on the surface. No soil could ever form. It will either be flung away by the rotational speed, or be blown away by the tremendous wind leaving just the unobtanium core. If the unobtanium is not smooth, it's possible there could be catchments in its ripples.

As stated, this is simply not possible for the planet as a whole. In principle, you could get zero gs at the equator, but nowhere else.

Take the Earth as an example. The orbital period in LEO is about 90 minutes. So, if the Earth were to have a 90-minute day, the equator would be at zero gs. However, the poles have no centrifugal force associated with them, and surface gravity would be 1 g, dropping with latitude.

As a result, the other answers are correct about atmosphere loss, except that it would only occur in the vicinity of the equator, and therefore would take a bit longer than if the surface field were zero over the whole planet.