So I'm creating a sci-fi book, and I'd like to know a way to calculate how fast would the moon need to spin to create a certain amount of gravity, and how much energy would be required to get it to spin that fast. I couldn't find any information on this on google. If you do answer, please provide the calculator/equations you used to calculate the mentioned information.

Okay, I see some people misunderstood my question. I know artificial gravity can be "generated" by rotation. What I meant was, how fast (RPMs) would the moon need to spin in order to achieve the Earth's gravity for any underground habitats. So that anyone living in those underground habitats would experience gravity similar to Earth's. Something similar to hypothetical rotating space stations, except in this case, the entire moon would spin, so that anyone living in it's underground habitats would experience Earth-like gravity.

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    $\begingroup$ Rotation creates centrifugal forces, which oppose gravity, right? So, faster rotation, means smaller apparent gravitational acceleration on the surface. So.. I am not really sure I understand your question. Unless the structures in these moon are designed for upside-down gravity? $\endgroup$ Dec 22, 2021 at 21:24
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    $\begingroup$ that's not how gravity works, spin only lessens gravity and the moon is already spinning as slow as it can. $\endgroup$
    – John
    Dec 22, 2021 at 21:26
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    $\begingroup$ The question itself is nonsensical. The formula for gravity is the following. Fg=(G*m1*m2)/(d^2). Fg is the force of gravity. G is the gravitational constant, a fixed value. m1 is the mass of the first object. m2 is the mass of the second object. d is the distance between the objects. Notice how there is no mention of speed or rotation in this equation. Neither has anything to do with gravity. If you want to increase the gravity, either make one of the objects more massive or decrease the distance. If this is not the answer you want, you might need to be more specific with your question. $\endgroup$ Dec 22, 2021 at 21:31
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    $\begingroup$ Re the edit - if the moon is spun to treat it as a big O'Neill cylinder then any underground habitats will not be underground. As soon as the simulated spin-gravity exceeds the Moon's actual gravity (about 1/6 Earth's) then the Moon will start flying apart. There won't be any ground to be "under" long before the spin reaches 1.2G. $\endgroup$ Dec 22, 2021 at 21:40
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    $\begingroup$ It seems that you have a misunderstanding of "artificial gravity 'generated' by rotation". It's not spelled out explicitly in the answers, so I'll state it here: this "artificial gravity" goes outwards from the center of spacecraft, not down toward the center as you seem to be assuming. You're walking on the "ceiling" or possibly "walls" of rotational "gravity" systems. $\endgroup$
    – Ben
    Dec 23, 2021 at 23:24

6 Answers 6


$$ F=\frac{mv^2}{r} $$

That's your equation. Let's flesh it out.

m is the mass of the human, in kilograms. Let's call it 75kg (around 165lb).

F is the sum of the weight of the human on earth and the weight of the human on the (non-spinning) moon, in Newtons. This is because centripetal force due to a spinning planet (or moon) will act in a direction opposite that of gravity from that planet (or moon). Call it 735N in earth-weight, for just over 120N moon-weight (close enough). F = 855N.

r is the radius of rotation, in meters. The radius of the moon is around 1738100 meters at the equator, where we'll be building our base, for maximum velocity. r = 1738100 meters.

Now we just have to solve for velocity:

$$ \frac{Fr}{m} = v^2 $$

$$ \sqrt{\frac{Fr}{m}} = v $$

Substitute in our variables:

$$ v = \sqrt{\frac{855\mathrm{N} * 1738100\mathrm{\ meters}}{75kg}} $$

And solve:

$$ v = 4451 \mathrm{\ meters/second} $$

The circumference of the moon is 2×π×radius, or 10920804 meters. So, our moon would complete one revolution every 10920804/4451 seconds = 2453 seconds. That's roughly once ever 40.89 minutes.

That's not so unreasonable, right? But here's the catch. Anything at the equator of this moon on the outer surface, going at the speed of the equator, will be launched out into orbit. That means moon dust, rocks, spacecraft, everything. The moon will disintegrate, and the underground base with it, if you can even find a way to build the base in the first place (maybe before speeding up the rotation). The escape velocity of the moon is only 2380 meters/second, and the equator is clocking 4451. So, no more moon.

The solution to this is to make the moon stay together by some incredible binding force of essentially magical strength. If every molecule was immovably bound to every other, you could have your base. Whether that works for your story, I don't know.

Note: You'll run into this problem on any planet of any size, when you try to make its centripetal force greater than its gravity. The equatorial speed will rise over escape velocity, and the planet will disintegrate. That's why artificial gravity of this sort typically only comes up in space stations, which aren't kept together by gravity at all.

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    $\begingroup$ 2453 seconds is about 40 minutes, not 40 hours. You might want to revise the conclusions after that $\endgroup$
    – L.Dutch
    Dec 23, 2021 at 14:22
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    $\begingroup$ @L.Dutch I've edited it myself. No need to revise the conclusion: it's obvious that if the Earth was rotating in 40 minutes rather than 24 hours the centripetal force would be much higher. The rest of the answers stands. $\endgroup$
    – Rekesoft
    Dec 23, 2021 at 16:47
  • $\begingroup$ @L.Dutch Duh, thank you! Late night post, my bad. I knew something seemed off about that unit conversion. $\endgroup$ Dec 23, 2021 at 18:15
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    $\begingroup$ @Michael Even very strong steel has a breaking length of only about 30 km (meaning that a 30 km long cable made out of it would break under its own weight if subject to one g of gravity). Compare that to the moon's radius of about 1700 km. So a spinning steel moon would break apart, too. $\endgroup$ Dec 24, 2021 at 19:40
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    $\begingroup$ @Michael Of course, then we'd be orbiting the "moon," haha! $\endgroup$ Dec 24, 2021 at 19:56

Such a high rotational rate would cause the moon to shatter.

Not only would the rotational rate have to exert enough centrifugal force to cause people to experience 1.2 G while on the ceiling of a cavern, it would have have to over come the moons natural gravity.

At that speed objects such a rocks and lunar regolith to leave the surface and fly off into space. It would probably be high enough to cause the lunar bedrock to break apart and also fly off into space.

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    $\begingroup$ This. "probably be high enough" means definitely high enough to destroy the crust, mantle, and core. Earth would face a mass extinction from the debris (just a 50 millionth of the moon is about a 10-km diameter sphere). Don't you hate that pesky square-cube law? $\endgroup$ Dec 23, 2021 at 8:21
  • $\begingroup$ @KevinKostlan I don't think it would bombard Earth, the mass is still there in orbit. We'd just get a really nice set of rings consisting of moon rock and bits of moonbase citizens. $\endgroup$
    – Ben
    Dec 23, 2021 at 23:19
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    $\begingroup$ @Ben The mass is still there in orbit, yes, but each piece in a different orbit, some of them round, some highly eccentric with perigee <6e6m. $\endgroup$ Dec 24, 2021 at 5:02
  • $\begingroup$ The moon shattering won't happen - because the energy needed to speed it up enough will melt it anyway if it's done on a reasonably fast time scale. $\endgroup$
    – Dan Hanson
    Dec 25, 2021 at 3:11


I want a giant spinning habitat on the moon! If we can't spin the whole moon that fast without having the rocks on it lift off (which makes sense, I grudgingly accept) we will construct a habitat on the surface of the moon and have that spin. Now, how fast?

from https://rechneronline.de/g-acceleration/centrifuge.php

screenshot from Gforce calculator page

So if my habitat makes 0.025 revolutions in a minute that means 1 revolution in 40 minutes. The circumference of the moon is 3,476,000 meters. That means in the habitat we are going 86900 meters in a minute. That is 5214 km/hour.

Oh that is fast you say. Too fast, oh my. Things will break. The ship canna take much more. Pish posh I tell you! This is the moon! There are no pesky bugs to splat on the windshield. And we have a maglev track. The rotating habitat will be spinning like a fashionable belt about the midsecton of the moon.

You will have to get up to speed before you get on. There will be special places to do that.

  • $\begingroup$ The first step, the moon has to spin fast enough to give everything just below the surface an acceleration that equals escape velocity. Then, the spin needs to be increased sufficient to create the required artificial gravity.. $\endgroup$ Dec 23, 2021 at 0:06
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    $\begingroup$ @KerrAvon2055 - The 0.2G is the moon's gravitational pull. Then the remaining 1 is earth gravity. Passengers in the spinning moon belt experience 1 G as they walk on the ceiling. $\endgroup$
    – Willk
    Dec 23, 2021 at 0:17
  • $\begingroup$ Mea culpa - previous comment withdrawn $\endgroup$ Dec 23, 2021 at 2:31
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    $\begingroup$ @PaŭloEbermann - you don't get the same scenic view out the windows. $\endgroup$
    – Willk
    Dec 23, 2021 at 19:21
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    $\begingroup$ A track all the way around the moon to get enough speed for 1 g?!? That's completely insane! I love it. How soon can we get started building it? $\endgroup$ Dec 26, 2021 at 14:40

An intuitive explanation

If you're spinning the Moon, you want apparent gravity to point away from the core, into space. That's fine, except that it's not only your habitat which experiences that apparent gravity. Rocks on the surface also "want to fall", and if things are falling away from the core ... the rocks will "fall" into the sky.

That becomes a runaway process, in fact - anything which "falls away" reduces the mass of the Moon, decreases its gravitational pull, and lets more material pull away faster. (This is a mix of escape velocity and the Roche limit.) See also this.

Such a disintegration was a plot point in part of the Long Earth series.

There are two interesting ways to handle this. One is to glue the surface together, but at that point you might as well build your own habitat - even the risks are the same. The other is to core the Moon and put your habitats inside, then have them rest on the stationary rock above to not fall into space, which supports the rock at the same time. You'll want a frictionless set of rails. The precise dimensions will affect your needed speed, but that's given by the centrifuge-calculation answers.


Wayfaring Stranger gives you the math on why your idea is impractical (impossible, without super materials).

A more practical way to use rotation to increase apparent gravity on the moon, would be to have rotating habitats in the shape of a frustum (cylinder with one end wider than the other; much wider in this case).

Basically you would embed a misshapen O'Neil Cylinder in the crust of the moon (narrow end down) so that the combination of the force of gravity from the Moon and centripetal from from the rotating habitat align and result in an apparent 1G within the habitat.


So this has kind of already been addressed but here we go: The problem with the question is that centrifugal force is the opposite of gravity, in that increasing the centrifugal force of the moon would lessen gravity. In order for this to work, the people in the caverns would need to be standing on the ceiling of those caverns.

If you think of references to this in media such as Interstellar, Cowboy Bebop etc the part of the space station that has gravity is the outer wall but only on the inside, with a person's feet pointing toward space and their head pointing to the rotation point. This makes sense too, take a string/belt, hold it over your head and spin it - the centrifugal force pushes outwards.

The other part as mentioned by others is the stress placed on the moon's surface. Needing to bind the surface together would probably be the first issue to be addressed.


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