For accuracy, it’s nice to know how fast a rotating circular (or toroidal) space station would have to be rotating to simulate Earth-like gravity for its occupants. After some searching, I finally dug up a concise and relevant Wikipedia article that helps explain exactly that.

The rotating wheel space station wiki says:

Occupants of the station would experience centripetal acceleration according to the following equation,

\[a = − ω ^ 2 r\]

where \(ω\) is the angular velocity of the station, \(r\) is its radius, and \(a\) is linear acceleration at any point along its perimeter.

Notice that the right-hand side is negative – this is because gravity generally pulls us down, so the typical acceleration we experience is actually \(-9.81 \mathrm{m/s^2}\).

The goal here, for me, is to check the required angular velocities required to adequately spin stations of a certain radius. Is a station with a radius of \(100\mathrm{m}\) even plausible? Let’s find out!

For reference, \(100\mathrm{m}\) is slightly larger than the fictional Serenity ship, which is \(82\mathrm{m}\) in length.

First let’s solve for \(ω\), which will give us radians per second, the standard unit of angular velocity.

$$ a = − ω ^ 2 r \\ - a = ω ^ 2 r \\ - a / r = ω ^ 2 \\ \sqrt{- a / r} = ω \\ ω = \sqrt{- a / r} $$

So if we plug in our values for \(100\mathrm{m}\) (and Earth gravity):

\[ω = \sqrt{- a / r} \\ ω = \sqrt{\frac{- (-9.81\mathrm{m/s^2})}{100\mathrm{m}}} \\ ω = \sqrt{\frac{9.81 \mathrm{m/s^2}}{100\mathrm{m}}} \\ ω = \sqrt{0.0981 \mathrm{/s^2}} \\ ω = 0.313 \text{ radians per second}\]

Almost there! But what I really want to know is how many seconds per revolution that is. This is called the period and gets a new variable, \(T\). Remember there’s \(2 π\) radians in a revolution, of course, so we just need to divide, and then invert:

\[ω = \frac{2 π}{T} \text{ (standard formula)} \\ T = \frac{2 π}{ω} \\ T = \frac{2 π}{0.313} \text{ seconds per revolution} \\ T = 20.1 \text{ seconds per revolution}\]

To get the original formula in terms of seconds per revolution, we can do that too:

\[ω = \sqrt{- a / r} \text{ (angular velocity in radians per second)} \\ T = \frac{2 π}{\sqrt{- a / r}} \text{ (period in seconds)}\]

So for our \(100\mathrm{m}\)-radius station, it would have to rotate once every 20 seconds to simulate Earth’s gravity. That’s not too bad.

What about for a larger station, like the (also fictional) Deep Space 9 station, which is about \(1450\mathrm{m}\) in diameter?

\[T = \frac{2 π}{\sqrt{- a / r}} \\ T = \frac{2 π}{\sqrt{\frac{9.81\mathrm{m / s ^ 2}}{725\mathrm{m}}}} \\ T = \frac{2 π}{0.116\mathrm{/s}} \\ T = 54.0 \text{ seconds per revolution}\]

So it takes longer (which makes sense), but not linearly longer. The Deep Space 9 station is 7.25 times bigger than the first one, but only needs to spin 2.69 times slower. This is because the radius is in a square root in the denominator; \(\sqrt{7.25}\) is actually 2.69! This means for a station with four times the radius, the required angular spin speed will drop by half. Let’s confirm this just to be sure: we’re expecting a station with radius \(400\mathrm{m}\) (\(100 × 4\)) to require about \(40.2\mathrm{s}\) (\(20.1 × 2\)) per revolution, based on our first station calculation.

\[T = \frac{2 π}{\sqrt{- a / r}} \\ T = \frac{2 π}{\sqrt{\frac{9.81\mathrm{m / s ^ 2}}{400\mathrm{m}}}} \\ T = \frac{2 π}{0.157\mathrm{/s}} \\ T = 40.1 \text{ seconds per revolution}\]

Cool, well, that answers that!

Linear Velocity

One more random thought – I saw somewhere that ships would want to dock in the center of the spinning ring because the outside would simply be moving too fast to dock with. How fast would that be, exactly? Let’s consider our smallest station, the \(100\mathrm{m}\) one.

\[ω = \sqrt{- a / r} \\ v = ω × r \text{ (standard formula)} \\ v = r\sqrt{- a / r}\]

Eh, \(r\) is in there twice but I don’t think it’s reasonable to factor that out further.

So let’s plug in our value for \(r\):

\[v = r\sqrt{- a / r} \\ v = 100\sqrt{9.81 / 100} \\ v = 31.3\mathrm{m/s}\]

That’s roughly \(113 \mathrm{km/h}\) or \(70.0 \mathrm{mph}\).

Let’s try our other two stations:

\[v_{400} = 400.\sqrt{9.81 / 400.} = 62.6\mathrm{m/s} = 225 \mathrm{km/h} = 140. \mathrm{mph} \\ v_{725} = 725\sqrt{9.81 / 725} = 84.3\mathrm{m/s} = 304\mathrm{km/h} = 189 \mathrm{mph}\]

Each of these are somewhat fast, by land speed standards – in excess of 100 mph. If you planned to dock ships on the outside of the ring, you’d need to take the following into account:

  • Ships would need to have their docking ports on the side of the ship, perpendicular to their main means of thrust (as opposed to the front or back). While it’d be trivial for a ship to reach 300 km/h going straight forward, going 300 km/h sideways, if the docking port was on the front of the ship, seems less attainable.
  • Depending on the mass of the station and the mass of the ship, docking on the outer ring would shift the center of gravity (and center of spin) and could introduce wobble. Then again, using the fictional Serenity and Deep Space 9 entities for comparison, the Firefly is 265 metric tons, while Deep Space 9 is 45 million metric tons1 – 170,000 times larger. So in that particular case, the effect would likely be negligible on a day-to-day basis.
  • The simulated gravity we’ve worked so hard to create would be working against you – gravity would be slightly higher than Earth-normal outside the radius of the station, and then you’d have to transport the cargo to whatever other part of the station it needs to go, again likely in normal gravity.
    • If, instead, the ship had docked in the middle of the ring, cargo could then be unloaded and gently “fall” into the desired section of the outer ring – assuming the outer ring was connected to the center with spokes.
  • On the plus side, the outer ring can accomodate far more ships simultaneously than the center of the ring could accomodate (which would only accomodate two ships – one on the top of the ring, one on the bottom.)
    • A compromise could be possible: if ships could instead dock with spokes instead of the hub, it would accomodate more ships, but would still require some amount of thrust to align while inside the ring.
    • A more advanced design might have movable docking ports on the station: dock at the center hub, and then the station would slowly move the docking port (with ship attached) down one of the spokes to allow the next ship to dock. Then you’d be limited by the number of spokes – each spoke could accomodate two ships (one each for the top and bottom). A four-spoke station could accomodate eight ships simultaneously.
  • Docked ships could benefit from the simulated gravity, but their docking port would have to be in a particular position. This position would probably be on the top of the ship, with the bottom of the ship pointing directly outwards, away from the center of the station.
    • While this would be fairly trivial when docking on the outside of the ring, docking near the hub this way would be more difficult. Instead of docking the ship directly to the station, there’d likely need to be an intermediate docking arm that was perpendicular to the station. But then this would rotate with the station as well (making it hard to dock with), unless it was rotating to track the ship, bu then docking and stopping the rotation would apply torque to the station, and…station design is complicated.

1: 45 million metric tons seems like quite a lot, even for a made up number. Is it gravitationally significant? The formula for gravity is \(F = G\frac{m_1 m_2}{r^2}\), which if we plug in the values \(6.67 × 10^{−11} (\frac{265,000 × 45,000,000,000}{725^2})\) yields \(1.51 N\). If we plug that into \(a=F/m\) (Newton’s second law of motion) we get \(1.51 / 265,000 = 5.70× 10^{−6} m/s^2\), which is…quite tiny, actually. 0.00006% of Earth’s gravity. So, no, clearly not gravitationally significant.