Showing My Math


So here’s the deal. I’m working on a new project and I want to make sure at least some of the details have actual technical grounding. I’m okay with a little handwavium, it’s probably unavoidable, but I want to at least have some grounding in reality. Problem is, I’m not all that great at actual technical grounding, as the last physics class I took was non-AP physics in high school where I barely got a B. So I might occasionally make these posts, I might make just this one, in an attempt to crowdsource some of my equations. The questions I have are whether I’ve got the right equations, and whether I’m using them correctly, and to also play around with LaTeX a little. But mostly the first two things.

So here’s today’s problem: Given a cylindrical space ship with an internal radius of 6km, how fast must it be rotating to create a centripetal acceleration equivalent to earth gravity for someone standing on the inside surface? I didn’t know any of these equations, but found them at this rather helpful forum post. First, we must find the speed at that 6km point that would produce an acceleration of 9.8m/s²:

a=\frac{v^2}{r}
a=9.8 \frac{m}{s^2}
r=6000 m
9.8=\frac{v^2}{6000}
9.8*6000=v^2
58800=v^2
v=242.49 \frac{m}{s}

That number sounds awfully damn fast, but consider the speed of rotation of the earth at sea level on the equator is roughly 465 m/s. Next step, at least what I’m assured is the next step, is converting this into radians/second:

W=\frac{v}{r}
v=242.49
r=6000
W=\frac{242.49}{6000}=0.0404

Finally this can be converted to revolutions per minute. The conversion formula I found is:

1 \frac{rad}{s} = \frac{60}{2pi} = \frac{30}{pi} rpm
0.0404 \frac{rad}{s} = \frac{30*0.0404}{pi}=\frac{1.212}{pi}=0.386 rpm

Therefore the ship is rotating at a rate slightly faster than once every three minutes. What I didn’t expect is that, since the rate of rotation is a constant, centripetal acceleration increases linearly from the axis of rotation. I’m so accustomed to formulas for gravity having squares all over the place, but this isn’t, strictly speaking, gravity. It’s an acceleration equal to gravity. So at half the distance from the axis of rotation, we can work backwards with W as a constant…

W=\frac{v}{r}
0.0404=\frac{v}{3000}
v=0.0404*3000=121.2\frac{m}{s}

a=\frac{v^2}{r}
a=\frac{121.2^2}{3000} = \frac{14689.44}{3000} = 4.9 \frac{m}{s^2}

Which is equivalent to half gravity.

My next trick will be to find a formula that describes the rate of descent for a body falling through linearly increasing gravity. That’s less likely to come up in-story, but more for my own curiosity.

Edit: Some further poking around (which, I’m ashamed to say, has mostly been at Wikipedia so far) suggests that 2rpm is about the maximum rotation that most humans can adjust to with no ill effects, so my rotation of nearly 1/6 that rate is shockingly safe in and of itself. So that’s good to know. Now if only it didn’t have a “citation needed” tag.

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  1. avatar

    #1 by John McMullen on March 22, 2013 - 2:08 pm

    Oddly enough, I used to keep a set of equations like this on a page in Geocities, but lost it when Geocities closed. (Oh, wait: apparently Oocities, which I haven’t heard of, keeps an archive of Geocities. The PDF is at http://www.oocities.org/area51/chamber/2838/fm_wbcookbook.pdf )

    Given that someone in a space station is essentially weightless (falling around the primary), and the illusion of gravity is caused by the deck under them slamming up against them at 9.8 meters per second, there might be some question about whether they really “fall” on the space station. (And I ask this in all seriousness: my degree was in biology, and we didn’t have to take much physics for it.)

    Obviously, something happens: you’re moving around the circle at some tremendous speed, and when you are let loose, you travel in some tangent to your course: coriolis force instead of falling. But I lack the knowledge to say whether the effect is almost identical to falling in a gravity field or whether there’s a noticeable difference.

    It might be that instead of falling, you slam into the opposite end of the tube you’re falling down and then you begin some process that’s essentially the same as falling. Or maybe not; I am not educated in the right way to know.

    • avatar

      #2 by DLThurston on March 22, 2013 - 2:19 pm

      I keep hitting the same point. Not really understanding how centripetal acceleration works, having instead only found some formulas for computing it, I have no idea where the force lies either. That’s something else I’m going to need explained to me. When someone jumps into the air, what brings them back down? A ball released just outside the axis of rotation, what initially pushes it towards the outside. Clearly something or rotation wouldn’t be such a common trope for replacing gravity. But it might as well be invisible gremlins for all I know.

    • avatar

      #3 by DLThurston on March 22, 2013 - 2:22 pm

      Looks like you’re actually pretty close with the “slamming up against them” notion. Per wikipedia:

      A rotating spacecraft will produce the feeling of gravity on its inside hull. The rotation drives any object inside the spacecraft toward the hull, thereby giving the appearance of a gravitational pull directed outward. Often referred to as a centrifugal force, the “pull” is actually a manifestation of the objects inside the spacecraft attempting to travel in a straight line due to inertia. The spacecraft’s hull provides the centripetal force required for the objects to travel in a circle (if they continued in a straight line, they would leave the spacecraft’s confines). Thus, the gravity felt by the objects is simply the reaction force of the object on the hull reacting to the centripetal force of the hull on the object, in accordance with Newton’s Third Law.

      That’s still a little Greek to me, but I’m getting closer.

  2. avatar

    #4 by John McMullen on March 22, 2013 - 2:17 pm

    (Oh: the rotational velocity to rpm conversion seems right; I went at it in a different way, calculating the circumference of a circle of 6 km radius and then figuring how long it takes to make one rotation. Allowing for rounding errors, it’s 0.386 rpm.)

    Using my version of the equation, I get the same rotational speed.

  3. avatar

    #5 by Day Al-Mohamed on March 23, 2013 - 2:06 am

    It may be slightly “odd” that the blind person is the visual learner BUT I thought this an interesting illustration of some of what you’re talking about and seeing it on a real human being even in a variant form was really illuminating.

  4. avatar

    #6 by Day Al-Mohamed on March 23, 2013 - 2:10 am

    If you skip the weird text and fancy pages and get to the professor’s lecture, this is awesome. I’m just cranky that I couldn’t find the whole lecture.

    • avatar

      #7 by DLThurston on March 23, 2013 - 1:11 pm

      Ask and ye shall receive. The portion excerpted in the above video starts around 26:30

  5. avatar

    #8 by Day Al-Mohamed on March 23, 2013 - 2:14 am

    Btw…just wanted to say, I LOVE NASA videos. :)

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