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Q & A: Rotating space station story

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Most recent answer: 04/24/2013
Q:
For a story I'm writing I have a space station about 500 meters long with a diameter of about 100 meters. How does one figure out how fast the station has to rotate (correct term?) around it's long axis so that it has 1 Gravity around the circumference inside? Thanks.
- David (age 65)
Oceanside, CA, USA
A:

Hello David,
Newton's second law states that F = m a or, equivalently  a = F/m,   where m is the mass of an object,  a  is the acceleration  and F is the force acting on the body.   Now on earth, the force of gravity produces an equivalent acceleration, g = 9.8 meters per sec2.   So what you have to do for your space ship is to have the rotational acceleration at the circumference equal to g.  For an object traveling in a circle the acceleration is given by  a = (2πf)2 R where f is the rotational frequency and R is the radius.  Putting it all together you get: f = sqrt(g/R)/2π.
The answer is independent of the length of the ship.  Good luck on your story.

LeeH

 

Hi David,

I just saw your question, and wanted to add a word of warning.

As you may know, observors in a rotating reference frame feel a force called the Coriolis force, which gives rise to an acceleration a = 2*v × ω, where v is the velocity of the object and omega is the rotation rate of the reference frame. On the earth, for example, this force deflects projectiles, and can have a significant effect on missiles or airplanes. 

In your space station world, ω is 0.44 radians/second, much larger than the earth's 0.000073 radians/second. So, the Coriolis force is correspondingly over 6000 times greater! If you fired a bullet perpendicular to your space station's axis of rotation, it would accelerate sideways at over 40 times the force of Earth gravity, and person running at top speed around the circumference of your station would feel a sideways force half as strong as gravity!

Another fact to keep in mind is that the gravity will depend highly on the distance from the station's axis. For example, someone 10 meters up from the "surface" (say, on the third story of a house) would feel a decrease in effective gravity of ~20%. In fact, a professional baseball pitcher could throw a baseball so high that it would reach the axis, at which point it would feel negative effective gravity and be attracted towards the opposite side of the space station (assuming he compensated correctly for the Coriolis force).

If you aren't careful, these effects might spoil some of your plot. If you are careful, it might make for some excitement of its own! ;)

Cheers,

David Schmid


(published on 04/24/2013)

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