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How fast (linear speed) would the rim of each craft need to be travelling at?

In the film the astronaut, Dave Bowman, is seen jogging around the inside of the Discovery One cabin. Assuming that he can jog at 12 km/h will it have any effect on the way he feels? Will the direction of his jogging have any effect?

Assume that Discovery One and the space station had diameters of 17 m and 600 m respectively. Take g as 9.8 m s-1.

Use the equation for centripetal acceleration and rearranging:

 Equation.

But

 Equation.

So

 Equation.

For the space station r = 300 m, so

 Equation.

Or about 35 seconds per revolution.

For Discovery One however:

 Equation.

Or about 6 seconds per revolution.

For the speed of the rim:

 Equation.

For the space station:

 Equation.

And for Discovery One:

 Equation.

This is quite a low speed and leads to an interesting effect. Dave Bowman runs at 12 km/h which is approximately 3.3 m s-1 and this is an appreciable fraction of the rotational velocity. If he runs in the direction of rotation he will have a net speed of 12.46 m s-1 but if he runs against the rotation it will only be 5.79 m s-1. Putting these speeds back in to the equation we can see that his accelerations whilst running are:

 Equation.

Or about 1.86 g and when running against the rotation:

 Equation.

Or about 0.40 g. Even worse David Bowman’s height is an appreciable fraction of the radius of rotating section. Assuming that he is around 1.85 m tall then the centre of his head (and so the balance organs in his inner ears) will be around 1.7 m above the deck and so only 6.8 m from the centre of rotation. Putting these figures back in to the equation gives accelerations of 1.59 g and 0.28 g respectively. Whilst nobody has yet attempted to put an artificial gravity centrifuge in to orbit yet it is clearly more involved that just installing a giant hamster wheel.

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