Gravity Variation Along a Rotating Space Station

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SUMMARY

The discussion focuses on calculating the perceived gravity along the spokes of a rotating space station, similar to the design in "2001: A Space Odyssey." The centripetal force equation, F = mω²r, is used to determine that gravity increases linearly from the center to the outer ring. Specifically, at 25 meters from the ring, the gravity is 0.75g; at 50 meters, it is 0.5g; at 75 meters, it is 0.25g; and at 99 meters, it is 0.01g. This linear relationship is crucial for understanding gravitational variation in rotating systems.

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enzi
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This may seem like a very simple question but given my lack of knowledge in the area, I'm stumped:

If we have a rotating "space station" a la 2001: A Space Odyssey where the outer ring is at normal Earth gravity, what is the factor to determine how much gravity would be at various points along the "spokes"? So let's say the "spoke" is 100 metres long, what would the gravity be 25 metres from the ring, 50 metres from the ring, 75 metres from the ring, and 99 metres from the ring (1 metre from the hub)?

Thanks in advance for your help on this and I apologize for being so dense that I can't figure this out on my own! :)
 
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In terms of angular speed ω, the centripetal force needed to keep an object of mass m moving in a circle of radius r is F = mω2r. So the perceived "gravity" along a spoke increases linearly from the center (where it is zero) to the rim.
 
Linearly! That one I wouldn't have guessed. So we can safely say that 25 metres from the ring we have .75g, 50 metres from the ring we have .5g, 75 metres from the ring we have .25g, and 99 metres from the ring we have .01g. Thanks!
 

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