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Question: How fast would something spin to create earth like gravity from centrifugal force?

(Posted by: satsumo on 2009-06-21 13:03:28)

In the 2001 movie, a space station is shown spinning. I assume the concept is to simulate gravity by using centrifugal force. How fast would something need to spin to create Earth like gravity in this way? Does it depend on the radius of the object?

Posted by: Andrew S on 2009-06-21, 13:14:58

Yes, it depends on the radius. Centri_petal_ force is given by mv^2/ r, where m is in kilograms, v is in metres per second and r is in metres. As you can see the force needed is proportional to the mass but that is the same as for gravity. In practice the radius needs to be fairly large, both to ensure than the "gravity " does not vary too significantly with a small change in height and also to keep the RPM down to a level that does not induce dizzyness.

Posted by: campbelp2002 on 2009-06-21, 13:31:33

I did a rough calculation and figured a 1,000 foot diameter wheel (the 2001 space station looks about that big to me) would have to do about 2 RPM (one rotation in 30 seconds) to make 1G. That is about how fast it looked in the movie. This is not by chance. They had science consultants to get as realistic as possible, and I am sure they calculated the correct rotation rate and size and show it that way in the movie on purpose. Larger diameter can rotate slower. An 8,000 mile diameter space station would need to make one rotation in 90 minutes. I pick 8,000 miles because that is Earth's diameter. If Earth rotated once in 90 minutes instead of once in 24 hours, objects at the equator would be weightless.

Posted by: anordtug on 2009-06-21, 13:44:34

Rotation do not crate gravitation. Gravitation is a unique force created by mass.

Posted by: Slowfinger on 2009-06-22, 05:49:13

Principle behind that is fairly simple. Inside a room that spins fast, like park ride described here en.wikipedia.org/ wiki/ Rotor_(ride) centrifugal acceleration presses you against the inner wall. To simulate gravity, centrifugal acceleration needs to be equal to acceleration of gravity g. acf = g v^2 / R = g where v is linear velocity on the rim and R is radius of a wheel. v = 2 pi R / T where T is period of rotation i.e. time needed to make one turn, T= 2pi sqrt (R/ g) and since g= 9.81ms^-2, T= 2 sqrt(R) This means if radius of a wheel is 100m, it should make one turn in every T= 2 sqrt(100)= 20s to simulate the effect of gravity.

Posted by: Meklar on 2009-06-22, 08:06:37

It depends on the radius. The equation is as follows: a= v^2/ r where a is the accelertion, v is the true speed of rotation (NOT the angular speed) and r is the radius from the center of the rotation. For Earth gravity we set a= 9.81m/ s^2, take any arbitrary r we like (depending on the size of the space station in question) and then solve for v as follows: v= √(ar)