New answers tagged centrifugal-force
Ignoring the Coriolis effect, which is velocity dependent. I get for an x-y Cartesian coordinate system, with the y component toward the North Pole and x component pointing out from the equator at the position you are measuring the effective g=G , the following (in ordered pair vector representation); G= ...
The technique would work, but whether you would use it would depend on many many details. Rotating frames of motion come with all sorts of counter intuitive bits, so lets look at it from an inertial frame outside the station. We perceive the station as rotating. The velocity vector of any object on the station is a tangent, so we see the result of every ...
Assume the vehicle is already there before the space station is built. So it is floating. If it is floating slightly above space station ground, this does not change if the station starts to spin (neglecting its acceleration due to air friction). In the reference frame of the station it will move with the velocity of the outer cylinder. If you get in the ...
You're presumably talking about objects dropped away from the equator. At the equator it causes a lower effective gravity. Away from the equator it causes a deflection towards the equator.
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