How fast would a space station have to spin to reproduce Earth’s gravity?

-  1783  -  How fast would a space station have to spin to reproduce Earth’s gravity?  This review goes through the math for centripetal force in a rotating space station.
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-----------------  1783  -  How fast to spin to reproduce gravity?
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-  Jeff’s question:  How fast would a satellite ( space station ) have to spin to reproduce gravity the same as felt on the surface of the Earth?
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-  This is a simple problem of calculating Centripetal Force.  Here’s how:
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-  Say we have a space station that is a hollow circular tube that has a radius to the outer wall of 50 meters.  We rotate the orbiting tube at some constant speed.  What speed is needed so the astronaut walking on the outside wall weighs the same as he does back home?
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-  The acceleration of gravity on the surface of Earth is 32 feet per second per second.  Regardless of mass all things fall at that same acceleration:
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-----------------------------  a  =  32 ft/sec^2
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-  When an object in circular motion is at a constant rotating speed it is constantly accelerating because its velocity is constantly changing direction.  Speed is the absolute magnitude , but , velocity is a vector with magnitude and direction.
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---------------------------  Speed  =  “v”
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---------------------------  Radius  =  “r”
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-  The vector of acceleration is “a”
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--------------------------  a  = - v^2 / r
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-  The minus sign means the vector is pointing outward for the center of the circular motion.  Newton’s second law of motion is Force  =  mass * acceleration.
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--------------------------  F  =  m * a
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---------------------------  a  =  v^2  /  r
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------------------------  F  =  - m* v^2  /  r
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-  The astronaut in the space station standing on the outside wall of the tube has a mass “m”
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-  The space station radius is “r”  (50 meters)
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-  The speed of rotation is a constant  = “v”
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------------------------  a  =  m * v^2 / r
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-  On the surface of Earth the acceleration of gravity is called “g”,  G-forces.  Therefore:
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------------------------  a  =  m * g
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-  Put the two equations equal to each other:
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------------------------  m * g  =  m * v^2  / r
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-  This equation  reduces because mass cancels from both sides of the equation.  All mass falls at the same acceleration.  Drop an iron bowling ball and a plastic bowling ball from the tower of Pizza and they both hit the ground at the same time. So, would a feather, but you would have to remove all the air.
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----------------------------  g  =  v^2 / r
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----------------------------  v^2  =  g * r
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-  Converting “g” to the metric system:  g  =  32 feet / sec^2  =  9.8 meters / sec^2
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------------------------  v^2  =  9.8 m / s^2  *  50 meters  =  490 m^2 s^2
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-----------------------  v  =  22 m / sec
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-  That is about 50 miles per hour.  22 m/sec  *  mile /  1610 m  * 3600 sec / hour  =  49.5 miles / hour.
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-  That is spinning pretty fast.  I think astronauts will need to get used to operating in a little less gravity.  Work would be a lot easier.  Requires less energy to move around.
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-  Jeff, does that answer your question?
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