Artificial gravity is produced in a space station by rotating it, so it is a noninertial reference frame. The rotation means that there must be a centripetal force exerted on the occupants; this centripetal force is exerted by the walls of the station. The space station in Arthur C. Clarke's 2001: A Space Odyssey is in the shape of a four-spoked wheel with a diameter of 150 m. If the space station rotates at a rate of 2.45 revolutions per minute, what is the magnitude of the artificial gravitational acceleration provided to a space tourist walking on the inner wall of the station?

Question

Amber Bass

Physics

24 March, 00:13

Artificial gravity is produced in a space station by rotating it, so it is a noninertial reference frame. The rotation means that there must be a centripetal force exerted on the occupants; this centripetal force is exerted by the walls of the station. The space station in Arthur C. Clarke's 2001: A Space Odyssey is in the shape of a four-spoked wheel with a diameter of 150 m. If the space station rotates at a rate of 2.45 revolutions per minute, what is the magnitude of the artificial gravitational acceleration provided to a space tourist walking on the inner wall of the station?

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Nathan Wall · Answer 1

centripetal acceleration = ω² R

ω is angular speed of rotation and R is radius of circular path

ω = 2π n

n is no of rotation per sec

= 2.45 / 60

ω = 2π x 2.45 / 60

=.2564 rad / s

centripetal acceleration = ω² R

=.2564² x 150/2

= 4.93 m / s²

gravitational acceleration provided = 4.93 m / s²