The magnitude of the artificial gravitational acceleration provided to a space tourist walking on the inner wall of the station is 5.6 m/s²
Explanation:
Given-
Diameter, d = 185 m
radius, r = d/2 = 185/2 = 92.5 m
Frequency, f = 2.35 rev/min = 2.35/60 rev/sec
Magnitude of gravitational acceleration, a = ?
Centripetal acceleration, a = v²/r -1
and v = 2πrf
Substituting the value of v in equation 1
a = (2πrf)²/r
a = 2πrf X 2πf
a = 4π²rf²
a = 4 X 3.14 X 3.14 X 92.5 X 2.35 X 2.35 / 60 X 60
a = 5.6 m/s²
Therefore, the magnitude of the artificial gravitational acceleration provided to a space tourist walking on the inner wall of the station is 5.6 m/s²
