Futurists project large space stations which would rotate so as to use centripetal acceleration to simulate gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments such as the current International Space station. (a) If the space station were 200. m in diameter, what velocity would produce an “artificial gravity” of 9.80 m/s2 at the rim? (b) What is this value in revolutions per minute?
To calculate the velocity required to produce an "artificial gravity" of 9.80 m/s² at the rim of the rotating space station, we can use the formula for centripetal acceleration:
a = (v² / r)
where:
a is the centripetal acceleration
v is the velocity
r is the radius (half of the diameter)
(a) Velocity calculation:
Given:
a = 9.80 m/s²
r = 200 m / 2 = 100 m
Rearranging the formula, we can solve for v:
v = √(a * r)
Substituting the given values:
v = √(9.80 m/s² * 100 m)
v = √(980 m²/s²)
v = 31.30 m/s
Therefore, a velocity of approximately 31.30 m/s would produce an "artificial gravity" of 9.80 m/s² at the rim of the rotating space station.
(b) To convert this value to revolutions per minute (rpm), we need to know the circumference of the rotating space station. The circumference is equal to the distance traveled in one revolution.
Circumference = π * diameter = π * 200 m
To convert from meters per second (m/s) to revolutions per minute (rpm), we need to convert the velocity to meters per minute (m/min):
31.30 m/s * 60 s = 1878 m/min
Since the circumference is measured in meters, we can set up a proportion to find the value in rpm:
31.30 m/s = 1878 m/min
1 revolution/sec = x revolutions/min
x = (1878 m/min) / (31.30 m/s)
x ≈ 59.96 rpm
So, the velocity of 31.30 m/s would be approximately equal to 59.96 revolutions per minute (rpm) in the rotating space station.