One problem for humans living in outer space is that they are apparently weightl

Question

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates "artificial gravity" at the outside rim of the station.

Part A

If the diameter of the space station is 900 m , how many revolutions per minute are needed for the "artificial gravity" acceleration to be 9.80m/s2?

Part B

If the space station is a waiting area for travelers going to Mars, it might be desirable to simulate the acceleration due to gravity on the Martian surface (3.70m/s2). How many revolutions per minute are needed in this case?

Explanation / Answer

A)

The circumference can be used to obtain the radius of the space station

2R = 900 m

R = 143.312 m

The acceleration a for a rotational motion in terms of angular velocity

a = g(in this case) = v^2/r = r^2

= sqrt(g/r) = 0.2615 radians/sec

The number of revolution required is the frequency of 2 rotation in a minute

N = 60sec/(2) = 0.2615 60/(6.28) = 2.5 rev/min

B)

We adopt the same procedure as we did for Part A except that the acceleration in now amars = 3.7 m/sec2

= sqrt(a/r) = 0.1606 radians/sec

N = 60sec/(2) = 0.1606 60/(6.28) = 1.5363 rev/min

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