A rotating space station has radius 1310 m, measured from the center of rotation to the outer deck where the crew lives. What should the period of rotation be if the crew is to feel that they weigh one-half their Earth weight?
v^2/R =9.81/2
but v = 2 pi R/T
so
4 pi^2 R^2/(T^2 R) = 9.81/2
T^2= (8/9.81) pi^2 R
To determine the period of rotation required for the crew to feel one-half their Earth weight, we can use the concept of apparent weight experienced in a rotating reference frame.
In a rotating space station, the crew experiences a centrifugal force outward due to the rotation, which counteracts the force of gravity. This apparent outward force reduces the crew's weight. The crew will feel one-half their Earth weight when the outward centrifugal force equals half the force of gravity.
To calculate the period of rotation, we can apply the following formula:
Fc = 1/2 * mg
Where Fc is the centrifugal force, m is the mass of an object, and g is the acceleration due to gravity.
In this case, the mass of the crew doesn't affect the calculations, so we can ignore it. We'll use the Earth's acceleration due to gravity, which is approximately 9.8 m/s^2.
Fc = 1/2 * mg
Fc = 1/2 * (m * g)
Fc = 1/2 * (m * 9.8)
Fc = 4.9 * m
Now, the centrifugal force can be related to the acceleration due to the rotating motion:
Fc = m * (ω^2 * r)
Where ω is the angular velocity in radians per second, and r is the radius of the rotating space station.
Applying these equations, we have:
4.9 * m = m * (ω^2 * r)
The mass (m) cancels out on both sides of the equation, leaving us with:
4.9 = ω^2 * r
Solving for the angular velocity (ω), we get:
ω^2 = 4.9 / r
ω = sqrt(4.9 / r)
Now, the period of rotation (T) is related to the angular velocity (ω) by the equation:
T = 2π / ω
Substituting the value of ω, we get:
T = 2π / sqrt(4.9 / r)
T = 2π * sqrt(r / 4.9)
Now we can substitute the given radius of the space station (r = 1310 m) into the equation to find the period of rotation (T):
T = 2π * sqrt(1310 / 4.9)
T ≈ 2π * sqrt(267.35)
T ≈ 2π * 16.358
T ≈ 102.65 seconds
Therefore, the period of rotation for the crew to feel one-half their Earth weight in this rotating space station with a radius of 1310 m should be approximately 102.65 seconds.