Space station designed like a donut outer radius 460 m with what period must the station rotate so that a person sitting in the outward wall experiences artificial gravity
To determine the period at which the space station must rotate in order for a person sitting in the outward wall to experience artificial gravity, we can use the formula for centripetal acceleration:
a = (v^2)/r
Where a is the acceleration, v is the velocity, and r is the radius of rotation. The acceleration is equal to the gravitational acceleration on Earth (9.8 m/s^2) in this case.
Since the person sitting on the outward wall of the space station should experience artificial gravity, the acceleration should be equal to 9.8 m/s^2.
Let's assume the person is at a distance of 460 m from the center of rotation (the outer radius). Hence, the radius (r) is 460 m.
We can rearrange the formula to solve for velocity (v):
v = sqrt(a * r)
Plugging in the values, we get:
v = sqrt(9.8 m/s^2 * 460 m) ≈ 67.85 m/s
Now we can use the formula for period (T) in terms of velocity (v) and radius (r):
T = (2 * pi * r) / v
Plugging in the values, we get:
T = (2 * pi * 460 m) / 67.85 m/s ≈ 43.14 s
Therefore, the space station must rotate with a period of approximately 43.14 seconds for a person sitting in the outward wall to experience artificial gravity.