Question

In designing rotating space stations to provide for artificial-gravity environments, one of the constraints that must be considered is motion sickness. Studies have shown that the negative effects of motion sickness begin to appear when the rotational motion is faster than approximately 1.65 revolutions per minute. On the other hand, the magnitude of the centripetal acceleration at the astronauts' feet should equal the magnitude of the acceleration due to gravity on earth. Thus, to eliminate the difficulties with motion sickness, designers must choose the distance between the astronaut's feet and the axis about which the space station rotates to be greater than a certain minimum value. What is this minimum value

Answer 1

In this case, the minimum value for the distance between the astronaut's feet and the axis of rotation should be approximately 97.24 meters

Step-by-step explanation:

To determine the minimum value for the distance between the astronaut's feet and the axis about which the space station rotates, we need to consider two factors: the threshold for motion sickness and the requirement for centripetal acceleration to equal the acceleration due to gravity on Earth.

1. Motion Sickness Threshold: Studies have shown that motion sickness begins to appear when the rotational motion exceeds approximately 1.65 revolutions per minute. This means that the space station should rotate at a speed slower than this threshold to avoid inducing motion sickness in the astronauts.

2. Centripetal Acceleration: In order to simulate the effects of gravity on Earth, the magnitude of the centripetal acceleration at the astronauts' feet must be equal to the acceleration due to gravity on Earth. This ensures that the astronauts feel a similar force to what they experience on Earth.

To calculate the centripetal acceleration, we can use the formula:

Centripetal acceleration = (angular velocity)² * radius

Now, let's assume that the minimum value for the distance between the astronaut's feet and the axis of rotation is "r" (in meters). We can use this value to calculate the required angular velocity.

Since we want the centripetal acceleration at the astronaut's feet to equal the acceleration due to gravity on Earth, we can equate the two:

(angular velocity)² * r = 9.8 m/s²

To find the minimum value of "r," we need to solve for it. Rearranging the equation, we have:

r = (9.8 m/s²) / (angular velocity)²

Since we know that the rotational motion should be slower than 1.65 revolutions per minute, we can convert this to radians per second. One revolution is equal to 2π radians, and there are 60 seconds in a minute, so:

1.65 revolutions per minute * (2π radians / 1 revolution) * (1 minute / 60 seconds) ≈ 0.1725 radians per second

Now, substituting this value into the equation for "r," we get:

r = (9.8 m/s²) / (0.1725 rad/s)² ≈ 97.24 meters

Therefore, the minimum value for the distance between the astronaut's feet and the axis of rotation should be approximately 97.24 meters. This ensures that the astronauts experience an artificial-gravity environment without the negative effects of motion sickness.

Answer 2

The minimum radius for a rotating space station to provide an artificial gravity of 9.80 m/s² at an angular velocity of 0.173 rad/s without causing motion sickness is approximately 327.87 meters.

Step-by-step explanation:

To find the minimum radius for a rotating space station that provides an artificial gravity of 9.80 m/s² without causing motion sickness, we must consider the maximum permissible rotational speed of 1.65 revolutions per minute, as well as the desired centripetal acceleration. First, we convert the rotational speed to radians per second:

1.65 rev/min × (2π rad/rev) × (1 min/60 s) = 0.173 rad/s

Next, the formula for centripetal acceleration (a_c) is:

a_c = rω²

where r is the radius of the circular path and ω is the angular velocity. Rearranging this formula to solve for r gives:

r = a_c / ω²

Given that the desired acceleration is equal to gravity on Earth (9.80 m/s²):

r = 9.80 m/s² / (0.173 rad/s)²

r ≈ 327.87 m

Therefore, the minimum radius of the space station should be approximately 327.87 meters to provide an artificial gravity of 9.80 m/s² without inducing motion sickness due to excessive rotational speed.