A rotating space station has radius 1380 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?

To determine the period of rotation required for the crew to feel one-half their Earth weight, we can equate the gravitational force on the crew on the rotating space station to one-half their Earth weight.

The gravitational force experienced by an object is given by the equation:

F = (G * M * m) / r^2

Where:
F is the gravitational force,
G is the gravitational constant (approximately 6.67430 × 10^-11 N(m/kg)^2),
M is the mass of the Earth (approximately 5.972 × 10^24 kg),
m is the mass of the crew,
and r is the radius of the rotating space station.

We want the gravitational force on the crew to be one-half their Earth weight. Therefore:

(G * M * m) / r^2 = (1/2) * (g * m)

Where:
g is the acceleration due to gravity on Earth (approximately 9.8 m/s^2).

Simplifying the equation:

(G * M * m) / r^2 = (1/2) * (g * m)

(G * M) / r^2 = (1/2) * g

Solving for the period of rotation:

T = 2 * pi * sqrt(r / g)

Where:
T is the period of rotation (in seconds),
r is the radius of the rotating space station (in meters),
and g is the acceleration due to gravity on Earth (approximately 9.8 m/s^2).

Substituting the given values:

T = 2 * pi * sqrt(1380 / 9.8)

Calculating the value:

T ≈ 2 * pi * sqrt(140.82)

T ≈ 2 * pi * 11.86

T ≈ 74.38 seconds

Therefore, the period of rotation for the crew to feel one-half their Earth weight is approximately 74.38 seconds.

To find the period of rotation of the space station required for the crew to feel that they weigh one-half their Earth weight, we need to consider the centripetal force and gravitational force acting on the crew.

The centripetal force required for an object to move in a circle of radius "r" with a speed "v" can be calculated using the formula:

F_c = m * a_c

where F_c is the centripetal force, m is the mass of the object, and a_c is the centripetal acceleration.

The centripetal acceleration can be found using the formula:

a_c = v^2 / r

where v is the linear velocity of the object.

The gravitational force acting on the crew can be calculated using the formula:

F_g = m * g

where F_g is the gravitational force and g is the acceleration due to gravity.

For the crew to feel that they weigh one-half their Earth weight, the centripetal force must be equal to half of the gravitational force. So we can equate these two forces:

F_c = F_g / 2

Since F_c = m * a_c and F_g = m * g, we can rewrite the equation as:

m * a_c = (m * g) / 2

Now we can substitute the expressions for a_c and rearrange the equation:

m * (v^2 / r) = (m * g) / 2

Simplifying the equation:

v^2 / r = g / 2

v^2 = (g * r) / 2

v = sqrt((g * r) / 2)

The linear velocity "v" can be found using the equation above. The period of rotation "T" can be calculated using the formula:

T = 2 * π * r / v

Now we can substitute the expression for "v":

T = 2 * π * r / sqrt((g * r) / 2)

Using the given values, the acceleration due to gravity on Earth is approximately 9.81 m/s^2. Substituting this value, along with the radius of the space station (1380 m), into the equation, we can calculate the period of rotation "T".