Problem 48 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A proposed space station consists of a circular tube that will rotate about its center (like a tubular bicycle tire), Fig. \(5-44 .\) The circle formed by the tube has a diameter of about \(1.1 \mathrm{~km} .\) What must be the rotation speed (revolutions per day) if an effect equal to gravity at the surface of the Earth \((1.0 g)\) is to be felt?

Step-by-Step Solution

Answer

The space station must rotate at approximately 1827 revolutions per day to simulate Earth's gravity.

Step 1: Understand the Problem

We need to find the rotation speed of the space station in revolutions per day such that the centrifugal force experienced by people inside feels like Earth's gravitational force (1g). We know that the diameter of the station is 1.1 km.

Step 2: Convert Diameter to Radius

First, recognize that the radius of the station's circle is needed. Divide the diameter by 2: \[ r = \frac{1.1 \text{ km}}{2} = 0.55 \text{ km} = 550 \text{ m} \]

Step 3: Determine Required Centripetal Acceleration

To simulate Earth's gravity, the centripetal acceleration must be equal to \( g = 9.8 \text{ m/s}^2 \). We express centripetal acceleration as \( a_c = \omega^2 r \), where \( \omega \) is the angular velocity.

Step 4: Solve for Angular Velocity

Set the centripetal acceleration equal to Earth's gravitational acceleration and solve for \( \omega \): \[ \omega^2 \cdot 550 = 9.8 \] \[ \omega^2 = \frac{9.8}{550} \] \[ \omega = \sqrt{\frac{9.8}{550}} \approx 0.133 \text{ rad/s} \]

Step 5: Convert Angular Velocity to Revolutions per Second

Find the number of revolutions per second by dividing \( \omega \) by \( 2\pi \): \[ f = \frac{0.133}{2\pi} \approx 0.02116 \text{ revolutions per second} \]

Step 6: Convert Revolutions per Second to Revolutions per Day

Multiply revolutions per second by the number of seconds in a day (86,400 seconds): \[ \text{revolutions per day} = 0.02116 \times 86400 \approx 1827 \text{ revolutions per day} \]

Step 7: Conclusion

To produce a centrifugal acceleration equivalent to Earth's gravity, the space station must rotate at approximately 1827 revolutions per day.

Key Concepts

Space Station Design

In space station design, creating an environment that mimics Earth's gravity is a primary concern for the health and well-being of astronauts. By simulating gravity, researchers can help prevent the negative physical effects of prolonged weightlessness, such as muscle atrophy and bone loss. One innovative solution is to design a rotating space station.
Rotating spaces mimic gravitational pull using centrifugal force, similar to how clothes move to the perimeter of a spinning dryer. The force keeps astronauts "grounded" against the station's interior surface.
Here's how it works:

• The space station is designed in a circular shape, often likened to a giant wheel or donut.
• As it spins, centrifugal force is directed outward, away from the wheel's center.
• To feel 1g—or Earth-like gravity—inside the station, the rotational speed must be just right, calculated specifically as its radius and gravitational requirements.

Rich understanding and design of such stations can effectively replicate the sensation of gravity found on Earth, offering more comfort and practicality for long-duration space missions.

Centripetal Acceleration

Centripetal acceleration is the acceleration that makes a body follow a curved path. It points towards the center of the circle along which the body moves. For a space station aiming to provide Earth-like gravity through rotation, achieving correct centripetal acceleration is key.
This is due to the need for the same pull, or sensation, as being on Earth, which is achieved through controlling the acceleration to match the gravitational pull (9.8 m/s²).
To break this down further:

• This force is dependent on both the speed at which the object is travelling and the radius of the path.
• Mathematically, it is expressed using the formula: \[ a_c = rac{v^2}{r} = ext{or alternatively} \quad a_c = ext{} \omega^2 r \] Where:
  • \( a_c \) is the centripetal acceleration.
  • \( v \) is the linear velocity.
  • \( r \) is the radius of curvature.
  • \( \omega \) is angular velocity, which is often used in scenarios involving rotation, like our space station.

Careful calculation of these parameters ensures that the space station gives the intended gravitational experience, crucial for successful long-term inhabitation of outer space.

Angular Velocity Calculation

Understanding angular velocity is critical in the context of calculating the spin required for a rotating space station. Angular velocity defines how fast something rotates or orbits, measured in radians per second (rad/s). In our problem, we calculated the angular velocity needed to achieve Earth-like gravity for the station occupants.
Here's a concise breakdown of why that's vital:

• It accounts for the rate of rotation as perceived from a fixed point (center of the station, in this case).
• To find it, we resolved the centripetal force equation, substituting the desired acceleration (9.8 m/s², Earth's gravity) to find \[ \omega^2 = \frac{9.8}{550} \]
• The square root of the solution gives the angular velocity, \( \omega \), approximately 0.133 rad/s.
• This informs us of how quickly each segment of the circular space station moves, allowing further conversion to more practical units for everyday terms, such as revolutions per day.

Accurate calculation of angular velocity is thus pivotal in designing space stations that consider both safety and human comfort by mimicking Earth's gravity through rotation.