What is the relationship between the radius of a ring in a rotating space station and the strength of the artificial gravity generated?

Question

Suppose engineers built a rotating space station similar to Space Station V from the film 2011: A Space Odyssey, but with multiple concentric rings where astronauts can live.

    ____________
   /  ____c___  \
  /  /  __b_  \  \
 /  /  /  a \  \  \
/  /  /      \  \  \
\  \  \      /  /  /
 \  \  \____/  /  /
  \  \________/  /
   \____________/

Ring a is 10 meters from the center, ring b is 20 meters from the center, and ring c is 30 meters from the center.

• If the station is set to rotate at a speed where the gravity within ring b is identical to Earth's gravity and the rings rotated together, would the gravity in rings a and c be significantly stronger or weaker?
• If the concentric rings could rotate freely, would rings a and c need to rotate faster or slower than ring b to ensure that the gravity at each ring had identical strength?

Answer 1

The acceleration into the "ground" that is experienced at a radial distance $r$ from the axis of rotation (the center of the space station) is given as:

$$a= \omega^2 r$$

where $\omega$ is the angular velocity of the space station (i.e. how many times it rotates per second). This is called a "centrifugal force" (not to be confused with "centripetal force" though they are intimately related). You can see that acceleration is linear with respect to how far you are from the axis of rotation.

A good way to demonstrate that this equation makes sense intuitively is to tie a ball to the end of a rope. If you hold the rope very close to the ball and begin to swing it around in a circular path, it is relatively easy to hold on to. If you increase the length of the rope, you can feel that it becomes more difficult to hang on.

Back to your space station example. If we set the angular velocity of the space station so that at $r=20$ meters the acceleration is equivalent to the acceleration due to gravity on Earth, it is easy to solve for the accelerations at the other two locations:

Since ring A is $10$ meters $= \frac{1}{2}(20)$ meters away from the axis, the acceleration there will be half of what it is at B, or $4.9$ meters/second/second. Ring C is $30$ meters $= \frac{3}{2}(20)$ meters away from the axis, so the acceleration there will be $1.5$ times what it is at B, or $14.7$ meters/second/second.

If the concentric rings were allowed to rotate independently (ignoring the fact that this would create problems when trying to get from one ring to another) then from our equation above we can see that ring A would need to rotate $\sqrt{2} \approx 1.414$ times faster than ring B, and ring C would need to rotate $\sqrt{\frac{2}{3}} \approx 0.816$ times slower for them all to have the same centrifugal acceleration.

Answer 2

Let's assume that a particular ring of radius $r$ is spinning at angular speed $\omega$, then a person of mass $m$ in this ring will experience a centrifugal force $F_c$ given in magnitude by $$ F_c = m r\omega^2 $$ and pointing radially outward.

Important Aside. The centrifugal force, which points radially outward, is what's termed a "fictitious force" in classical mechanics because it is not the result of interactions with another object. Instead, it is merely an apparent force that results from making observations from a non-intertial frame of reference. However, if a person were actually walking along the outside of one of the rings, then he would experience the apparent gravity by virtue of the contact force between his feet and the ring.

If this person were to apply Newton's second law in this rotating frame and treat this apparent outward force as he would a gravitational force on the surface of the Earth where the force is given in mangitude by $mg$ where $g$ is the acceleration due to gravity, then he would find that $$ mg = m r \omega^2 $$ and therefore, he would experience an effective gravitational acceleration which is a function of $r$ and is given by $$ \boxed{g_\mathrm{eff} = r\omega^2} $$ So we see that the effective acceleration due to gravity in a given ring increases linearly with the radius of the ring and quadratically with its angular speed.

I just saw that someone else also wrote an answer that takes it from here, so I hope this helps!

Cheers!

Answer 3

The illusory force of gravity is created by the centripetal force.

See Centripetal Force

Stated as follows:

$F_c=\frac{mv^2}{r}$

Be careful with this though - because the constant of motion is the angular velocity $\omega$.

Consider the expression after substituting $v=\omega r$:

$F_c=m\omega^2r$

Now you can see what happens to the force with increasing radius and thus you should be able to deduce at what angular velocities your independent rings should rotate.

Answer 4

So at the end of all this genius guy talk 99% of every person that read everything still has no clue of what it means to simulate gravity by spinning in space. So maybe this will help, imagine a swimming pool 20ft deep and the pool has a metal cover over the top of it and submerged in the 20 ft of water is a huge ball that is about 10 feet in diameter and your inside the ball standing up, if the ball began to spin at a medium speed you would NOT be spinning as long as you were walking in place. Gravity Simulation Achieved! Now, imagine everything I just said, but this time make everything, the pool and the ball much much bigger. Now you would NOT have to walk in place, you could just walk about as the ball spins, but on the inside you are NOT spinning, you are walking!