Some people watch science fiction shows to be entertained, and that's fine. But you can also use them to find excellent physics problems. How about the recent Netflix movie *Stowaway*? It's about a crew traveling to Mars and—guess what?—there's an extra person aboard the ship! (I mean, that's not really a spoiler. It’s right in the title.)

But in order to get to the fun physics, I might have to actually reveal a small spoiler later. You've been warned.

The coolest part of the show is the ship they use to get from Earth to Mars. It's basically a crew compartment connected to a counterweight by very long cables. The two parts then rotate around a common center of mass as it moves through space. Why would you do this? It's a real way to make artificial gravity. In fact, lots of sci-fi shows, from *2001: A Space Odyssey* to *The Martian* to *The Expanse**,* have spacecraft that rotate.

Artificial Gravity

How does that rotation make artificial gravity? Let's start by imagining a human sitting in a stationary chair on the surface of the Earth, perhaps just like you are doing now. There are two forces acting on the person. First, there is the downward-pulling gravitational force—this is equal in magnitude to the mass of the human (*m*) multiplied by the gravitational field (*g* = 9.8 newtons per kilogram). The other force is the upward-pushing force from the chair on the human's butt. Since this upward force is perpendicular to the surface of the chair, we call it a "normal" force. This normal force must be equal in magnitude to the gravitational force so that the net force is zero and the person stays at rest.

Humans don't really feel forces. What we think of as a force is really just some type of compression in our bodies. You can imagine a person as being sort of like a giant spring. If you want to squish a spring, you actually need *two* forces pushing both ends of the spring together. In the case of a human sitting on a chair, these two forces are the normal force and the gravitational force, squishing them from opposite directions.

But what if there isn’t a gravitational force and a normal force? Can we get this kind of compression even if we’re not on Earth, being pulled down by its gravity?

For this thought experiment, let’s replace our human with a spring connected to a single mass. As we know, if I place this mass-and-spring combo on a floor on Earth, the gravitational force pulls down on the mass and the floor pushes up. The combination of these two forces will compress the spring. But let’s say we try this on the floor of a ship in outer space, where there is no gravitational force. I can create this compression by instead making the floor accelerate upwards. The mass must push back on the spring, compressing it just as gravity would. Maybe this diagram will help.

Yes, we humans can't tell the difference between a gravitational force and an acceleration. (This is Einstein's equivalence principle.) But what does this have to do with rotation? Well, if you are moving in a circle, you are accelerating. Acceleration is defined as the rate of change of velocity—that means that any change in velocity would result in an acceleration. But velocity is a vector quantity, which means that you could change speed to change velocity *or* you could change direction. Either would be a change in velocity.

For an object moving in a circular path, it's constantly changing direction and thus always accelerating. The direction of acceleration is pointed towards the center of the circle of rotation. That means that you can use this acceleration to simulate gravity. Here's a rough sketch of how it would work for a rotating spacecraft like the one in *Stowaway*.

In the case of a rotating ship, it's actually easier to think about the angular velocity of the whole thing instead of the linear velocity of each part. The angular velocity (we use ω) is the rate that the angle of orientation changes in units of radians per second. With that, the acceleration of a human depends on two things: the angular velocity and the radius of the circle. Since this acceleration is just like the gravitational field (*g*), we can calculate the apparent gravity as:

There are two ways to increase the artificial gravity. You can spin faster (increase ω) or make a bigger circle (R). Now you can see why the ship uses two objects (the spacecraft and a counterweight) that are connected by cables; together, the whole thing spins like a baton. By using cables, the circular radius of rotation can be quite large without having to build a gigantic spacecraft.

Analysis of Gravity in *Stowaway*

OK, let's go over the details of this show’s rotating ship. Near the beginning, the commander says: "Artificial gravity is climbing. It looks like we will hit close to 5 g's."

What the heck does that mean? A “g” is a measure of artificial gravity where 1 g is equal to 9.8 meters per second squared—the value of gravity on the surface of the Earth. So the commander is saying that gravity on the ship is now *five times* that, which means that everyone onboard would be five times their normal weight on Earth. I'm going to go ahead and call this a mistake. Surely, that would make it impossible to move around. I bet she really meant to say "0.5 g's"—half the gravitational force on Earth.

Now for the fun part. Let's actually measure the rotation rate of the spacecraft. Fortunately, they have several shots showing the whole thing spinning. I can then use video analysis with the Tracker app to plot the angular position as a function of time. The slope of this data will be the angular velocity. Here's what I get:

The slope of this line is 0.1024 radians per second. If that's the angular velocity, then what circular radius would you need to get an artificial gravity of 0.5 g?

Now, we just need to solve for R:

Let's just make it simple: To achieve an artificial gravity of 0.5 g, you’ll need a radius of 450 meters and a spacecraft-to-counterweight distance of twice that (900 meters).

Just for fun, the Wikipedia page lists the tether distance at 450 meters. This would give a rotational radius of 225 meters. Using the same angular velocity, the astronauts would have an artificial gravity of just 0.25 g's.

I mean, that's not terrible. In fact, the gravitational field on Mars is 0.38 g's, so this would be almost good enough for the astronauts to prepare for work on Mars. But I'm going to stick with my artificial gravity of 0.5 g's and a tether length of 900 meters.

What Would It Be Like to Slide Down a Tether?

Without going into too much detail, let's consider what would happen if an astronaut was going to climb one of the cables from the spacecraft to the counterweight on the other side for some reason. Maybe life's just better on the other side—who knows?

When the astronaut starts up the cable (I'm calling "up" the direction that’s opposite the artificial gravity), physics dictates that they will feel the same apparent weight as the other astronauts on the spacecraft. However, as they get higher on the cable, their circular radius (their distance from the center of rotation) decreases, making the artificial gravity also decrease. They would keep feeling lighter until they got to the center of the tether, where they would feel weightless. As they continued their journey to the other side, their apparent weight would start to increase—but in the opposite direction, pulling them toward the counterweight at the other end of the tether.

But that's not very exciting for a movie. So here is something very dramatic instead. Suppose an astronaut starts near the center of rotation with very little artificial gravity. Instead of slowly climbing "down" the tether, what if they just let the fake gravity *pull* them down? How fast would they be going when they get to the end of the line? (This would sort of be like falling on Earth, except that as they "fall," the gravitational force would increase as their distance from the center does. In other words, the farther they fall, the greater the force on them.)

Since the force on the astronaut changes as they move down, this becomes a more challenging problem. But don't worry, there's a simple way to get a solution. It might seem like a cheat, but it works. The key is to break the motion into tiny pieces of time.

If we consider their motion during a time interval of just 0.01 seconds, then they don't move very far. This means that the artificial gravity force is mostly constant, since their circular radius is also approximately constant. However, if we assume a constant force during that short time interval, then we can use simpler kinematic equations to find the position and velocity of the astronaut after 0.01 seconds. Then we use their new position to find the new force and repeat the whole process again. This method is called a numerical calculation.

If you want to model the motion after 1 second, you would need 100 of these 0.01 time intervals. You could do this calculation on paper, but it's easier to make a computer program do it. I will take the easy way out and use Python. You can see my code here, but this is what it would look like. (Note: I made the astronaut larger so you could see them, and this animation is running at 10X speed.)

For this slide down the cable, it takes the astronaut around 44 seconds to slide, with a final speed (in the direction of the cable) of 44 meters per second, or 98 miles per hour. So, this is *not* a safe thing to do.

But wait! What if they slow down their "fall" by holding onto the cable with their hand and producing some friction? It's difficult to determine just how much of a backward-pushing force they could produce, so I'm just going to estimate that it would be about 25 percent of their weight on Earth at around 180 newtons (40 pounds). Of course, at the start of the motion, this frictional force would be greater than the artificial gravity. And that means they wouldn’t start to slide at all. So, let’s say they apply the frictional force only after they are moving at 10 meters per second.

With that, here is a plot of the speed as a function of time for both the no-friction and friction slides (for comparison):

In the case of the frictional force, the astronaut reaches the end of the tether with a speed of 23.7 m/s (53 mph). That's still pretty fast for a safe landing.

Also notice that the time is around 30 seconds. The speed curve with friction just flattens out there. Why? That's the point at which they reach 10 m/s. However, the frictional force is greater than the artificial gravity at this location. That means they are able to reduce the amount of friction so that the net force is zero and they slide at a constant speed.

However, after about 40 seconds, they get too far. At that time, their position on the tether produces a gravitational force that's greater than friction. This means their net force is now pushing them *away* from the center of rotation and increasing their speed. And that means they're going to have a significant impact when they hit the spacecraft at the other end.

There's Another Force to Consider

In the analysis above, I assumed that the astronaut would be able to stay on the tether as they moved down. But that might not be the case. In the reference frame of the rotating spacecraft, there is another force to think about: the Coriolis force. In short, this is a sideways force in a rotating frame due to an object’s motion toward or away from the center of rotation. We can understand this force by considering the astronaut as they move away from the center of rotation. Imagine that they are going from position 1 to position 2 in the diagram below.

In order to stay on the cable, the astronaut has to have the same angular velocity (ω) in both positions 1 *and* 2. Since they have the same angular velocity, the astronaut will make one complete revolution in the same amount of time for both positions. But for position 2, they have a longer distance to go, since their circular path is longer. This means that as they move from 1 to 2, their linear velocity (v) has to also increase. How do you increase a velocity? With the Coriolis force. The direction of this force will always be perpendicular to the cable and will have a magnitude of:

In this equation, *m* is the mass of the astronaut, *v*_{r} is the velocity of the astronaut in the direction of the cable (which does not include their motion due to the rotation of the spacecraft), and ω is the angular velocity of the reference frame (the rotating spacecraft).

Let's consider the two sliding cases (with and without friction) and calculate the sideways force due to the Coriolis force. Here is a plot of these forces (in both cases) as a function of distance from the rotational center of the spacecraft. Just so it's easier to understand the magnitude of these sideways forces, I'm using units of g's (where 1 g = the astronaut's weight on Earth).

For the no-friction case, things get out of control. By the end of the motion, the Coriolis force is almost 1 g. That means the astronaut would need to hold on with their hand just to support their full weight. Their body would also swing out and away from the cable, although not completely horizontally.

But things are different when they use friction to slow down. Yes, they're still going fast—but the Coriolis force depends on their speed moving away from the center of rotation. Since the friction prevents their speed from getting super fast, the Coriolis force is lower in magnitude—at a value of around 0.5 g's by the time they reach the end of the cable. It would still be difficult to hang on, but at least it's plausible.

Motion of a Falling Object

Let's say (just for fun and not for any reason related to the plot) that the astronaut is carrying some object with them as they slide down the tether. When they reach the end, they hit the spacecraft and fall over. Oh no! That unidentified object rolls over the top of the spacecraft and "falls" away. What would the motion of this "falling" object be like? Would it look just like an object dropped from the top of a tall building?

Well, no—that wouldn't happen, since once it leaves the rotating spacecraft there won't be an artificial gravity force acting on it. It should move with a constant velocity. But in what direction?

How about I just model both ways and you tell me which one looks better? Here is option A. I'm going to assume the object just barely falls off the rotating spacecraft.

Now for option B. Notice in this case, the object moves away from the rotating spacecraft in a straight line with respect to the background.

Which one looks better, and which one is better for physics? In physics, the correct motion is option B. When the object "falls off," it has a component of velocity moving away from the center of rotation, as well as a component of velocity tangent to the direction of rotation. This is like a rock in a sling. Once it leaves the sling, it will continue to move along at a constant velocity in the direction it was traveling.

But from a cinematography perspective, option A might look more dramatic. It looks like the object is just barely "floating" away, and like an astronaut could almost be tempted to jump off the spacecraft to try to fetch it. And there’s another nice thing about option A: If you were an astronaut watching the object, you could continue to see it in your field of view, since it's still right by the spacecraft. Of course, this couldn't actually happen in the real world, because there’s no force that would keep pushing the “falling” object to move around in a circle.

OK, you might be able to guess that the producers went with option A. Honestly, it's wrong—but I'm fine with that. It's a visual element of the story, and telling a story is what the film is all about.