Everything You Need to Know About Death Star Orbital Physics
A fully armed and operational physics lesson about ‘Return of the Jedi’
It might seem silly to look at the physics from a movie that was released more than 30 years ago, but here we are. Yes, Star Wars: Return of the Jedi was first in the movie theaters back in 1983. So, why does it even matter? Well, first, it matters to me because I’m a huge Star Wars fan. Second, it seems like the Death Star from Return of the Jedi (or at least pieces of it) shows up in the upcoming Star Wars: The Rise of Skywalker.
So, let’s think about the physics of the Death Star II. Here is a quick refresher: The emperor is building a new (and slightly bigger) Death Star. Hopefully, this time it doesn’t have the vulnerable ventilation shaft — but who knows. While it’s under construction, the Death Star orbits the planet moon of Endor. You see, Endor has this bitchin’ shield generator that the Imperials project from the moon’s surface to protect the space station.
That’s it for the plot — now for the physics. I’m going to assume that the Death Star is in a geostationary orbit around Endor. That means that the space station is not flying under its own power, but is instead just moving due to a gravitational interaction. The geostationary part means that the orbital angular velocity of the Death Star is the same as the rotational angular velocity of Endor. The result is that the Death Star always appears in the same part of the sky relative to Endor.
But how does that work? How do you make something appear to stay in the same place as it orbits a planet? Let’s start with just a basic orbit. There are two big ideas you need to really understand orbital motion. The first is the momentum principle. It looks like this:
The momentum principle says that the nature of forces and motion is that a net force changes the momentum of an object, and the momentum is the product of mass and velocity. The other idea is the model for the gravitational interaction. This holds that there is a gravitational force between any two objects with mass. In particular, this gravitational force decreases in magnitude as the two objects move farther apart. Here is the mathematical model for the gravitational force.
Some important notes for this gravitational force:
- m₁ and m₂ are the masses of the two interacting objects.
- r is the distance between the centers of the two objects.
- The “r hat” (r with a pointy hat on it) is called a unit vector. You don’t have to worry about this, but I just want to be technically correct.
- Finally, G is the universal gravitational constant. It’s tiny. It has a value of 6.67 x 10⁻¹¹ Nm²/kg².
Now we are ready for some serious physics. Let’s say the two interacting objects are the Death Star and the planet moon, Endor. The mass of Endor is so large (even compared to the Death Star) that the gravitational interaction doesn’t really change it. The Death Star is near Endor and moving along with some momentum. The gravitational force from Endor causes a change in this momentum. If we make a smart choice of the magnitude and direction of the Death Star’s momentum, we can get it to move around Endor in a circle.
Here you can see the Death Star at position 1 and then 2 with a change in momentum. Even though it didn’t change in speed, the direction changes — and that’s still a change in momentum. See, orbital physics isn’t so difficult — right?
But let’s change some stuff up. What happens if you move the Death Star farther away from Endor, but you still want it to move in a circular orbit? At a greater distance, the gravitational force is weaker. That means it will produce a smaller change in momentum. But also with a bigger circular orbit, the momentum doesn’t need to change direction so much to stay in a circle.
How about this — an animation instead. Suppose there were a large object orbiting the Earth. I can easily create a numerical model to show the motion around the Earth. A numerical model solves a problem by breaking it into many short time intervals and then making some assumptions during those intervals. If you really want to see it, you can look at (and modify) the code for this animation.
Okay, there’s a bunch of stuff going on here. Let me make some important points.
- This is a view from above the North Pole of the Earth so that we can see the rotation of the Earth.
- This animation is not in “real speed” or it would take a whole day for the Earth to rotate. But to help notice the rotation, I put a “shield generator” on the surface. It’s that yellow dot on the Earth.
- There are two orbiting objects. The first one is at a lower altitude. Notice that it takes less time for this one to complete an orbit.
- The goal is to have an object far enough away such that the time it takes to orbit is the same as the time it takes for the Earth to rotate.
I’ll skip the math, but it’s sort of useful to show the relationship between the angular velocity ω (in radians per second) and the orbital distance as measured from the center of the Earth.
This says the object will have a lower angular velocity with increased distance from the planet. But wait! It also depends on the mass of the planet. That’s important.
Suppose we want a geostationary orbit for the Earth. We know the mass of the Earth and we also know how long it takes to rotate, thus we know the angular velocity. I guess I should say something about sidereal vs. synodic rotations — if I don’t, some dork (like me) is going to point out my error. The Earth does not take 24 hours to rotate. It takes 24 hours for the sun to go from the highest point in the sky (noon) until that position happens again. This is called a synodic day.
But during these 24 hours, the Earth is also orbiting the sun. So it actually takes a little less time to rotate — the sidereal day is 23 hours and 56 minutes (approximately). This is the time needed to calculate the angular velocity of our Death Star. (Or maybe you just want a normal, non-planet-destroying satellite.)
Finally, here is that animation, along with the code.
The object rotates with the same angular velocity as the Earth such that it stays right above the “shield generator.” But the shield generator has to be on the equator of the Earth. That’s the only way the angular velocities of the two objects (Earth and Death Star) can exactly match up. This is why all those satellite TV dishes you see on people’s houses point to the south. That’s where the equator is and that’s where geostationary objects orbit.
But you want to talk about the planet moon of Endor. Me too! If the Death Star is in a geostationary orbit (technically, it should be called an endostationary orbit, because of Endor), then there are three factors:
- The mass of Endor
- The angular velocity of Endor (related to the length of a day) — assuming sidereal and synodic days are close
- The orbital radius
If I know two of these things, I can find the third. Let’s start with this first view of the orbiting Death Star as seen in the rebel attack briefing.
From this image (along with a Death Star diameter of 160 km), the Death Star orbits at a distance of 1.564 x 10⁶ meters. This puts the radius of Endor at 1.09 x 10⁶ meters. But wait. Here is another shot from that same scene.
In this case, the Death Star is farther away, with an orbital distance of 1.564 x 10⁶ meters. I assume they just moved it closer to the planet in the animation for effect. I’m going with the bigger value. So now I have one of the three values.
Of course, I can’t get the angular velocity of Endor from this scene. What fool would make the animation rotate in “real speed?” That wouldn’t look cool. If you are someone like Admiral Ackbar, you surely want to have your Death Star animation look cool. Especially in front of a Jedi like Luke Skywalker. You gotta impress. But that means we don’t know the angular velocity.
That’s fine. We can get the mass of Endor. What? Yup. Here’s how it works. We have seen people on the surface of Endor. They move around just like they would on Earth (because it was filmed on Earth). This means the gravitational field on the surface of Endor is the same as on Earth. The gravitational field is just the force per unit mass — on Earth this has a value of 9.8 newtons per kilogram. (We use the symbol “g.”) Since the gravitational field depends on the mass and radius of the planet, I can solve for the mass because I already know the radius by comparing it to the known size of the Death Star II.
With my known values, I get an Endor mass of 1.76 x 10²³ kilograms. Note: This gives Endor a smaller radius than the Earth’s moon, but also more than twice the mass of the moon. Just in case you are keeping track.
But that’s it. I have the orbital radius and the mass of Endor. With that, the angular velocity for a stationary orbit would be 3 x 10⁻³ radians per second. This would put the length of an Endor day at 34 minutes long. That’s not a very long day. Maybe that’s what makes the Ewoks short — they don’t get enough sleep.
Now for some homework and preemptive comments:
- There are a few times in the movie that you see both Endor and the Death Star. For instance, you can see the Death Star from the surface of Endor when it explodes and you can also see both Endor and the Death Star when the rebels arrive. As a homework assignment, you can use these to estimate the orbital distance for the Death Star.
- How far away would the Death Star orbit if Endor had a day of 24 hours? What would that look like?
- Is it possible that the Death Star is not in orbit, but rather flying? Yes. Estimate the mass of the Death Star (good luck with that one) and then calculate the force needed to keep it in that position.
- Wait. If Endor makes one complete rotation in just 34 minutes, then the effective surface gravity (especially at the equator) will be quite lower because of the centrifugal force. (Don’t worry, I used that term correctly.) That means that Endor’s mass could actually be higher and still produce an effective surface gravity of 9.8 N/kg such that it wouldn’t have to rotate as fast for the Death Star orbit. Oh, this is complicated. Maybe I will circle back to this later.
- I just had an idea. What if Endor has a reverse rotation and a reverse orbit? Remember, it’s a moon. That means that it is orbiting a planet that orbits a star. Most planets and moons rotate in the same way they orbit, but it’s possible to be spinning “backwards.” Is it possible to have a 34-minute period but have a much longer synodic day with some type of backwards orbit?
- Don’t you have something better to do than waste your time on physics and Star Wars? Nope.