Whoa! Is That Death Star Wreckage on That Planet?

Near the end of the teaser for Star Wars: The Rise of Skywalker, there’s something very interesting—towering chunks of a Death Star, rising like mountains from a blustery sea. Oh, there’s no doubt. It’s a Death Star. We can argue about which Death Star, the one from Episode IV or the one from Episode VI, but I won’t accept a non–Death Star answer.

One thing I like about movie trailers are these little hints about the plot. Why are there pieces of a Death Star on this planet? What planet is this? Why are the heroes here? Are they looking for something or just out sightseeing? We’ll find out when the movie is released in December.

But there is one question we can answer, without spoilers, using only our knowledge of physics: How could the wreckage of a blown-up Death Star get onto this planet? Is it possible to “fall” from outer space? And if so, would there be anything left for space tourists to see?

Trust the Force

Let’s start with a simple case. Imagine a universe containing a single, stationary planet and a big hunk of space junk. Now, even if the thing is really remote, there’s still a gravitational tug on it. Any two objects with mass will be drawn toward each other by a gravitational force. (Including you and the screen you’re reading this on, but luckily it’s pretty tiny.)

As this equation shows, the magnitude of the gravitational force depends on the mass of each object (m1 and m2) and the distance (r ) between them.

Illustration: Rhett Allain

Here, G is the universal gravitational constant, with a value of 6.67 x 10–11 N×m2/kg2. It’s “universal,” so it should work even in a galaxy far far away. See the r2 in the bottom? This says the attractive force between two things declines with the square of the distance. Double the distance and the gravitational force is only a quarter as strong.

So a remote object has a very small force on it—but it’s never actually zero. There’s no cutoff line. And even a tiny force will change the momentum of the object. That inches it closer to the planet, where the gravitational force is a tiny bit stronger. As a result, the object gradually speeds up and eventually hits the surface. So far so good.

Sizing Up the Impact Velocity

But how hard does it hit? To answer that, it’s easier to think in terms of energy, rather than gravitational force. Think of the planet and space object as a system, where the total energy in the system is constant—there’s no external force acting on it. There’s two types of energy we need to consider here, kinetic energy and gravitational potential energy.

Kinetic energy is the energy an object has by virtue of being in motion, and it depends (like momentum) on its speed (v) and mass:

Illustration: Rhett Allain

Gravitational potential energy (U ) depends on the masses of both the planet and the object, as well as the distance between their centers. OK, technically there isn’t a potential energy—there is only a change in potential energy. However, we can define the potential energy with respect to an infinite distance away and get the following expression:

Illustration: Rhett Allain

As you can see, that’s pretty similar to our equation above for FG. Why is there is a minus sign on this one? Well, if the gravitational potential energy at an infinite distance is zero and the potential energy decreases as the object gets closer, the potential must be negative. So as the kinetic energy increases, the potential energy declines to keep the sum constant.

With these two energies, we can calculate the speed of the Death Star fragment when it hits the planet. A few notes:

  • I’m assuming it starts infinitely far away with zero initial velocity.
  • The size of the planet matters, not just the mass, because the “falling” object stops at the surface, a distance from the planet’s center equal to its radius.
  • The planet is stationary. That’s not realistic, but a moving planet’s kinetic energy is pretty constant, so this simplification actually doesn’t really matter.

So, if the energy of the system is constant and the object starts far away, then its speed at the surface of the planet (I’ll call this distance R ) will be:

Illustration: Rhett Allain

This is pretty interesting. Notice that the impact velocity depends on the mass of the planet (mp), but not the mass of the object. Also, when we derive this using energy, there is no time variable. It doesn’t matter how long this “falling” process takes—and it would take pretty long if you started infinitely far away—the answer is the same.

(In fact, it doesn’t even matter which way the thing is going. You can use this same equation in reverse to compute “escape velocity,” the speed at which you’d have to launch something off the planet for it to keep going and never come back.)

Now let’s plug in some numbers and see what we get. I’m using the mass and radius of Earth, but I’m attaching the code here, so you can easily change the values to some other planet—like, say, Endor? Just a thought. Click the pencil icon on the window to modify the code.

This is bad. A fragment of the Death Star, even starting at a velocity of zero, would hit Earth with an impact velocity of 11,176 meters per second—25,000 miles per hour. That’s damn fast. I don’t care what futuristic materials it was made of, the pieces would be obliterated. Things on the ground would be messed up too.

What if we use something easier—like the moon? I listed the size and mass of the moon in the code above. If you change the values to those of the moon, you still get an impact velocity of more than 2,000 meters per second (4,474 mph). Still no good.

Kind of a Drag

Wait, what if there’s an atmosphere to slow it down? When an object moves through a gas, there’s an air drag force that pushes in a direction opposite to its motion. Now, I’m not going to lie—at high speeds this gets really complicated. But we can get a rough estimate of the effect with this simple equation:

Illustration: Rhett Allain

This says air drag is proportional to the square of the velocity (v ). So, yeah, the faster something goes, the more important air drag becomes. Here ρ is the density of the air (1.2 kg/m3 at Earth’s surface). A is the cross-sectional area of the object. C is a drag coefficient that depends on its shape—let’s go with 1.0, which is on the high side. (A cube would have a drag coefficient of 1.05.)

We also need to know the height of the atmosphere … Er, there is no height. Atmospheres don’t end, they just fade away. But that’s not simple. Let’s just use a height of 20 km—about twice as high as a passenger jet flies—and assume the air density is constant. It’s not, but I’m erring on the side of overestimating air drag.

Finally, we need to know the mass and size of the Death Star fragment. No problem: If this is Death Star 2 from Return of the Jedi, it has a radius of 100 kilometers, according to Wookieepedia, so a piece of it will be smaller than that. I’m going with a square slab 5 km long on each side and 1 km thick. To get the mass, I’ll assume it has a density similar to that of a ship floating on water, about 500 kg/m3.

Now I’ll just plug these values into the equation for Fair , along with the final speed of 11,176 m/s, to estimate the maximum air drag force. Then I can use that to calculate the minimum impact velocity with air drag. Here is my code, so you can try different assumptions:

With these parameters, the piece will hit the atmosphere 20 km out and then decrease in velocity with an acceleration of 140 m/s2. That gives us a new impact speed of 10,923 m/s, or 24,400 mph. Yeah, not really much of a difference.

It’s pretty obvious that having an atmosphere on our planet isn’t going to help land any Death Star debris intact. Oh, I know. I made some crazy-bad assumptions. But even if I could bring the impact speed down to half of what we got for a tiny orb like the moon, it would still be 1,000 meters per second. That’s just too fast.

What does this mean? It means those awesome shards of the Death Star didn’t just fall there. Someone had to move them. You know, with a giant tractor beam.

Just to be clear—if the movie comes out and acts like this Death Star debris fell down from space, I’m still going to enjoy the show. I know that Star Wars isn’t real anyway. I really do. But that won’t stop me from poking some fun at the filmmakers’ wishful physics.

More Great WIRED Stories

Read More