## Whoa! Is That Death Star Wreckage on That Planet?

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:

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/m^{3} 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/m^{3}.

Now I’ll just plug these values into the equation for *F*_{air} , 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/s^{2}. That gives us a new impact speed of 10,923 m/s, or 24,400 mph. Yeah, not really much of a difference.

**Original Source**