FREEDOM AND SAFETY
AS OBI-WAN SAYS when he drops in on Grievous: Hello there! Today is Star Wars Day. (May the 4th be with you.) Which means I get to post another physics analysis of a scene from one of the Star Wars movies. Last year, I looked at the acceleration of Jedi in all of their jumps - including Jar Jar, because why not?
This time it’s The Rise of Skywalker. The Final Order wants to teach everyone in the galaxy a lesson. So, on orders from Emperor Palpatine, a Xyston-class Star Destroyer fires a super powerful beam from space and blows up the planet Kijimi. Just like that.
I know what you’re thinking: How much energy would it take to blow up a planet? Of course, it’s just an academic question. I’m sure you’re not a Sith lord with bad intentions, so I’ll show you how to figure this out. But even if this is not a real thing, it’s still fun to calculate.
To start, we need to estimate the speed of the planetary shards as they’re blasted into space. We can do that with the Tracker video analysis app. The idea is to pick out a few specific pieces and map their position in each frame of the video.
This position is measured in pixels, but we can convert it to distance by scaling it to a known object in the scene. Then we can get time data from the frame rate - 24 frames per second in this case. Assuming the scene is filmed at regular speed (i.e., not slow motion), we know that each frame represents 1/24th of a second. With position and time data, we can the compute the speed.
To fix the distance scale, I’m going to use the size of Kijimi itself. How big is this planet? Who knows? I'll just say it has a radius of 1 K, where K = the radius of Kijimi. Yes, it seems silly to define the unit in terms of the thing we’re measuring, but we do that all the time in science. (Before people knew the actual distance from Earth to the sun, they set it equal to 1 “astronomical unit.”) Don't worry, it’ll work out in the end.
There is one more issue. We can really only measure the speed of stuff moving perpendicular to the camera - i.e., in the picture plane. Why? Suppose a chunk is sort of angling toward the camera. In each frame, it would move slightly to the side and get slightly bigger. But if I only plot its pixel position, I’ll underestimate the distance traveled and thus the speed.
With that in mind, I picked three fragments that start at the edge of the planet (as seen from the camera) and travel outward in different directions. The Tracker app then gave me this plot of distance traveled (the radial position of each object as measured from the center of the planet) versus time:
You can see that they mostly plot out as straight lines, and the slope of each line (change in position/change in time) is the radial velocity in units of K per second. The green and blue objects have very similar speeds of around 0.3 K/s. The red one starts off at 0.24 K/s then falls to about 0.08 K/s. That’s probably an error by the software; it’s hard to track objects in a field with a bunch of other stuff flying around.
I also looked at some pieces in a later shot and found a speed there of around 0.4 K/s. Since different stuff is moving at different speeds, I'm just going to go with 0.3 K/s as a rough average.
Is that fast? Well, it depends on the value of K. If this planet was the size of Earth, then K is 6.37 million meters. Using this to convert the speed units, I get a debris velocity of 1.9 million meters per second. That is super fast. But it’s still only about 0.6 percent of the speed of light (300 million m/s) - which is good, since weird stuff happens when objects approach the speed of light.
Of course, the speed would be even higher if the planet’s radius is larger than Earth’s. Is that likely? Well, in our solar system, Earth is the largest rocky planet that people could walk around on. Planets like Jupiter are much, much larger but don't have a nice surface with rocks to shoot out when they blow up.
Outside our solar system, most known exoplanets are gas giants like Jupiter, with low densities suggesting they’re not rocky. However, there are some terrestrial planets out there. The biggest one, Kepler-20b, has a radius 1.87 times that of Earth. Using that to scale the video would give a debris velocity of 3.5 million m/s. Still well below the speed of light.
Now we can answer your question. Let me start with three rough approximations. Let’s say the planet is the same size as Earth, with a radius of 6.37 million meters. Let’s also use the mass of the Earth, 5.972 × 1024 kg, and assume a uniform density (which is not true).
Finally, let's assume that all of the planet is ejected with an average velocity of 0.5 million m/s. That’s much slower than my measurement. Why go with a lower speed? Well, the stuff I tracked was probably the fastest debris, since it was on the leading edge of the explosion. Also, I want to be conservative in my estimate of the energy demand.
With this average speed estimate, I can now calculate the total energy of the explosion as the kinetic energy (K) of all the flying fragments. (Sorry, I guess I'm using K as a symbol for two different things.) This kinetic energy is a function of m, the total mass of the planet, and the average velocity (v) at which the pieces are traveling:
Using the mass of the Earth and the lower estimate of debris speed, 0.5 million m/s, I get an energy of 7.465 x 1035 joules. To put that in context: If you grab a physics textbook off the floor and put it on a table, that takes around 10 joules of energy. This is just like that, except with 35 more zeros after it. Yes, it’s a big number.
Power is defined as the rate that energy changes:
If the energy is measured in joules and the time in seconds, then the power would be in units of watts. So let's look back at the video and estimate the time it takes the Star Destroyer to deliver all this energy to the planet. I give it a time interval of about 10 seconds.
By the way, this is not a laser weapon, even though Wookieepedia calls it a “superlaser.” If it were a laser, it would be invisible. You can see laser beams on Earth because the light reflects off dust particles and stuff. In space, there would be nothing to scatter the beam and nothing would get to your eye. You’d just see the planet suddenly blow up.
Anyway, with a time of 10 seconds and an energy change of about 7 x 1035 joules, this would imply a power of 7 x 1034 watts.
For comparison, imagine you could use all the radiation from the sun. That would be pretty hard, since it shines in all directions. You’d have to surround it with a giant spherical solar panel, like a Dyson sphere. But let's just say you could. The sun has a total power output of 3.8 x 1026 watts.
Yes, that means the Star Destroyer is more powerful than our sun by a factor of over 100 million (108). To put it another way, the Final Order has the power of hundreds of millions of suns on each of those starships. That's a daunting enemy.
But you know what Han Solo would say: “Never tell me the odds.”
Rhett Allain is an associate professor of physics at Southeastern Louisiana University. He enjoys teaching and talking about physics. Sometimes he takes things apart and can't put them back together.