In preserving with my long tradition of executing some variety of physics analysis for Star Wars Working day (May perhaps the fourth Be With You), I rejoice this calendar year by hunting at the Death Star II in Return of the Jedi.
In a pivotal scene, the Rebels mount an all-out offensive from an great Super Star Destroyer in the fight about Endor. The attack on the bridge sends the ship crashing into the Death Star. That’s about all you have to have to know.
I do not know why the Destroyer crashed after a fighter hit the bridge. Surely a vessel so vast would have a backup bridge. But realizing why is not my career. In its place, I’ll emphasis on search at the time between when the crew missing manage of the Destroyer until eventually the time it hit the Death Star.
Allow me commence with two important specifics. To start with, the dimension of the Super Star Destroyer. I will not try out to estimate this, I’ll just use values from other authorities. Wookipedia pegs the Executor-course Star Destroyer at 19 kilometers long. The 2nd important depth: Death Star has a diameter of a hundred and sixty kilometers.
Following, there are two matters I have to have to identify from the film. To start with, the altitude (about the Death Star) of of the Star Destroyer. Honestly, this is instead tough. We under no circumstances get a clear shot of the two the spacecraft and the Death Star. Actually, the ideal look at displays the Death Star as witnessed from the bridge of the Star Destroyer just prior to it crashes.
It is completely feasible to use the clear curvature of the area of the Death Star to estimate the altitude of the starship. Even so, it nonetheless would be an estimate, and it would consist of some relatively sophisticated calculations (granted, there is a chance that it is not much too tricky and it’s my fault). In its place enable me just say that the Death Star appears to be like comparable to Earth as witnessed from the orbital altitude of the Intercontinental Place Station. I am heading to presume that the ratio of altitude to “planet” radius is the exact same for the two situations. That usually means:
Making use of an ISS altitude of three hundred km and and Earth radius of 6,370 km, I get a Star Destroyer altitude of 3.seventy seven km. Of course that just cannot be appropriate given that the length of the Executioner-course Star Destroyer is 19 km. Ok, let’s just maintain off on this value for now.
Now enable me emphasis on the affect speed. If I use the length of the Star Destroyer, I can get the place of the spacecraft in each and every body applying Tracker Video Evaluation. Searching at just the motion towards the Death Star, I get the subsequent plot.
The slope of this plot displays an affect speed of 3.five km/s (seven,829 mph). Of course, that’s really speedy.
The genuine query remains—why is it relocating so speedy? There are 3 possible answers:
For the reason of this analysis, I am heading to presume the collision was thanks only to the gravitational interaction. If that’s the scenario, I can use this to estimate the mass of the Death Star. I will make a couple much more assumptions:
Given that we do not definitely care about time through this motion, we will use the Get the job done-Electricity Theory. This states that the total perform carried out on a process is equivalent to its alter in strength. If I consist of the two the Destroyer and the Death Star in the process, there will be no perform and the strength will consist of the two kinetic strength and prospective strength. I can compose this as:
Just one important take note. Technically, the two the Death Star and the Star Destroyer will have variations in kinetic strength. Even so, if we presume the mass of the Death Star is significantly better than the spacecraft it will have a negligible alter in velocity. From the perform-strength equation, I can address for the mass of the Death Star (I am contacting that mass-2).
G is the gravitational constant (6.sixty seven x 10-11 N*m2/kg2) and v2 is the velocity of the Star Destroyer on affect. Making use of my value for the affect velocity and the commencing and ending place (with regard to the centre of the Death Star), I get a mass of 2.seven x 1022 kg. That would give it an ordinary density of one.25 x 10seven kg/m3. If the Death Star was good metal, it would have a density around 8,000 kg/m3.
Of course, this leads to some complications. To start with, how do you get the density that superior? Possibly the Death Star has a super dense internal core—maybe. Also, this massive mass alongside with the rather tiny dimension would make the gravitational subject on the area very massive, 28.seven times more substantial than the Earth’s area subject. Next, the earlier mentioned plot of place displays a virtually constant velocity. With this massive of a mass, the Star Destroyer would have a non-constant acceleration as it moved closer to the Death Star.
Of course, I know there is a different clarification for the massive calculated mass of the Death Star. The other purpose is that Star Wars is just a film and the Star Destroyer crashes because it is a design controlled by individuals. Honestly, I am Ok with this explanation—because it’s definitely correct.
Given that there have been tales about how experts like to the suck the pleasurable out of anything, enable me make a handful of factors.
Oh, you believed this would all be about by now? Mistaken. The physics under no circumstances ends. Verify out the Super Star Destroyer correct after it loses manage.
Let us initial search at the rotation of this crashing starship. If I mark the bridge as the stage of rotation, I can get the subsequent plot of angular place vs. time.
From the slope of this plot I get a virtually constant angular velocity of .159 radians/2nd. Huge deal, correct? Of course, it is. Given that the Star Destroyer is rotating, anyone aboard is relocating in a circular path. To go in a circle, you have to speed up. This centripetal acceleration relies upon on the angular velocity (ω) and the radius of the circle. The magnitude of this acceleration can be penned as:
The angular velocity for this Super Star Destroyer is not super large—however, the circular radius for the Imperial crew at the entrance of the vessel will have a super massive radius. If I approximate this radius at about 15 km, I can work out the centripetal acceleration with a value of 379 m/s2 or 39 G’s. That could possibly not be a superior ample acceleration to destroy you outright, but the crew possibly handed out prior to the collision with the Death Star. I guess that is for the ideal. Who would want to witness that crash and the explosion of the Death Star?
Now you can see the value of a wonderful physics analysis of Star Wars. It’s not just to stage out errors, but to supply much more that means to the plot. Following time you watch Return of the Jedi, just think of all the helpless Imperials caught in the doomed spacecraft.
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