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How did gravity worked on the Death Star?
How did gravity worked on the Death Star?
46 comments
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It's well established in the Star Wars universe that all you need to create gravity is a floor. Take, for example, any scene from within any of the space ships. Gravity is never a problem.
Of course, a deep chasm also seems to create gravity, as seen in the first movie when Luke and Leia swing from one ramp to another to escape the stormtroopers chasing them.
Regardless, it's easy to see from the blueprints that the layout is stacked like your first image.
Edit: upon closer examination it turns out it's both. The plans show three 'concentric surface decks' that apparently work like your second image. So I guess the answer is 'it depends on where you are in the death star', and, I guess, which way 'down' is where you are.
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Since the sequels had space "bombers" dropping unguided bombs by just opening a hatch and letting them go, you only need to have a vaguely identifiable "down" for gravity to work...
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You see, Star Wars unguided bombs work via the buoyancy principle, as outlined by our current Flat Earth Scholars.
/s
Haha but for real, in the original trilogy, I think there are a handful of scenes where TIE Bombers are shown dropping bombs, but its on things like pretty large asteroids or planets, and in at least the early games like TIE Fighter, the bombs are basically just replaced with things sort of like torpedoes: very slow, no or slow guidance/tracking, only useful against large slow targets.
Whilst the bomber scene in 8 is silly, I think its mainly silly that its like a bomb bay with a fucking open hole, like a ww2 bomber, but the crew are not wearing pressure suits and oxygen masks.
Even if there is one if those hangar forcefield type things... thats a component which can obviously fail, so you'd think the crew would be fully suited up the whole time.
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I don't know why people have a problem with this. The bomb bay had gravity, just like all parts of every other ship, big or small. If you drop the bombs while they are inside of the ship, and they fall out of a hole (and we've seen big access holes in ships before) then once they are in space they will continue with inertia.
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But what if the Deathstar wore pants...
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Big enough nerd to know the canonical answer was both. The exterior of the Death star all had gravity pointed down towards the middle of the station. However once you got past this defensive layer artificial gravity was oriented like the left side.
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It's Star Wars so most likely nr 1.
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If it's a big as a little moon it'll have his own weak gravitational pull
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It's no moon
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I'm curious if it would be similar to that of the moon, less than, or moreso.
It is made entirely of metal which is super dense and has a lot of mass, but a lot of it is also open space I imagine. Curious if the smaller amount of dense metal would have a larger or smaller mass than a solid moon.
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Wait. I just realized energy also creates a gravitational pull, and the death star’s whole thing is destroying a planet right? That’s got to take a huuuge amount of energy because the explosion has to massively overcome the gravity holding the target together.
A quick google search says you'd need 10^32 Joules to blow up the earth. E=mc2 so dividing that energy by the speed of light squared gives about 1.1e15 kg of equivalent mass which is relatively small compared to earths mass (6e24) but still large.
For reference, if the radius of the Death Star was 1000m you’d get about 5.2m/s2 acceleration from just that energy in its core.
But if the Death Star is able to blow up multiple planets, then the energy it has to have on hand goes up. So if the Death Star contains enough energy to blow up 5.4 billion planets, then just that stored energy would have nearly equivalent “mass” to the earth.
But gravitational acceleration is inversely proportional to distance squared. So since the Death Star is small, you wouldn’t need that much energy to get earth gravity. If we assume the Death Star has about a 160km radius, then you’d only need enough stored energy to blow up ~45,000 earths to get a surface gravity of 9.1m/s2.
This gravity would increase as you got closer to the core or whatever part stores all that energy. But if you spread that energy out a bit you could probably extend how large the earth-like gravity range in the station would be.
The mass of the structure itself would contribute to the gravity too so that 45,000 is probably an overestimate.
TL;DR: From rough math in my head, assuming a radius of 160km, point mass, and ignoring the mass of the structure, you’d only need to store ~5e19 J of energy in the Death Star to get earth like gravity on the surface. That is approximately the amount of energy required to blow up 45,000 earths
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Just like earth! With magnets! GOSH.
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Or like the picture on the right but with the people standing on the inner surfaces instead of the outer ones? Centrifugal force from rotation is the only way we know how to make "artificial" gravity, although you can imagine the comical scene if it had to stop rotating in order to fire.
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It could spin around the cannon’s axis pretty easily, I imagine.
Also it would need to spin in layers because the outside would be moving a lot faster than the inside. For the size of it that actually probably wouldn’t matter all that much since the closer layers wouldn’t need to be inhabited by anyone but much more resilient droids.
One thing for sure is that even with such afancy technology they probably wouldn’t have directional floor gravity producers. The stacked version would then need to have gravity generators at the very bottom and it would get weaker as you go up which would be…strange.
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It being a sphere should make walking around very uncomfortable. Every single place is a ramp, and the gravity is never the same anywhere.
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And this is Star Wars is much closer to fantasy than sci-fi…
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Mix of both probably. Star Wars tech seem to swap gravity direction easily and smoothly. Like the gun chambers on the Millennium Falcon.
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since death star is capable of delivery a blast with high energy density, its core might be a nuclear fusion or anti matter power plant. maybe the mass there generates sufficient gravitational force.
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The main weapon of death star was powered by kyber crystals, the same crystals that powers a Jedi's lighsaber.
That said, the death star also needs power for various other stuff not only the planet destroying mega cannon, so it's easy to assume there's also a powerful power plant for powering rest of the station.
There was also 123 "highly radioactive" generators powering it's propulsion system and hyperdrive.
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Lol it's probably magic from a gnome in a weird helmet
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Could also be spin gravity, with the feet pointing outwards.
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How would the elevators work if it was like 2nd picture?
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The same way they do on Earth. Up towards the outside, down towards the center. Of course, most of the elevator shafts would only go down partway.
But -- since we see artificial gravity in Star Wars, you could say that all the rooms in the Death Star are laid out as haphazardly as the toys in a kid's box. It wouldn't be the silliest thing in the franchise, after all.
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