The gravity is negligible. The official sizes of the Death Stars have been 120 - 900 km in diameter according to rebel scale. For comparison Earths moon is ≈35000 km in diameter and it’s gravity is 1/6 of earth’s. On top of that, the Death Stars are mostly hallow, being a metal framework, instead of solid rock.
Comment on How did gravity worked on the Death Star?
_haha_oh_wow_@sh.itjust.works 1 month agoAny chance you know the canon explanation of how they counteract the gravity generated by the Deathstar’s mass?
MyTurtleSwimsUpsideDown@fedia.io 1 month ago
nadiaraven@lemmy.world 1 month ago
the Death Stars are mostly hallow
Our Death Star, who art in heaven, Hallowed be thy name, Thy empire come, thy will be done, On Alderaan as it is in heaven, Give us this day our daily rations, And forgive us our rebellion, As we forgive those who rebel against us, And lead us not to the light side, But deliver us from the Jedi, For thine is the empire, and the unlimited power, and the dark side forever, Amen
Gestrid@lemmy.ca 1 month ago
If Lemmy had gold, I’d give it to you.
RecluseRamble@lemmy.dbzer0.com 1 month ago
Earths moon is ≈35000 km in diameter, and it’s gravity is 1/6 of earth’s.
Off the a factor of 10. The Moon has a diameter of almost 3500 km (Earth’s circumference is about 40,000 km, so your diameter would make the Moon larger than Earth).
However, the Death Star being hollow still means you’re probably right about gravity being negligible.
MyTurtleSwimsUpsideDown@fedia.io 1 month ago
Whoops. Good catch! so about 4-30 times the size of the Death Star. That would mean the gravity of the Death Star is at most 1/24th that of earth’s if it were solid rock, and my math is correct. That’s at the surface, though. As you go inside, gravity will decrease until you reach the center where there will be no gravity at all because all the mass of the space station is pulling you away from the center equally. (assuming a uniform mass distribution).
g ≈ M/r^2
V ≈ r^3.
uniform density: ρ for simplicity’s sake
M = ρV
—> g ≈ ρr where r is the distance from the center of the death star, but no further than the surface
finley@lemm.ee 1 month ago
It doesn’t create enough gravity because it isn’t massive enough to create gravity. They use gravity deck plating.
MudMan@fedia.io 1 month ago
Even if it was massive enough, if they can keep people sticking to the ground in a tiny ship they can surely counteract the gravity of a space station.
Also, most of their spaceships have wings. We're thinking about this way too hard.
cyborganism@lemmy.ca 1 month ago
They don’t all have wings. Only the X-Wing and Imperial transport ships have actual wings, and we’ve seen them fly through atmospheres.
MudMan@fedia.io 1 month ago
Well, yeah, but we've also seen the ones that look like a hamburger patty fly through the atmosphere (and, in fact, outmaneouver the winged ones). Clearly that's not what they're for.
dwindling7373@feddit.it 1 month ago
They are clearly not wings meant to create aerodinamic lift.
MudMan@fedia.io 1 month ago
Well, no, they're meant to make the pew-pew laser fights look like a film about airplane dogfights. So yeah, way overthinking it.
finley@lemm.ee 1 month ago
You’re here discussing it too, man
MudMan@fedia.io 1 month ago
Oh, yeah, no, but that's because I'm a nerd.
KoboldCoterie@pawb.social 1 month ago
Technically everything with mass creates its own gravitational field; most things just aren’t massive enough for it to be detectable.
Ageroth@reddthat.com 1 month ago
One of my favorite science facts: Because of how the strength of gravity diminishes as you get further away and stronger as you get closer, when you approach to within arms length of another person (approx 1m) the gravitational attraction between the two masses of your bodies can exceed the gravitational attraction between your body and the sun at any given time.
finley@lemm.ee 1 month ago
When the inverse square law and weak gravitational forces meet
shutz@lemmy.ca 1 month ago
Yeah, but no one can escape the gravitational field of your mom.
(Sorry, couldn’t resist, as I half expected your comment to end with a “your mom” joke)
ramble81@lemm.ee 1 month ago
“Gravity deck plating”… okay that makes sense. So basically each floor has its own gravity generation to orient you to it. They’re all placed “bottom to top” to work like a building but it’d be possible to put one in at a 90-degree angle for say maintenance work.
Tlaloc_Temporal@lemmy.ca 1 month ago
There are crew walkways (I think they even have handguards!) along the beam path of the superlaser, so there are definitely at least a few small decks at different angles.
intensely_human@lemm.ee 1 month ago
Like in the Falcon
gravitas_deficiency@sh.itjust.works 1 month ago
Artificial gravity and inertial compensators are pervasive (if relatively handwaved/unexplained) in the SW universe
Tlaloc_Temporal@lemmy.ca 1 month ago
Hoverpads have fully replaced wheels and datapads have fully replaced paper, even in the poorest most remote communities.
thejml@lemm.ee 1 month ago
How much gravity would the Deathstar’s mass provide? I feel like it would be very small considering it has no real massive central solid or liquid core.
_haha_oh_wow_@sh.itjust.works 1 month ago
It’s the size of a moon and made from metal: It’s definitely generating some gravity (even a small amount of mass generates gravity) but I guess whatever tech they use to generate gravity overcomes it.
lolcatnip@reddthat.com 1 month ago
It’s the size of a very, very small moon, and mostly hollow.
Kusimulkku@lemm.ee 1 month ago
I should call her
Khrux@ttrpg.network 1 month ago
Yeah the fact it’s called a small moon is slightly deceptive to us because our moon is absolutely huge as far as moons go. The natives of the SW universe would be used to much much smaller moons.
For reference, our moon is 3475km across and the death star is 150km across, so it’s diameter is 23 smaller. It’s also weighed at about 900million tonnes or 9*10^14kg.
If I’m right (which I’m likely not). g=(GM)/r² or g=(6.66710^-119*10^13)/75².
That’s a gravity of 1.086x10^-5m/s² or if I round with pure disrespect for physics, 100,000 times weaker than earth’s gravity. Essentially it’s totally negligible compared to their artificial gravity. Hell, I don’t even think a marble on the floor would overcome it’s own grip and roll towards the center of the space station.
My maths is almost certainly wrong somewhere here, I failed it badly.
frezik@midwest.social 1 month ago
So small that a natural body of that size probably wouldn’t be massive enough to hold a spherical shape. DS1 was a little smaller than 128 Nemesis, which isn’t spherical. Maybe if it were made of something extremely dense, it would be, but you’re not likely to find a natural spherical object that small.
Now that I think of it, this puts the “that’s no moon” scene in perspective. Luke is a country bumpkin who just calls it a moon, but Obi-wan has an idea of its size (perhaps from glancing at the Falcon’s scans, since size and distance is hard to judge by eye), and knows a natural object couldn’t be that spherical.
butter@midwest.social 1 month ago
It wouldn’t need to generate gravity.
Acceleration “down” would be enough.
Scubus@sh.itjust.works 1 month ago
Only if it was undergoing constant acceleration, which we know it to be incapable of.
MotoAsh@lemmy.world 1 month ago
I mean, those are equivalent forces. Gravity doesn’t actually exist as a separate force, just like acceleration isn’t a magical force appearing from nowhere.
Agent641@lemmy.world 1 month ago
I ran the numbers, and it says about 0.00473m^2^2