Comment on How did gravity worked on the Death Star?
MyTurtleSwimsUpsideDown@fedia.io 3 months agoThe 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.
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.
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
nadiaraven@lemmy.world 3 months ago
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 3 months ago
If Lemmy had gold, I’d give it to you.