Comment on How did gravity worked on the Death Star?
RecluseRamble@lemmy.dbzer0.com 3 months agoEarths 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 3 months 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