cross-posted from: https://feddit.org/post/1104168
Any chance you know the canon explanation of how they counteract the gravity generated by the Deathstar’s mass?
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.
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.
It wouldn’t need to generate gravity.
Acceleration “down” would be enough.
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 Idiameter, and its 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.
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
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 mostly filled with air 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
It’s not massive enough to create its own gravity. They use gravity deck plating.
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.
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.
Technically everything with mass creates its own gravitational field; most things just aren’t massive enough for it to be detectable.
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.
“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.
Artificial gravity and inertial compensators are pervasive (if relatively handwaved/unexplained) in the SW universe
