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How did gravity worked on the Death Star? (feddit.org)
submitted 2 years ago by to c/science_memes@mander.xyz
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cross-posted from: https://feddit.org/post/1104168

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[–] 61 points 2 years ago* (last edited 2 years ago) (2 children)

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.

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  • [–] 37 points 2 years ago* (1 child)

    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

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  • [–] 8 points 2 years ago* (1 child)

    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.

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  • [–] 2 points 2 years ago* (last edited 2 years 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

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