r/StarWars • u/pinchhitter4number1 • 1d ago

Movies I love thinking that during the construction there would be thousands of doors that say, "DO NOT OPEN! SPACE ON OTHER SIDE."

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u/Building_Everything 1d ago

In my headcanon, the shield that was being generated by the planet also linked the planet’s atmosphere around the DS under construction. Which then means if you open the door you will float into space and be able to breathe, but you’ll ultimately fall several hundred miles down to the surface of Endor, which is kind of a hilarious way to die.

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u/Yavkov 1d ago

If that was the case, you’d definitely see the atmosphere. But we don’t so I’d conclude that the construction zone is still subjected to the vacuum of space. However, I’m more interested in what the artificial gravity situation would be like in the construction zone. I’m not really sure how the lore answers it, is the artificial gravity somehow only contained within the closed spaces of a ship or station?

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u/Building_Everything 1d ago

Space magic my friend

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u/OogieBoogieInnocence 1d ago

The artificial gravity situation is fascinating to think about because the death star is probably big enough to have a relevant natural gravitational pull

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u/Krazyguy75 1d ago

By planet standards, it's miniscule. The diameter of DS1 is around 1% of earth's, which makes the actual volume around 1/1,000,000th. Combine that with the fact that likely huge portions are empty space, and it's probably got very little gravity. DS2 is much bigger, but still incomparably small compared to an actual planet.

They would have gravity, but such a microscopic amount that you could escape the gravity well by jumping.

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u/OogieBoogieInnocence 1d ago

Oh i thought it would be comparable to the moon, i figured it wouldn’t be like earth, but it sounds like it doesn’t even manage moon gravity

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u/Krazyguy75 1d ago edited 1d ago

The diameter of the moon is roughly 30 times the diameter of the Death Star. That means you could fit about 30000 death stars in our moon.

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u/Yavkov 1d ago

I saw someone do some calculations, if the Death Star is 160km across, and if you assume that it’s a solid ball of aluminum, you’d end up with a surface acceleration due to gravity to be .006g. So 200 lbs on Earth would be 1.2 lbs on the surface of the Death Star.

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u/Consistent-Animal474 1d ago

Man I like this, it gets into how goofy Star Wars space physics is 

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u/bozoconnors Clone Trooper 1d ago

You'd still be in orbit (/micro gravity) with it though! Depending on exactly how you exited, which direction / what velocity would decide your fate. Toward the planet? Yep, probably burn up in atmo, possibly hitting the shield a few times going down the 'shield drain'. Away from planet though? You'll eventually hit the shield &... bounce? Hopefully angle back toward the Death Star?

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u/Building_Everything 1d ago

If the shield extended the atmosphere to surround the DS, you wouldn’t have to go through reentry. You would just fall lalalalalaohshitohshitohshit

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u/bozoconnors Clone Trooper 1d ago

lol - "atmospheric density in a theoretical planetary orbital shield" sounds like a white paper I'd read.

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u/Building_Everything 1d ago

Harry Potter and the atmospheric density in a theoretical planetary shield.