I’m thinking of getting into the planet-destroying business, and I figured I’d borrow some inspiration from one Mr. Lucas. You want to destroy something planet-sized, you need a really big weapon. So of course, you build a space station the size of a moon.
So you build your giant, spherical space station. Much of it is effectively hollow, to accommodate the giant laser and room for the crew and the great big bottomless pits that all evil lairs must have. It’s still pretty massive, though, because it’s really big and made out of metal. It’s big enough that it’s going to have some gravitational pull of it’s own - certainly not a lot (according to people who are nerdier than I am, it’s roughly 1/20th the size of our moon), but it’ll have something.
Now, I’m not quite up to the Empire’s technological level myself, so I haven’t quite perfected artificial gravity. But assuming I keep the same general size, shouldn’t my Death Moon have enough gravity itself such that I could “walk” along the outer surface of it? I mean, obviously, it’ll be awkward, given the need for a space suit and I imagine the extremely low gravity, but it would be somewhere between just floating and, say, the bunny-hopping Neil Armstrong et al did on our moon, right?
Secondly, how will this gravity effect the layout of my living space? There’s something so appealing about the spherical appearance, but obviously I’ll need to divide the interior into decks and provide things like floor for my minions. Since the government shut down my last project, I didn’t get to finish my experiments with building interior walkways through the center of the earth. Assuming I just go ahead and divide it into vertical decks parallel to the diameter of the circle, the way the Empire did, what’s going to be the effect of the station’s own gravity on that?
Well, it looks like the death star is about 1/20 the diameter of the Moon (165km versus 3,400 km). Which means it’s 1/8,000th of the volume of the moon.
And the death star is going to be mostly hollow (well, full of air), so even if the steel takes up nearly 1/10 of the total volume, the effective gravity of the deathstar is going to be something like 1/100,000th of the Moon’s.
That’s going to be pretty negligible. I’m afraid ‘floating’ is the operative description for anyone on the surface.
Which makes the second part of the question kind of moot. So for fun let’s assume you’ve got a quite a few million tons of unobtanium to build it out of, which is like a million times denser than steel (and strong enough to build a hollow station with the gravity of the Earth). Now we’re at something like Earth gravity at the outer surface of the station.
In this case, spherical decks are absolutely the way to go. Gravity at the outer deck will be pulling straight toward the center. Gravity on the spherical deck below that will also be pulling straight toward the center, but a little less strongly, decreasing with each deck until at the absolute center of the station, there’s no gravitational pull at all. (So maybe abandon the spherical layout towards the middle, when gravity is low enough to ignore and spherical decks would be too curved). If you built a deck parallel to the ‘equator’, it would feel tilted anytime you weren’t along the axis from ‘pole’ to ‘pole’. Not too bad up near a pole, where it would just feel like the deck curved up a little at each edge, but at the outer edge of the equatorial deck, gravity would be pulling sideways to the floor.
I always wonder why Lucas didn’t have the Death Star spin. Then he would at least have some plausible excuse for gravity on some portions of it. He could have had the Death Ray fire out the axis at the north or south ‘pole’.
Why? They clearly have some sort of artificial gravity, or else they’d be floating around on the Millenium Falcon.
The alternative is to believe that they simply ignored all gravity-related questions. Philistines would then point out that the gravity situation when they were inside the Giant Space Worm on the Bigger Asteroid in The Empire Strikes Back made no sense whatsoever (was gravity pulling sideways?), but we True Believers simply ignore such comments, preferring to ask questions about flying intestinal parasites like Mynochs.
Because, in the version of the Death Star that I’m building I prefer not to waste my budget on an artificial gravity system when I can have it for free (power required to induce spin notwithstanding).
If you’re still committed to building spherical Death Star, rather than changing to a cylinder (the Death Rod?), you’re going to have a lot of wasted real estate inside. I’d spring for the Gravity Generator, myself.
That just brings up the question: “Just how many people do I need to run this thing anyhow?” It’s not a colony ship. I would think that the surface area on just the first deck in from the inner pressure hull would hold plenty. The rest of the interior volume can be inhabited with mechanical things that don’t require gravity.
Given Quercus’s calculation that the gravity will be pretty minimal, this shouldn’t be much of an issue. Mostly floating will still be mostly floating.
UncleRojelio said:
CalMeacham said:
UncleRojelio said:
But you didn’t ask about your choice to design your own Death Star, you asked about Lucas’s choice and him having a plausible excuse. Well, he had a plausible excuse - artificial gravity.
Build a shell surrounding an artificially-created black hole. A black hole with the Earth’s mass is less than an inch wide. You can use that for a power source, too.
Blowing up a planet may look cool and have a lot of shock value, but in a Universe with thousands of years of history, you’d be better off throwing a couple of thousand nukes at it, killing every living thing on it (Exterminating a population is still a lot of shock value) and then allowing it to rest for a couple of hundred years so that you can eventually repopulate it.
I think your calculations are somewhat off here. If we assume that the death star is 10% iron and the rest air, we get that the density of the deathstar is about 1/4 that of the moon. The volume is 1/8,000, as you say, and so the mass will be about 1/32,000 that of the moon. However, since the radius is 1/20th and gravitational attraction decreases with radius according to 1/R^2, the actual gravity of the deathstar will be about 1/80th that of the moon, or 0.02m/sec^2 So still pretty low but not 1/100,000th
A Death Rod would only be appropriate if I were a male with a very tiny penis, and I am in fact female with no penis.
Nonsense! You people clearly have no idea how to rule with an iron fists: the lack of hand rails helps weed out the weak, as does forcing your crew to resort to cannibalism.
This is an idea I like.
Quercus shall be rewarded with protection when my plan comes to fruition, as that’s pretty much I was looking for. That always kind of bugged me in the films, since you don’t really get any visual sense of scale, just “Hey, it’s kinda moon-sized”, which I always took as having some sort of gravity, but then you’ve got the decks running perpendicular to the poles, which made me a little intellectually uncomfortable. Good to know that it’s just entirely silly, not mostly silly.
[ul]
[li]Take a rocket the size of the space shuttle.[/li][li]Accelerate it to say 20% of the speed of light.[/li][li]Smash rocket into planet[/li][li]Profit(?)[/li][li]Planet destroyed.[/li][/ul]
