Yeah, but in the immediate confusion upon coming out of hyperspace I think Obi-Wan was the only one capable of immediately accepting that the “asteroid field” was, in fact, the demolished Alderaan. No one else could really process the reality of the situation, Han outright denies the possibility. No reason for Luke to think they’re at the demolished remains of an ex-Alderaan, no reason to call that thing a moon.
But, out of curiosity, does anyone know the official IAU stance on what to call that which was formerly a satellite of a now no longer existent planet? Does it still get to be called a moon?
It’s hard to believe a galaxy spanning empire would fall due to the death of a single individual and a couple of capital ships. It must have hundreds of billions, maybe trillions of soldiers and loyal citizens. Which I guess is basically the premise of episode 7.
A Death Star does seem impractical, but I think of it as a future version of the Nazi “super weapons” that didn’t really do much. Terror weapons, mostly. I’m sure the Empire has a MIC pushing for absurd weapons platforms that don’t really apply to the current enemy, just like we do.
The gravity generated by the Death Star would be next to nothing, completely useless to help a person walk around normally.
According to most sources, the Death Star is about 100 KM across. By way of comparison, Mimas, the moon of Saturn that looks creepily like the Death Star, is four times wider and therefore eight times the volume. To add further convenience to the comparison, in addition to looking like the Death Star, Mimas is the smallest thing known to be rounded due to its own gravity.
But that gravity is REALLY small. The surface gravity is less than a hundredth of Earth’s. The Death Star’s would therefore be smaller than that, since it’s smaller. Furthermore, the Death Star is way less massive than a moon of similar size would be, because unlike Mimas, it’s not a solid rock. It’s mostly empty, in fact. The great majority of the Death Star’s volume appears to be air. So the gravity would be super tiny.
The Death Star is really super duper big by the standards of spacecraft, but by the standards of planetary objects it’s tiny.
I think there’s a kind of pop culture understanding of bodies broken into planets, moons, asteroids, meteors, dust - things are more about size than placement. Certainly more so when this was written. By a guy who thought parsecs were a unit of time. So it’s too big to be an asteroid but too small to be a planet - ergo a moon.
Yeah, wrong, but pop culture science understanding often is.
I don’t have the decent cite that was thrown at me at the time, but it was on this board that it was shown to me that the explanation for the use of the word parsec was just a straw-grasping after the fact explanation and that it truly was a mistake on Lucas’ part to begin with.
That said, it was a truly artful straw-grasping after the fact explanation. Always made perfect sense to me.
The Empire had been in place less than 20 years, so I’m not sure how many loyal citizens there were. Scared ones, yes. There was also not a lot of succession planning, so who was going to run the place with Papatine and Vader both dead? Sith have but one dark-side minion, I can’t see Palpatine giving a lot of power to anyone else.
I’ve only read the first EU series, but I got the impression that Palpatine made use of the force to keep his troops in line. Without it, the fleet would fragment. The celebration at the end of the new version of Ep. VI certainly makes it seem that the inhabitants of Coruscant weren’t afraid with him gone.
Oh, sure but every little bit helps. And even a slight sideways pull would be somewhat disorienting. The floors would never feel level. And when the 127th Storm Trooper Catering Corps hosts a miniature golf game against a detachment of mechanics from the 823rd AT-AT Expeditionary Force, the home team would have a tremendous advantage.
Okay. It’s Luke who calls it a moon. He has never even been off-planet before. I think that’s good enough reason for his mistake, even though it’s not an unreasonable supposition, all things considered.
Personally, my favorite fanwank for the parsecs thing is that Han is an idiot, or at least doing a convincing job of imitating one. You’ll notice that Obi-Wan rolls his eyes at the boast.
I think implicit in the ability to create artificial gravity is the ability to negate naturally occurring gravity.
I like to picture Darth Vader participating in the tournament, and just really blatantly using Force telekinesis to cheat, and nobody having the balls to call him on it.
What gravity the Death Star had would not only be minuscule, but fairly evenly applied. I mean, for any point inside the sphere, there is going to be mass all around you. The imperceptible tug from the fusion core will be off-set by the imperceptible tug from the hyperdrive, etc. None of it would have any effect worth noting.
No, the thing I want to know is the actual mechanics of getting the artificial gravity to center at the south pole, and does said gravity increase as you move down deck levels.
I don’t think that’s right. Unless you’re right in the middle, from most points within a sphere, there’s going to be more mass in one direction from you than in another, and consequently, higher gravity.
Ah, now this part I do know. We worked through the math in a Physics class in college, and I found it fascinating.
There are a few assumptions involved. The math works on a perfect sphere of uniform density. You can analyze it as a collection of concentric, hollow spheres (again, like the layers in an onion). The overall gravitational pull is the sum of the pulls to each of the shells.
From a point outside of a shell, you are attracted to each part of the shell according to the law of gravitation. If you sum it all up, it’s exactly the same force as if the entire mass of the shell was condensed to a single point at its center. For a point inside a shell, the parts of the shell pulling you in different directions perfectly cancels out. And here’s the really fascinating bit, that goes for anywhere inside the shell, not just at the the center.
So the overall attraction to a spherical body is the sum of all the shells closer to the center than you are. So, there’s the shell you’re standing on, all the shells within it, add up the mass, consider the mass to be concentrated at the center, and use the equation for gravity to figure out the force. As you get closer to the center, there’s less mass to consider (which would lessen the force) but you’re getting closer (which would increase the force).
Now, the Death Star is not of uniform density, but if my understanding of the math is correct, as long as each shell is uniform it shouldn’t matter. It’s just the total mass inside of the shell you’re standing on. I think a dense core would tend to make the situation worse for those who were close (relatively speaking) to it.