Would the Death Star affect a planet's tides?

[Wayneguard]

I don't actually think the Death Star orbits anything, so I doubt it.

"you can see here the Death Star orbiting the forest moon of Endor."

...

 

[Lovely Mixture]

Everything that needs to be said has been said. The keyword here is Mass.

 

[Guffe]

I don't know. But I doubt they'd need to care about the tides or water in general and start kissing the emperors ass pretty freakin' quick.

 

[RufusMcLaser]

To summarize all the good posts so far, it depends on the actual mass of the Death Star (I don't have that figure) and its distance from the body of water (we'll assume the notional planet has H20-based oceans rather than, for instance, molten lead) in question.
And yes, do submit this as a What If to Randall at xkcd.

 

[ssgt splatter]

Well the Death Star is so massive that it has its own natural gravity so I think it would cause some tidal fluctuation.

 

[Spade Lead]

Alright all you science-inclined Escapists, I have a question that I've been rattling around in my mind for a while.

In the Star Wars movies the Death Star is supposed to be big. Really, really big. Big enough that it could be mistaken for a moon when it's near a planet. But when I thought about this, a rather interesting question sprang to mind: if the Death Star is large enough to be mistaken for a moon, would it affect the tides of a planet when it came close to it? I realize that tides are normally caused by the moon, so would the addition of a moon-sized object so close to a planet cause the normal patterns to get screwed up? Or would it be either too small or too far away to make a difference?

The Death Star, some 160 Kilometers in diameter, is still mostly empty space. While it is HUGE, the effects on the tides of say, Endor, would actually be minimal. That is because, while there is fucktons of Duralloy armor and stuff, in the end, it is mostly empty space at the core for the reactor, incredibly large hangar sections, living spaces for the million + crew members, and consumables storage. If a planet had NO MOON, yes, it would gain a tidal surge from the Death Star, but really, why would you orbit a planet anyway? Zip in, blow it up, zip back out. (Zip is used ironically).

Edit: Got my stats wrong, it was 160, not 140, and Diameter, not Circumference.

 

[Souplex]

I don't think it comes close enough before firing to affect tides.
Mars is pretty big, and it doesn't affect tides.

 

[Spade Lead]

If it were the right kind of orbit around a planet, then yes. I don't think they ever specify exactly in the movies where the death star(s) were, exactly, though. I think they were in orbit of stars, rather than planets, though.

We never see the Death star go into hyperspace though, and the rebels seem close enough to launch an attack with what can be assumed to be short-range craft... the implication here is presumably that the Empire built their superweapon right next to the rebel base they spent the entire film looking for, and just didn't notice.

Yes, in the Alderaan system, right next to the Red Giant star we never see until the final scene. They MENTION traveling through hyperspace IN THE MOVIE.

 

[artanis_neravar]

If it were the right kind of orbit around a planet, then yes. I don't think they ever specify exactly in the movies where the death star(s) were, exactly, though. I think they were in orbit of stars, rather than planets, though.

We never see the Death star go into hyperspace though, and the rebels seem close enough to launch an attack with what can be assumed to be short-range craft... the implication here is presumably that the Empire built their superweapon right next to the rebel base they spent the entire film looking for, and just didn't notice.

All of the Rebel "wings" are hyperspace capable and are long-range fighters/bombers/interceptors. The death star is also seen in the Alderaan system before the rebels sensors pick it up entering the Yavin system.

snip

It did have a strong enough gravitational pull to draw the Executor into it.

 

[DaWaffledude]

You should check this out: http://www.theforce.net/swtc/holocaust.html

Very good read about the aftermath of the second death start destruction and its affects on Endor!

A-chem

http://starwars.wikia.com/wiki/Endor_Holocaust

 

[Spade Lead]

snip

It did have a strong enough gravitational pull to draw the Executor into it.

Actually, no. The Executor spun out of control and DOVE into the SECOND Death Star, which was over twice the size of the first Death Star. The Executor was in no position to "Fall" into the Death Star, as it was actually cruising no where near the Death Star while engaging the rebels. The whole point of the Imperial fleet was to draw the rebels AWAY from the Death Star to lure them into it's engagement envelope. The Death Star has a range of 3 Million Kilometers. If the ships are 20 feet away, it won't be able to target them. It needed the space to crush the rebel capital ships. Executor went what we on the navigation team call NUC, Not Under Command. Usually, vessels NUC end up running aground, whether their engines are unable to be stopped, pushing them onto shoal waters, the helm is not responsive, causing them to travel in generally whatever direction the motors and rudder are steering when you lose command, or the engines die and the water pushes them aground. That is all that happened. She lost helm control and "Ran Aground" on the Death Star.

 

[Fappy]

Not trying to be mean or anything but it seems like the threads are starting to get lazy. If only we had more gender and drowning threads...

Just awhile ago we were drowning in gender threads.

 

[Blue_vision]

The Death Star is apparently about 160 km in diameter. This is compared to the Earth's moon's diameter of 1,700 km. So, the moon has over 100 times the volume of the Death Star, and is likely considerably denser (it seems like most of the Death Star's interior is made up of corridors, so a sizeable proportion of its volume would be gaseous.) So let's say that the moon is 150 times as massive as the Death Star. Using this (conservative) estimate, the Death Star would have to be 5 times as close to the earth as the Moon to have a similar effect on the Earth's tides, or about 70,000 km from the surface.

Of course, when we look at the Alderaan scene from Episode IV, we see that, from the Death Star, Alderaan seems to be only marginally larger than the Earth in the famous "earthrise" photo. This means that the biggest view of Alderaan from the Death Star could only have been about twice as big as the Earth from "earthrise", which was shot on the moon (assuming that the image of Aleraan being destroyed was what a person would see if they were seated right in front of the Death Star.) Since we know Alderaan is of similar size to Earth, we can calculate that the Death Star would have been about 170,000 km away from Alderaan, 2.5 times as far as it would need to be to have the same tidal effects on the Earth as the Moon, well outside of the realm of error in estimation.
At this distance, the Death Star would have about 1/15 of the gravitational effect on Earth as the Moon would, and tidal effects would be greatly diminished.

We can assume that if the Death Star's laser worked at this distance of 170,000 km away, it would be unnecessary for it to ever be any closer to a planet, as doing so would put it at greater danger of being damaged by enemy defences. Based on these assumptions, we can find that the Death Star would have a greatly diminished effect on the Earth's tides. Seeing that the inhabited moons of the Star Wars universe appear to have the same surface gravity as Earth, we can assume that the same is true of all moons which the Death Star would be orbiting; at the best, it could have about a magnitude difference in tidal effects as compared to the Earth-Moon system.

EDIT: I see that the discussion has shifted towards the Second Death Star. My argument is invalid.

 

[Spade Lead]

EDIT: I see that the discussion has shifted towards the Second Death Star. My argument is invalid.

Yeah, that kind of took me by surprise, too. Like, a LOT.

 

[Samantha Burt]

The original Death Star? Its technical specs list it as 1.9x10^13 kg. That's about a thousandth the mass of Phobos - really the most pathetic moon (its escape velocity is only about 28mph). The DS2 is about 10 times greater in diameter, so it could affect a small moon, such as Endor's, but it wasn't even finished, and the moon is largely lacking in surface water, anyway.

 

[F'Angus]

It it passed by a suitable planet then possibly. Although it is fairly hollow with all the corridors so I don't know if that would affect anything.

 

[Spade Lead]

The original Death Star? Its technical specs list it as 1.9x10^13 kg. That's about a thousandth the mass of Phobos - really the most pathetic moon (its escape velocity is only about 28mph). The DS2 is about 10 times greater in diameter, so it could affect a small moon, such as Endor's, but it wasn't even finished, and the moon is largely lacking in surface water, anyway.

But Phobos is an ACTUAL moon, and thus High Density. The Death Star 2 would have been, at completion, still a very low density object, according to your calculations, only 1/100th the mass of Phobos. Not counting the implied probability of STAR DESTROYER Docking bays (a feature never shown in the first Death Star). That actually further reduces the mass of the station as a whole.

 

[snekadid]

As people have said - the actual mass of the Death Star is crucial. It's big, but also hollow, and we don't know how dense the materials it's made from are.

It also depends on how far away from a planet it is. Gravitational pull diminishes at (roughly) the square of the distance a body is from another, so it would need to be good and close.

Lastly, it would also matter if it was orbiting around the planet or stably hovering over a fixed point. Tides are caused by the moon travelling over the surface of the planet, pulling water with it slightly. This wouldn't happen if it was above the same point in the sky all the time. In that case, initially the death star would draw water to it when it arrived over a planet, but after that water levels wouldn't change.

You're forgetting that the Death Star has to be moving since the planet is constantly moving, not just revolving but rotating around the systems sun. Even if the Death Star was able to hold perfect relative position to the planet while it Rotated, for your scenario to work it would also have to perfectly orbit at the same rate and position that the planet revolves to have that effect on the waves.

In all likely hood the water would rise as the area closest to the DS passed in sight of it and then settle as it revolved away, the degree of which would of course be reliant on the previously mentioned mass and density.

captcha: i'm blushing, THE CAPTCHA! IT LIES!!!!!

 

[BeeGeenie]

No. 'cause it got blowed up. *ba-dum tish*

but if it hadn't been blown up, then yes, it would have exerted a small gravitational force on any planet it got close to. Whether it would have been noticeable? Probably not. It depends how close it is, and by the time it got close enough, the planet in question might not be there anymore.

 

[dvd_72]

It all depends on the mass of the death star. Despite it being smaller than our current moon (as was mentioned here) it would be made of primarily metal. Primarily hollow metal, so I'm not sure how these factors would balance out. Best I can do for you is that it is possible. sorry.