If the death star was coming into orbit around a planet, I think the least of its worries would be what time high tide would be
Would the Death Star affect a planet's tides?
- Thread starter Lord Garnaat
- Start date
Yeah, even if a ship does have artificial gravity the pull of it shouldn't affect much, I have seen the formula to calculate how far it should be effective. However as you stated ships doesn't seem to be affected by the relatively small gravity field that has to be there meaning it doesn't really do much even close by. It would probably have helped a lot if it actually did have a stronger gravity field since it would make it all but impossible to fly near it without getting pulled in and that would make the hatches (which shouldn't be there) impossible safer.Lucane said:
Well, it seems like we've got a conclusive answer here.
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.
And yes, do submit this as a What If to Randall at xkcd.
Well the Death Star is so massive that it has its own natural gravity so I think it would cause some tidal fluctuation.
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).Lord Garnaat said:
Edit: Got my stats wrong, it was 160, not 140, and Diameter, not Circumference.
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.Daverson said:
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.Daverson said:
It did have a strong enough gravitational pull to draw the Executor into it.Spade Lead said:
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.artanis_neravar said:
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.
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.
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.Samantha Burt said: