The Death Star is a huge space station, literally the size of a moon. Now I know you can park TIE fighters and shuttles and even craft as big as the Millennium Falcon inside of the station. But when you see this photo:

Death Star 1 and a star destroyer

Where exactly do you park a Star Destroyer at the Death Star? In the first episodes of the series, we've seen that the older version of star destroyers could actually land on the surface of a planet. While I'm certain that this newer version doesn't normally do that, would the Death Star actually have a landing bay for a vehicle this size?

Once again, where do you park a Star Destroyer once it's at the Death Star?

  • 40
    The photo's a tad misleading. The Star Destroyer is pretty far away from the Death Star; a side-by-side comparison would show that the Star Destroyer is tiny in comparison. I'm still not sure if large enough landing bays exist. – HDE 226868 Feb 28 '15 at 16:23
  • 22
    @Chris: What would be the point of docking a Star Destroyer inside the Death Star? Routine personnel travel is handled by shuttles and small craft, as is material transport, and routine maintenance is handled by the Star Destroyer's onboard repair staff. Literally everything that smaller craft need to land/dock to deal with is a solved problem for capital ships (like Mon Cal cruisers and Star Destroyers). – Jeff Feb 28 '15 at 16:41
  • 4
    The way I see that photo, I assume perspective such that the distance from the "camera" to the star destroyer is much less than the distance from the star destroyer to the death star. There is not enough information in the image to infer much beyond that. The slight fuzziness of the death star's surface detail and difference in lighting (star destroyer is well lit but overlaps the shadowed portion of the death star) leads me to infer likely relative distances such that the death star must be vastly larger than the star destroyer. – Anthony X Feb 28 '15 at 16:55
  • 9
    jeff Russell's Starship Dimensions site has all the relative sizes you need for comparison. It looks to me like there would be no problem docking even a Super Star Destroyer inside a Death Star. – Joe L. Feb 28 '15 at 16:58
  • 3
    Just to fully clarify the Star Destroyer/Death Star relative sizes: youtube.com/watch?v=dS12p0Zqlt0 – Paul D. Waite Feb 28 '15 at 19:34

First, I find it hard to believe that those are scaled similarly. The Imperial Star Destroyers (both the first and second class) are 1,600 meters long. Essentially, they are about a mile long. You could run from one end to the other in less than 10 minutes.

The Death Star was the size of a small moon. The first Death Star was a rough sphere 120 kilometers in diameter. In other words, a Star Destroyer's length was 1.3% of the Death Star's diameter. A docking bay capable of holding a Star Destroyer would need to be enormous. Let's assume the Star Destroyer is a rectangular block 1600 meters long, 800 meters wide, and 800 meters high (which it isn't, it is significantly smaller than that) and that a docking bay needs to hold two of those with sufficient space for clearance and maintenance hardware. Let's say this results in a bay that is 1800 meters long, 2400 meters wide, and 1000 meters high. That is a volume of 4,320,000,000 cubic meters (4.32 cubic kilometers).

That sounds HUGE, right? It really isn't, compared to the internal volume of the Death Star. The Death Star has an 60,000 meter (60km) radius. Plugging that into the calculations for the volume of a sphere, we find that it has a volume of 904,779 cubic kilometers.

That's means the theoretical bay would be less than 0.0005% of the internal volume of the Death Star.

So you can see, there could be LOTS of places they could put a docking bay for a Star Destroyer.

That said, there's no real need for them to have a docking bay for ships that size.

The purposes of docking bays is to allow maintenance, restocking of provisions/fuel, personnel movement, etc for small craft. You park the Millennium Falcon, Chewie can climb up on top and mess about with hardware, you can attach feeder hoses to restock the water reserves, reclaim waste, refill consumables, etc. People can walk onboard or leave. All of this is needed with small ships, where space is as a premium.

Star Destroyers don't need to dock to accomplish this. All of their routine maintenance is done without docking. They were designed such that anything they can repair out of onboard capabilities can be done without needing to dock. They have personnel who are trained to work EVA in those rare cases where external repairs are needed, and they have machine shops and engineers who can produce most of the parts they could need.

Refueling is not something that has been addressed. We know (according to Wookieepedia) that the ships are powered by an 'I-a2b solar ionization reactor', which is referred to as seeming like a miniature sun. This reactor uses fully half of the volume of the ship, and the fuel it consumes is unknown. The Imperial Star Destroyer is listed as carrying 'consumables' for 2 years for its standard crew complement (according to a rulebook published in 2014 for a licensed game). Given that, we can assume that the ship was expected to not need to take on consumables for about a year at a time. It would be reasonable to assume that it refueled when it restocked.

Star Destroyers that aren't in need of severe maintenance would find it easy to orbit the Death Star (as it could any moon-sized object). Those which did need serious maintenance would likely have been sent to a dedicated repair facility, such as Kuat Drive yards (where many were constructed).

TL;DR: Your picture is way off scale, the Death Star is easily big enough to have docking bays big enough for Star Destroyers to land within, but it's unlikely that anyone would build them in because why would they need to land Star Destroyers within the Death Star?

  • 6
    Also, I don't think the picture is off scale, it's just that we see a Star Destroyer approaching the Death Star, but it's still a long way off. And also, how did that surplus r end up in my other comment? – SQB Mar 1 '15 at 19:48
  • 2
    Even if you did want to dock a Star Destroyer with the Death Star, you could do it in any number of ways. "Nose-in" docking, landing on the surface, low orbits etc... – Stephen Mar 2 '15 at 4:52
  • 4
    This ship is too big! If I walk, the movie will be over! – BBlake Mar 2 '15 at 12:54
  • @SQB: Math corrected to account for DS1's correct radius, as well as a minor calculation error I made. – Jeff Mar 2 '15 at 16:27
  • @SQB I think the extra 'r' is because there were only two Death Stars, and the both got destroyed. So clearly, the Empire didn't have enough of them. There was a dearth of Death Stars. – KSmarts Mar 2 '15 at 16:30

A Super Star Destroyer crashing into the Death Star

As you can see, even a Super Star Destroyer is much smaller than the Death Star.

Of course, this was the second Death Star, which was just a tad bigger than the first one — a diameter of 160 kilometres vs 120 kilometres.

Your picture shows a Star Destroyer approaching the Death Star, but it seems to be still a long way off, which is why it looks relatively big, compared to the Death Star.

  • 146
    This is a very liberal interpretation of the word "Park". – Valorum Feb 28 '15 at 20:20
  • 19
    @BMWurm, just being pedantic here, but when you increase the radius (or diameter) of a sphere by a factor of 33%, you increase the volume by a factor 137% (more than twice, but still not near 5 times) – Boluc Papuccuoglu Feb 28 '15 at 23:03
  • 3
    Also, for reference - the Executor was 19,000 metres long. The Imperial-I class in episode IV was "only" 1,600 metres long - making a the Executor more than 12 times length of an Imperial-I – HorusKol Mar 2 '15 at 4:39
  • 5
    Actually the picture in your answer does not seem at scale. If the depicted ship is 19km long (as claimed by @HorusKol), the part of the Death Star at the left edge of the picture would be roughly 60km from impact. Now, since the radius of the Death Star is 80km, at 60km distance the curvature should be pronounced significantly. All I see in that picture is a flat expanse of techno-jumbo. – M.Herzkamp Mar 3 '15 at 15:16
  • 8
    Lucas really messed up the scale factors in the scene shown in this post. If you assume a 19km length for the Executor, based on the measurements of the curvature in this picture, the death star would have a diameter of about 11,300km; just a smidge smaller than the Earth. – Stewbob Mar 3 '15 at 15:45

The DS-1 Orbital Battle Station (also known as Death Star) did not have any specific docking bays for Star Destroyers. It had numerous docking bays for smaller ships, like TIE-class fighters and bombers, or landing shuttles, along its equator in a 1 km wide trench (source: Wookieepedia article on the Death Star).

They didn't need a special docking bay for larger ships, because nearly all Star Destroyers had their own docking bays, so they could simply shuttle over. Also, consider that a docking bay of that size would require a massive forcefield to keep the air in. (source: Original Trilogy footage).

In addition, the way physics work is that it would be very difficult for a Star Destroyer to stay airborne in the artificial gravity the Death Star has without inflicting massive structural damage to the docking bay. This is because, for the Star Destroyer to stay airborne, it would have to exert a force equal to its own mass on the floor. According to https://archives.swc-empire.com/a/index.php?title=Imperial_Star_Destroyer_I, an Imperial I Star Destroyer can weigh 6.4 million T. You would need a VERY strong docking bay floor to support that kind of weight.

However, the DS-1 does have a number of dry docks. I am not sure how canon these are, since this whole new Canon/Legends division has me confused, but here is an image from https://starwars.fandom.com/wiki/Death_Star

Star Destroyer in docking bay

  • 5
    This all sounds very plausible. That said, it also sounds like fan-theory. Can you back any of this up? – Valorum Feb 28 '15 at 21:18
  • @Richard I shared some of my sources on this matter. I also removed the biggest speculation part about the technology. – Nzall Feb 28 '15 at 22:45
  • 2
    Except that we see ships nearly as large parked on the ground at Coruscant. When you've got gravity-manipulation tech, the actual mass of an object is largely irrelevant. – Valorum Mar 1 '15 at 0:29
  • You are getting your descriptions and therefore your units wrong. Thrust(force) != mass. The weight (force) of a Star Destroyer may be 6.4 megaTonnes x m/sec^2 but it's mass would only be 640 kiloTonnes. Since the gravitational acceleration of DS would be <1/10 of Earth's, assume it weighs 640 kiloTonnes x m/sec^2 or less on DS. – Jim2B Mar 1 '15 at 6:29
  • @Jim2B I think Nate is assuming an earth-like artificial gravity in the landing bay, not the natural gravitation of the death star (which would be much lower, as the DS is not massive as a real moon). – Paŭlo Ebermann Mar 1 '15 at 16:47

The Death Star had mooring towers for docking large spaceships like Star Destroyers.

Mooring tower on the surface of the Death Star

  • Welcome to SFF.SE. Good find (+1)! I added an image to improve the answer (link-only answers are discouraged), but the addition of the image provides visual evidence here on SFF.SE. – Null Mar 17 '15 at 17:32
  • You may wish to add a note that the referencing for this is from a non-canon source, namely an online promotional game called "Death Star Designer" – Valorum Mar 17 '15 at 18:29
  • Isn't that tower in the picture the location of the Emperor's throne room? Definitely not a mooring tower. – RichS Nov 13 '19 at 5:28

The movies never address the question, however the series background FAQ might. In the movies, we do however see multiple times shuttles landing on the death stars to transport resources, troopers and dignitaries (specifically Vader and Palpatine).

So i would say that it is far more likely that star destroyers do not dock and instead use shuttles to transport between the ships.

However as mentioned above, the death star would have adequate space for thousands of star destroyer docking bays, so it is not at all out of the question.


From what I recall of the original “Star Wars” movie trilogy, Star Destroyers would only ever have been “parked” around the Death Star, not inside it, even though I suppose there would be plenty of room to park a whole fleet of them inside.


First, as @SOB's picture points out, the Star Destroyers are still noticeably smaller than the Death Star, and so one could potentially fit inside.

Second, we can see that the Millenium Falcon manages it by a simple application of landing struts.

enter image description here

While a Star Destroyer could potentially have landing gears to do the same, evidence is inconclusive at best on whether or not they do - and leans towards 'they don't'. However, 'older' models do in fact have landing struts, which were designed-on in the prequels that came out after ANH-ROJ.

They could also dock by hanging, which is how TIE fighters do it in some canon, but there's even less evidence to support this theory.

So while there are some ways in which a Star Destroyer could dock internally with the Death Star, none seem to be shown in any canon.

  • 5
    -1 for a speculative ("it's not difficult to imagine") answer. +1 for hand-drawn circles. – KSmarts Mar 2 '15 at 16:32
  • @KSmarts Speculative but based on real ship models. I know this answer would be improved with something as simple as the design documents for the Star Destroyer (or fan-made ones, or lucas-made ones) but I don't know where those are. – Zibbobz Mar 2 '15 at 17:23
  • 4
    -1 for failing the Square-Cube Law. Landing struts don't scale well. – Shadur Mar 3 '15 at 13:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.