In the movie Star Wars - Episode VI: Return of the Jedi, the Death Star 2 orbits around the moon of Endor.

Incomplete Death Star 2 in orbit around Endor

According to the Death Star 2 wiki page on Wookiepedia, the space station did not have functioning ion engines nor a functioning hyperdrive system:

enter image description here


Something as big and heavy as the Death Star 2 would quickly be pulled down by gravity if it did not have any means of propulsion.

How did the Death Star 2 stay in orbit above Endor when it did not have functional ion engines nor a functional hyperdrive system?

  • 74
    I think the key words are right there "in orbit."
    – Buzz
    Commented Jan 5, 2023 at 2:18
  • 93
    How does the Moon stay in orbit without engines?
    – Jon Custer
    Commented Jan 5, 2023 at 2:22
  • @JonCuster, that's a good point. The Earth and the Moon are gravitationally balanced as far as I understand it.
    – user143126
    Commented Jan 5, 2023 at 2:28
  • 57
    Yes, that is what ‘in orbit’ means.
    – Jon Custer
    Commented Jan 5, 2023 at 2:53
  • I’ve deleted a whole bunch of comments that all pretty much repeated the same information. Let’s keep the comments here to a minimum now.
    – TheLethalCarrot
    Commented Jan 7, 2023 at 13:14

3 Answers 3


Repulsorlift generators

This is addressed by the canon reference book Star Wars: Complete Locations (2016). While the second Death Star was being constructed, it was at first held up by a repulsorlift generator located on the Endor moon itself:

Destructive Might

The Death Star II is not parked in a naturally synchronous orbit above the Forest Moon, so remaining over one point on the moon's surface requires a considerable uplift force against Endor's gravity. Initially, the station was supported by a repulsorlift field projected from the same ground facility that would eventually supply the station with its defensive shield. Tales told by Ewok shamans relate that the extra weight on the moon's crust had dramatic side effects, including massive groundquakes, land that shifted and buckled, and lakes that spilled out of their natural basins.

(Source: Complete Locations (2016), page 167)

However, by the time of Return of the Jedi, the repulsorlift generators are located on the Death Star itself. The illustration on this page says that "Arrays of temporary repulsorlift generators fill engineering sectors" of the Death Star II. Presumably these were intended to only be necessary until the Death Star was finished and mobile.

  • 35
    That's so dumb.
    – OrangeDog
    Commented Jan 5, 2023 at 11:34
  • 3
    @OrangeDog I find this somewhat reasonable, assuming limitations of the shield generator. The Death Star must remain directly above the shield generator, and must do so by being in stationary orbit. But if the shield generator can't project a shield out to an orbit that high, the Death Star may maintain an artificially low stationary orbit by constantly accelerating upward. An object doesn't require propulsion to maintain an arbitrary stable orbit, but maintaining geosynchronous orbit at the "wrong" altitude does. Commented Jan 5, 2023 at 17:09
  • 5
    @OrangeDog There are any number of occasions in Star Wars where vessels remain in one place over a planet without being in orbit and without the use of any sort of thrust: At the start of Revenge of the Sith, where the separatist fleet is just hanging there, and the one battleship plummets when it loses power; the start of The Last Jedi, where the Imperial ships arrive in sight of the base and just hang there and a bomber can drop gravity bombs; Rogue One, where clearly the Star Destroyer is floating over Jedha City. Commented Jan 5, 2023 at 18:12
  • 10
    @OrangeDog it seems in line with the rest of the behavior of ships in Star Wars. From what we see in the movies, I don't think that universe is running on the Newtonion or Keplerian mechanics we know and love in the real world. I think the reference to "synchronous orbit" is what's out of place here. Commented Jan 5, 2023 at 22:52
  • 8
    @OrangeDog whatever a "repulsorlift" is, it's efficient enough that you can have something like landspeeders or hoverbikes that hover all the time. If the cost is that low (like the cost of leaving an LCD watch always-on), why wouldn't you make decisions like this?
    – fectin
    Commented Jan 6, 2023 at 13:56

The Death Star is in a geostationary (endostationary?) orbit over Endor.

As has been pointed out many times in the comments, the Death Star can simply be in orbit. Orbits do not require any power to sustain them, though a low orbit might degrade due to very small amounts of atmospheric drag.

Others have pointed out that the Death Star does not move relative to the surface of Endor. This might be an optical illusion, we don't see the Death Star for any appreciable period of time, but it also seems to need to stay directly above its shield generator.

A geostationary orbit is one where its orbital period is the same as the rotation period of the planet. On the Earth, that's an orbital period of 24 hours and requires a height of about 36,000 km above the surface or 42,000 km from the center of the Earth.

I've calculated the height of the Death Star above the center of Endor in two ways, from Wookiepedia and from on-screen. I get from 18,000 km (Wookiepedia) or 18,800 km to 38,000 km (on-screen). To make this stationary above the shield generator, Endor would rotate once every 18 hours (Wookiepedia) or 19.6 hours to 2.3 Earth days (on-screen). Not unreasonable.

I have the math and on-screen canon to prove it! Buckle up.

We can do the calculation from Wookiepedia and then check with what is seen on screen.

Calculating From Wookiepedia

To know how high above Endor the Death Star has to be to be geostationary, we need to know...

  1. The mass of Endor.
  2. Length of a day on Endor.

Length of Day on Endor.

Wookiepedia states 18 hours.

Mass of Endor

This is pretty easy. We take the equation for gravity: g = GM/R^2 (G is the Gravitational constant, which is equal to 6.67×10^−11 N m^2/kg^2, M is the mass of Endor, R is the radius of Endor) and solve for M. M = gR^2/G. Now we need to know the radius and gravity of Endor.

Wookiepedia describes its gravity as "light". We see people walking around normally on Endor, so let's say it's about 90% of Earth's gravity. Earth's gravity is 9.8 m/s^2 so Endor is 8.82 m/s^2.

Wookiepedia says Endor has a radius of 2450 km or 2,450,000 m.

We plug that into our equation: M = 8.82 m/s^2 * (2,450,000 m)^2 / 6.67×10^−11 m^3/kg s^2 and we get 7.9e23 kg. For comparison the Earth has a mass of 6e24 kg, so the answer for the smaller and less dense Endor is reasonable.

Height of the Death Star above the center of Endor

In order for the Death Star's orbital period to match Endor's 18 hour rotation, we need to solve the orbital period equation for height.

Orbital period is 2pi * sqrt(h^3/GM) where M is the mass of Endor and h is the height above its center. When we solve for height we get cbrt( (period/(2pi))^2 * GM ). Solving with a little Ruby program ...

include Math

G = 6.67e-11
M = 7.9e23

period = 18 * 60 * 60
puts cbrt((period/(2 * Math::PI))**2 * G * M)

We get about 17,762,921 m. Let's call it 18,000 km.

Calculating from the Movie

Height of the Death Star above the center of Endor

To keep this answer manageable, I made a separate question and answered it there.

Depending on the estimate of the angular diameter of Endor, I had to eyeball it, we get 18,800 km to 38,000 km. The lower bound is pretty close to the Wookiepedia numbers!

Orbital Period of the Death Star

Orbital period is 2pi * sqrt(h^3/GM) where M is the mass of Endor and h is the height above its center. Using the high on-screen value for the distance ...

  • Height of the Death Star: 38,000,000 m
  • Mass of Endor: 7.9e23 kg
  • G: 6.67e-11 Nm^2/kg^2
include Math

G = 6.67e-11
M = 7.9e23
H = 38_800_000

puts 2 * Math::PI * sqrt(H**3 / (G * M))

Plugging that all in: 2pi * sqrt(38,000,000^3 / 6.67e-11 * 7.9e23) we get about 203,000 seconds or 56 hours or 2.3 Earth days.

Using the low value for the distance, 18,800 km, we get 70557 seconds or about 19.6 hours. Pretty close to the Wookipedia 18 hours!

Maybe it's dumb luck it turned out reasonable, but I like to think the special effects people worked this all out.

What about when the Death Star is viewed from Endor?

When Luke surrenders, we're treated to a view of the Death Star from the surface of Endor. It's huge and looming. It is a common sci-fi trope to make things in orbit appear outlandishly large from the surface for visual drama. Let's check how the on-screen depiction of the Death Star compares to the Earth's Moon anyway.

Death Star as viewed from Endor's surface. It's large and looming in a dark sky.

The Earth's Moon has an angular diameter of about .5 degrees. For Death Star to appear that large it would need to be about 18,000 km away from the surface which, again, is very close to our other calculations!

In addition, objects close to the horizon appear larger than when they are high in the sky: the Moon Illusion. Using a high zoom lens "flattens" the image makes things seem larger and closer together. These two things in conjunction would make the Death Star seem quite large in the sky.

  • 2
    This would mean that the distance between the Death Start II and Endor was more than 10 times the radius of Endor. However, in the picture posted in the questions it seams that the distance between Death Star II and Endor was much less than that (maybe < 2x radius_of_endor between surface of Endor and Death Star II). Commented Jan 5, 2023 at 21:52
  • 3
    @MiloP Ha, I totally missed that Wookipedia lists the rotation period! If we try to match 18 hours we get 18,000 km. Pretty good! As for it using Star Wars magic rather than orbital mechanics, I'm with OrangeDog.
    – Schwern
    Commented Jan 5, 2023 at 22:52
  • 1
    @daniel.heydebreck 1) The picture in the question is not from the movie, Endor is much larger in the movie. 2) We don't know how far away we are from the Death Star in that picture, that can change the apparent size of the Death Star considerably. 3) We have no frame of reference to estimate their apparent diameters. 4) You can't eyeball it. For example, the Sun is 500 times more distant from us and 400 times larger than the Moon. Yet they are equal size in the sky.
    – Schwern
    Commented Jan 5, 2023 at 23:40
  • 1
    An endostationery orbit is one which consumes pencils and erasers as it orbits, versus an exostationery orbit, which produces pencils and erasers. Commented Jan 7, 2023 at 2:35
  • 1
    @RobertF Good question. Technically, yes. The difference between the two sides is 0.0029 m/s^2. To put this into perspective, an 80 kg person would feel about .23 N of force or less than it takes to type on this keyboard. The Death Star is quite safe from being torn apart.
    – Schwern
    Commented Jan 8, 2023 at 0:45

Propulsion is not required to be in orbit.

A body is in orbit around another if it is moving fast enough in an appropriate direction. The only force acting on it is gravity, which is constantly accelerating it towards the larger object (and vice-versa, but that doesn't really matter when one is significantly larger) and keeping it in orbit.

Once the Death Star 2 was placed in orbit - or rather the pieces of it as it was constructed - no further power is needed to keep it there.

  • 7
    The Death Star does not appear to be in an orbit that would sustain it.
    – Valorum
    Commented Jan 5, 2023 at 11:48
  • 2
    @OrangeDog propulsion is required to keep that low stationary orbit. Commented Jan 5, 2023 at 17:27
  • 3
    @OrangeDog - Orbits can be decaying; en.m.wikipedia.org/wiki/….
    – Valorum
    Commented Jan 5, 2023 at 17:28
  • 7
    @YaroslavKornachevskyi Note that they are around the "Forest Moon of Endor", not around the planet of Endor itself. Endor is a large, unihabitable gas giant. The moon is much smaller, so the distance required for a GEO orbit is probably much less. (Granted, it still seems to have Earth-like gravity, but then, so does every planet and moon and even asteroid they ever visit in the franchise, so I guess you just gotta let that one go.) Commented Jan 5, 2023 at 17:32
  • 3
    @HorusKol According to Wookiepedia, both the planet and the moon, and apparently the entire star system, are all called Endor, though the planet may also be called Tana (apparently only by the Ewoks). Presumably the moon is the only habitable body in the system so everyone just conflates them? Commented Jan 5, 2023 at 21:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.