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

Question

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

https://starwars.fandom.com/wiki/DS-2_Death_Star_II_Mobile_Battle_Station

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?

Answer 1

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.

Answer 2

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.

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.

Answer 3

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.