I have read the Death Star (the first one) had a diameter of about 186 km, which made me think about whether it could somehow host a satellite or moon due to its size, just like some dwarf planets do.
-
4Everything can have a satellite, even a comet. It depends on 3 things: mass of the primary body, mass of the satellite, and the distance between them (also velocities, but that's not so important in this case). I'd guess you could have a Star Destroyer orbiting Death Star on a very low orbit, though I wasn't abe to find the mass of the Death Star.– GallifreyanJan 12, 2017 at 8:38
-
3@Gallifreian - The mass of the orbiting body appears in both the centripetal and gravitational forces, so it drops out. It does (negligibly) influence the position of the COM of the system.– AdamantJan 12, 2017 at 9:37
-
3@Valorum - Even if that can be demonstrated, it is not necessarily straightforward. Besides, it's probably false, in theory - at sufficient speed, a object can orbit even a fairly small body. Star Wars ships likely cannot go that fast, but it is not obvious from the answers to the other question.– AdamantJan 12, 2017 at 10:57
-
2@Valorum - an atmosphere is a collection of gaseous molecules, which is quite different from a collection of solid or liquid molecules. Molecules of gas are constantly in motion, meaning their velocity may be significantly higher than the escape velocity, due to heating. In any case, Adamant is a physicist, he knows better.– GallifreyanJan 12, 2017 at 12:31
-
2This is not off-topic, as it is not asking for speculation based on real-world science.– GallifreyanJan 12, 2017 at 14:55
|
Show 14 more comments
3 Answers
In theory - yes
As pointed out by Adamant, anything can have a moon, if the moon moves with the right velocity at a right distance. Below I will try to provide some calculations that support this.
In practice - hardly
According to Newton, the gravitational force between two objects depends on their mass and the distance between them. We want our moon to orbit the Death Star due to the gravitational attraction between them, so the former has to be in uniform circular motion. This tells us that the feasibility of such system also depends on the linear (tangential) speed of the moon.
Long story short, after equating two formulas mentioned above, we obtain the following expression:
v is the velocity of the moon, M is the mass of the Death Star, G is the universal gravitational constant, and r is the radius of the orbit. I leave the derivation to reader
Since we can tune the velocity and the radius of the orbit as we want, we only need to find the mass.
This website draws an analogy between the Death Star and a modern warship, and calculates its mass to be equal to 1.1 x 1015 metric tonnes.
Let's assume our body is orbiting at an altitude of 160 kilometres from the station's centre of mass. Then the speed would have to be 0.6 metres per second. While feasible, this speed is so low, the system wouldn't be very stable. On the other hand, if the moon orbits at 85 kilometres (only 5 kilometres above the hull), then its speed would be 0.9 metres per second, which is better.
Overall, the Death Star could have an object of negligible mass orbit it at a low altitude and a low speed.
answered Jan 12, 2017 at 16:20
Since a satellite can be anything that is gravitationally bound in an orbit around an object, yes, the Death Star could (easily) hold a satellite. Even objects much smaller like a comet nucleus can hold a spacecraft in orbit.
Basically anything below the mass of the Death Star itself can become a satellite if it moves close enough to be caught in an orbit by its gravitational field. For that it must not be too fast nor must it be too slow.
Just the fact that the Death Star has (and needs) an own propulsion system makes it lose all satellites once it accelerates fast enough to overcome the gravitational force between the satellite and itself and thus leave the satellite behind as soon as it moves under its own power.
In principle, any object can orbit any other object if there aren't any other large enough objects that are too close to them; the gravitational attraction pulls them towards each other, and if they're moving at the right lateral speed relative to each other, they keep "missing" each other and circling around each other instead. It might be a pretty slow orbit, but it would still be an orbit.
However, if there are other objects nearby, their influence can be sufficiently large that it disrupts the orbit. For example, if we could get the Space Shuttle into deep space, an astronaut could orbit it (very slowly). But when the Space Shuttle is in Low Earth Orbit, the gravitational force from the Earth on the astronaut is much, much larger than the gravitational force from the Space Shuttle, and so the astronaut effectively just orbits the Earth in parallel with the Space Shuttle instead.
This concept is formalized by the concept of the Hill sphere of a celestial object, which says that if a large object of mass M is orbited by a planet/moon of mass m at a distance of a, then satellites of the smaller planet m must have an orbital radius r less than about
To actually carry this any further, we need to have rough estimates of the masses of the planets that the Death Star orbited. For example, at the Battle of Yavin, the Death Star orbited Yavin at about 3 planetary radii. Yavin is a red gas giant, so let's estimate that it has approximately the same properties as Jupiter: its mass M is about 1.9E27 kg, and given the known size of Jupiter, its orbit would have a radius of 210,000 km. As a rough estimate, we can suppose that the density of the Death Star was about 1/3 that of aluminum, and given a diameter of 186 km this works out to a mass of about 3.0E18 kg. Thus, any satellites of the Death Star during the Battle of Yavin would have had to orbit at a distance of less than
(210,000 km)*(3.0e18 kg / 3*1.9E27 kg)1/3 ≈ 170 km.
Thus, it would be possible, but the radius of the orbit couldn't be more than about double the radius of the Death Star itself.
