It may appear trivial to dissect the physics of a film released over three decades ago, but here we are. Indeed, Star Wars: Return of the Jedi hit theaters in 1983. But what’s the significance? For one, I'm a dedicated Star Wars enthusiast. Additionally, elements of the Death Star from Return of the Jedi are rumored to appear in the forthcoming Star Wars: The Rise of Skywalker.

Let's delve into the physics surrounding the Death Star II. To recap briefly: the Emperor is constructing a new, slightly larger Death Star, presumably without the exposed ventilation shaft — but who can say? During its assembly, this space station orbits the moon of Endor, which is equipped with an impressive shield generator that protects the station from the surface of the moon.

Now, turning our focus to the physics. I’ll operate under the premise that the Death Star maintains a geostationary orbit around Endor. This means the station isn’t maneuvering under its own propulsion; instead, it moves due to gravitational interactions. The geostationary aspect indicates that the orbital angular velocity of the Death Star matches the rotational angular velocity of Endor, ensuring it remains visible at a consistent point in the sky relative to Endor.

But how does this occur? How can an object appear stationary while orbiting a planet? Let's begin with the fundamentals of orbital mechanics, which hinge on two key concepts. The first is the momentum principle, expressed as follows:

The momentum principle states that a net force alters the momentum of an object, where momentum is defined as the product of mass and velocity. The second concept pertains to gravitational interaction, which asserts that every two mass-bearing objects exert a gravitational force on each other, diminishing as the distance between them increases. Here’s the mathematical representation of gravitational force:

Key points regarding gravitational force include: - m? and m? signify the masses of the two objects. - r represents the distance between their centers. - The "r hat" denotes a unit vector, a technical detail worth noting. - G symbolizes the universal gravitational constant, a minuscule value of 6.67 x 10?¹¹ Nm²/kg².

Now we’re prepared for some serious physics. Consider the two interacting bodies: the Death Star and Endor. Given that Endor's mass dwarfs that of the Death Star, the gravitational interaction barely alters its momentum. As the Death Star approaches Endor, the gravitational pull modifies its momentum. By wisely adjusting the magnitude and direction of the Death Star’s momentum, we can achieve circular motion around Endor.

In the illustration, the Death Star transitions from position 1 to position 2, demonstrating a change in momentum. While speed remains constant, the direction shifts — a valid alteration in momentum. See? Orbital physics isn't so daunting after all!

Let’s alter the scenario. What if the Death Star is positioned farther from Endor, yet still aims for a circular orbit? At increased distances, gravitational force weakens, leading to a smaller change in momentum. However, with a broader circular orbit, less directional change is needed to maintain circularity.

How about exploring this through animation? Imagine a substantial object orbiting Earth. A numerical model can easily simulate its motion, breaking the problem into shorter time intervals and making specific assumptions. If you’re curious, you can view and modify the code behind this animation.

Several important observations arise: - The view is from above Earth’s North Pole, illustrating its rotation. - The animation is not in “real time,” as that would require an entire day for a complete rotation. A “shield generator” is marked on the surface to aid in noticing the rotation. - Two objects orbit: the first at a lower altitude, completing its orbit faster. - The objective is to position an object far enough away so that its orbital period matches Earth’s rotation.

Skipping the math, it’s beneficial to illustrate the relationship between angular velocity (in radians per second) and orbital distance from Earth’s center.

This suggests that as distance increases, angular velocity diminishes. However, it also relies on the planet's mass, which is crucial.

If we seek a geostationary orbit for Earth, we know its mass and rotation duration, allowing us to calculate angular velocity. A quick note on sidereal vs. synodic rotations: the Earth doesn’t take precisely 24 hours to complete a rotation. It takes 24 hours for the sun to return to the zenith (noon), termed a synodic day.

During this period, Earth orbits the sun, resulting in a sidereal day of approximately 23 hours and 56 minutes — the duration needed to determine the angular velocity of our Death Star (or any conventional satellite).

Lastly, here’s the animation along with the code.

The object maintains the same angular velocity as Earth, positioning it directly above the “shield generator.” To achieve this, the generator must be located at Earth’s equator, ensuring that the angular velocities of both Earth and the Death Star align precisely. This explains why satellite dishes are typically pointed southward; that’s where geostationary objects orbit.

Now, shifting our focus back to Endor. If the Death Star is in a geostationary orbit (technically an "endostationary" orbit due to Endor), three critical factors come into play: - The mass of Endor - The angular velocity of Endor (linked to day length), assuming sidereal and synodic days are nearly equivalent - The orbital radius

Knowing any two of these parameters allows us to calculate the third. Here’s an initial view of the orbiting Death Star from the rebel attack briefing.

From this image (alongside a Death Star diameter of 160 km), we can deduce that it orbits at a distance of 1.564 x 10? meters. This estimates Endor’s radius at 1.09 x 10? meters. However, here’s another frame from the same scene.

In this instance, the Death Star appears farther away, maintaining an orbital distance of 1.564 x 10? meters. I’ll adopt the larger value for our calculations.

Of course, deriving Endor’s angular velocity from this scene is implausible. Who would animate it in “real time”? That wouldn’t be visually appealing. If you’re Admiral Ackbar, your Death Star animation must dazzle, especially when in the presence of Jedi like Luke Skywalker.

No worries; we can ascertain the mass of Endor. How? Well, we’ve observed inhabitants on Endor’s surface, moving similarly to those on Earth (since filming occurred on Earth). This indicates Endor’s gravitational field mirrors that of Earth, with a force of 9.8 newtons per kilogram (symbolized as “g”). Given that gravitational fields depend on mass and radius, I can derive the mass using the known radius in relation to the Death Star II’s size.

Using the known values, I calculate Endor’s mass at 1.76 x 10²³ kilograms. Note that this results in a radius smaller than that of Earth’s moon, yet it has more than double the moon’s mass.

That’s it! With the orbital radius and Endor’s mass determined, the angular velocity for a stationary orbit would be 3 x 10?³ radians per second, suggesting an Endor day lasts just 34 minutes. That’s quite brief — perhaps that’s why Ewoks are so small; they don't get sufficient rest.

Now, here are some thoughts and homework assignments: - There are scenes in the movie showcasing both Endor and the Death Star. For instance, when the Death Star explodes, it’s visible from Endor’s surface. Use these moments to estimate the Death Star's orbital distance. - What would be the Death Star’s orbital distance if Endor had a 24-hour day? What implications would that have? - Could it be possible that the Death Star isn’t in orbit but rather flying? Try estimating its mass (good luck with that) and calculate the necessary force to maintain that position. - If Endor completes a rotation in just 34 minutes, the effective surface gravity (especially at the equator) would likely be lower due to centrifugal force. This means Endor could possess a greater mass while still achieving an effective surface gravity of 9.8 N/kg, which would lower the rotation speed for the Death Star's orbit. This is complex; I may revisit this later. - Consider the idea of Endor having a reverse rotation and orbit. Since it’s a moon, it orbits a planet that orbits a star. Most celestial bodies rotate in the same direction they orbit, but backward spinning is possible. Could a 34-minute rotation coincide with a significantly longer synodic day? - Lastly, don’t you have better things to do than ponder physics and Star Wars? Nope!