Probably not unscathed.
The Death Star, the Galactic Empire’s planet-destroying superweapon, is an iconic symbol of sci-fi power. But if this colossal space station were to orbit Earth, would it remain in one piece? Let’s compare the Death Star to Earth and explore the immense gravitational forces at play.
To understand the stresses on the Death Star in Earth orbit, we need to consider tidal forces. Imagine the Death Star as two halves: one facing Earth and the other facing away. The hemisphere closer to Earth experiences a slightly stronger gravitational pull than the far hemisphere. This difference in gravitational force, known as tidal force, stretches the Death Star.
Let’s assume the Death Star orbits about 300 km above Earth, a typical low Earth orbit altitude. With a hypothetical radius of 80 km, the bottom of the Death Star would be roughly 270 km from Earth, and the top would be about 330 km away. Even this relatively small difference in distance creates a noticeable disparity in gravitational acceleration.
We can estimate the acceleration difference using the formula for gravitational acceleration, $g = GM/R^2$. The slight change in distance ($Delta R$) between the near and far hemispheres results in a differential acceleration. This difference manifests as a force trying to pull the Death Star apart. Approximating the force, we find it’s roughly proportional to $g frac{Delta R}{R} m$, where $g$ is Earth’s gravitational acceleration in low orbit (approximately 10 m/s²), $R$ is the orbital radius, $Delta R$ is the difference in distance to Earth across the Death Star, and $m$ is the mass of a Death Star hemisphere.
Considering a density of 1 gram per cubic centimeter for the Death Star (similar to water, for simplicity), its massive size (radius of 80km) would give it an enormous mass, around $10^{18}$ kg. The calculated tidal force translates to a stress within the Death Star. Our estimations suggest a tension of about 2.5 million Pascals.
How does this stress compare to material strength? Steel, a common structural material, has a tensile strength significantly higher, in the range of hundreds of millions of Pascals. At first glance, it seems the Death Star, if constructed from something akin to steel, might just hold together. However, our calculation indicates the stress is only about two orders of magnitude below steel’s strength. This relatively small safety factor means the Death Star would experience considerable deformation in low Earth orbit. It wouldn’t violently rip apart, but it would likely flex and distort noticeably under these stresses.
Furthermore, we should consider the Death Star’s own gravity. The gravitational attraction between the two hemispheres acts to pull them together, counteracting the tidal forces. While significant, our estimations show this self-gravity is not strong enough to overcome the tidal stresses entirely; it contributes a force roughly comparable to the tidal force, but ultimately, structural integrity is crucial for the Death Star to maintain its shape.
In conclusion, while the Death Star might not be instantly pulverized by Earth’s gravity, it would face substantial stress in low Earth orbit. Tidal forces would cause significant deformation, and its structural components would be under immense tension. Therefore, while not completely destroyed, the Death Star wouldn’t have a comfortable time orbiting Earth and certainly wouldn’t remain perfectly spherical. It would require a robust internal structure far exceeding typical steel to withstand these gravitational challenges without noticeable and potentially critical deformation.
