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Calculation time!

Key assumptions. Wookiepedia lists 120 km for the diameter of the first Death Star [1]. Density of steel is around 8000 kg/m^3 [2]. As a wild guess, I'll assume a uniform sphere with about 1/4 that density (i.e., rho = 2000 kg/m^3), to account for open space etc.

Surface gravity [3] is given by g = (4pi/3) * G * rho * r = 0.03 m/s^2 (i.e., about 0.3% that of Earth).

Escape velocity [4] is given by v = sqrt(2 * g * r) = 60 m/s = 134 mph

This is faster than I would have guessed, but I would expect that to be in reach for a good trebuchet. It's reasonable to assume the compactor is required to load compressed trash into such a device.

[1] https://starwars.fandom.com/wiki/Death_Star/Legends [2] https://en.wikipedia.org/wiki/Steel#Properties [3] https://en.wikipedia.org/wiki/Surface_gravity#Relationship_o... [4] https://en.wikipedia.org/wiki/Escape_velocity#Calculation



Do you think whatever artificial means they use to create gravity on the Death Star would affect the escape velocity?




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