Total eclipse of the Death Star

Happy Eclipse day!

Congratulations to everybody who is lucky enough to live in the eclipse path, or who made the effort to get under the shadow of the moon! I hope it was grand; I was on the wrong side of the planet this time, so I have had to enjoy it via the internet.

Of course, the Internet likes to have fun, so along with the various actual eclipse photos (which range from cool to spectacular), there have been some pictures replacing the black disk of the moon with the DeathStar.  Long time readers of this blog will know that this Lounge has a great view of imaginary spacecraft in orbit; the Death Star fits into that category nicely. So with a bit of basic math and physics, we can calculate the conditions under which the Death Star can eclipse the sun, as viewed from here on Earth.

But first, we need to define our Death Star. I won’t dig too far down into the seedy underbelly of Srat Wars fandom, but a oft repeated figure for the size of the Death Star is a diameter of 100 miles, which yields an 80km radius. As for the density (which we’ll need later for reasons I don’t want to spoil), we will go with 800kg/m3. This is the density of something that is 10% steel and 90% air, which would give it the same general construction as modern naval vessels. This makes the Death Star slightly more dence than pure ethanol, but substantially lighter than the beer which fuels this blog.

In order to eclipse the sun, the Death Star needs to subtend a larger angle of sky than the Sun. For the sake of simplicity, we will call the sun angle 0.5 degrees, or 30 minutes of arc (it actually varies slightly, as the Earth’s orbit is elliptical, and the eccentricity of this orbit changes between 0 and 6 percent depending on where in the Milanković cycle we are). So, given a 80 km radius, the Death Star can eclipse the sun if it is closer than 80/sin(0.25deg)= ~18,300 km.

This is much farther than near Earth orbit, but much closer than geosynchronous orbit (about 36,000 km altitude). It is also, of course, about 21 times closer than the Moon, which is about 21 times larger than the Death Star.  However, it means that if the Death Star was in Geosynchronous orbit (to ‘hover’ over a target, for example), it would not eclipse the sun; it would block out at most a quarter of the light, which would be barely noticeable by people down below.

On the other had, if the Death Star was in low Earth orbit, like the International Space Station, it could easily eclipse the Sun. An 80km radius space station only 360 km up would be huge from the point of view of an observer directly underneath, blotting out more than 25 degrees of arc in the sky as it zoomed past at 8 km/sec (or one diameter every 20 seconds). However, it isn’t clear if the Death Star could fly this close to our planet.

The orbital velocity of a satellite around the Earth, in meters per second, is sqrt(GM/R), where G is the Gravitational Constant (6.67E-11 m³kg^-1s^-2), M is the mass of the Earth (6E24 kg), and R is the radius of the orbit IN METERS (not km). So with an orbital radius of 6700 km (329km above the mean surface), the orbital velocity is 7728 m/s. The problem for the Empire is that the Death Star has a radius of 80km, so the guys sitting in the gun turrets facing the Earth only have an orbital radius of 6620 km. Thus they will be orbiting at 7775 km/s, 47 m/s faster than the space station. For people who live in the real world, that’s a 105 miles per hour, or 165 km/hour difference. Smashing your troops against the walls at a hundered miles per hour is going to impede their ability to fire their super laser, and it is possible that even the structural integrity of the Death Star would be under threat this close to the Earth.

Back here in Science Land, we call the closest that a satellite can get to a planet without being torn apart by this sort of differential orbital speed the Roche Limit. The Roche Limit determines the closest approach a satellite can orbit a planet without being torn apart. Technically, the Roche limit only applies to objects held together by gravity- e.g. with no tensile strength. Steel, the purported structural material of the Death Star, has substantial tensile strength- this is why it’s used for everything from bicycle spokes to suspension bridge cables. But even if the space station is held together by the tensile strength of the steel, that will be little comfort to everything and everyone that isn’t tied down; even if the Death Star could survive inside the Roche limit, the occupants wouldn’t. So in order to know if a fully operational Death Star can eclipse the sun, we need to calculate the Roche Limit, and determine whether it is closer or farther than the maximum eclipse distance of ~18,300 km calculated at the top of this blog post.

The Roche limit equation is d = R (2 rho_M/rho_m)^1/3, where

R is the radius of the Primary, rho_M is the density of the primary, and rho_m is the density of the moon. And the assumption we are using is that the density of the death star is 0.8 g/cc or 800 kg/m³ (a bit less than my second beer).

 

As for the other numbers, the Earth’s radius is 6371km, and the earth’s density is 5500kg/m³. So the Roche limit for the Death Star is 15,263 km.

This is closer than the maximum eclipse distance of 18,300 km (which is a distance, not a radius, so you can add up to 6371 more km for an equatorial eclipse viewer), so there is a range, albeit a fairly well restricted range, in the orbital radius of roughly 16,000 to 24,000 km where the Death Star is far enough from Earth to not be tidally disrupted, but still close enough to blot out the sun. But it wouldn’t be blotted out for very long. A 16000 radius orbit has an orbital velocity of 5 km/s. So even with an equatorial observer only 10,000 km away, where the shadow is largest, at 72 km wide, totality would last less than 15 seconds. This is almost the exact elapsed time from Tarkin’s “Fire when ready” to weapon discharge. And Bonnie Tyler wouldn’t even have time to get a little bit tired of listening to the sound of her tears.