You may remember that last year there was a tongue-in-cheek petition to the White House to have the US build a Death Star in order to boost the economy. That petition was responded to in the most wonderful way. But besides the absurd cost and the stupidity of actually having a planet-destroying space station, there is the wonder of the limits on feasibility. Not just engineering, but what does the physics suggest. I am hardly the first person to look at the question, but the ones I have found online seem incomplete or have some small errors. So, using some basic physics and modeling, let’s see what we can say about the power of the Death Star. And given Darth Vader claims that it is nothing compared to the power of the Force, then we can put some lower limits on what the Force can do.

Now, the key thing your Death Star needs to do is deposit enough energy into a planet to cause it to explode. So we need a way to define “blow up the planet” so we can create a physics model for that. Given what is seen in the film and fulfills our intuitions, the best way to think about it is putting enough energy into a planet so all of its matter is no longer (gravitationally) bound into a single object. This would mean that any given piece of the Earth will fly off and never return to reassemble the planet. Thus, we need to determine how much energy there is bound up in the Earth. In this case, we need to consider the gravitational potential energy. To make the calculations and modeling simple, let’s assume that the Earth is of constant density, ρ (even though this is not true, it will have a minimal effect on the final calculations, all things considered). We will also consider the Earth to be a perfect sphere, again to make the calculations easy. For a given piece of matter with mass m and another piece of matter with mass M that are distant r away, the gravitational potential energy (U) is

The negative sign indicates it is a potential due to an attractive force (in this case, gravity). Now, we need a way of modeling pieces of the Earth bounded to the remainder of the Earth. To do this, it is best with calculus. So we need to determine what will be a infinitesimal piece of mass for the Earth (dm) and use that to set up what will be its infinitesimal piece of potential energy (dU). Since we have spherical symmetry, let us use thin sheets of surface area of the Earth. The surface area of a sphere is 4πr^2, so if we take a surface that is infinitesimally thin (dr), then

 

dm = 4πr^2 dr

As for the mass of the rest of the Earth below that infinitesimal region, that would be the remaining volume times the density. Since the volume of the sphere will be 4π/3r^3, then

M = 4πρr^3/3

We can now plug this all into the potential energy equation and integrate from 0 to the radius of the Earth, R. Given all that, along with the average density of the Earth being the total mass of the Earth divided by the spherical volume of the Earth, then

U = -3M^2/5R

where M is the mass of Earth, and R is its radius. This means that you will need to pump at least this much energy into the mass of the planet so that it will no longer be gravitationally bound. Additional energy will give the matter additional kinetic energy. To repeat, this is a minimum amount you would need to properly blow up a planet of any given mass and radius.

So, let’s plug in for the Earth, And we get

~2.2*10^32 Joules (J)

That amount of energy is not something to easily grasp. For some perspective, the atomic weapon dropped on Hiroshima in 1945 had an energy level of around 10^14 J. We would need about 10^18 (a billion, billion) such bombs to compete with the Death Star. If you go by all the energy produced in all of the electrical power plants on Earth each year (~10^20 J), we would need to run for a trillion years, which is longer than the age of the universe. And I doubt we have the coal and other fossil fuels to run that long. For a most astronomical energy source, the Sun has a luminosity of about 3.8*10^26 Watts (W). That means that the Sun will produce as much energy as the Death Star fires into a planet if you let the Sun shine for a week. But the movie shows the blast being charged and released in a matter of seconds. So, we have some major power.

From where is the Death Star going to get that much energy? Fusion isn’t going to do that, given that the fusion core of the Sun is larger than a Death Star and it needs a week just to power up a single shot. According to sources such as Wookieepedia, the energy source is likely hypermatter using a hypermatter annihilator. Apparently you get hypermatter from hyperspace, and it is said to by “tachyonic”. This is all made-up Sci Fi jargon, but let’s consider the key point of any form of matter: it’s really energy. Einstein’s E = mc^2 tells us that we can convert matter to energy. So, how much matter do we need? Our unit is that of the kilogram (no matter what type of matter it is), so we cans we that we need

~2.4*10^15 kg

This is on the order of the mass of Mount Everest. Given that Death Star I is only 120 kilometers across, it seems unlikely it can have Mount Everests of space being taken up. However, given that there isn’t much said about hypermatter, and given that it supposed to be used by all sorts of ships to go into hyperspace, we can make some educated guesses about its physics. Because of space limitations, perhaps hypermatter is a boson, like photons and the Higgs, rather than a fermion. The importance of that is that multiple bosons can occupy the same space, while fermions (which is normal matter) cannot. In that case, it is less difficult to store huge quantities of hypermatter without needing to have another super-station just filled with the stuff. You can also see from the designs out there, you aren’t going to fit Mount Everest into the reactor very easily.

Before continuing, there is something else that hasn’t been made explicit: this is the minimum amount of energy required to do this task. In particular, the calculations based on the needed amount of energy from hypermatter was assuming all of the energy went into blowing up the planet. But such 100% energy efficiency is not possible, even with advanced technology. The Star Wars universe also supports this. Remember that the original Death Star was destroyed by firing torpedoes into an exhaust port. Why does a station need exhaust unless not everything goes into the operation? This would appear to be a heat exhaust from firing the superlaser. We cannot guess how efficient the system on the Death Star is, but physics requires it to be less than 100%. If it’s anything like the efficiency seen in modern nuclear power planets, then we are only looking at perhaps 10% energy efficiency. That means we will need ten times as much energy as the initial calculation required. But for simplicity, let’s just take this minimum energy calculation, because things are only going to get worse.

One final detail that I have not seen considered is the recoil from firing the superlaser. It is actually the case that light (in the form of photons) has momentum, much like a bullet. For our lives, that momentum is puny, but the push of light from the Sun is something NASA engineers have to consider when they are monitoring probes and how their trajectories differ from just considering gravity. The momentum (p) in light is easy to calculate.

E = pc, or p = E/c where c = speed of light.

The reason this matters is going to be the conservation of momentum. Before the Death Star fires, we assume it is initially at rest. After it fires, the momentum of the light will be headed toward the planet, and the Death Star will need to have an equal magnitude of momentum in the opposite direction so that the total momentum after firing is still zero. This is why a gun recoils after firing a bullet: the bullet has forward momentum, so the gun needs backward-pointing momentum to balance it out. Since the gun is much more massive than the bullet, the gun doesn’t move that fast.

To figure out the recoil and thus how fast the Death Star is moving backward, we require the definition of momentum for massive objects and the mass of the Death Star. Because the energies are so huge in this situation, we really ought to use the relativistic version of momentum:

Here, γ is the relativistic factor from special relativity and is proportional to the velocity of an object; at rest, it has a value of 1, and as an object approaches the speed of light it goes to infinity (for note: when v ~ 86% of the speed of light, γ = 2, so you need to go really fast to have a notable gamma-factor). As for the mass of the Death Star, I cannot find any data on that subject. So, let’s make some assumptions and make the Death Star as massive as possible. Let’s assume it is a solid sphere of iron. This will certainly be an exaggeration of its real mass, since there is space of people to walk around and for equipment and tubes, etc. But this exaggerated mass will mean that the recoil speed of the Death Star will be an underestimate; if it’s already bad, it only gets worse. So, given that Death Star I had a diameter of 120 km, and the density of iron is ~7800 kg/m^3, the absurdly-large mass of the Death Star would be ~7.1*10^18 kg.

Let’s now take all the info, and assume the lowest amount of energy needed to fire and destroy the Earth, and assume the most massive Death Star possible, we find that the recoil speed of the Death Star must be…

~100,000 m/s

(I will note that this did not require the correction for relativistic speeds, but if you want to redo the calculation with a more reasonable amount of mass, then you probably will.)

So this superlaser will be able to fling a moon-sized station to an amazing speed. But this isn’t the worst part. The last issue is that this happens pretty darn fast. The film shows the superlaser taking about 5 seconds to discharge. That means the Death Star will have its momentum change occur in 5 seconds, going from zero to a third the speed of light in 5 seconds. We are looking at over 2000 gs of acceleration. Everyone on board the ship is dead. Dead, dead, dead! Just by firing the superlaser. I’m not even considering structural integrity, just the fact that a human body cannot undergo that level of acceleration and live.

Remember, this is also a best-case scenario. We considered using the minimum amount of energy with 100% energy efficiency and fired for an absurdly-mass Death Star. Make any of these numbers more “realistic” and the problem is only worse.

So, if the Galactic Empire were to build such a station, not only would it require huge sums of cash and materials never before heard-of, not only would it be a horrific weapon, but it would effectively self-destruct just by using its primary weapon. It is the ultimately genocide-suicide weapon. In other words, a pretty stupid idea. And you have an exhaust port barely 2 meters wide that is an easy target for X-wing fighters, and you let your station get blown up? Not even Jar Jar is that incompetent.

So kids, don’t build the Death Star. Now, building the Enterprise, that’s another story…

(Note: if you see any mistakes, please feel free to say so in the comments.)