End of the Earth II - Blaze of Glory

In honor of earth day today, many bloggers are posting things about how to save the earth, or retrospectives on earth days past. We here at The Virtuosi decided, what better way to celebrate the earth than to figure out how to destroy it? So that is exactly what we intend to do. This post will focus on the destruction of the earth by a laser beam. This is a familiar concept. Whether it is Marvin the Martian or the Death Star, destroying planets with lasers (or threatening to) is a common theme. We will be considering two questions today. The first is fairly obvious: how much power would the death star need to destroy the earth? The second relates to a topic of continuing interest of mine: how much would the death star recoil upon firing? First, the power we would need. Most estimates of the death star’s power seem to rely on simply overcoming the gravitational binding energy of the earth. We’re going to assume the earth is a uniform density. We can easily calculate the gravitational binding energy to be $$U_G=\frac{3GM_{earth}^2}{5R_{earth}}$$ $$U=2.3\cdot 10^{32} J$$ However, looking at the video of the death star blast, it looks like the earth was more than just gravitationally unbound. It looks like we’ve actually managed to atomize some of the constituent molecules. Let us assume that about half of the earth gets atomized. We’re also going to assume that the earth is made entirely of iron (not too bad an assumption). In a lattice, iron will have \~6 bonds, and the energy of each bond will be on the order of \~2 eV. We can calculate the number of iron atoms, N, needed to give the mass of the earth by: $$N=\frac{M_{earth}}{m_{iron}}$$ $$N=3\cdot 10^{50} atoms$$ We assume half of these have all of their bonds broken, this gives an energy required to break the bonds of $$E \approx 3N(2eV)=2.88\cdot 10^{32} J$$ This gives us a total destruction energy of $$E_{tot}=5.2\cdot 10^{32} J$$ Analyzing the video of the firing of the first death star tells us that the laser fired for about 4s, so this is a power of 1.310^32 W! As an aside, that much energy is about the total energy output of the sun over a day. That means the impulse this laser delivered (momentum per second, or force) to the death star must be the power over c, the speed of light (see my earlier post for a discussion of laser gun recoil). This is a force of 4.310^23 N. We now need to estimate the death star mass. According to confidential sources the death star had a diameter of 150 km. The death star is made of metal, but it also has a lot of empty space inside of it. We’ll go ahead and assume it has the density of water. This gives is a mass of $$M_{ds}=\tfrac{4}{3}\pi r^3\rho=2.4\cdot 10^{13} kg$$ A quick glance at our numbers reveals that we’re going to need special relativity here! Otherwise we’d be accelerating this thing well past the speed of light. We have a total momentum change of 1.710^{24}kgm/s. This is a relativistic momentum, which we can solve for v. The algebra is a tiny bit ugly, but it turns out that velocity in terms of relativistic momentum is $$v=\sqrt{\frac{p^2}{m^2+p^2/c^2}} \approx c$$ It turns out that the death star would be moving so close to speed of light as to not matter. That’s fast! I think we can safely say that this laser recoil would be noticed!