The Death Star: A How To (v. 0.9)

David Woolsey

This is a fanciful exercise in using the maxim: "Imagine your well intended invention in the hands of a ruthless South American dictator." Or, in this case, The Evil Empire!


When I saw the movie Star Wars for the first time I thought that the weapon on the Death Star could never be built. I mean it just seemed completely impossible to generate and transport enough energy to destroy an entire planet as depicted. But, inspired one day by an informal discussion among the graduate students in the Holzapfel Lab of the gigantic particle accelerator imagined by Sugawara, Hagura, and Sanami I'm now not so sure it could not be built.

Articles To Reference

<> Destruction of Nuclear Bombs Using Ultra-High Energy Neutrino Beam by Hirotaka Sugawara, Hiroyuki Hagura, and Toshiya Sanami.


We discuss the possibility of using the ultra- high energy neutrino beam (about 1000 TeV) to detect and destroy the nuclear bombs wherever they are and whoever possess them.


<> Substructure of the inner core of the Earth by J. Marvin Herndon.


The rationale is disclosed for a substructure within the Earth's inner core, consisting of an actinide subcore at the center of the Earth, surrounded by a subshell composed of the products of nuclear fission and radioactive decay. Estimates are made as to possible densities, physical dimensions, and chemical compositions. The feasibility for self-sustaining nuclear fission within the subcore is demonstrated, and implications bearing on the structure and geodynamic activity of the inner core are discussed.

Here, roughly, is how to build the Death Star's planet busting weapon. An ultra-high energy neutrino beam, such as proposed by Sugawara, Hagura, and Sanami (S H & S) could be built and used to initiate a fission chain-reaction in the Uranium mono-Sulfide (US) planetary core of the Earth in exactly the same way that the beam would be used (by S H & S) to initiate fission reactions in weapon stock piles. The rough issues and details are something like as follows.

Necessary Parts

These are some of the parts needed for the "device".

Neutron Reflector

A neutron reflector to contain fission neutrons within the bomb's chain reaction zone would not be necessary. The 8 km dia core is big enough that it would take 24 microseconds for light to transit. Thus it will take 240 microseconds for the very fastest neutrons to transit it. The fission chain time constant is of the order of microseconds. The number of generations needed for a high yield bomb is about 57 or so. There would be insufficient time for even the fastest neutrons to escape the bomb core and so no reflector would be needed.

Tamping Jacket

As for a tamping jacket, you could literally not get more tamping material on Earth.

High Pressure and Density

High pressure and density? Got that too. Pressure at the Earth's core is 3.6 Mbar (360 GPa). The density at the inner core is 23,000 kg/m^3. (Compare to an average density of the Earth of 5,500 kg/m^3.) At present I am unsure of how this compares to the pressure achieved in an implosion type fission bomb but I'll bet it's not too different.

Calculation of Bomb Yield

But what about the yield? Is it enough to blow up the whole world? To know this we need a calculation of the gravitational binding energy of the Earth and an estimate (probably optimistic) of the explosive yield of the Earth's Uranium core.

Escape velocity from Earth is 10 km/s. That leads to a specific binding energy requirement of U = (0.5 * m * v^2) / m = 50e6 [J/kg].

The mass of the Earth is

M_earth = volume * average density = ((4/3) * pi * r^3)*(5,500 [kg/m^3])
= ((4/3) * pi * (6.4e6 [m])^3) * (5.5e3 [kg/m^3])
= 6.0e24 [kg]

E_bind <= U * M_earth = 50e6 [J/kg] * 6.0e24 [kg] = 3.0e32 [J]

However, the total gravitational binding energy of the Earth is less than the specific potential energy times the mass (because as you remove mass the escape velocity goes down -- but the calculation above does supply an upper bound).

The exact calculation for the gravitational binding energy of the Earth goes (check me, please) like this

E_bind = integral from r=0 to r=R of U * dm where U = G * M / r = G * (5.5e3 [kg/m^3]) * (4/3) * pi * r^2 and dm = (5.5e3 [kg/m^3]) * 4 * pi * r^2 * dr.


E_bind = integral from 0 to R of G * (5.5e3 [kg/m^3])^2 * (16/3) * pi^2 * r^4 * dr
= G * (5.5e3 [kg/m^3])^2 * (16/15) * pi^2 * R^5 = 2.3e32 [J]

The energy yield from the 8 km ball of Uranium should be something like Y = 0.005 * M_u * c^2 where

M_u = ((4/3) * pi * (4e3 [m])^3) * (26e3 [kg/m^3]) = 7.0e15 [kg]

and the factor of 0.005 is the mass fraction converted to energy by fission. So we have

Y = 0.005 * M_u * c^2 = 0.005 * 7.0e15 [kg] * 9.0e16 [m^/s^2] = 3.2e30 [J]

Y / E_bind = 3.2e30 [J] / 2.3e32 [J] which is only about 1.5% of the energy needed to completely blow up the world and separate the fragments to r = infinity. It would still make quite a mess of things though. Just imagine what would happen if the surface of the Earth jumped up one or two percent of the Earth's radius and then fell back down.

Now, this weapon would only work against worlds with Uranium (or Thorium) cores. But since those are probably the only ones with long lasting (~5 G year) geothermal heat they would be the only ones with permanent tectonic activity. Some current theories on the evolution of life assert that tectonic activity is necessary for life to evolve and persist. In other words, the weapon would only work against "living" planets -- hence the name Death Star is far more appropriate than it might have otherwise been.

So, now we know where the Asteroid Belt came from. ...because long, long ago (but not too far, far away) the quantity of fissionables was higher. When the Earth formed (4.3 G year ago) the proportion of U_235 (half life 0.76 G year) to U_238 (half life 4.5 G year) was about 25/75 which is closer to "bomb grade". More importantly though, the quantity of fissionables was about (or, rather, at least) three times higher than at present.

Now we know that the Death Star would NOT need to supply anything like all the energy needed to blow up an entire planet. It would only be supplying a "catalyst" for the reaction in the form of a high energy neutrino beam.

Among the open questions I still have is whether the fission products from U_238 fast fission includes enough neutrons to build a chain reaction. I know that U_238 will undergo fission when hit with fast neutrons but, again, I need to look closer into the neutron output.

The chain reaction question important because the beam of neutrinos, and resulting hadron shower, passing through the inner core would only lower the critical mass in the path of the beam. In the path of the beam we'd get a chain reaction limited only by the energy and power of the beam (and we can make up anything we want to for this -- what the heck, that's what Sugawara, Hagura, and Sanami are doing). If it'll work on warheads it could be made to work here. But whether the rest of the surrounding core takes off in a chain reaction will depend on what happens outside the beam. And the answer to that depends on the neutron yield of U_238 fast fissions. What we want is a "flame front" or "shock front" to propagate out and away from the beam path.

More to come...

Page first created Monday, July 11, 2005
Page last modified Thursday, July 21, 2005 11:11 AM
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