# Can Earth become a micro Death Star?

Can we harvest the fission energy inside of Earth and produce a energy beam powerful enough to destroy Near Earth Object such as an asteroid? Is this technology probable in the next couple of centuries?

• Maybe we could channel that energy but there is no nuclear fission inside Earth. – Vincent Apr 18 '15 at 5:14
• The Death Star has a max diameter of 900 km. The Earth. Has a diameter of 12,000 km. I only point this out because of the use of micro in the title. There are lots of other day ways to power an energy beam that don't involve messing with the core. – AndyD273 Apr 18 '15 at 6:21
• An obvious question coming to mind: If it's ground-based, wont syncing be an issue due to planetary rotation : imagine the best place to fire from is China but the weapon is built in Alaska. Besides, and incoming enemy can then just make sure they're always on the other side of the planet. We'd then need to place probably 3 "guns" around the planet. – Nikhil VJ Jan 20 '18 at 5:00

• The temperature of the earth core comes in part from radioactive decay.
• This temperature can be turned into useful power by geothermal power plants.
• Using this power for a laser or particle weapon is thinkable but impractical in the next decades. SDI projects were chemical lasers, or bomb-pumped lasers, or various kinetic interceptors.

• Since we're on the worldbuilding site, what would be the reason to take geothermal power and not, say, nuclear power or wind power? Perhaps a geothermal plant would be easier to hide in a cave, as long as you can deal with waste heat. Pipe it into an area of hot springs?
• A non-chemical laser like a solid-state laser sounds like a good idea if you can do the tech. No specialized fuel required, just plug it into the power grid on your base.
• Why would you mount the weapon on the surface of Earth and not on a station in orbit? First, direct access to the civilian/commercial power grid. No need to generate the power on site, just shut down the rest of the country when you have to fire. Next, a surface installation can be hidden underground. The enemy can't map them from long range.

Would these reasons balance the obvious benefits of mounting the weapon in space, like no atmosphere to shoot through? Depends on the technology in a couple of centuries.

FOLLOW-UP regarding Samuel's comment: The laser will not vaporize the atmosphere. It might have negative environmental consequences, but whoever makes the firing decision would balance those drawbacks against the drawbacks of not firing on an incoming asteroid or enemy battle fleet. Against a natural threat like an asteroid, a space-based system is almost certainly the better idea.

• Word choice: One cannot vaporize a gas. One could heat it some more or turn it into a plasma. I guess that's what you mean.
• Firing the laser beam through the atmosphere means that some energy will be lost from the beam, in addition to any waste heat from the laser itself. Bad for the efficiency of the weapon, bad for the atmosphere.
• A planet is big. Unless the laser has actually planet-shattering power levels, like the original Death Star, it won't be big enough to affect an entire planet.
• Did you consider that firing the laser from the surface of the Earth might vaporize the atmosphere? – Samuel Apr 18 '15 at 7:37
• @Samuel Related --- what-if.xkcd.com/13 – Rmano Apr 20 '15 at 11:09

This sort of engineering can be done without turning the planet into a "death star", although some of the engineering is likely to take several years to a century of R&D.

Method one is to simply create a squadron (or even a wing) of kinetic kill vehicles powered by the ORION pulse drive.

From NextBigFuture:

http://nextbigfuture.com/2009/02/nuclear-orion-home-run-shot-all-fallout.html

*Nuclear explosive propelled Interceptor for deflecting objects on collision course with Earth. Johndale Solem, Los Alamos, proposed unmanned vehicle. No shock absorber or shielding. The pulse units were 25kg bombs of 2.5 kiloton yield for 100G acceleration of a 3.3 ton Orion. So an unmanned nuclear Orion can survive very high G forces. A single 25 kiloton yield would accelerate 3.3 tons to 1000Gs. A 2.5 megaton yield would accelerate 330 tons by 1000Gs. 25 megaton yield would accelerate 3,300 tons by 1000Gs. The highest acceleration had 0.4 seconds between charges so to get up to speed two or three charges might be needed to get 1.2 seconds of acceleration. Earth escape velocity is 11.2 km/s. 1000Gs is 9.8km/s**2.*

The projected performance is pretty impressive as well:

http://nextbigfuture.com/2009/02/unmanned-sprint-start-for-nuclear-orion.html

"Launch against a 100 meter chondritic asteroid coming at 25 km/sec. 1000 megatons if it hits. Launch when it is 15 million kilometers away and try to cause 10000km deflection. A minimal Orion weighing 3.3 tons with no warhead would do the job. 115 charges with a total of 288 kiloton yield. Launch to intercept in 5 hours. Ample time to launch a second if the first failed."

Being able to deliver a gigaton of energy 15 million kilometres away within 5 hours of launch should make most people pause when considering how to conduct warfare in space.

If Ravening Beams of Death is more your style, then consider the section in Atomic Rockets on high energy lasers: http://www.projectrho.com/public_html/rocket/spacegunconvent.php

The ultimate laser weapon would be a monster FEL "xaser" emitting a beam of coherent x-rays (focused through a diffraction grating) at targets out to one light minute away. Consider for a moment that the Moon is a bit farther than one light second from Earth, having one of these monsters in orbit would mean you have effectively controlled space almost 60 times farther than the distance from the Earth to the Moon!

Again, the description:

"Laser guru Luke Campbell thinks it not impossible to make an x-ray laser which does NOT require a nuclear device to pump it. In theory a Free Electron laser can produce any wavelength. It is possible approximate an x-ray lens by having the rays make glancing blows off dense materials. Bottom line is an x-ray laser is technologically very challenging, but if you manage to make one you have an Unstoppable Death Ray of Stupendous Range.

Let's take a 10 MW ERC pumped FEL at just above the lead K-edge. This particular wavelength is used because lead is pretty much the heaviest non-radioactive element you can get, and at just above the highest core level absorption for a material you can get total external reflection at grazing angles - so no absorption or heating of a lead grazing incidence mirror. We will use a 1 meter diameter mirror. The Pb K-edge x-ray transition radiates at 1.4E-11 m. This gives us a divergence angle of 1.4E-11 radians. At 1 light second, we get a spot size of 5 mm, and an intensity of 5E11 W/m2. Looking at the NIST table of x-ray attenuation coefficients, and noting that 1.4E-11 m is a 88 keV photon, we find an attenuation coefficient of about 0.5 cm2/g for iron (we'll use this for steel), 0.15 cm2/g for graphite (we'll use this for high tech carbon materials) and 0.18 cm2/g for borosilicate glass (a very rough approximation for ceramics). Since graphite has a density of 1.7 g/cm3, we get a 1/e falloff distance (attenuation length) of 4 cm. Iron, with a density of 7.9 g/cm3, has an attenuation length of 0.25 cm. Glass, density 2.2 g/cm3, has an attenuation length of 2.5 cm.

At 1 light second, therefore, the beam is depositing 2E12 W/cm3 in iron at the surface and 7E11 W/cm3 at 0.25 cm depth; 1.2E11 W/cm3 in graphite at the surface and 5E10 W/cm3 at 4 cm depth; and 2E11 W/cm3 in glass at the surface and 7E10 W/cm3 at 2.5 cm depth. Using 6E4 J/cm3 to vaporize iron initially at 300 K, we find that iron flashes to vapor within a microsecond to a depth of 0.9 cm. The glass, assumed to take 4.5E4 J/cm3 to vaporize (roughly appropriate for quartz) will flash to vapor within a microsecond to a depth of 4 cm within a microsecond. Graphite, at 1E5 J/cm3 for vaporization, will flash to vapor to a depth of 0.7 cm within a microsecond (the laser performs better if we let it dwell on graphite for a bit longer, we get a vaporization depth of 10 cm after ten microseconds).

Net conclusion - ravening death beam at one light second.

Now lets look at one light minute. The beam is now 30 cm across. This is much deeper than the attenuation length in all cases, so we will just find the radiant intensity and the equilibrium black body temperature of that intensity. We have an area of 7E-2 m2, and an intensity of 1.4E8 W/m2. You need to reach 7000 K before the irradiated surface is radiating as much energy away as heat as it is receiving as coherent x-rays. The boiling point of iron is 3023 K, the boiling point of quartz is 2503 K, and the sublimation temperature of graphite is 3640 K. All of these will be vaporized long before they stop gaining heat. At this range, the iron is subject to 5.6E8 W/cm3 at the surface, the graphite to 3.3E7 W/cm3 at the surface, and the glass to 5.6E7 W/cm3 at the surface. Using the above values for energy of vaporization, we get about 0.1 milliseconds before the iron starts to vaporize, 0.8 milliseconds before the glass starts to vaporize, and 3 milliseconds before the graphite begins to vaporize (because of its long attenuation length, once it begins to sublimate, graphite sublimates rapidly to a deep depth, while you essentially have to remove the iron layer by layer).

Net conclusion - still a ravening death beam at one light minute.

What about at one light hour? The beam is 18 meters across. The equilibrium black body temperature is 900 K. This is well below the melting point of most structural materials. Ten megawatts, however, is a lot of ionizing radiation. Any unhardened vehicle will be radiation killed at these ranges.

Luke Campbell

However, he goes on to note that in order to boost electrons to the velocities required for an X-ray free electron laser, you will need an acceleration ring approximately one freaking kilometer in diameter. So this X-ray laser would only be suitable for exceedingly huge warships, orbital fortresses, and Death Stars.

Since the time he wrote the above, Luke Campbell has reconsidered the use of lead grazing incidence mirror. Now he favors using diffraction."

In either case, you do not need the entire energy budget to output of the Earth to gain the effect that you desire, simply brute force engineering. For political and practical reasons you would probably want to mount devices like ORION pulse interceptors and massive "xaser" weapons on the Moon or in orbit around the Earth (this would also resolve some engineering issues like heat dissipation or allowing the laser beam to engage the target without being defocused or diffused by passage through the atmosphere).

No (and yes)

The Job
A "Death Star" type weapon is one that disrupts the target body. The amount of energy you must supply to disrupt the body is called its gravitational binding energy. Earth's gravitational binding energy is $2.24 \times 10^{32} J$.

Using the formula $E = mc^2$ and plugging in Earth's binding energy and the speed of light, we get:

$$2.24 \times 10^{32} = m(3 \times 10^8)^2 \rightarrow \frac{2.24 \times 10^{32}}{9 \times 10^{16}} = m$$

$$m = 2.5 \times 10^{15} kg$$

or you need total energy equal to total energy conversion of $2.5 \times 10^{15} kg$ of matter to disrupt an Earth sized target.

YES
The Answer is "Yes" because if we look at the inventory of materials available in the Earth, assume 100% efficiencies, that we can extract the materials, and "burn" them all at the same time, we do in fact have enough energy reserves to do the job.

Doing it with Antimatter
Divide that by two (half of it is antimatter and half is matter) and you get the minimum amount of energy conversion required to power a beam to disrupt a planet. The number I get is $1.25 \times 10^{15} kg$ of antimatter to the target.

Putting this another way, you need a solid chunk of antimatter about the mass of the asteroid 433 Eros which is roughly 20 km in diameter.

The Earth actually has some antimatter in orbit around it. However, there is only about $1.6 \times 10^{-10} kg$ of antimatter there. Which means we only need about $1 \times 10^{25}$ times as much antimatter as we have to do the job.

So "No" we can't do it with antimatter.

Doing it with Fusion
Less than 0.4% of the mass in a fusion reaction is converted to energy, so divide the mass requirement above by 0.4% and you get the following:

$$m = \frac{2.5 \times 10^{15} kg}{0.004} \rightarrow m = 6.25 \times 10^{17} kg$$

The mass of Earth's entire hydrosphere of $1.4 \times 10^{21}kg$. Only 1/9 of that mass is hydrogen, which means Earth's hydrosphere contains $1.5 \times 10^{20} kg$ of hydrogen.

Putting this another way, you need a body of pure hydrogen and its isotopes about the mass of the asteroid 243 Ida which, if you used Solid Hydrogen's Density of $0.086 g/cm^3$, would be more than 85 km in diameter.

So "Yes", if we fused all of Earth's hydrogen at the same time, it would provide enough energy to disrupt 240 Earth massed planet.

Doing it with Fission
Fission releases less than 1/4 the amount of energy as fusion per unit starting mass. So use the numbers from above and multiplying by 4, I get

$$m = \frac{2.5 \times 10^{15} kg}{0.001} \rightarrow m = 2.5 \times 10^{18} kg$$

You will need approximately this much starting mass as fuel for the fission reaction necessary to disrupt an Earth sized planet.

Putting this another way, you need a body of pure fissionable ($U_{233}$, $U_{235}$, & $Pu_{239}$) and precursor ($Th_{232}$ & $U_{238}$) isotopes about the mass of the asteroid 243 Ida which if it was the same density as Eros would be roughly 120 km in diameter.

The Earth's inventory of Uranium and Thorium isotopes is:
$Th_{232} = 2.19 \times 10^{14} kg$ Used in breeder reactors to make $U_{233}$
$U_{233} = Negligible$
$U_{235} = 2 \times 10^{14} kg$
$U_{238} = 6 \times 10^{16} kg$ Used in breeder reactors to make $Pu_{239}$
$Pu_{239} = Negligible$

So most of our fissionable inventory is in the relatively stable $U_{238}$. We would have to convert all of this into $Pu_{239}$ in a breeder reactor before using.

So "No", if we did this with fission, we have only have 1% of the amount necessary to disrupt an Earth massed planet.

NO
The Answer is "No" because even if we harvest all the energy resources, the inefficiencies of the weapon would fry us if we tried.

The laser
Most lasers that we humans have used throughout the last 50 years have very poor energy conversion efficiencies (something under 20%). Meaning if we used all that energy to form a "Death Star Beam", we would deposit four times more energy on Earth (through heat losses) than we would deliver at the pointy end of our beam.

The Upshot of this is, with conventional lasers (solid state, gas dynamic, fluid, organic etc.) we'd disrupt the Earth with our waste heat long before we disrupted our target.

Recent laser breakthroughs have lead to laser types with theoretical efficiencies that are much higher. These are diode pumped solid state lasers (DPSSL) and free electron lasers (FEL).

Some DPSSL have a theoretical maximum efficiency above 80%

FEL have a theoretical maximum efficiency above 90% (the reference doesn't discuss efficiencies, this is a number I recall reading but don't remember where).

The DPSSL, isn't suited for emitting such powerful beams, however, the FEL has no limits on frequency or power of the beam. If we used this beam, we could deposit more energy on target than the Earth received.

The problem is that 10% of $2.24 \times 10^{32} J$ is still a rather large $2.24 \times 10^{31} J$ of energy. This is equivalent to frying the Earth with 16 minutes of all of the Sun's power output. This is enough to vaporize all water on the planet 96 times.

Definitely not a healthy thing to do.

This stuff scales linearly with mass (but not diameter), so you'd get the following for other Solar System bodies:

Target  Disrupt      The energy left on Earth will
Body's  Mass
Mass    (kg)
Earth   6x10^24      Boil oceans 96x, we have enough fusionables to do this.
Mars    6x10^23      Boil oceans 9.6x
Europa  5x10^22      Boil oceans 1 time, we have enough fissionables to do this.
Haumea  4x10^21      Boil 10% of the oceans
Tethys  6x10^20      Boil 1% of the oceans


So "No" even if we "just" disrupted a mass about 1% of our Moon's mass, we'd still kill every living thing on the planet.