Could a 'Death Star' really destroy a planet?

Jan 19, 2012 By Ray Sanders, Universe Today
The Death Star. Image Credit: Wookieepedia / Lucasfilm

Countless Sci-Fi fans vividly remember the famous scene in Star Wars in which the Death Star obliterates the planet Alderaan.

Mirroring many late night caffeine-fueled arguments among Sci-Fi fans, a University of Leicester researcher asks the question:

Could a small moon-sized battle station generate enough energy to destroy an Earth-sized planet?

A paper by David Boulderston (University of Leicester) sets out to answer that very question. First, for the uninitiated, just what the heck is a Death Star?

According to Star Wars lore, the DS-1 Orbital Battle Station, or Death Star, is a moon-sized battle station designed to spread fear throughout the galaxy. The image above shows the Death Star as it appeared in Star Wars Episode IV: A New Hope (1977). The Death Star’s main weapon is depicted as a superlaser capable of destroying with a single blast.

Boulderston claims that it is possible to estimate how much energy the Death Star would need in order to destroy a planet with its superlaser. There are a number of assumptions made, however, in order to come up with the energy requirement.

For starters, Boulderston assumed that Alderaan did not have any sort of planetary “deflector” shield. A second assumption is that the planet is a solid body of uniform density – essentially ignoring the complex interior of planets, due to lack of information on Alderaan itself. Using the idealized sphere model based on ’s mass and diameter, it was possible to determine the gravitational binding energy of Alderaan, using a simple equation of:

U= 3GMp2

——
5Rp

Where G is the Gravitational Constant (6.673×10-11), Mp is planet mass, and Rp is the planet’s radius. Using Earth’s mass and radius, the required energy comes out to 2.25 x 1032 Joules. Using Jupiter’s data, the energy required goes up to 2 x 1036 Joules.

Boulderston asserts that (according to lore) the Death Star is powered by a ‘hypermatter’ reactor, possessing the energy output of several main-sequence stars. Given that the power output of our Sun is about 3 x 1026 Joules per second, it’s a reasonable assumption the Death Star’s reactor could power the superlaser.

Despite using a simplified model of a planet, Boulderstone states the simplified model is reasonable to use since the Death Star’s main power reactor has the energy output equal to several main-sequence stars. Even if Earth’s exact composition were used in the equation above, the required energy to destroy a planet would only be affected by a few orders of magnitude – well within the Death Star’s power budget.

Boulderstone reiterated that the required to destroy a Jupiter-sized planet would put considerable strain on the Death Star. To destroy a planet like Jupiter, all power from essential systems and life support (no re-routing from the auxiliary EPS conduits – that’s a Star Trek hack!) would be required, which is not necessarily possible.

Boulderstone’s conclusion is that the Death Star could indeed destroy Earth-like planets, given its main power source. While the Death Star could destroy an Earth-sized planet, a Jupiter-sized planet would be a tough challenge, and the Galactic Empire would need to resort to using a Suncrusher to destroy stars.

If you’d like to read Boulderstone’s paper, you can access it at: physics.le.ac.uk/journals/inde… article/view/328/195

Explore further: Thermoelectric power plants could offer economically competitive renewable energy

add to favorites email to friend print save as pdf

Related Stories

Binary star system found by following gamma-ray signal

Jan 13, 2012

(PhysOrg.com) -- To find a binary star system, which is where two stars are in close proximity to one another, astronomers have traditionally relied on pure luck. They’d first start studying what would look like a single ...

New planet discovered in Trinary star system

Jul 14, 2011

Until recently, astronomers were highly skeptical of whether or not planets should be possible in multiple star systems. It was expected that the constantly varying gravitational force would eventually tug ...

Star blasts planet with X-rays

Sep 13, 2011

A nearby star is pummeling a companion planet with a barrage of X-rays a hundred thousand times more intense than the Earth receives from the Sun.

Recommended for you

User comments : 37

Adjust slider to filter visible comments by rank

Display comments: newest first

DaFranker
3.6 / 5 (5) Jan 19, 2012
If the maths involved in determining the energy required to destroy a planet in the manner shown in Star Wars were really that simple, we'd have figured out how to make fusion reactors long ago. Gravity is not the only factor here, since gravity is not all that "holds the planet together".

Nice of him to bring up the idea and publish this article though - I assume he wrote this in his free time, and we could benefit from more people doing things like this with their free time.
btb101
3.3 / 5 (3) Jan 19, 2012
some people have way too much time on their hands.

its good work though. lol.
TrinityComplex
5 / 5 (4) Jan 19, 2012
They explain what the Death Star is and then end the paper with an ambiguous reference to the Suncrusher. That's a little cruel.
jonah_arnot
not rated yet Jan 19, 2012
Yeah, what is up with that last line? I thought we were talking about crunching planets, not stars. Maybe the author is trying to indicate that destruction of a planet like Jupiter would require the amount of destructive capability that it would take to destroy a star, but if that's the case, what evidence do they provide to back it up? Or maybe it's a simple error on the part of the author, who just meant to put planet instead of star. Oh well, the article was moderately informative at least. Yay Star Wars!
that_guy
3.1 / 5 (8) Jan 19, 2012
@dafranker -
Idiot.
The article states that the math is idealized due to unknowns.
You may be right that some factors may have been missed, but this is just an exercise.

You say that if it "were so easy, then we'd already figured out how to make fusion reactors..."

No. We have figured out the approximate energy of the sun, how it generates energy, it's radius, etc. Some of these estimates are very easy to make. That doesn't mean that it gives us a method how to make energy.

All this guy does is show that the lore of the deathstar is consistent with physics in this one narrow area, that the entire output of several stars is enough energy to destroy a planet. DRRRRR!!@!!

1 star for you, 5 for everyone else.
nkalanaga
5 / 5 (2) Jan 19, 2012
Actually, they did give the evidence. The binding energy of Jupiter is about 10,000 times that of Earth, so it would take a weapon with 10,000 times the energy output to destroy it.

The Sun's energy output would require 750,000 seconds to generate enough energy to destroy the Earth. "Several stars" is rather vague, since the don't specify the stars. Assuming the Sun is average, and ten stars that size, it would require 75,000 seconds to generate the energy, or less than a day.

If we're talking A, B, or O stars, it could be an hour or less.

However, even at that rate, it would take months, if not years, to produce enough energy to destroy Jupiter, so the Death Star wouldn't be an effective weapon there.
Eikka
2.3 / 5 (3) Jan 19, 2012
Reminds me of this quote:

The great logician Betrand Russell (or was it A.N. Whitehead?) once claimed that he could prove anything if given that 1 plus 1 equals 1. So one day, some smarty-pants asked him, "Ok. Prove that you're the Pope." He thought for a while and proclaimed, "I am one. The Pope is one. Therefore, the Pope and I are one."


Take fanciful assumptions and you can extrapolate anything.

That said, there's still the problem that there's ten orders of magnitude difference between the power output of the sun, and the assumed energy it takes to destroy a planet. Doesn't make much difference if there's several of them.

10 to the 10th power seconds is still 300 years before the Death Star is fully charced and ready to fire.
yyz
5 / 5 (1) Jan 19, 2012
So how would the Death Star compare with the Doomsday Machine (remember, with the pure neutronium hull): http://en.wikiped..._Trek%29
javjav
4 / 5 (1) Jan 19, 2012
This is not practical. If the empire wanted to destroy our planet, it would require trillions of times less energy to simply point their superlaser to the moon in front of it's orbit direction, to slow down our dear satellite and let it smash our planet.
Seeker2
1 / 5 (2) Jan 19, 2012
If we should get in the path of some death ray coming out of a black hole it could be curtains. I think I heard there is actually one in the Andromeda galaxy with us in its sights.
nkalanaga
5 / 5 (6) Jan 19, 2012
Eikka: It would take a little less than 10 to the 6th power seconds for Earth (10^26 vs 10^32), and if they had big enough capacitors, that could be done. How big a capacitor can one put in hyperspace?

On the other hand, rendering the planet uninhabitable would be even simpler than Javjav's idea. Given Star Wars technology, put one of their gravity drives on 10 km diameter asteroid, accelerate it to half the speed of light, and aim it at the planet. You'd splatter magma over the entire surface of the planet. The planet might still be in one piece, but there certainly wouldn't be any survivors!

In fact, any drive suitable for manned interstellar travel would be capable of destroying a civilization.
ziphead
4.5 / 5 (8) Jan 19, 2012
Lets just hope Mythbusters do not get to see this article.
Cave_Man
1 / 5 (1) Jan 19, 2012
How do they assume the laser energy is going to interact with matter, how big is beam size, all you need is to put a hole through the planet and later pressure from displacement or collapse would probably be enough to crumble. What would be interesting is to find out if the large scale mathematics would be anything like the movies...would the planet crumble? explode? split into chunks?

idk, a marble going 90% the speed of light would have some bad effects on the habitability of earth.
barakn
4.2 / 5 (5) Jan 20, 2012
Gravity is not the only factor here, since gravity is not all that "holds the planet together"
Take, for example, water. It takes 2x458.9 kJ/mol = 917.8 kJ/mol to dissociate all the molecular bonds in a mole of water. The energy required to lift a mole of water from the Earth's surface to infinity (with zero leftover kinetic energy) is, using U=-GMm/r
6.673 E-11 m^3/kg/s^2 * 5.9742 E24 kg * .018 kg/mol / 6,371,000 m = 1,126 kJ/mol. This may not seem much larger than the molecular dissociation energy, but consider that you don't need to atomize the planet. If you blow it up so that the average fragment size is a meter cubed, you leave the vast majority of the molecules alone whole but still destroy the planet.
nkalanaga
5 / 5 (2) Jan 20, 2012
For those who are really interested:
http://qntm.org/destroy
"Things Of Interest: Blog: Geocide:
How to destroy the Earth"

Barakn is right. The problem is that the mass of the Earth has to be not only broken up, but accelerated to escape velocity. Otherwise it will fall back together, and you'll still have a planet, although not a habitable one.
plasticpower
1.8 / 5 (5) Jan 20, 2012
We can already destroy the whole planet.. or at least all life on it. With nukes.
zz6549
not rated yet Jan 20, 2012
A more important question is: is it possible to generate the required energy given the size of the Deathstar?

Using his lower estimate of 2.25E32 J, and assuming perfect mass->energy conversion, 2.5E15 Kg of matter would be required.

The first death star was 160km in diameter. Assuming it was solid steel (density of 7.85E12 Kg/Km^3), it would weigh approximately 1.7E19 Kg.

The density of hypermatter is unknown, but if it had an equal density to steel, it would be quite possible to store 10000 shots worth of it within the Deathstar.

Moebius
2.1 / 5 (14) Jan 20, 2012
What a bunch of crap. You make the assumption that the death star produces enough power to destroy a planet and then deduce that it in fact can destroy a planet because it has enough energy. Also I don't recall them identifying the weapon as a laser.

My assumptions would be that it's a hyper efficient weapon and can be powered with a nine volt battery. The death star was in fact the size of a moon to house the thousands of people required to run the bureaucracy involved with an empire that has too many weapons and is all too willing to use them. Sound familiar?
mosahlah
1.2 / 5 (5) Jan 20, 2012
I think what they really should be using to destroy a planet is a large CO2 generator. Then the planet will simply bake itself into oblivion. I came up with that idea from reading the mainstream media.
Blakut
not rated yet Jan 20, 2012
What does explode mean? Not taking all pieces of matter from here to infinity, but throwing them so hard that they won't ever return. It's a big difference...You have to take POWER into account, not just energy.
nkalanaga
5 / 5 (4) Jan 21, 2012
Mosahlah: Won't work. If all of the carbon on Earth was converted to carbon dioxide, we'd have an atmosphere like Venus. Venus is still there, and while uninhabitable, shows no sign of significant structural damage.
Seeker2
4 / 5 (6) Jan 21, 2012
I think what they really should be using to destroy a planet is a large CO2 generator. Then the planet will simply bake itself into oblivion. I came up with that idea from reading the mainstream media.
We actually have a huge methane generator under the arctic tundra. When it melts we may need oxygen masks if we can keep it from igniting.
Seeker2
1 / 5 (3) Jan 21, 2012
If it does ignite sayonara.
nkalanaga
5 / 5 (4) Jan 22, 2012
Methane is only explosive between about 5% and 15% in our atmosphere. There isn't enough methane in the arctic and oceans to reach that level.

On the other hand, methane is a much stronger greenhouse gas than carbon dioxide, so things could get uncomfortably warm if too much is released.
TychoCraterCafe
1 / 5 (1) Jan 22, 2012
I think it's more likely that your could shake a planet apart using Tesla-like resonance than hitting it with a mega-laser :) Interesting to think about though.
Isaacsname
not rated yet Jan 22, 2012
Wouldn't that be a fair bit of thrust ?
daqman
1 / 5 (2) Jan 22, 2012
I agree with one of the earlier posters. The energy required to "destroy" a planet is much less than that quoted in the article. The paper calculates the energy to completely dissociate the planet so that all parts are given escape velocity. In principle all you have to do is put most of the planet into orbit around it's center of gravity. This takes much less energy but delivering that much energy in the form of a beam is a problem.

A missile made from a one kilometer on a side cube of anti-iron should do the trick. This would annihilate with the equivalent amount of molten iron in the planet's core and release enough energy to move the bulk of the planet into orbit around itself.

Of course finding that much anti-iron it a problem.
Graeme
not rated yet Jan 22, 2012
The efficiency of conversion would have to be extremely high otherwise even only a tiny fraction of waste energy in the moon sized object would destroy the deathstar as well.
It is one thing to make a laser 10% efficient, and perhaps extremely hard to make one 90% efficient, but making a laser convert power at 99.999999% efficiency I am sure has never happened. Only fiction however, so it can be as unrealistic as they want.
jsa09
5 / 5 (2) Jan 22, 2012
I am a little confused. There is a mathematical equation of the energy required to destroy an Earth-like planet with no definition at all as to what is meant by "destroy".

I would think that the equation is out by several orders of magnitude but then my definition of "destroy" is probably different than theirs.

One needs to define ALL the terms before one can generate an equation.
ThanderMAX
not rated yet Jan 24, 2012
In StarTrek , they could re-route energy even from toaster and grinders :P
antialias_physorg
not rated yet Jan 24, 2012
There is a mathematical equation of the energy required to destroy an Earth-like planet with no definition at all as to what is meant by "destroy".

They use the gravitational binding energy
http://en.wikiped...g_energy
Which is an upper limit to the energy you would need.

Gravitational binding energy assumes that particles are loose and that you pull them so far apart that they do not coalesce again.

A real planet does not consist of loose particles (and is therefore easier to pull apart because you don't need to pull all particles away from each other but just however many chunks you have blown it into. On the other hand you have to employ some energy to get those chunks to come loose from each other - but that is less than getting the stuff to fly off into space)

Additionally it would be OK to have the parts orbit each other afterwards through gravity. For all intents and purposes it would still be "destroyed". So the energy needed is less.
antialias_physorg
not rated yet Jan 24, 2012
A missile made from a one kilometer on a side cube of anti-iron should do the trick. This would annihilate with the equivalent amount of molten iron in the planet's core

It wouldn't get that far because it would already annihilate with the first cubic kilometer it meets on the surface.

Antimatter reacts with any matter - not just matter that has the same chemical properties (it reacts on a proton-antiproton/electron-positron/neutron-antineutron level).
GSwift7
1 / 5 (1) Jan 24, 2012
I do love Star Wars, and the new MMO is great, but trying to calculate the amount of energy needed is futile. Alderan is weak and therefore easy to destroy. You should all join the Dark Side; we have cookies.

Oh, and who's your daddy? I am.
FrankHerbert
3.3 / 5 (12) Jan 24, 2012
Shouldn't we really be worried about the power of a Stone Burner required to destroy an Earth-sized planet and also the distance one must be from it to keep one's eyes from being fried out of their sockets by J-rays? That is unless you don't need eyes :)
GSwift7
1 / 5 (1) Jan 24, 2012
What I want to know is "what's an aluminum falcon?"

http://www.youtub...d3QWsyk0

lolz
Meyer
1 / 5 (1) Jan 25, 2012
We actually have a huge methane generator under the arctic tundra. When it melts we may need oxygen masks if we can keep it from igniting.

Fortunately heat rises so the ocean floors are barely affected by atmospheric temperatures, and most of the hydrate deposits have 5-10 degrees of leeway. If a large quantity did melt somehow (e.g. by an ice age reducing the mass and pressure of the oceans), the first thing to worry about would be oxygen depletion in the ocean from methanotrophs consuming the dissolved methane, as was observed in some parts of the Gulf of Mexico after the oil spill. What little that escapes into the atmosphere will probably not be enough to pull us out of the ice age.
Xbw
1.8 / 5 (5) Jan 30, 2012
The number of comments on this article shows where our real priorities are hehe.

I'd like his next article to be on the doomsday machine from Star Trek TOS.

edit: Oops here is a linky. http://en.wikiped...eries%29

Please sign in to add a comment. Registration is free, and takes less than a minute. Read more

Click here to reset your password.
Sign in to get notified via email when new comments are made.