Built on Facts

Standing on the edge of Niagra Falls you can watch the water pour over. Falling down the gravity of the earth, it exchanges its potential energy for kinetic energy by picking up speed. Some of that energy is extracted by turbines and lights the homes and businesses of Yankees and Canucks alike. Some of that energy is used to pump water up into water towers to maintain the water pressure which those same people use to cook and clean.

Move with a gravitational field, get energy. Move against it, lose energy.

Now let’s say you wanted to take apart the earth. Yes, the whole thing. You want to grab each bit of dirt and pull it out of the earth’s gravity and move it out to deep space. How much energy in total would it take? There’s some practical reasons for wanting to know. The engineers who built the Death Star would need for their superlaser to deliver at least that much energy. And what could be more practical than that?

But figuring out exactly how much energy requires some finesse – figure out how much energy it takes to remove a kilogram worth of earth, and suddenly the earth has one fewer kilogram worth of gravity. So the next kilogram won’t take quite as much energy to remove, and so on. We’re going to have to do some thinking about how to get around this. We’ll start with the equation that tells you the potential energy of an object in the gravitational field of a uniform spherical mass. The potential energy is just the energy required to pull the mass entirely out of the gravity of the sphere. Here G is the gravitational constant, M is the mass of the uniform sphere, m is the mass of the object in the sphere’s field, and r is the distance between the mass and the center of the sphere:

Now here’s my plan to get figure out how to take the changing mass of the sphere into account. Imagine you scrape off a very thin but uniform layer off of the entire earth, like removing the peel from an apple. The surface area of a sphere is 4 pi times its radius, so the volume of that layer is going to be 4 pi times the radius times the thickness of that layer. Its mass is its density times its volume. In the laconic language of math, that means:

Where the greek letter rho is the density of the layer, and dr is the thickness. I choose to call the thickness “dr” so that the calculus-fluent can at this point nod their heads sagely and recognize my line of attack. Now we’ll do the same thing for M, the mass of the sphere. Clearly it’s just going to be the volume of the sphere (which is 4/3 pi r^3) times its density:

Now that’s the energy required to remove one layer. All that remains to do is add up the successive energy required to remove all those infinite numbers of successive infinitely thin layers. Sounds difficult. But that’s why God created Newton, who created calculus. We can add up the total energy to remove all the layers like this:

That’s just the energy equation with M and m replaced, and an integral sign in front of it noting that we’re adding all the layers from the center of the earth at r = 0 to the normal radius of the earth r = R. Perform the integral and get:

But the density rho is itself just the mass of earth divided by its volume, which is just M/(4/3 pi R^3). Plugging that in and reducing:

And that’s our final answer, which is also called the gravitational binding energy of a sphere. For the earth, it works out to around 2.2 x 10³² joules.

That’s a preposterously huge amount of energy! It’s a solid week of the sun’s entire power output. Dumping it in about a single second, as required to blow up Alderaan, is a very, very impressive feat. Doubly so when you take into account the fact that the binding energy is just enough to dissociate the planet into a diffuse cloud. If you want to actually blow the thing up into pieces flying out at many times escape velocity, you need much more energy.

Darth Vader told us that the power to destroy a planet is insignificant next to the power of the Force. Either he’s being a typically mystical ex-Jedi offering wisdom along the lines of the pen outmatching the sword, or the Force must be pretty dang powerful.

Comments

#1 Nemo
February 4, 2009

Or maybe Alderaan is a lot smaller than it looked. Happens in movies sometimes.
#2 Jacob
February 4, 2009

Wookiepedia (yes, it exists!) lists the diameter of Alderaan as 12,500 km – almost exactly the same as Earth’s. Seems like the approximation holds, although it could of course be less dense.
#3 Cannonball Jones
February 4, 2009

I only remember about 3 people in the whole Star Wars saga coming from Alderaan, maybe it was really tiny as Nemo said. Like the size of a caravan.
#4 Daniel
February 4, 2009

These same calculations were done years ago by Dr Curtis Saxton of Star Wars Technical Commentaries, and Michael Wong on stardestroyer.net . Clearly great minds think alike, though for a truly accurate measure you would need to see how fast the debris was moving after the explosion, this is just a lower limit otherwise.

I don’t suppose you will speculate on the origins of the “ring” anytime soon, will you?
#5 Matt Springer
February 4, 2009

Hmm. The ring is tricky. Maybe an obscure angular-momentum effect, sort of the inverse of black hole jets?

(Of course a cynic might say that they originate in Star Trek. But I was never a Trek fan anyway.)
#6 Spaceballs
February 4, 2009

This all assumes that the Death Star destroys with its own energy, rather than creating a chain reaction in the target itself.
#7 Karl Withakay
February 4, 2009

This was #7 (and technically also #4 and #3)on livescience.com’s Top Ten Ways to Destroy the Earth.

http://www.livescience.com/technology/destroy_earth_mp-1.html

They calculated you would need 2.5X10^12 metric tons of antimatter to produce the 2.24X10^32J (assuming 100% efficiency) . Isn’t that off by a factor of two? Wouldn’t your need half that much antimatter, and the same amount of matter for it to annihilate with since BOTH the reacting matter and antimatter are converted into energy? The TOTAL amount of mass (matter + antimatter) needed to be converted into energy is 2.5X10^15 kg, right?
#8 Karl Withakay
February 4, 2009

To put the energy requirements in perspective, that’s a solid antimatter asteroid about the size and mass or Mars’ smaller moon Deimos.
#9 t
February 4, 2009

stumbleupon ftw.
nice math im just starting learning the finer applications of calculus, and found myself thinking when reading the introduction that it should be used.
Just a thought though, couldn’t some of the energy present on the planet, or even the mass (with mass-energy equivelance) work towards lowering the energy needed by the laser?
#10 Aaron
February 4, 2009

Now, of course your calculations hold correct if you want to absolutely destroy a planet (essentially vaporizing it), but what if we just wanted to functionally destroy it? Say, thoroughly wipe out all multicellular life? I would think it would be sufficient to liquiefy, vaporize, or otherwise destroy the surface of an earth-size planet to the depth of – oh – a mile or so?
#11 Uncle Al
February 4, 2009

-2.2 x 10^32 joules, E=mc^2, c = 299,792,458 m/s… 2.4×10^15 tonnes of energy. How much did the Death Star mass?

The actual gravitational binding energy of the Earth skews from not being a homogenous body,

http://www.splung.com/kinematics/images/gravitation/variation%20of%20g.png
black curve; red is homogeneous body.
http://www.typnet.net/Essays/EarthGrav.htm
bottom; gee increases with depth to 50% radius
http://www.typnet.net/Essays/EarthBind.htm
2/3 down

Average density: -2.242×10^32 J, 2.495×10^15 kg
Empirical density: -1.711×10^32 J, 1.904×10^15 kg

That second number seems a bit suspect. With most of the Earth’s density being near its center one might expect greater negative binding energy (not a less negative number) for the greater mass having fallen deeper.
#12 Dave
February 4, 2009

to #10

That wouldn’t look quite as impressive would it? And we wouldn’t get that satisfying boom that you know we’d hear when you explode a planet. ^^
#13 dWj
February 4, 2009

The “chain reaction” suggestion (#6) requires that there be some potential energy in the earth — presumably you’re imagining chemical potential energy, though perhaps you’re thinking something nuclear. Me, I’m thinking bulk modulus; is the earth springy? At equilibrium, the lower layers are compressed enough to push up with the force with which they’re being pushed down; if there’s compressibility, that means some energy is released as it expands. On no real basis whatsoever, I’d imagine the potential energy of compression is a couple orders of magnitude smaller than the gravitational binding energy; this is still an impressive amount of energy, though.
#14 The Science Pundit
February 4, 2009

Nitpick:

The surface of a sphere is 4 pi times its radius squared. I see you corrected the mistake by the time you got to the integral, so it didn’t affrect your result.
#15 The Science Pundit
February 4, 2009

affrect? I need to proofread my comments better.
#16 Karl Withakay
February 4, 2009

Maybe the chain reaction is a stangelet catalyzation deal.
(Method #9 for destroying the Earth.)
#17 Joshua Zelinsky
February 4, 2009

Praxis exploding was in Star Trek 6 which was years after Star Wars so if anything the inspiration was the other way around.
#18 pjb
February 4, 2009

Joshua:
If I’m not mistaken, the original Star Wars footage had no ring. I think it was added when the trilogy was remastered in the 90s.
#19 Eric Lund
February 4, 2009

Of course, you have assumed in this calculation a uniform density for the Earth, which isn’t quite true. The outer layers of the Earth are rock, which is less dense than the iron core. That should reduce the energy requirement by some factor of order 1.

#12 is correct that there is some compression due to the weight of the material above. This is why there is a solid inner core surrounded by a liquid outer core: at some depth the pressure pushes the iron back into the solid phase despite the elevated temperature. But I agree with his intuition that this is a small effect; the nonuniform density will produce a bigger correction.

I also agree with Dave (#11) that the point in a work of space opera fiction is to get, as Marvin the Martian would say, an Earth-shattering kaboom. In practical applications, however, vaporizing the surface as Aaron (#10) advocates would get the job done–but you would have to go a bit deeper because of topography. Somewhere between 10 km and 30 km, depending how thorough you want to be (30 km is the elevation difference between the summit of the highest mountain and the depth of the deepest ocean trench), would suffice.
#20 Ian
February 4, 2009

Spaceballs said:
“This all assumes that the Death Star destroys with its own energy, rather than creating a chain reaction in the target itself.”

You mean like this?

http://en.wikipedia.org/wiki/Concepts_in_the_Ender%27s_Game_series#Molecular_Disruption_Device
#21 Pimco
February 4, 2009

pjb wrote:
If I’m not mistaken, the original Star Wars footage had no ring. I think it was added when the trilogy was remastered in the 90s.

Inspiring no small amount of consternation in many filmgoers at the time.
#22 Matt Springer
February 4, 2009

Thanks, Science Pundit! I’ve fixed the missing 2.

A couple of other comments: yes, the explosion in the original version didn’t have the ring. It got added in the special editions in the 90s. Also, Han shot first.

As far as using energy that happens to be latent in the earth, that’s possible but I don’t think it’s going to be anywhere near the order of magnitude needed. In theory you could do nuclear fusion on all the atoms lighter than iron but I still don’t think it would be adequate. Energy production drops off pretty quick for elements heavier than hydrogen. Worse, a large fraction of the earth is iron, which is a dead end in terms of nuclear energy production.
#23 Pimco
February 4, 2009

Spaceballs@6: The eponymous planet-’sploder weapon in the Lexx TV series was implied to work this way.
#24 Greg
February 4, 2009

We may also want to consider that they have the physics to manipulate gravity itself as seen by the fact that even small ships like the falcon have standard gravity and don’t have to rotate or anything similarly. Also, the anti-gravity craft that luke and others use throughout the movies on planets suggest they can easily manipulate gravity. The Death Star may simply change the gravitational field of all the matter the planet is made of making it “explode” away from itself as seen in the movie. This may take far less energy than working against the planet’s gravity. Could the “laser” be an anti-graviton beam?
#25 Joshua Zelinsky
February 4, 2009

Ok, I guess I lost whatever geek cred I had with that comment. Before I go back to my little corner I’ll just point out that apparently a very minimal price estimate for the Death Star has been made:

http://i.gizmodo.com/5146010/death-star-costs-156-septillion-14-trillion-times-the-us-debt
#26 Karl Withakay
February 4, 2009

Greg, RE#23,

I’ll defer to a theoretical physicist on this, but I assume E would still = mc^2 no matter how you approached it.
#27 Karl Withakay
February 4, 2009

RE#23
Total mass-energy has to be conserved, no matter what method is used to do the work.
#28 Anonymous
February 4, 2009

“Also, Han shot first.”

nope, he didn’t. in the original Greedo did not shoot at all, so “first” does not apply. Han simply shot, period

[/nerdier than you]
#29 skyotter
February 4, 2009

nope, he didn’t. in the original Greedo did not shoot at all, so “first” does not apply. Han simply shot, period

that’s mine, sorry. i didn’t mean to be Anonymous =)
#30 Jedispy
February 4, 2009

Another thing you have to consider is how quickly the reaction occurred before the explosion. A laswer (assuming the superlaser is actually based on laser technology) would need to bore down into the crust in order to cause the detonation (see the logic provided in Deep Impact and Armageddon for what I’m talking about here). The fact that the entire reaction is completed in roughly 1 second means that the superlaser outputs enough energy to pass through the entire radius of the planet. Them’s be some big energies I tell you whut
#31 Realist
February 4, 2009

It’s just a movie, you losers.
#32 Griffin
February 4, 2009

Seems overly simplified. You are treating the planet as a stationary body with uniform consistency, that is somehow being targeted over the entire surface. I see no accounting for the energy due to the existing motion of the planet (rotation and revolution). You are also assuming that the DS beam is energizing the planet evenly. Since it was a narrow beam, it would only be affecting a localized area, certainly not the entire surface, especially not the far side of the planet. Blowing off a sizable chunk from one side would cause fractures through the crust and shock waves through the core which would help to tear the planet apart as it rotates.
#33 Offended
February 4, 2009

And you’re just a boring fuddy duddy #30. Get over yourself and let us have our fun.
#34 Glaivester
February 4, 2009

I must admit that the explosion of Alderaan has always struck me as unsatisfactory – at least after the firs viewing or so. The laser shots, the planet goes boom – I alwyas thought it would be more satisfying if the laser hit the planet, kept shooting for a second with a vibration noise rapidly increasing into a shriek, and then the planet explodes. Perhaps there was not enough money for special effects to fund any explosion more than what was shown, but a three second delay before the explosion, with the laser hitting all the time would have been possible.
#35 Anonymous
February 4, 2009

Never understood why they didn’t just take a decommissioned Star Destroyer, aim it at a planet, and kick on the hyperdrive. Hyperdrive is FTL, so there’s got to be a moment (the impact) where it’s at least traveling at C; infinite mass object slamming into something with significantly less mass than “infinite”, good to go.
#36 Joshua Zelinsky
February 4, 2009

Anonymous, hyperdrives only allow you to travel faster than the speed of light in hyperspace. Once you drop out of hyperspace you resume at the same vector you entered in at (or something like that).
#37 frog
February 4, 2009

You don’t have to go through all that effort. Since the energy of this kind of system (as long as it’s spherically symmetrical) for a change outside of the sphere is exactly equal to the same change with all the mass at the center of the sphere, and the energy already takes into account the rest of the system (the 1/2 in the energy for these kinds of systems — same for Maxwell’s equations), you just have to calculate the energy gained to move all the mass from infinitely far away to the center (as long as you set it up to avoid the singularity — three dimensions is nice because it’s the only dimensionality which let’s you do that!)
#38 Star Destroyer
February 4, 2009

In response to #10: The Death Star does not need to melt the crust to a mile’s depth.

Imperial class Star Destroyers are designed for that purpose, and they have flown *escort* to the Death Star.
#39 EJ
February 4, 2009

“Perform” the integral??

An integral is not a procedure. You can evaluate an integral, but an integral is not something you perform.
#40 llewelly
February 4, 2009

I keep looking at your 2nd equation and not seeing the rho you’re referring to. Hard to believe I’m the only person to notice this omission. It comes back in equation 4.
#41 Matt Springer
February 5, 2009

Fixed. I probably dropped it when I fixed it the first time for the missing 2.

More comments: The rotational kinetic energy of the earth won’t help us much even if we could recover all of it. It’s a couple orders of magnitude smaller than the gravitational binding energy.

And ok, EJ, you perform an integration. You’re probably one of those mathematicians who get all bothered by the sloppy way physicists talk about limits too.
#42 Dax
February 5, 2009

Matt, this is great. I was thinking about doing my BS in Physics and now I am sure. You have illuminated it for me. Write more.
#43 carlos
February 5, 2009

you got it all wrong dude, the laser transports a tiny little chimp, who’s kind of a wacko and activates a damn powerful nuclear warhead on the core of the earth and destroys it big time, unfortunately the chimp knows he’s going to die, so before the lasers of the death star teletransports the chimp to the earth, Darth Vader and friends make a hug party to celebreate the chimp’s feat and honor him telling him that when he dies he will be received in heaven by 30 hot virgin female chimps, at the end, the chimp is happy. True story.
#44 Steve
February 5, 2009

This is a great start, though you must realize that the amount of energy calculated here is the absolute minimum.
It would be interesting to apply this amount of kinetic energy to an assumed number of particles of an assumed mass and see how fast they would travel away from the center of the planet. I’d also be interested to see what kind of mechanical bonds would need to be broken, as this is just the energy required to move a mass the size of the earth, not tear it apart.
#45 Azkyroth
February 5, 2009

We may also want to consider that they have the physics to manipulate gravity itself as seen by the fact that even small ships like the falcon have standard gravity and don’t have to rotate or anything similarly. Also, the anti-gravity craft that luke and others use throughout the movies on planets suggest they can easily manipulate gravity. The Death Star may simply change the gravitational field of all the matter the planet is made of making it “explode” away from itself as seen in the movie. This may take far less energy than working against the planet’s gravity. Could the “laser” be an anti-graviton beam?

On a related note, wouldn’t every atomic nucleus in a sample of matter essentially explode, sending its entire complement of protons flying outward at near-relativistic velocity, if you could find a way to momentarily neutralize or disrupt the strong nuclear force?

(Have any science fiction weapons using this concept been proposed?)
#46 siddharth
February 5, 2009

http://abstrusegoose.com/95

The trick is in obtaining an appropriate Lagrangian Field Density of “The Force”.
#47 ghostbusterbob
February 5, 2009

If you are reading down this far (I did) you are a geek and a nerd (as I am.)

When the Falcon came out of hyperspace, they were not greeted by a fine Alderaan dust, but what they thought was an uncharted asteroid field. The planet wasn’t completely disintegrated, but was just completely “blown away.” That’ll reduce your overall energy usage. Hey, Vader is greener than al gore!
#48 MRW
February 5, 2009

Re: #25

Greg, RE#23,
I’ll defer to a theoretical physicist on this, but I assume E would still = mc^2 no matter how you approached it.

Presumably, but what does that have to do with anything? E=mc^2 is nowhere in these equations.
#49 Troy
February 5, 2009

It wasn’t a simple laser. It was augmented by the dark side of the force. That’s way more power than needed considering the force (dark or light) allows the wielder unlimited access to the entire universes force energy instantly. The amount of force energy needed would not lessen the overall universal ‘force’ as that is infinite.

This why Vader had to be in or near the Death Star when it is using the force augmented laser.
#50 kevin
February 5, 2009

in the words of the immortal sage… joel hodgeson..

“if you are wondering how he eats and breathes and other science facts…(la la la)…then repeat to yourself ITS JUST A SHOW..I should really just RELAX”…

and if you dont get the reference… you are a mental turd anyway.
#51 kevin
February 5, 2009

oops.. i misspelled the name of the immortal sage…

joel hodgson…. please forgive me oh great one.
#52 kevin
February 5, 2009

granted, its entertaining.. but if we were to use all of our intellectual capacity in… say… oh, i dont know…. FIGURIING OUT HOW TO GET ENOUGH POWER OUT OF ALTERNATIVE FUELS, or A WAY TO HARNESS THE SUNS POWER INEXPENSIVELY, OR, HOW TO GROW FOOD, TO HELP US IN OUR “REAL SITUATION, RIGHT NOW” INSTEAD OF investing all of that intellectual capacity on solving the logistics of FICTIONAL STORIES …..that were never meant to be real, but to be allegories and a new mythology…… you might even be able to make enough money to move out of your parent’s basement…to a house of your own….then you can have more space to put all of your “ACTION FIGURES”.
i know… its a radical thought… but… the world is in the toilet.. how about putting some of that brain power to use for your society and stop putting your finger in your belly button and sniffing it.
#53 Karl Withakay
February 5, 2009

MRW RE#47
My point was that if you could produce some sort of antigravity beam, the amount of energy required to create a beam powerful enough to produce a reverse gravity filed sufficient to disassemble the planet would be the same as the amount of energy to blow it apart. I added #26 because after looking at it, I though #25 did a poor job of expressing my point.
#54 Lee Child
February 5, 2009

This is just sad. Kevin pretty much said it all.
#55 Kail
February 5, 2009

#51, kevin:

I assume you’re scouring the Internet posting similar rants in other fandom-related areas. My Little Pony modding communities, various fanfiction fora, alt.fan.TrickMyTruck, etc. That must take a lot of your time.

Because you wouldn’t just be picking on people who commented on a nifty back-of-the-envelope pop physics demonstration that referenced a popular movie so people’d have something to relate to the equations. Would you? That would seem… dickish.

Either way, this implies that you aren’t a 100% efficient cognition machine either, and you occasionally do things with your brain that aren’t “putting some of that brain power to use for your society.”

Anyway, good luck with moving “out of your parent’s [sic, possibly] basement.” HAND!
#56 Kail
February 5, 2009

I’m sorry I take all that back I’m an idiot
#57 Neil B ☺
February 5, 2009

@44: That’s a clever idea, and simply “disconnecting” the strong force for a moment wouldn’t even cost any energy input at all! It would be like unhooking a spring. That is essentially how atomic bombs work: the extra neutron entering a fissionable nucleus disrupts the normal strong force relation, and the mutual positive repulsion flings the nuclear fragments apart. This also shows that one must be careful about hidden assumptions such as, you must use extra input energy to make X happen.

But as annoying as Kevin@51 seems, consider getting inspired and put some time at least into practical offerings, you might come up with something good. I’m gonna try …
#58 Karl Withakay
February 5, 2009

Neil B ☺,

That’s kind of like saying, “I can light gasoline on fire which releases energy from the chemical bonds of the gasoline molecules, if only I could find a way to do the same thing with water molecules.”

It makes for decent science fiction, though.

Of course, if you’re Q, you can just change the gravitational constant of the universe for the planet and let it spin itself apart.
#59 Tulse
February 5, 2009

As an alternative (and noting that the planet seemed to come apart in “chunks”), would it require less energy to vaporize a hole to the centre of the planet, and then vaporize enough of the core to explode the planet from within due to the vapor pressure?
#60 Abelman
February 5, 2009

So, this laser heats up the whole of the planet causing it to explode, or what? I think it’s interesting to determine the energy necessary, but what, physically, is actually going on with the energy beam?

Does it slice through the planet affecting the core of Alderaan? Is the Death Start in a geosynchronous orbit around the planet? Perhaps what we see as a “beam” is actually a grouping of high-energy beta particles or something designed to create a chain reaction within the planets core.

Still, it does seem to happen a bit faster than science would allow.
#61 slarkin
February 5, 2009

wow, It’s a movie. It’s not real get over it….. Losers
#62 Matt Springer
February 5, 2009

Just for the record – disagreement is great, but if you’re one guy posting under several different names and complaining profanely about this being a “waste of time”, well I think you’re wasting your time.

Who’s more foolish the fool or the fool who follows him?
-Obi-Wan Kenobi. Who was on the Death Star at the time…

Also your comments won’t stay published regardless.
#63 Jørgen Hansen
February 5, 2009

Don’t question my faith. This really happended. It’s still happening. Stupidity has blinded humanity to the true power of the force. Rather than mocking the few of us able to tap, albeit miniscule, power of the force, you should all open your eyes and join us. We can eventually rebuild this lost connection to the force and make the world a better place.
#64 Karl Withtaky
February 5, 2009

Tulse , RE#58

The amount of energy to do the net work required is independent of the means used to do the work, so long as we are assuming 100% efficiency (or even the same efficiency for all methods considered) of the energy used.

Net work is independent of the path taken, and we are essentially talking about the net work required to disassemble the planet against the force of its own gravity.
#65 Stupu
February 5, 2009

just a little aside, you might want to change the M/(4/3 pi R^2) in your last step to M/(4/3 pi R^3). i spent a good 10 minutes scratching my head before i realised that wasnt the volume of a sphere… and good god, i have a physics exam in 4 days.

I’m taking this as a hint from the force that i should study a bit, because that was poor form for a math geek.

nice derivation all round though, i liked =D
#66 Tulse
February 5, 2009

Karl, I really should have known that…I am abashed.
#67 gond
February 5, 2009

We may also want to consider that they have the physics to manipulate gravity itself as seen by the fact that even small ships like the falcon have standard gravity and don’t have to rotate or anything similarly. Also, the anti-gravity craft that luke and others use throughout the movies on planets suggest they can easily manipulate gravity. The Death Star may simply change the gravitational field of all the matter the planet is made of making it “explode” away from itself as seen in the movie. This may take far less energy than working against the planet’s gravity. Could the “laser” be an anti-graviton beam?

You could go even one further by saying that the “laser” is a “graviton gun”, where it ADDS gravitons to the entire system, causing it to start to IMPLODE, then the gun is shut off, returning graviton levels to normal, creating the EXPLOSION we see in the film. Just a thought.
G-
#68 Tim S
February 5, 2009

At all the people that are talking about this being a waste of time and energy (#51 in particular), the mathematical exercise here wasn’t that complex. It seems to be about at the level as physics problems I’ve done in college. Problems that weren’t designed to take very long or exhaust all of a person’s intellectual energy. I really doubt that anyone here actually does spend all their time and energy on these musings and some people like to apply what they know to all sorts of hypothetical questions, for example I had a final with a problem about a cow rolling down an inclined plane(Assuming the cow is spherical).
#69 dude
February 5, 2009

#18 is partially correct… I agree with you that the calculations are skewed because they assume a uniform distribution of mass, however, this would increase the energy requirements since the outer layers would contain less mass, the energy required to remove each subsequent layer would be lessened by less.

#23 is correct… and in support of this theory, I submit… They had tractor beam technology (the Falcon was trapped in their tractor beam)… This obviously shows a mastery of gravitational manipulation… Also, you would notice that the laser beams change direction (they come out at an angle and change to leave the Deathstar perpendicular to its axis).

#31 and #46 are correct, you don’t need to turn the planet into dust, you just need
enough concusive force to break it apart. The Falcon dropped out of hyperspace into what they thought was an asteroid field.

You don’t need to strip the planet apart layer by layer, just “blow up the damn [planet] Jean-Luc”.
#70 Jeff bwles
February 5, 2009

Uhh, I think that the original article nearly hit it on the head: That’s a preposterously huge amount of energy! It’s a solid week of the sun’s entire power output. Dumping it in about a single second, as required to blow up Alderaan, is a very, very impressive feat.

In other words, somehow make a small “sun” erupt in the planet’s core, and its energy will be enough to blow the thing apart.

Causing the core of the planet to turn into a sun (“go nuclear”) with pulses or other-world matter-to-energy technology, would do it. Boom. Big boom. Big fast nasty boom.
#71 Greg
February 5, 2009

#66
Love that Idea. Especially if you focus the beam to create the conditions of a neutron star or even black hole in the center of the planet. I think that having a neutron star/small black hole for an instant then letting it return to normal would shoot out the material at sufficient velocities to get the explosion.
#68
I also want to point out that the falcon (and smaller craft) can create enough artificial gravity to make it seem like 1G for everyone and everything on board. If this is done with gravity manipulation/ gravitons then it shows that they can produce huge quantities of energy in very small spaces and/or can manipulate gravity/other forces in ways which are just incredible. Consider that the falcon can produce gravity, go full throttle, and travel through hyperspace and still have plenty of energy to run the lights and holographic chess boards as well as its own laser weapons. Another exercise might be just to estimate the amount of energy the Falcon can produce, then we can discuss what the DS might be able to make.
#72 Benny
February 5, 2009

I just read the book “The Death Star”. According to the book the Death Star used a “hypermatter” reactor. And the beam acted to send normal matter into hyperspace. Or some such. So there is more going on than just blasting a planet with a laser beam. I KNOW this is actually possible because this is science fiction and anything is possible here. After all….Superman can fly!
#73 Azkyroth
February 5, 2009

At all the people that are talking about this being a waste of time and energy (#51 in particular), the mathematical exercise here wasn’t that complex. It seems to be about at the level as physics problems I’ve done in college. Problems that weren’t designed to take very long or exhaust all of a person’s intellectual energy.

On the other hand, even elementary algebra is likely to exhaust the intellectual capacity of the sort of person who shows up on a science blog to whine about people analyzing the portrayal of physics in movies.
#74 Zach
February 5, 2009

Why go for the “big kaboom?” It’s not like sound travels through a vacuum.

#66, the particles would not explode after gravity was returned to normal, because there is still an attractive force between them. Maybe if the increase in gravity increased the density so much that it spontaneously fused and went nova, but it would not explode from a return to normal gravity.

Also, the existence of The Force breaks down most of the laws governing our universe. Who’s to say the rest of the laws of physics need not apply? Perhaps it is only coincidental that there appears to be some sort of attractive force, and that there is humanoid life that resembles our own? Also, how’d they get the camera there to film the whole thing?

@Trolls: Go get a life. If you’re not going to enjoy the mental exercise, then YOU are wasting YOUR time, and ours.
#75 Anonymous
February 5, 2009

I didn’t feel like looking through all posts to see if someone mentioned this, but the novel Death Star mentions that the energy blast knocks a sizable portion of the target into hyperspace, meaning that little of the original planet is left to be blown apart.
#76 chris
February 5, 2009

Tim S@67 – Additionally, I would also like to point out that if said poster thinks that we are spending too much “intellectual capacity” on these equations then maybe said poster is quite a bit below the average intelligence of those discussing this matter cogently and coherently – in other words, “kevin” is dumb.
#77 Aresen
February 5, 2009

Actually, I always assumed that the Death Star simply suppressed the strong nuclear force in a calculated volume at the core of the target, causing the baryons to spontaneously decay, yielding 900+ mev each.
#78 John
February 5, 2009

Heathens!

May His Noodliness have mercy on you even though you mock Him!

Can I just say that no-one has been called a Nazi yet?

No I’m sorry there isn’t time!
#79 LiThiuM56
February 6, 2009

Ok.. check out Star Wars: Forces Unleashed… eventually you get to the insides of the Death Star. It’s totally awesome! I could see it making sense. And lets be honest, I’m sure they didn’t use a ‘LASER’. Who know what kind of crazy beam came out of that sucker
#80 Paul
February 6, 2009

Physics + Star Wars = Always interesting
#81 Ben
February 6, 2009

RE # 21 actually the energy gained from nuclear reactions on the planet would be more than sufficient. Consider, the average binding energy of a iron nucleus is about 8 MeV per nucleon (proton or neutron). The average binding energy of the rest of the elements falls somewhere below this number with the majority being between 7 and 8 MeV. For the sake of argument let us say that it is on average a 0.1 MeV/nucleon difference (although it is probably closer to .5 MeV). As you pointed out, iron cannot contribute, so let us suppose that 99% of the earth is composed of iron. This leaves us only 1% of the earths mass being converted to iron in nuclear reactions. This gives us about 3.5*10^49 (The mass of electrons is insignificant) nucleons or 3.5*10^48 MeV generated. Converting to joules gets us on the order of 5*10^35 joules as the LOW end of possible energy to be obtained. This is several orders of magnitude more than is necessary to overcome the force of gravity holding the planet together. Thus it is feasible that the Death Star somehow causes a nuclear chain reaction. (in truth doing so would still involve a prohibitive amount of energy, but it would be several orders of magnitude less than attempting to directly overwhelm gravity) Also I want to point out to everyone who wants loud explosions: there would in fact be no sound no matter how fabulous the explosion. Sound does not carry in space.
#82 Ben
February 6, 2009

and #51, this problem took me 15 minutes to solve and write an answer to. Now I’m going to spend four hours doing me real homework for one of my Nuclear Engineering classes. It will consist entirely of problems like this and they will all be solved by 50 other students tonight, they have been solved by thousands of students before me, heck they were solved 50 years ago. We do theoretical, FICTIONAL problems so we can solve some of the way harder problems you mention. It is not a waste of time.
#83 Sambob
February 6, 2009

Just a thought, was it not said that the force runs through everything? Wouldn’t something that is part of everything have a hell of a lot of power? So that would make sense.

Also I really like it, might really be useful if I am planning to build a Death Star anytime soon.
#84 y
February 6, 2009

what causes the spontaneous symmetry braking?
the explosion is not spherically symmetric?
#85 ben
February 6, 2009

fist thing all you would need to blow up a planet would be a anti gravitron beam that would give all matter that it touched an anti gravitron radiation effect so that anything within the radiation field would be contaminated by the radiation to put out its own radiation just like contaminents from a nuke.
#86 Azkyroth
February 6, 2009

fist thing all you would need to blow up a planet would be a anti gravitron beam that would give all matter that it touched an anti gravitron radiation effect so that anything within the radiation field would be contaminated by the radiation to put out its own radiation just like contaminents from a nuke.

Radiation does not work that way.
#87 kevin
February 6, 2009

i love this post, but must point out, that the death star is massive, confusable with a small moon. making it fly would require an impressive amount of energy, so i do not think they would have problems finding the energy to destroy a planet.
#88 Mike
February 6, 2009

Well, given they have light speed travel, using OUR laws of physics to solve this problem is silly at best. I propose whatever they caused the core to phase out of existence, perhaps by somehow creating a singularity in the center of the planet, or by bending space-time to “teleport” the core… away somewhere else. The resulting sudden drop in gravity from the sudden loss of density could easily cause a planet to dissociate the way Alderaan did.

My 2-cents.
#89 llanitedave
February 7, 2009

Keep in mind that this all happened a long time ago in a galaxy far, far away.

At the time the expansion of the universe was slowing down. Adding all that dark energy to the system in order to destroy Alderaan is what got the acceleration started.
#90 Glaivester
February 7, 2009

Elaborating on the point made by Benny (#71):

Here is a quote from Star Wars: Death Star by Michael Reeves and Steve Perry (2008 Del Rey Mass Market Edition (paperback)), page 186:

“well, the hypermatter reactor was capable of generating an energy burst equivalent o the total weekly output of several main-sequence stars.”

Someone in the Lucasverse did the math.
#91 pwe
February 8, 2009

@ #89
Ok, so we only need to know, what kind of energy source was used, that could store enough energy for that in an object the size of the earth’s moon.
#92 Tim Gaede
February 9, 2009

For a slightly better approximation of the gravitational binding energy, assume that density (instead of being uniform) is a linear function of distance from the center.
The derivation takes a bit longer but does not require any more advanced math:

(2*G*π^2 * R^5 / 315) (13*ρ_c^2 + 65*ρ_c*ρ_s + 90*ρ_s^2)

where ρ_c is the density at the center and
ρ_s is the density at the surface

It took a lot of cancellation and simplification to derive but you can see it reduces to

(16*G*π^2 * R^5 / 15) ρ^2

when ρ_c = ρ_s = ρ

It means you’ll need a bit more energy to blow up a typical planet than had uniform density been assumed.

(NOTE: The mass of a planet is (G*π R^3 / 3) (ρ_c + 3*ρ_s)
#93 Maxmanta
February 9, 2009

In response to #35, every point in normal space has a corresponding point in hyperspace. Anything with mass exists in hyperspace as a “shadow” (to put it in SW terminology).

Given the dynamic nature of space (in that everything is in constantly in motion in relation to everything else) hyperspace computers are kept up-to-date with the most recent, known locations of masses in space (relative to the vehicle they are navigating), and used to calculate the safest trajectory through hyperspace to keep collisions with hyperspace shadows at a minumum. Still, hyperspace travel carries at least a small amount of risk.

As a payperson, I always assumed the Death Star beam excites the subatomic particles in its target’s atomic assemblies to the point where their individual kinetic energies overcome the binding nuclear force holding them together.

Do I pass? Am I in? Am I in?
#94 Maxmanta
February 9, 2009

Correction–I am a LAYperson, not a PAYperson.

I don’t even know what a payperson is.
#95 Ben
February 9, 2009

#93, while that is a good idea, it would in fact require INCREDIBLE amounts of energy to overcome the binding nuclear forces. Don’t forget that the nuclear force is actually the strongest force and gravity is the weakest, gravity just acts over a larger distance. The kinetic energy required to dissolve a iron nucleus is a 8-9 MeV per nucleon, which works out to more than 10^40 joules for the whole planet. It would require much less energy to overcome the gravitational force. In addition if the planet was reduced to subatomic particles there would be no asteroid field, rather there would be a hydrogen nebula.
#96 smerky
February 9, 2009

Are you absolutely sure the Death Star’s weapon was a laser? Sure it was a beam of some sort, but maybe it was a gravitational weapon that destroyed the atomic bonds between the atoms that made up the planet and allowed it to just fall apart. If it somehow resonated with this binding force it could get away with much less energy than you have discussed. It was obviously a combination of 8 smaller beams that combined to make the larger one. You notice how the smaller beams were pulsating? They could each have been modulated by several signals resonant with the various major elements comprising the planet’s crust.
#97 Sbark
February 9, 2009

One alternative that I haven’t seen proposed is to deliver anti-electrons sufficient to eliminate all the electrons on the planet. Would the force from the conversion of the matter to energy combined with the force of the proton repulsion be enough to get the job done?
#98 JJNash
February 10, 2009

And then, as the engineers on the Death Star take a moment to do their congratulation toast, someone says “ohhhh, crap, exploding planet shockwave in two minutes. Didn’t think of that. Say, did we build enough shielding?”
#99 Jet Black
February 10, 2009

The mean temperature of the planet as it explodes seems to be quite low. Looking at the image given and assuming a black body spectra in the yellow portion, even the insides only seem to be about as hot as the surface of the sun (about 6000 deg)
#100 Proscriptus
February 10, 2009

Well, I ain’t no scientifer, jus’ a plain ol’ nerd, but the way I sees it y’all are making at least one fundamental assumption that limits your options, which is use of the word “beam” in the singular. I postulate that the emanation from the Death Star is in fact two (or more) tightly collimated beams which intersect at a planned point deep within Alderan and Vadar and co. are, in fact, CROSSING THE BEAMS! I said never to do that, Ray!
#101 Jom
February 10, 2009

Interesting article yet im lost with all its technicalities
#102 Luke Skywalker
February 10, 2009

star wars is better than star trek by farrrrrr.
#103 Mr. Sulu
February 10, 2009

Shutup, not even close you loser
#104 po
February 10, 2009

“the ability to destroy a planet is insignificant next to the power of the force
#105 Maxmanta
February 12, 2009

The twitch twilek woman’s lekku does more for me than a thousand Orion slave girls.

Star Wars roolz.
#106 Maxmanta
February 12, 2009

I meant ‘the twitch OF A twilek woman’s…”

I should really learn to proofread.
#107 Maxmanta
February 12, 2009

“fist thing all you would need to blow up a planet would be a anti gravitron beam that would give all matter that it touched an anti gravitron radiation effect so that anything within the radiation field would be contaminated by the radiation to put out its own radiation just like contaminents from a nuke.”

Thank you, Mister LaForge. That will be all.
#108 Mike E.
February 13, 2009

I wish that the people like #51 would take a step back and realize a few truths. One, it is only Science Fiction; until someone actually does it. Two, most of today’s technology that we take for granted, in the 1960′s and 70′s were considered Science Fiction.

Here is a list of examples:
Robby to Robot (Lost in Space) — ASIMO from Honda
Star Trek Communicator — Cell Phone anybody?
Computers in almost every one were huge (Trek, SW, etc etc) — We have Laptops, Desktops, even Cell Phones that are able to process much more and more efficiently.
Artificial Intellegence — Take a look at the AI projects, in particular Alice.
Transporter — Slashdot has been covering this one for a while

The whole point of this is that through Science Fiction, we allow ourselves to dream of possibilities. It is through that dream, that the real discoveries in science are found.

“Genius is one percent inspiration and ninety-nine percent perspiration” — Thomas Edison.
#109 Pricey-Mon
February 18, 2009

All those calculations are based on using current technology. Its a worthless argument really isn’t it? Using todays technology to blow up something that is not real based in the future. Man, you guys need to get out a bit more.
#110 John Seal
February 19, 2009

I am breaking the mold a little bit here by asking a question of location. Not about where Alderan was…but by where I can find this kind of discussion on Men in Black II. Serlina destroys 4(?) planets in a variety of unique and dynamic ways. I would like to see some “geeky’ theorycrafting on that as well.
As my theoretical mathematical skills are lacking, and I just like to read you smart people’s ideas…maybe I can persuade some of you to expand into a new movie? (I wonder if that makes me geekier?)
Thanks in advance! ^^
#111 Matty
April 19, 2009

Just wondering about your derivation, did you treat G as a constant when you did the integral? I always thought G=9.81 but only at sea level. or am I thinking little g and you’re using the G that relates the attractive force between two bodies?

Thanks for the entertainment none the less.
#112 An Idiot
April 19, 2009

nvm you’re using G I should have double checked you’re equation
#113 Anonymous
May 18, 2009

Where did the explosion come from if there is no oxygen in outer space for the fire to burn?
#114 Deepak
July 30, 2009

Nice calculations – I had worked out the formula to approximate the energy required to create/destroy a planetary mass before, but never considered it from an Earth-destroying or Star Wars perspective.

Some more useless calculations to further the theme:

I wanted to calculate the recoil velocity of the death star after lasing.

Wookipedia (http://starwars.wikia.com/wiki/Death_Star) gave the diameter of the first death star (the one that took out Alderaan) as 160 km, giving a radius of 8^10^4 metres.

The mass of the death star is not given. However, there is this bit in that article that states: “The incredible energies harnessed by the station combined with its great mass gave the Death Star magnetic and artificial gravitational fields equal to those found on orbital bodies many times greater in size.”

OK, we can sort of work with that. I have to neglect the “magnetic” part of it and just assume that gravity is the only thing acting here. If magnetic fields had any major role, it will really throw the calculation off, but nothing in the show led me to believe they were being held down by a massive magnetic field acting on metallic boots. So let’s just talk about gravity here.

Let’s say the local gravitation field on the surface of the death star is equal to that on Earth (g = 9.81 m/s^2).

g = GM/(R^2)

That gives M = 9.4*10^20kg.

The momentum (p) of the superlaser beam would be given by:

E = pc

Since we’ve already worked out E to be 2.2 * 10^32 J, we get

p = 7.3*10^23 Ns

The other issue is whether there’s a significant mass deficit in the death star after lasing.

Using E = mc^2, the mass deficit corresponding to 2.2*10^32 J of energy is about 2.5*10^15 kg, which is on the order of a millionth of the initial mass of the death star, and hence can be safely neglected.

Hence the recoil velocity (v) of the death star after firing can be calculated with:

p = mv (no need to consider special relativistic corrections, because we’re dealing with v << c)

giving v = 780 m/s or about 2800 km/hr.

This assumes a g value equal to Earth’s gravity. But even if the death star were to have the gravity of Jupiter on its surface (2.58g), that would still give around 1085 km/hr as the recall velocity, still pretty darn quick.

This is a lower estimate, assuming (as you did), that the energy is only sufficient to disintegrate the planet into a diffuse cloud. The estimate would naturally be much higher in the scenario of blowing the planet up into smithereens, with every particle being imparted with a speed much higher than the escape velocity.
#115 Deepak
July 30, 2009

Sorry, this line: “p = mv (no need to consider special relativistic corrections, because we’re dealing with v”

should read:

“p = mv (no need to consider special relativistic corrections, because we’re dealing with v much less than c.”
#116 porno izle
January 12, 2010

In response to #35, every point in normal space has a corresponding point in hyperspace. Anything with mass exists in hyperspace as a “shadow” (to put it in SW terminology).
#117 MeNotYou
February 16, 2010

http://www.stardestroyer.net/Empire/Tech/Beam/DeathStar.html

The original article can be found in the above link.
It was apparently written by a Michael Wong in 1998.
It’s copyrighted material.

Is Michael Wong dead?
His last update on the above linked site was in 2004.
#118 porn teen tube
February 17, 2010

hyperspace. Anything with mass exists in hyperspace as a “shadow” (to put it in SW terminology)
#119 penis büyütücü
March 24, 2010

Just wondering about your derivation, did you treat G as a constant when you did the integral? I always thought G=9.81 but only at sea level. or am I thinking little g and you’re using the G that relates the attractive force between two bodies?
#120 sikis
June 7, 2010

just wondering about your derivation, did you treat G as a constant when you did the integral? I always thought G=9.81 but only at sea level. or am I thinking little g and you’re using the G that relates the attractive force between two bodies?
#121 Bang!
September 11, 2010

I always thought that the laser idea was pretty lame. Why not just fire a ball of lead at near cee? Provided that the slug was the size of a mountian, it would work. Mabey the laser destroys the planet AFTER a gravity feild generated
by the Death Star weakens it? Either thought would probably make a more interesting movie, though… At least in my opinion.
#122 dizi izle
December 15, 2010

would probably make a more interesting movie, though… At least in my opinion.
#123 Ghost
February 17, 2011

I love the math and thought put into this chain, but I don’t think we will ever really know until we need to blow up a real-life planet.
#124 tütüne son
March 26, 2011

The estimate would naturally be much higher in the scenario of blowing the planet up into smithereens, with every particle being imparted with a speed much higher than the escape velocity…
#125 honorphysicssucks
May 8, 2011

Hey I’m doing a project on why star wars physics doesn’t compare with real physics (for example:space is vacuum meaning their shouldn’t have been any cheesy engine noises) can anyone help me brainstorm. PS AP physics sucks
#126 God
May 8, 2011

………………………..
#127 adr
July 9, 2011

this is the utmost lower since it only negates the gravitational potential energy for all the mass….. hence it would go nowhere after that much power is supplied in this little perfect world we create where energy distributes instantaneously over an entire body. Normally tho with 2 times 10 to the 32 power you would accelerate several layers of crust and mantle to relativistic speeds and leave a large molten asteroid in the middle. If u want total obliteration ud need to add kinetic energy ,preferably relativistic for a big boom to the sum he has stated
#128 Weldon
July 12, 2011

That’s pretty interesting. Never really thought about the energy required to actually blow up the world. You see things like that in movies and most of the time keep on going about your day to day life.

honorphysicssucks, how big of a list do you want? OOH and if the life forms in star wars were so advanced, why did some of them live in deserts with out an A/C. Come on you have a space ship but can not afford a nice A/C?
#129 Lorcan Collins
July 25, 2011

NERD!!!!
#130 R-3P0
November 25, 2011

What if the superlaser simply reverses the charges of the sub-atomic particles in the planet, effectively converting them into anti-matter which will annihilate with the surrounding atoms.

Karl Withakay mentioned the anti-matter neccessary to destroy an Earth-sized planet would be “about the size and mass or Mars’ smaller moon Deimos” If the Death Star converted a narrow beam across the entire diameter of Alderaan into anti-matter via “particle charge reversal” than I suppose that could easily generate the sufficient energy to scatter the entire mass of the planet.
#131 schonovic
February 23, 2012

hypothetically a concentration of lasers on one point con open a point of hyperspace.
perhaps the superlaser places a specific amount of the core into hyperspace for a small amount of time exploding the entire core into a large momentum, close to the speed of light. the relative kinetic energy from droping out of hyperspace could rip the planet appart.
#132 omnomnom
March 29, 2012

Too low. Read this: http://stardestroyer.net/Empire/Tech/Beam/Alderaan.html
#133 Saitir
May 13, 2012

Just throw a Nimitz class aircraft carrier (about 100,000 tonnes) at 0.999999999999999c at the planet and it’ll generate fairly close to that much energy (2.01 * 10^32 joules). I’ll let someone else figure out how to accelerate something that size to that speed though!

See RKV (http://en.wikipedia.org/wiki/Relativistic_kill_vehicle) and using Einstein’s kinetic energy equations.