Estimate the power of the laser on board of Darth Vader's Death Star

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Discussion Overview

The discussion revolves around estimating the power of the laser on Darth Vader's Death Star, specifically its capability to vaporize an Earth-sized planet in approximately one second. Participants explore various theoretical and mathematical approaches to calculate the energy required for such an event, considering factors like the mass of the Earth, heat of vaporization, and the dynamics of shockwaves.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants suggest using the mass of Earth and the heat of vaporization of rock to estimate the energy needed for vaporization.
  • Others argue that the rapid destruction would generate a significant shockwave, necessitating a large amount of energy to propagate through the planet.
  • A participant calculates that converting the mass of Earth (approximately 6 x 10^24 kg) to gas would require around 10^31 J of energy.
  • Another participant compares the energy output of the Death Star to that of a supernova, suggesting that the Death Star would need to harness energy on a massive scale.
  • Some participants introduce Buckingham's pi theorem as a method for dimensional analysis, proposing an equation to estimate the energy based on the radius of the blast and the density of Earth.
  • There are discussions about the creation of ultrarelativistic plasma during the energy release and questions regarding the laser's frequency and focusing methods.
  • One participant notes that the energy calculated is close to the gravitational binding energy of Earth, raising further questions about the implications of such energy levels.
  • Some participants express a desire to simplify the problem for educational purposes, suggesting a more straightforward thermodynamic approach.
  • There is a debate about the destruction of Alderaan, with differing opinions on whether it was completely vaporized or left with debris.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the exact calculations or methods to estimate the laser's power. Multiple competing views and approaches are presented, and the discussion remains unresolved.

Contextual Notes

Participants acknowledge various assumptions, such as the heat of vaporization values and the dynamics of shockwaves, which may influence their calculations. The discussion also touches on the complexities of energy dissipation in different states of matter under extreme conditions.

Who May Find This Useful

This discussion may be of interest to those studying physics, particularly in areas related to energy calculations, thermodynamics, and theoretical applications of physics in popular culture.

Count Iblis
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The laser is able to vaporize an Earth sized planet in about a second.
 
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Then you just need the mass of Earth and the heat of vaporisation of rock ( or the crust)
 
mgb_phys said:
Then you just need the mass of Earth and the heat of vaporisation of rock ( or the crust)

We need to do more than that. The Earth has to be destroyed in just a second. When a huge amount of energy is delivered to the surface a big schockwave develops. This has to travel through the entire planet in a second.
 
Speed of sound through Earth is only about 8-10km/s so to put a mach 1000 hypersonic shockwave through Earth is going to take some serious energy into the crust on one side.
Still if you intending to convert 6x10^24kg of rock to gas that's going to take quite a bit of energy anyway leaving aside the dynamics ;-)
 
allright, the mass of the Earth is 5.97x10^24

I'd like to see this solved. It would also be a neat problem to show to my physics teacher.
 
leftyguitarjo said:
allright, the mass of the Earth is 5.97x10^24

I'd like to see this solved. It would also be a neat problem to show to my physics teacher.

Ok, let us estimate: latent heat of evaporation, say 1000 J/g (see http://en.wikipedia.org/wiki/Latent_heat) or 1 MJ/kg. That brings us to about 6 10^30 J, or let's say, 10^31 J.

Now, a supernova has an energy output of something (if I remember well) like 10^42 J.
So if the death star runs on supernovas, one supernova has more than 100 billion "bullets" that can blow up the earth...
 
Or 2x10^15 megatons, which is somewhere around 200 billion times the nuclear warheads on earth.
 
Russ - you've obviously been planning this, should we be worried?
 
(1000th post!)

There is a neat method for doing problems like this that doesn't get taught to undergraduates as I feel it shoud be:

http://en.wikipedia.org/wiki/Buckingham_π_theorem

It's called Buckingham's pi theorem, where pi refers not to Archimedes's constant in this case but to the defacto appearance of the greek character in the notation. Anyway, Buckingham's theorem formalizes the use of dimensional analysis to generate relevant equations out of no where. When I saw this question I thought of the classic example whereby the energy released in the trinity test of the atomic bomb was estimated before the actual value was declassified. Fortunately Wikipedia has this example:

http://en.wikipedia.org/wiki/Buckingham_π_theorem#The_atomic_bomb

The equation we get is:

R = (\frac{E t^2}{\rho})^{\frac{1}{5}}

This equation describes the expansion of the radius of the blast as a function of time in terms of the energy of the blast and the density of the environment. If we use the average density of the earth, the radius of the earth, and a destruction time of 0.5 seconds (maybe someone with a stop watch and a new hope DVD can do a better estimate) I calculate a total energy of 2*10^32 Joules, which would be over a trillion megatons!
 
  • #10
This is such a nerdy thread.

What about the power ouput of captain kirk's gun when set to stun?

btw, Crosson thanks for posting that, I hadn't heard of Buckingham's pi theorem, I'm finding it very interesting
 
Last edited:
  • #11
It is very difficult problem. A screen of ultrarelativistic plasma will be created in the first moment and diisipation of energy will be mainly in this plasma screen

At such power densities vacuum will be olso plasma of particle-antiparticle

And what is the laser freaquency and width of the ray? How do we focus it?
 
  • #12
Minich said:
It is very difficult problem. A screen of ultrarelativistic plasma will be created in the first moment and diisipation of energy will be mainly in this plasma screen

At such power densities vacuum will be olso plasma of particle-antiparticle

And what is the laser freaquency and width of the ray? How do we focus it?

Perhaps it could be focussed using a black hole as a gravitational lens?
 
  • #13
Crosson said:
(1000th post!)

If we use the average density of the earth, the radius of the earth, and a destruction time of 0.5 seconds (maybe someone with a stop watch and a new hope DVD can do a better estimate) I calculate a total energy of 2*10^32 Joules, which would be over a trillion megatons!

Interesting that that number works out to very close to the gravitational binding energy of the Earth (2.24*10^32 J)
 
  • #14
Minich said:
It is very difficult problem. A screen of ultrarelativistic plasma will be created in the first moment and diisipation of energy will be mainly in this plasma screen

At such power densities vacuum will be olso plasma of particle-antiparticle

And what is the laser freaquency and width of the ray? How do we focus it?

Well.. since this is for a High school class (im giessing since leftyguitarjo said to show his teacher), I think it could be simplified, couldn't it? I mean just not overthought.

The original Q is to vaporize the Earth - so use the melting point and heat of vaporization of Rock and the mass of the Earth as your data. Then its a simple thermo problem.

When you find the total energy needed, that is the power since the time taken to output that much energy is 1 second.
 
  • #15
Assume increased pressure as you go down into the Earth's mantle almost 11 times atmospheric level and that the mantle is a plastic moving solid mainly peridotite rock composition, the outer core is liquid iron also under heavy pressure and inner core solid iron.
 
  • #16
Alderaan was not evaporated. There was a lot of debris left.
 
  • #17
yeah but you don't know how large Alderaan was, might have been slightly larger than earth.
 
  • #18
An estimation using these numbers gives:

<br /> P=2*10^{32}W,\;<br /> &lt;S&gt;=\frac{\epsilon_0c}{2}E^2, \;W/m^2,\;<br /> P=A*&lt;S&gt;\;<br /> A=\pi R^2,\;<br /> R=R_{earth}/100,\;<br /> E_{ion,Hydro}=13eV/0.5 Angstrom<br />

This gives that the electric field over the bohr-radii is abot 13 times bigger than the "break-down" of an hydrogen atom (E_{ion,Hydro}=13eV/0.5Å). reosonable?
 

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