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

  • #1
1,827
7

Main Question or Discussion Point

The laser is able to vaporize an Earth sized planet in about a second.
 

Answers and Replies

  • #2
mgb_phys
Science Advisor
Homework Helper
7,744
12
Then you just need the mass of Earth and the heat of vaporisation of rock ( or the crust)
 
  • #3
1,827
7
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.
 
  • #4
mgb_phys
Science Advisor
Homework Helper
7,744
12
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 ;-)
 
  • #5
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.
 
  • #6
vanesch
Staff Emeritus
Science Advisor
Gold Member
5,007
16
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...
 
  • #7
russ_watters
Mentor
19,028
5,187
Or 2x10^15 megatons, which is somewhere around 200 billion times the nuclear warheads on earth.
 
  • #8
mgb_phys
Science Advisor
Homework Helper
7,744
12
Russ - you've obviously been planning this, should we be worried?
 
  • #9
1,222
2
(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:

[tex] R = (\frac{E t^2}{\rho})^{\frac{1}{5}} [/tex]

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
neu
220
1
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
85
0
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
1,827
7
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
Janus
Staff Emeritus
Science Advisor
Insights Author
Gold Member
3,427
1,117
(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
167
0
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, couldnt it? I mean just not overthought.

The original Q is to vaporize teh 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
DrClapeyron
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
Borek
Mentor
28,137
2,648
Alderaan was not evaporated. There was a lot of debris left.
 
  • #17
5
0
yeah but you dont know how large Alderaan was, might have been slightly larger than earth.
 
  • #18
An estimation using these numbers gives:

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

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?
 

Related Threads for: Estimate the power of the laser on board of Darth Vader's Death Star

Replies
8
Views
2K
Replies
6
Views
733
Replies
19
Views
1K
Replies
3
Views
9K
  • Last Post
Replies
11
Views
5K
Replies
54
Views
15K
Replies
3
Views
823
  • Last Post
Replies
3
Views
714
Top