Archive for September, 2007

Laser Thruster Physics

September 17, 2007

With the recent article about the prototype laser thruster, there has been discussions of using a larger, more powerful thruster to reach Mars in a week or less. By my basic calculations, though, there are a pair of problems with this.

The basis of the laser thruster is that light carries momentum as well as energy, and thus by conservation of momentum, a laser has some (usually miniscule) recoil. For our hypothetical trip to Mars, let us assume we maintain a constant 1 g = 9.80 m/s^2 thrust (accelerating to halfway point and decelerating beyond that). Then our velocities remain non-relativistic. Let M be the mass of our vessel. We have from Newton's Second Law that F=Ma=dp/dt (p is momentum). As this is produced by the equal and opposite reaction from the momentum flowing out with the laser, and as the relationship for momentum of light is p=E/c, we have Ma=dp/dt=d/dt [E/c]=(1/c)*dE/dt, and dE/dt is our laser power, P, so P=Mac.
For a vessel of about the same mass as the Apollo 11 moon mission, about 47000 Kg, we get a laser power of about 138 Terawatts. For comparison, the world's current most powerful laser, the NOVA laser used in fusion experiments at Laurence Livermore National Laboratory, produces about 100 TW. So there's a bit of scaling up needed.
Even then, we have another problem you need to figure out how to store the energy needed to power the laser and how to feed it in that quickly. The laser must be operated continuously for the trip, so we estimate with: 138 TW * 1 week=1.38*10^14 J * 604800 s = 8.35*10^19 J, which is equivalent (via E=mc^2) to about 930 Kg of mass via total conversion.
As nuclear fission converts about 0.1% of the total mass to energy, we'd need a fission reactor with at least 930000 Kg of uranium/plutonium fuel, about 20 times the total mass of our vessel! For solar panels, the solar constant (the intensity of radiation from the sun at the distance of the earth) is about 1366 W/m^2. With the average solar panel having about 14-16% efficiency, this gives a needed panel area of 6.7*10^11 m^2 = 670,000 km^2 = 260000 mi^2. In other words, one needs solar panels the size of Texas!

Edit: And fusion isn't very feasable either. The deuterium-tritium rection converts about 0.36% of mass to energy, so we'd need about 260000 Kg of heavy hydrogen, still more than 5 times the mass of our vessel.

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September 11, 2007


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