Find the Ideal Rocket for Interstellar Travel

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SUMMARY

The discussion centers on optimizing rocket propulsion for interstellar travel, emphasizing the balance between fuel mass conversion to energy and reaction mass ejection. It highlights that the most efficient propulsion method is light propulsion using matter-antimatter fuel, which allows for maximum momentum gain per weight of fuel. The relationship between energy and momentum is established through the equations E=mc² and p=mv, illustrating the complexities of kinetic energy calculations during fuel consumption. The conversation concludes that while light propulsion is fuel-efficient, optimal energy efficiency varies with target velocity.

PREREQUISITES
  • Understanding of the energy-momentum relation in physics
  • Familiarity with the principles of rocket propulsion
  • Knowledge of kinetic energy calculations
  • Concepts of matter-antimatter reactions and gamma radiation
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  • Research the principles of light propulsion and its applications in rocket design
  • Study the effects of redshift on energy calculations in relativistic contexts
  • Explore the feasibility of matter-antimatter fuel systems for space travel
  • Investigate advanced propulsion concepts such as photon sails and their efficiency
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Aerospace engineers, physicists, and researchers interested in advanced propulsion technologies and interstellar travel optimization.

mrspeedybob
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Suppose I am building a rocket for interstellar travel. I want it to be able to go as fast as possible for a given fuel load.

Since the momentum of my rocket will equal the momentum of the fuel I eject from the back of it, it behooves me to eject the fuel with as high a velocity as possible. However, momentum is proportional to velocity but kinetic energy is proportional to velocity squared so the higher the velocity of my exhaust the less energy efficient my rocket is. Since the energy must be obtained by converting some fuel mass into energy I set about trying to figure out the best use of fuel. What portion of the fuel mass should be converted to energy and what portion should be shot out the back as reaction mass?

During my research I came across this... http://en.wikipedia.org/wiki/Energy–momentum_relation. This seems to say that the greatest momentum my fuel will ever have is when I convert it all to energy and emit it from the rear of my rocket as massless radiation. Did I understand that properly?

Assuming that I did, The equation for momentum of energy given by the article is E=pc. since E=mc2 the momentum of a given mass of fuel converted to radiation should be...
mc2=pc
mc=p
Now this appears startlingly Newtonian. p=mv which in this case is p=mc.

Ok, so now suppose I have a 1,000,001 kg star ship and I begin my journey by burning 1 kg of fuel. My (now 1,000,000 kg) star ship should accelerate to about 300 m/s. But now my ship has a kinetic energy of about 45 gigajoules (relative to my former frame of reference). Where did it come from? Apparently if I wound up with 45 GJ, the energy I directed rearward should have been 45 GJ less then the energy equivalent of 1kg. But if the energy I directed rearward was less, then my 45 GJ number will be less also. How to I calculate how much of the kg of energy I emit from the rear, and how much is added to my star ships kinetic energy?
 
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First of all, yes, light propulsion is the most efficient use of fuel. But it requires matter-antimatter fuel. If you have that, and you find way to actually direct the produced gamma radiation, you have the most efficient engine.

In terms of your conundrum with extra energy, you need to realize that from rest frame, in which ship is gaining energy, the light radiated by the ship for propulsion is getting red-shifted, and therefore, has less energy than in the ship's frame.

P.S. By "most efficient", I mean most fuel-efficient. That is, highest gain of momentum per weight of fuel. It's not most energy-efficient. Optimal energy efficiency will depend on your target velocity.
 

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