- #1
mrspeedybob
- 869
- 65
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?
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?