3 ms·

You are right, this is plain wrong. You always want to have an exhaust velocity as high as possible. The energy you have to impart to the exhaust is 1/2 mv^2,

You are right, this is plain wrong. You always want to have an exhaust velocity as high as possible. The energy you have to impart to the exhaust is 1/2 mv^2, but you get m*v momentum. So small exhaust mass at very high speed will give you the same momentum, and cost you much less energy than bigger mass and less speed.

btilly 16y ago
Thus leading to ion drives. http://nmp.nasa.gov/ds1/tech/ionpropfaq.html http://nmp.nasa.gov/ds1/tech/ionpropfaq.html
alphaBetaGamma 16y ago
Sorry, this I'm completely wron... I mean: I was testing if you read the comments before upvoting. The 1/2mv^2 vs mv points to an exhaust speed as low as possible, to get as much momentum as possible from as little energy expenditure as possible. Now, if your exhaust comes from the outside world then that makes perfect sense. And this is why you have turboprop and turbofan engines: you slow down the exhaust of the turbine, speed up outside air, and do a favorable energy/momentum trade. But in space you have to carry the mass you will exhaust. You don't have much choice: all the energy you get from burning fuel will be transfer to kinetic energy of your exhaust (in the reference frame of the rocket).
- lutorm 16y ago
  Yeah, you're right. A bit of algebra reveals that your delta-V per MASS of fuel (which is a frame-independent quantity) goes as the (propellant specific chemical energy)^.5. However, it is also true that if you calculate the delta-V per unit propellant ENERGY used, it decreases with the propellant specific energy. Since you don't really care about how much energy you carry but only about how much fuel you need to use per delta-V, it's always better to have higher exhaust energy. (As long as you are using a chemical propellant, i.e. the mass carrying the energy and the reaction mass are the same.) The statement in the article about "propulsive efficiency" is total bogus. First: the relative kinetic energy of the rocket and the exhaust IS a frame-dependent quantity, so it's meaningless. Second, any reasonable measure of efficiency, like the delta-V per fuel energy, reveals that the "optimal" value is an exhaust velocity of zero. (Optimal in the sense that it's more energy efficient, but only because the delta-V goes to zero slower than the energy. It's still a useless optimum, because it results in no delta-V at all...)

btilly 16y ago

Thus leading to ion drives. http://nmp.nasa.gov/ds1/tech/ionpropfaq.html http://nmp.nasa.gov/ds1/tech/ionpropfaq.html

alphaBetaGamma 16y ago

Sorry, this I'm completely wron... I mean: I was testing if you read the comments before upvoting. The 1/2mv^2 vs mv points to an exhaust speed as low as possible, to get as much momentum as possible from as little energy expenditure as possible. Now, if your exhaust comes from the outside world then that makes perfect sense. And this is why you have turboprop and turbofan engines: you slow down the exhaust of the turbine, speed up outside air, and do a favorable energy/momentum trade. But in space you have to carry the mass you will exhaust. You don't have much choice: all the energy you get from burning fuel will be transfer to kinetic energy of your exhaust (in the reference frame of the rocket).

lutorm 16y ago
Yeah, you're right. A bit of algebra reveals that your delta-V per MASS of fuel (which is a frame-independent quantity) goes as the (propellant specific chemical energy)^.5. However, it is also true that if you calculate the delta-V per unit propellant ENERGY used, it decreases with the propellant specific energy. Since you don't really care about how much energy you carry but only about how much fuel you need to use per delta-V, it's always better to have higher exhaust energy. (As long as you are using a chemical propellant, i.e. the mass carrying the energy and the reaction mass are the same.) The statement in the article about "propulsive efficiency" is total bogus. First: the relative kinetic energy of the rocket and the exhaust IS a frame-dependent quantity, so it's meaningless. Second, any reasonable measure of efficiency, like the delta-V per fuel energy, reveals that the "optimal" value is an exhaust velocity of zero. (Optimal in the sense that it's more energy efficient, but only because the delta-V goes to zero slower than the energy. It's still a useless optimum, because it results in no delta-V at all...)