Ion Engine Q

[Rocket Man]

Rocket man you misunderstood my terms' date=' P represents power (FV or E/T), and you can be quite assured that power times time equals energy. You can also be quite assured that the basic laws of physics are the same on earth as they are in space, in other words my math applies to both the car and the ion rocket (conventional rockets are more complicated).

 

the problem with your logic is that the limiting factor here is the amount of power available to the rocket and not the force applied, also be sure you don't think of the rocket's reference frame as a valid one, as it is accelerating.[/quote']

 

i thought your p was momentum.

 

my refrence frames were on the rocket's centre of mass before and after it fires a particle, they dont move, they get replaced.

 

there is no difference between a conventional rocket and an ion rocket, an ion rocket uses electricity to accelerate a mass of propellant, while a conventional rocket uses gas pressure.

when a rocket exceeds it's exhuast velocity, the kinetic energy comes from the propellant being decellerated by the force from the engine.

at low velocities, a rocket is extremely inefficient, al most all of the energy goes into the propellant. at high velocities, energy is taken away from the propellant.

 

a rocket's power is measured in newtons not watts, F/M = A

a rocket's acceleration will be constant using a constant force as rockets do.

[CPL.Luke]

a conventional rocket is different as the mass of the fuel is a significant percentage of the rockets total mass, and so the rockets mass varies in time and this changes the math quite a bit. So for the sake of simplicity we should only talk about ion rockets.

 

 

[quote

a rocket's power is measured in newtons not watts]

 

thats like saying an electron's diameter is 5 ampere. Power has units of energy over time ie it can't be measured in newtons (which is a measure of force).

 

 

furthermore your math doesn't work because the reference frames before (A) and after (B) have experienced some change in momentum or impulse from the ejection of the particle, your model can't define any acceleration of the ship as that would invalidate the reference frames.

[Rocket Man]

i dont think you understand my refrence frames,

say the ship is stationary with a particle ready to fire, take a refrence frame.

the ship fires the particle and accelerates by a small amount. you still use the same refrence frame to determine the amount the ship has accelerated.

once the particle is out range to effect the ship's acceleration, you take a new refrence frame using the ship's center of mass along with the next particle to fire.

the final velocity can be obtained with the sum of the ejected particles' momentum.

 

ion rockets are no different to conventional ones. they still eject mass.

 

an ion rocket accelerates particles over a fixed distance to a fixed velocity relative to the engine. thus each particle has a set kinetic energy relative to the engine.

each particle will have a fixed mass and velocity relative to the engine and so, a fixed momentum relative to the engine just after it's ejected.

multiple particles in succession each with relative momentum will sum to give the ships final momentum.

 

an engine has a limited energy, but it is not applying force to a stationary mass. the mass it is applying force against has the same velocity as the engine.

 

energy = force X displacement

a car applies force to a stationary mass with increasing relative velocity, the displacement will increase per set time so the energy required to hold a constant force will increase

 

ANY rocket applies force against a mass with a constant relative velocity of 0

so the displacement is a constant, it does not increase with velocity.

it is the distance the rocket has to accelerate the propellant

 

when i said "a rockets power is measured in newton not watts", i was using the term in a trivial sense, a rocket will apply a constant force no matter what velocity it's travelling at using a constant amount of energy because the mass it is applying force against is stationary relative to the rocket.

[CPL.Luke]

the difference is in the mathmatics, The real definition of force is change in momentum with respect to time or dp/dt, now in a conventional rocket the mass changes significantly with respect to time, so the rate of change in momentum (force) becomes

 

F=(dM/dt)V+M(dV/dt)

 

not

 

F=MA

 

as far as I know the propellant mass in an ion rocket does not constitute a significant portion of the mass of the rocket and so F=MA works quite well, however if we wish to have a far more useful version of force for calculating the momentum of the rocket we should go back to dp/dt follow that through and you'll get

 

F=dm/dt V

 

where V is the velocity of the exhaust, and dm/dt is the rate of fuel consumption (in kg/time), integrating this with respect to time and we see that the final momentum of the rocket is going to be equal to m (the mass of the propellant) x V (the velocity of the propellant)

 

in other words

 

momentum=mv

 

however like I said before the rocket is limited by power so the force is not constant and the rate of fuel consumption is changed in order to have the maximum possible exhaust velocity.

 

 

^that is how you would do the problem out using momentum, and it is useful as it allows you to find what the rate of fuel consumption should be.

 

 

The problem with your reference frames is that you are in effect trying to define a force acting on the reference frame, and this invalidates it. It should be noted that the reason why saying a force is acting on a reference frame invalidates it is because you get non-sensical answers such as your saying the rocket experiences a constant acceleration under a constant amount of power, THIS DEFIES THE LAW OF CONSERVATION OF ENERGY.

[Rocket Man]

supposing you get a convetional rocket and strap a few extra hundred tonnes of payload to it, you will have roughly the same ratios as the ion rocket. the math will still be the same.

 

you still dont understand the frames, the force is applied to the propellant, the propellant has the same velocity as the rocket untill the particle is fired.

the refrence frame simply brings the relevant objects into focus

 

suppose you have a system to accelerate a particle to a velocity relative to the object firing it. have it stationary and measure the momentum the firing mechanisim gains after firing the particle.

 

then accelerate the entire loaded system to a high velocity and fire it backwards, is the momentum increase the same? all momentum laws say yes.

conservation of energy has no problem either, it applied a force over a fixed distance using a fixed amount of energy.

 

what's to say that the stationary system was not in motion and the accelerated one was not? that's what im getting at with the refrence frames.

all physical laws hold no matter what velocity you're travelling at.

 

the problem with your explanation is that once a particle is fired, it cannot hinder the rockets acceleration. each successive particle is fired with a relative momentum, even if the ships mass does not decrease, it will still have momentum equal to the sum of the relative momentum of each particle.

if each particle has a fixed relative momentum just after firing, the engine will experience a fixed impulse per particle it fires. thus a constant acceleration.

 

rockets do not defy conservation of energy simply because a very small percentage of energy is converted into kinetic energy. most of it goes into the propellant and is wasted. of several megajoules of stored energy, the final kinetic energy of a rocket will barely scratch the surface.

[swansont]

It's been a while since I derived the rocket equations, but I looked up the rocket speed equation and it seems familiar

 

v = w ln( 1 - mt/M₀)

 

for mass ejected at a rate m at a speed w relative to the rocket of mass M₀

 

That's not a linear equation, so its derivative is not constant (though it may look that way for small values of t)

[CPL.Luke]

Rocket man

 

you are right about the sum of the final momentum being the sum of the momentums of each particle fired, however you haven't considered the rate at which the particles are fired with respect to time relative to when the first reference frame, this is controlled by the amount of power the rocket has available to it.

 

also I found this post on a cornell forum

 

Note that for the accelerating car, there is a reaction

force that pushes the earth backwards; but since the delta

v of the earth is very small, and v^2 much tinier, the KE

change of the earth is negligible even though M of the earth

is large. In the rocket case the stuff that gets pushed back

(propellant) is much less massive, so its v is very large and

its KE can't be neglected. In other words, the physics of the

car scenario is really the same as that of the rocket, it just

has radically different numbers.

 

Total KE is the KE of rocket + KE of the ejected propellant.

At first (in the launchpad frame) the latter is by far the

greater, but as the rocket gains speed, the speed (relative

to the launchpad) at which the propellant is thrown backward

becomes less and less. If you calculate it out (an instructive

exercise) you find that the *total* KE increases linearly with

time, exactly as it should.

 

Btw you reach the same conclusion no matter what inertial frame

you do the calculation in. You will get different values for

the KE, of course, depending on frame.

[swansont]

It seems to me that at any instant, the rocket "sees" constant acceleration, by virtue of constant momentum of the ejected mass, in it's own frame. But the rocket is in an accelerating reference frame, and so this is not a valid application of Newton's laws if the elapsed time is not vanishingly small, and a ground-based observer will not see a constant acceleration.

 

The energy analysis bears this out, qualitatively. If the total KE of rocket + ejected matter is a conserved quantity, as CPL Luke's quote demostrates, the final speed of the ejected matter drops (as viewed from the launch pad) as the rocket speeds up. When v = w, the ejected matter is at rest and the transfer to the rocket is maximized. But what happens when the rocket exceeds the speed of the ejected matter? Now the ejected matter's KE starts increasing again, so the rate at which energy is transferred to the rocket begins to decrease.

[CPL.Luke]

swansont, Why would the transfer of energy from the propellant to the rocket be effected by the speed of the rocket? assuming the rocket ejects mass with a constant velocity, then the efficiency of the rocket should be constant, right?

[swansont]

swansont, Why would the transfer of energy from the propellant to the rocket be effected by the speed of the rocket? assuming the rocket ejects mass with a constant velocity, then the efficiency of the rocket should be constant, right?

 

 

The propellant is ejected at a constant speed wrt the rocket, and at a constant rate. But viewed from the launchpad frame, this means the propellant is ejected at differing speeds, and takes up a differing fraction of the total energy, depending on the rocket's speed.

[CPL.Luke]

alright, I played with that a bit and I see that your right, it's still a bit odd though,

 

espescially considering that the final momentum of the rocket is going to be equal to the mass of the propellant times the velocity at which its ejected, so in order to get the most momentum out of the fuel you would want to have the rocket eject the fuel at the highest speed possible, but if you want to have the most kinetic energy in the rocket you would want to have the rocket eject the fuel at such a speed such that its velocity with respect to us would be zero.

 

SO my question is then are the two equivelant? or is the extra energy lost as heat?

[swansont]

alright' date=' I played with that a bit and I see that your right, it's still a bit odd though,

 

espescially considering that the final momentum of the rocket is going to be equal to the mass of the propellant times the velocity at which its ejected, so in order to get the most momentum out of the fuel you would want to have the rocket eject the fuel at the highest speed possible, but if you want to have the most kinetic energy in the rocket you would want to have the rocket eject the fuel at such a speed such that its velocity with respect to us would be zero.

 

SO my question is then are the two equivelant? or is the extra energy lost as heat?[/quote']

 

The issue, I think, is that force is change in momentum, not change in energy, per unit time. The most efficient propellant, energy wise, would be something that has a huge mass but ejected extremely slowly. However, you are restricted in what you can build.

 

I think that the KE is a conserved quantity, once you've accounted for the efficiency of the system. That is, the thermodynamic efficiency of the engine at giving the ejected matter energy doesn't depend on the speed of the rocket; the rocket doesn't inherently heat up or cool down as a function of v. In a given frame, you will have a certain amount of KE available, and that amount will be split between the rocket and the propellant, with the ratio dictated by conservation of momentum. In the rocket's frame(s) that is a constant for mt<<M₀, but as I stated earlier, the rocket is accelerating, i.e. it keeps changing from one inertial frame to the next, and KE is not an invariant under those transformations.

 

So, because momentum and KE are not linearly related, the amount of KE imparted for a given change in momentum does depend on the momentum, and I don't think it has anything to do with energy being lost — all of the available KE should still be there; the "waste" is in how much you give up to the propellant, and that's unavaoidable when you restrict other conditions, like m and w (as I defined earlier) being constant.

[CPL.Luke]

so essentially what your saying is that while you can have more energy departed directly to the rocket if the velocity of the propellant is very low, but the rocket can gain a much larger velocity out of its given mass of propellant if it expends more energy and sends the fuel off with a larger velocity

 

so then the choice is with the designer of the rocket as to whether they want greater mass efficiency or greater energy efficiency, and I'm ignoring thermodynamical efficiency alltogether as it doesn't really effect this discussion.

 

am I getting what you mean right?

[swansont]

so essentially what your saying is that while you can have more energy departed directly to the rocket if the velocity of the propellant is very low' date=' but the rocket can gain a much larger velocity out of its given mass of propellant if it expends more energy and sends the fuel off with a larger velocity

 

so then the choice is with the designer of the rocket as to whether they want greater mass efficiency or greater energy efficiency, and I'm ignoring thermodynamical efficiency alltogether as it doesn't really effect this discussion.

 

am I getting what you mean right?[/quote']

 

 

I think so.

 

There are some practical considerations involved. The need to limit acceleration to a few g's, for example, which sort of dictates the continually ejected mass of a rocket. You can use a large gun to launch an object, pulling many g's, but you'd have to worry about they payload being damaged; that would be an example of the large "ejected" mass if you view the shell as ejecting the gun (and the earth) as the "propellant." You can certainly launch a shell for a lot less fuel than found in a rocket for the same payload mass.