Rocket Fuel Ejection: Intuitive & Math Explained

[jbriggs444]

This isn’t right; by mass ejection rate it is meant that there is some boundary ##\Sigma(t)## through which a mass ##r(t)## flows per unit time. Any clump of matter that you throw is of finite size (i.e. length parallel to velocity ##\sim l##) so no matter how fast you throw it, it still takes ##t \sim l/v## to eject the mass.

So we should stop teaching students about elastic and inelastic collisions and do away with the approximation that the interactions take negligible duration?

In any case, nothing requires you to redraw the system boundaries in a continuous fashion over time.

Edit: Spooky action at a distance with @Dale making the same point independently

 

[Dale]

This isn’t right; by mass ejection rate it is meant that there is some boundary ##\Sigma(t)## through which a mass ##r(t)## flows per unit time. Any clump of matter that you throw is of finite size (i.e. length parallel to velocity ##\sim l##) so no matter how fast you throw it, it still takes ##t \sim l/v## to eject the mass.

The boundary is not massive, and can move as fast as I like. The boundary merely defines the system. It can contain the rocket plus fuel at one instant and only the rocket at any subsequent time.

In any case, again, the physical condition needed is that the fuel be all ejected with the same velocity.

Edit: I see @jbriggs444 made the same point but faster!

 

[vanhees71]

So we should stop teaching students about elastic and inelastic collisions and do away with the approximation that the interactions take negligible duration?

In any case, nothing requires you to redraw the system boundaries in a continuous fashion over time.

Edit: Spooky action at a distance with @Dale making the same point independently

I don't understand this statement. For collisions we look at asymptotic states and don't care about the transient states at all. We prepare an asymptotic free state in the initial and then observe an asymptotic free final state.

 

[jbriggs444]

I don't understand this statement. For collisions we look at asymptotic states and don't care about the transient states at all. We prepare an asymptotic free state in the initial and then observe an asymptotic free final state.

So an instantaneous rate of change anywhere during the process is irrelevant. We very deliberately do not care about the details, only about the initial and final states. So we are in vigorous agreement.

How can one say that the incremental rate cannot be unbounded if you do not care about the incremental rate at all?

 

[Delta2]

So an instantaneous rate of change anywhere during the process is irrelevant. We very deliberately do not care about the details, only about the initial and final states. So we are in vigorous agreement.

How can one say that the incremental rate cannot be unbounded if you do not care about the incremental rate at all?

The instantaneous rate of change in collisions is only theoretical and doesn't affect the velocities of the final state. Here with the rocket if you assume an infinite instantaneous rate of change it affect the final velocity of the rocket.

 

[Dale]

Here with the rocket if you assume an infinite instantaneous rate of change it affect the final velocity of the rocket.

Yes, that is clear. What is being argued now is whether it is physical and sensical.

 

[jbriggs444]

The instantaneous rate of change in collisions is only theoretical and doesn't affect the velocities of the final state. Here with the rocket if you assume an infinite instantaneous rate of change it affect the final velocity of the rocket.

Nope. You have failed to grasp the perfectly physical process.

You burn the fuel, releasing energy.
You store the energy.
You package up the expended fuel.
You throw the expended fuel aft, using the stored energy.
You re-draw the system boundaries so as not to encompass the expended fuel.

The redrawing of system boundaries can be instantaneous. But that fact changes nothing about the physicality of the scenario. How and when to draw system boundaries is a free choice made when analyzing the scenario. One is not forced to make the redrawing instantaneous. But that is a convenient choice.

I am not suggesting that increasing the burn rate of a traditional rocket without bound suddenly changes the rocket behavior at some point. That would indeed be a non-physical and daft thing to say. I am not suggesting that the behavior at the limit is distinct from the limiting behavior. Such a suggestion could have no experimental support. I am suggesting that there are different ways to expend fuel that are not adequately modeled as continuous burns.

@Dale has pointed out that the key feature of this sort of impulsive burn is that the expended fuel all moves at the same velocity. That is a feature not shared by the normal continuous burn model.

 

[ergospherical]

In any case, nothing requires you to redraw the system boundaries in a continuous fashion over time.

The redrawing of system boundaries is instantaneous.

No, control volumes ##\Omega(t)## deform continuously with time. [in other words the mapping function ##\chi(\mathbf{X},t)## between the reference configuration ##\Omega_0## and deformed configuration is differentiable w.r.t. ##\mathbf{X}## and ##t##]. The rocket equation can then be derived by expanding ##\dfrac{d}{dt} \left[\displaystyle{\int}_{\Omega(t)} \rho \mathbf{v}(\mathbf{x},t) dV \right]## using the Reynolds transport theorem.

 

[jbriggs444]

No, control volumes ##\Omega(t)## deform continuously with time. [in other words the mapping function ##\chi(\mathbf{X},t)## between the reference configuration ##\Omega_0## and deformed configuration is differentiable w.r.t. ##\mathbf{X}## and ##t##]. The rocket equation can then be derived by expanding ##\dfrac{d}{dt} \left[\displaystyle{\int}_{\Omega(t)} \rho \mathbf{v}(\mathbf{x},t) dV \right]## using the Reynolds transport theorem.

No. You are just plain flat out 100% completely dead wrong here. I can change my definition of the system instantly at any time.

I am not bound by your requirements of differentiability and continuity.

 

[ergospherical]

No. You are just plain flat out 100% completely dead wrong here. I can change my definition of the system instantly at any time.

That is to say ##\Omega(t)## no longer varies continuously with time, in which case you no longer have a transport law for control volumes - then what is the point? Any infinities you get just due to discontinuously changing the control volume are not physically meaningful...

 

[Dale]

the key feature of this sort of impulsive burn is that the expended fuel all moves at the same velocity. That is a feature not shared by the normal continuous burn model.

Regardless of how fast the continuous burn is performed.

 

[Filip Larsen]

Instead of taking the time or burn rate limit of the rocket equation one can also consider that ejection of, say, half the total mass in one go will give the remaining mass (payload) same speed as the propellant, namely the ejection speed (##v_1 = 1## if we ejection speed to one). If now instead the same propellant is ejected as ##n## equal mass parts the total velocity ##v_n = \sum_{i=1}^n \frac{1}{2n-i}## quickly falls below 1 and has (as far as I can see) the rocket equation with mass factor 2 as limit, i.e. ##v_n = \ln(2)## for ##n \rightarrow \infty##. This seem like an "easy" argument for ejected all mass at once gives highest total velocity (i.e. no propellant is spend accelerating any other propellant). And this is actually the opposite conclusion as what the OP seeks.

Later: I can see why the above does not properly model rocket propulsion, because I have wrongly fixed the ejection speed relative to center of mass and not relative to the remaning mass ejecting the propellant. Correcting for that, keeping ##v_e## and using a total propellant ratio ##p## (where above text used ##p = 1/2##), I now get that ##v_n = \sum_{i=1}^n \frac{p}{n+1-p i} v_e## which starts much lower and now actually increase towards the rocket equation as ##n \rightarrow \infty## which is also what I would have expected and which supports the OP question.

 

[jbriggs444]

That is to say ##\Omega(t)## no longer varies continuously with time, in which case you no longer have a transport law for control volumes - then what is the point? Any infinities you get just due to discontinuously changing the control volume are not physically meaningful...

You are the one trying to push your inappropriate model. I am not obliged to use it.

If I want to model the expulsion of a discrete blob of expended fuel as the expulsion of a discrete blob of expended fuel, where is the harm?

 

[bob012345]

For the math, the Tsiolkovsky rocket equation is: $$\Delta v = v_e \ln \left( \frac{m_r + m_f}{m_r} \right)$$ where ##m_r## is the mass of the rocket without the fuel and ##m_f## is the mass of the fuel and ##v_e## is the exhaust velocity.

If you simply burn it all instantaneously then conservation of momentum gives $$\Delta v = v_e \frac{m_f}{m_r} $$

A series expansion of gives $$\ln \left( \frac{m_r+ m_f}{m_r} \right) \approx \frac{m_f}{m_r} - \frac{m_f^2}{2 m_r^2} + O(m_f^3)$$

So I get the opposite, that ejecting it all at once makes it go faster.

I believe the approximation is only valid if ## \frac{m_f}{m_r} < 1## however in most cases of interest ## \frac{m_f}{m_r} >1##.

But in either case $$\ln \left( \frac{m_r+ m_f}{m_r} \right) = \ln \left(1+ \frac{m_f}{m_r} \right) < \frac{m_f}{m_r}$$

Raising both sides to ##e##;
$$ \left(1+ \frac{m_f}{m_r} \right) < e^{\large(\frac{m_f}{m_r} )}$$

or taking the fraction to be ##x## for convenience where ##x>0##;
$$ \left(1+ x \right) < e^{\large(x )} ≈ 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}...$$

 

[ergospherical]

If I want to model the expulsion of a discrete blob of expended fuel as the expulsion of a discrete blob of expended fuel, where is the harm?

You are missing the point, which is that the rocket equation only goes over to the result of the collision problem in the limit of ##\delta t \rightarrow 0## if the mass ejected ##q\rightarrow 0##, which is guaranteed so long as the mass expulsion rate varies within finite limits.

 

[Dale]

I believe the approximation is only valid if …

Yes, I wasn’t too happy with that approximation either, but I went ahead with it because I didn’t have a better one and a quick plot showed that the sign of the difference was always the same regardless of the two masses.

 

[jbriggs444]

You are missing the point, which is that the rocket equation only goes over to the result of the collision problem in the limit of ##\delta t \rightarrow 0## if the mass ejected ##q\rightarrow 0##, which is guaranteed so long as the mass expulsion rate varies within finite limits.

Which suggests that the rocket equation is not the correct model for a scheme in which a bundle of fuel is expended for energy and ejected for reaction as a unit rather than having each incremental bit of energy tied separately to an incremental bit of reaction mass.

 

[ergospherical]

No they're both fine, since it's impossible to eject any finite amount of mass through a surface in zero time! [The problem does not lie with the rocket equation: this only depends on the initial and final masses, not how the mass was ejected.]

 

[jbriggs444]

No they're both fine, since it's impossible to eject any finite amount of mass through a surface in zero time! [The problem does not lie with the rocket equation: this only depends on the initial and final masses, not how the mass was ejected.]

The problem does, in fact, lie with the rocket equation. It does not correctly model the case under consideration.

 

[ergospherical]

The problem does, in fact, lie with the rocket equation. It does not correctly model the case under consideration.

This only happens when you eject a finite mass ##q## in the limit of zero time, in which case you have to do the integration as in #14. But since this is not a physically reasonable scenario, it's hardly a problem with the rocket equation!

 

[jbriggs444]

This only happens when you eject a finite mass ##q## in the limit of zero time, in which case you have to do the integration as in #14. But since this is not a physically reasonable scenario, it's hardly a problem with the rocket equation!

As I have pointed out multiple times already, it is a physically reasonable scenario. We need not do any integration whatsoever for a scenario that is adequately modeled as one or more discrete transitions, each of which conserves both energy and momentum.

 

[ergospherical]

As I have pointed out multiple times already, it is a physically reasonable scenario.

It's really not, as is the case for a lot of infinities that arise in physics. But I don't think I'm going to convince you at this point.

 

[jbriggs444]

It's really not, as is the case for a lot of infinities that arise in physics. But I don't think I'm going to convince you at this point.

Correct. You will not. Because you do not understand what I am saying. There is nothing fancy going on here. Nothing where one has to worry about integration at all. Discrete transitions, all of which conserve momentum and energy.

We need not concern ourselves with whether we model the transitions as taking zero or non-zero time. It is irrelevant to the initial and final states. It does matter whether the resulting expended fuel is physically present as an expanding cloud with lots of internal energy or is present as one or more co-moving blobs, each with minimal internal energy.

There is more to the proposed discrete scenario than just burning the fuel fast. How you divvy out the total energy in the unexpended fuel to the bits of resulting expended fuel makes a real measurable physical difference. You have to design those details into the rocket. The physical difference does not go away because you do not like the model that reflects the modified rocket design or because you have decided that there is only one possible way to divvy out the energy in the fuel.

Yes, essentially all rockets adopt a continuous burn methodology for what I assume are good and sufficient reasons. That does not mean that it is the only way to do things.

Edit: I should add that I respect you and would prefer not to engage in a discussion as... strident as this has become. When two intelligent individuals find themselves at loggerheads like this, my experience is that it often results from an underlying assumption that is not shared between the participants.

 

[bob012345]

As I have pointed out multiple times already, it is a physically reasonable scenario. We need not do any integration whatsoever for a scenario that is adequately modeled as one or more discrete transitions, each of which conserves both energy and momentum.

It would be interesting if you put some example numbers to your model to really see.

 

[Filip Larsen]

Just a quick comment for anyone that happened to have read my take on the limit in #47 (apparently Dale did) that I now think the relative speed used did not correctly model a rocket ejection process, and correcting for that I still get the rocket equation in the limit, but now from below instead of from above, which also reverses the conclusion in regards to the OP question such that it does indeed pays off to eject propellant over time rather than in one go (as expected).

 

[bob012345]

Instead of taking the time or burn rate limit of the rocket equation one can also consider that ejection of, say, half the total mass in one go will give the remaining mass (payload) same speed as the propellant, namely the ejection speed (##v_1 = 1## if we ejection speed to one). If now instead the same propellant is ejected as ##n## equal mass parts the total velocity ##v_n = \sum_{i=1}^n \frac{1}{2n-i}## quickly falls below 1 and has (as far as I can see) the rocket equation with mass factor 2 as limit, i.e. ##v_n = \ln(2)## for ##n \rightarrow \infty##. This seem like an "easy" argument for ejected all mass at once gives highest total velocity (i.e. no propellant is spend accelerating any other propellant). And this is actually the opposite conclusion as what the OP seeks.

Later: I can see why the above does not properly model rocket propulsion, because I have wrongly fixed the ejection speed relative to center of mass and not relative to the remaning mass ejecting the propellant. Correcting for that, keeping ##v_e## and using a total propellant ratio ##p## (where above text used ##p = 1/2##), I now get that ##v_n = \sum_{i=1}^n \frac{p}{n+1-p i} v_e## which starts much lower and now actually increase towards the rocket equation as ##n \rightarrow \infty## which is also what I would have expected and which supports the OP question.

Is ##p = \frac{m_f}{m_r}## mass of fuel over dry mass?

I got ##\Large \frac{v_f}{v_e} = \sum\limits_{k=1}^n \Large\frac{1}{n(1 + \frac{m_r}{m_f}) -k}##

which for ##n = 1## gives the simple mass ratio and for ##n →∞## approaches the rocket equation from above.

 

[Filip Larsen]

Is ##p = \frac{m_f}{m_r}## mass of fuel over dry mass?

No, not dry mass. In my expression ##p## is the fraction of total mass ejected (propellant or fuel mass) relative to the total mass of the rocket at start (propellant and payload/structure) which is also the reciprocal of the ratio used in the rocket equation. With your mass symbols it would be ##p = \frac{m_f}{m_r+m_f}##. It can of course be converted to any other fraction one would like.

 

[A.T.]

This only happens when you eject a finite mass ##q## in the limit of zero time, in which case you have to do the integration as in #14. But since this is not a physically reasonable scenario, it's hardly a problem with the rocket equation!

It seems that everyone agrees the rocket equation doesn't apply in the limit of zero ejection time for a finite mass. That is important to know, if you want to model something as an instantaneous separation. Whether you call it "a problem with the rocket equation" or just "limited applicability of the rocket equation" is just semantics.

 

[vanhees71]

It's clear that zero ejection time is impossible physically and the rocket equation provides a physically more realistic result though it is on the other hand of course also very much simplified, particularly in considering ##v_f=\text{const}##.

 

[A.T.]

It's clear that zero ejection time is impossible physically

This depends on the definition of "ejection time". Disagreement about that definition seems to be at the core of the argument here.

and the rocket equation provides a physically more realistic result

In practical terms, this depends on the scenario and the given information. For an explosion that separates a body into two pieces (rocket & exhaust) it won't give the correct rocket velocity.

 

[vanhees71]

Sure, if you apply a formula for a situation, for which it hasn't been derived, you cannot expect to get correct answers.

 

[bob012345]

It seems that everyone agrees the rocket equation doesn't apply in the limit of zero ejection time for a finite mass. That is important to know, if you want to model something as an instantaneous separation. Whether you call it "a problem with the rocket equation" or just "limited applicability of the rocket equation" is just semantics.

To me that seems like a red herring. Zero ejection time means infinite force. The OP concerned momentum change by continuous flow vs. single ejection. As a path to that equations with discreet chunks of ejecta from 1 to infinity were looked at. Which, if any, are correct? (I need to double check mine..)

 

[A.T.]

Zero ejection time means infinite force.

It can also mean that we don't care about the force and thus simply move the rocket boundary at some time point (separation) to exclude the ejected mass. When modelling collisions / explosions we do this all the time, and then apply conservation laws, without ever caring about the force.

 

[Filip Larsen]

I got ##\Large \frac{v_f}{v_e} = \sum\limits_{k=1}^n \Large\frac{1}{n(1 + \frac{m_r}{m_f}) -k}##

For ##n = 1## and, say ##m_f = m_r##, this gives ##v_f = v_e## which I take means both ejected fuel and remaining mass has speed ##v_e## in opposite direction, thus giving total speed of ##2 v_e## which is physically correct but does not model ejecting fuel "as a rocket" i.e. with relative velocity ##v_e##. If so this is the same mistake I made at first.

 

[bob012345]

For ##n = 1## and, say ##m_f = m_r##, this gives ##v_f = v_e## which I take means both ejected fuel and remaining mass has speed ##v_e## in opposite direction, thus giving total speed of ##2 v_e## which is physically correct but does not model ejecting fuel "as a rocket" i.e. with relative velocity ##v_e##. If so this is the same mistake I made at first.

I thought we were discussing throwing all the reaction mass off at once vs. continuous with a number of discreet steps as a bridge. For the case of the ratio ##\frac{m_f}{m_r} = 1##, it all depends on whether there is one reaction mass ##n = 1## or ##n > 1##. What you say is only true for equal reaction mass and dry mass and only one reaction.

In my equation, every reaction occurs in its own inertial reference frame different from the previous reaction reference frame by the additional delta##v## kick of the last reaction. The total accumulated ##v_f## is in relation then to the initial reference frame taken as at rest.