I Rocket Fuel Ejection: Intuitive & Math Explained

[fatpanda1]

TL;DR Summary

Why does ejecting fuel over time cause rocket to go faster than ejecting all at once?

Can you give both intuitive and mathematical explanation please.

 

[Baluncore]

Because if you eject it all at once it destroys the rocket structure, and the payload.

 

[Filip Larsen]

Why does ejecting fuel over time cause rocket to go faster than ejecting all at once?

In contexts or models where the rocket equation applies this is question is a bit misleading, so I am curious as to the context the question has been posed in? (There are quite common situations in the normal life of a launcher that is not modeled well by the rocket equation and which also gives a good answer to the question, but the question itself is just posed as if applicable to any situation which is not true).

 

[jbriggs444]

Summary:: Why does ejecting fuel over time cause rocket to go faster than ejecting all at once?

Can you give both intuitive and mathematical explanation please.

Because ejecting the fuel all at once and having a single clump of exhaust gasses then travel at its normal exhaust velocity relative to the rocket is an incorrect expectation.

If you eject a huge blob of fuel from a tiny rocket at a particular exhaust velocity relative to the rocket, it takes only a small amount of energy. You might well think of it as ejecting a tiny rocket from a huge blob of fuel. You wind up with a huge unburnt blob of fuel.

 

[anorlunda]

Amateur rockets come close. They expend their fuel in a small fraction of the flight duration.

 

[Vanadium 50]

Amateur rockets come close. They expend their fuel in a small fraction of the flight duration.

You don't want them to be really efficient.

You need them to be about a foot long +/- half an order of magnitude, so they can be easily built. You also don't want them flying more than a few hundred feet (say 500 feet +/- a half order of magnitude) because then recovery can prove difficult. So they need short burns.

 

[Dale]

Summary:: Why does ejecting fuel over time cause rocket to go faster than ejecting all at once?

Can you give both intuitive and mathematical explanation please.

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.

 

[berkeman]

Summary:: Why does ejecting fuel over time cause rocket to go faster than ejecting all at once?

Can you give both intuitive and mathematical explanation please.

In addition to the better answers above, as you shorten the duration of the burn, you increase the acceleration for that initial period during the burn. Depending on the payload, there is a practical limit to how high the acceleration can be. Even without humans on board, the electronic assemblies and electro-mechanical components have limited tolerance for high accelerations...

 

[vanhees71]

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 don't understand this argument. The Tsiolkovsky equation does not contain the time you take to exhaust the fuel. So in this approximation it doesn't matter how quickly you exhaust the fule. The change in velocity is always the same. So why should your 2nd formula apply for exhausting the fuel instantaneously and why shouldn't then apply the Tsiolkovsky equation as well?

The derivation of the Tsiolkovsky equation works as follows: You assume an arbitrary exhaust $\mu(t)$ (exhausted mass of fuel per unit time). Then assuming the relative velocity $v_f$ of the fuel to the rocket being constant yields the momentum balance
$$\mathrm{d}_t (m v)=-\mu(v_f-v)=\dot{m} (v-v_g),$$
i.e.,
$$m \dot{v}=-\dot{m} v_g$$
or
$$m \frac{\mathrm{d} v}{\mathrm{d} m}=-v_f$$
from which by separation of of the variables the Tsiolkovsky equation follows.

The main oversimplication seems to be the assumption that $v_f=\text{const}$.

 

[russ_watters]

Faster acceleration means less thrust(percentagewise) wasted supporting the weight of the rocket(when launching vertically).

 

[Dale]

I don't understand this argument. The Tiolkovsky equation does not contain the time you take to exhaust the fuel. So in this approximation it doesn't matter how quickly you exhaust the fule. The change in velocity is always the same. So why should your 2nd formula apply for exhausting the fuel instantaneously and why shouldn't then apply the Tsiolkovsky equation as well?

The 2nd formula is an instantaneous burn, meaning that all of the exhaust has the same final velocity. Even though the rocket equation doesn't explicitly contain the burn time, it is based on the assumption of a non-instantaneous flow, so the exhaust velocity varies as the rocket accelerates. So the final state in the two cases is different. I.e. in my formulation an instantaneous burn is not just a limiting case of a non-instantaneous burn with a very short burn time.

 

[vanhees71]

Hm, as I just added to my posting, the assumption underlying the Tsiolkovsky equation is $v_f=\text{const}$ (where $v_f$ is the velocity of the fuel relative to the rocket). If this is not the case you likely get a different result, but the standard Tsiolkovsky equation doesn't depend on the time needed to exhaust a given amount of fuel, $\Delta m=m_{\text{ini}}-m_{\text{end}}$. You simply get
$$\Delta v=v_f \ln(m_{\text{ini}}/m_{\text{end}})=v_f \ln [(m_{\text{ini}}/(m_{\text{ini}}-\Delta m)],$$
as you stated yourself above.

 

[A.T.]

Because ejecting the fuel all at once and having a single clump of exhaust gasses then travel at its normal exhaust velocity relative to the rocket is an incorrect expectation.

Based on that expectation, burning all at once would give maximal final velocity? Because that would give the exhaust the maximal momentum in the opposite direction?

If you eject a huge blob of fuel from a tiny rocket at a particular exhaust velocity relative to the rocket, it takes only a small amount of energy. You might well think of it as ejecting a tiny rocket from a huge blob of fuel. You wind up with a huge unburnt blob of fuel.

So the tiny rocket flies off before all fuel was burned, and the fuel energy goes into accelerating the exhaust in all directions, not just backwards?

 

[ergospherical]

My two cents - if a mass $q$ is ejected instantaneously at $t=0$, i.e. $m(t) = m(0) - q H(t)$, and at constant velocity $v_e$ relative to the rocket, then the differential equation describing the motion is\begin{align*}
m \dfrac{dv}{dt} = v_e q \delta(t)
\end{align*}where $\delta(t)$ is the Dirac delta. Thus the change in velocity is\begin{align*}
\int dv = v_e q \int \dfrac{\delta (t)}{m(t)} dt = \dfrac{v_e q}{m(0)}
\end{align*}where the integral is taken over any interval containing $t=0$. (The value of $m(0)$ is that of the rocket plus the initial fuel).

[This result is slightly different from the expression for an instantaneous burn given by @Dale].

 

[Filip Larsen]

Allow me to expand on my first reply above:
a) The ideal rocket equation indicates it doesn't matter how quickly you eject your propellant (as long as you do at same relative speed), the rocket will end up with the same total delta-V.
b) In situations where you have drag terms (i.e. atmosphere) for sections of the trajectory, maintaining a lower speed at those sections will (everything else being equal) give higher total delta-V. Add to this the practical limitation that even without atmosphere its not possible to enter a closed orbit around a planet from its surface using only one impulsive launch maneuver (but that only relates to the "shape" of the orbit, not the total speed).
c) For some orbital maneuvers the rocket will obtain the highest total delta-V if the maneuver is done impulsively (i.e. all at once) at the periapsis (see Olberth's effect for details).

So depending on the actual context the original question may be either misleading, true or false.

 

[DaveC426913]

The really short answer:

... ejecting all at once?

Because this is better known as 'the rocket esploded'.

It is left as an exercise for the OP to determine why this is inefficient.

 

[Filip Larsen]

Because this is better known as 'the rocket esploded'.

I think this is covered in reply #2

 

[DaveC426913]

I think this is covered in reply #2

I actually scanned for such replies before posting. I guess I wasn't as diligent as I should have been.

 

[Dale]

Hm, as I just added to my posting, the assumption underlying the Tsiolkovsky equation is $v_f=\text{const}$ (where $v_f$ is the velocity of the fuel relative to the rocket). If this is not the case you likely get a different result,

Exactly. For an instantaneous burn the velocity of the fuel relative to the rocket is undefined since the velocity of the rocket is discontinuous (of course it is all well defined in the limit, but I am not assuming a limiting case). So for an instantaneous burn (the way I assumed) the $v_e$ is the velocity of the exhaust gas relative to the inertial frame in which the rocket was initially at rest.

 

[ergospherical]

So for an instantaneous burn (the way I assumed) the $v_e$ is the velocity of the exhaust gas relative to the inertial frame in which the rocket was initially at rest.

In some ways it's more natural to take $v_e$ as the relative velocity after the burn because it gives the correct limiting behaviour as per #14 i.e. $m(0)v = (m(0)-q)(v+\Delta v) + q(v + \Delta v-v_e) \implies \Delta v = \dfrac{v_e q}{m(0)}$.

 

[Dale]

In some ways it's more natural to take $v_e$ as the relative velocity after the burn because it gives the correct limiting behaviour as per #14 i.e. $m(0)v = (m(0)-q)(v+\Delta v) + q(v + \Delta v-v_e) \implies \Delta v = \dfrac{v_e q}{m(0)}$.

That is interesting. I hadn’t looked at that.

 

[Halc]

The 2nd formula is an instantaneous burn, meaning that all of the exhaust has the same final velocity.

This seems wrong since the rocket also 'instantaneously' accelerates to its final speed, so it should be viewed as a limit as the burn time decreases to zero, which seems to be the same transfer of momentum to the exhaust regardless of the rate of burn. No matter how short you reduce the burn time, the exhaust velocity never approaches the same value.

In other words, it seems that given X amount of fuel and M rocket mass, in deep space already, you get exactly the same final speed regardless of the rate of burn, which is why burn time doesn't appear in that rocket equation. So the reason for the chosen rate of burn has more to do with engine efficiency and getting off the planet and other considerations.
Burn too slow and too much of the energy is used just countering gravity (a problem only if your rocket is pointed up, hardly the most efficient orientation). A sufficiently low power rocket will never leave the pad at all, at least not until its weight drops below its thrust. Hence the need for high G force at takeoff, but not so high when say leaving Earth orbit.

Faster fuel expenditure means larger engines, and that adds mass to the engines and, as has been pointed out in prior posts, greater structural mass needed to withstand the greater forces involved. Burns also are throttled back at first to prevent excessive speed while still in dense atmosphere. You always hear them talk about throttling up after a minute or so. No need to hold back if there's no atmosphere with which to contend.

Slow expenditure of fuel is indicated when already in space, which is why ion engines are optimal when time is not of the essence. If there's people on board, there's the mass of the life support to think about. Get there faster just so you don't have to cart along so many resources to keep your perishable payload intact.

 

[Dale]

so it should be viewed as a limit as the burn time decreases to zero

That is explicitly not what I assumed because then there is no difference. The OP can (should) clarify.

 

[ergospherical]

Nonetheless either approach is still valid, since as the burn time $\delta t \rightarrow 0$, then mass ejected $q \rightarrow 0$ for any finite mass ejection rate $r(t)$ and the change in velocity given by the rocket equation\begin{align*}
v_e \ln \left( \dfrac{m(0)}{m(0)-q} \right) = - v_e \ln \left( 1 - \dfrac{q}{m(0)} \right) = \dfrac{v_e q}{m(0)} + O(q^2)
\end{align*}reduces to the formulae in #14 and #20. Clearly a mass ejection rate $r(t)$ which does not remain finite is unphysical.

 

[Dale]

Unphysical, yes, but not nonsensical.

 

[jbriggs444]

That is explicitly not what I assumed because then there is no difference. The OP can clarify.

One possible interpretation would be the limit of pulsing $n$ chunks each of mass $\frac{m_\text{fuel}}{n}$ and each at exhaust velocity $v_e$. [Measuring exhaust velocity relative to the remaining payload velocity not relative to pre-burn craft velocity].

If you take the limit as n increases without bound, you wind up with the Tsiolkovsky equation. If you take the value for any finite n, including n=1 you get something less.

Taking the limit as a continuous mass flow rate increases without bound is silly because it's taking the limit of a constant.

Note that under this interpretation, we do not use all of the energy in the fuel. If we decide to use all of the energy in the fuel then the optimal (albeit impractical) approach is to burn it all at once and expel the payload like the bullet from a gun. Anything else leaves waste energy in the relative velocity of the scattering bits of exhaust gasses. You get delta v that scales with the square root of the fuel capacity rather than with the log of the fuel capacity.

 

[ergospherical]

Unphysical, yes, but not nonsensical.

I would say nonsensical too! That the mass ejection rate must vary only between finite limits (on physical grounds) guarantees that $q \rightarrow 0$ in the limit of an instantaneous burn, in which case the rocket equation reduces to the result of the collision problem.

 

[jbriggs444]

Based on that expectation, burning all at once would give maximal final velocity? Because that would give the exhaust the maximal momentum in the opposite direction?

Yes -- if we were to harvest all of the energy in the fuel and eject it as a single lump.

So the tiny rocket flies off before all fuel was burned, and the fuel energy goes into accelerating the exhaust in all directions, not just backwards?

The fuel energy goes -- somewhere. We do not care. The model just says we did not use it all. Maybe we are spraying unburnt fuel. Maybe the unburnt fuel burns and scatters. Whichever.

 

[Dale]

I would say nonsensical too! That the mass ejection rate must vary only between finite limits (on physical grounds) guarantees that $q \rightarrow 0$ in the limit of an instantaneous burn, in which case the rocket equation reduces to the result of the collision problem.

The collision problem is hardly non-sensical. Sorry, but I disagree.

 

[jbriggs444]

The non-sensical part is to consider any mass ejection rate $r(t)$ which is unbounded. If one does so then I don't think it should be a surprise when discrepancies arise.

Unbounded mass ejection rates are perfectly physical.

You burn all of the fuel and package it up into a canister ready for expulsion. You harvest the energy and use it to propel the canister, ejecting it a single impulsive *whoomp*. That's an inelastic collision.

Even if you decide to take a slow motion video and decide that the ejection was not instantaneous, it is still most easily analyzed as if it were.

 

[ergospherical]

Unbounded mass ejection rates are perfectly physical.

You burn all of the fuel and package it up into a canister ready for expulsion. You use the energy to expel the cannister so as to use up all of the energy harvested from the burning of the fuel. You eject it in a single impulsive *whoomp*. That's an inelastic collision.

Mhm, the mass ejection rate will still remain bounded...

Also, there's nothing to say you can't consider the idealisation in principle (see post #14), but it is not really surprising to me at least that it doesn't agree with the rocket equation for finite $q$.

 

[Dale]

Actually, as I think more about it, it isn’t even unphysical. The key point of the assumption is that all of the exhaust is going at the same speed. You can physically achieve that simply by throwing the fuel off the back of the rocket all together. The moment it leaves your hand is the moment where the mass ejection rate is infinite.

So neither unphysical nor nonsensical, but just not standard rocket design.

 

[ergospherical]

The moment it leaves your hand is the moment where the mass ejection rate is infinite.

This is the part that is not physically realisable - the duration of the release will be always non-zero hence the mass ejection rate always bounded.

 

[Dale]

This is the part that is not physically realisable - the duration of the release will be always non-zero hence the mass ejection rate always bounded.

Classically there will always be some instant where you stop touching the mass. That can be defined as the instant where the mass is ejected, giving an infinite mass ejection rate.

In any case, the exhaust will all have the same velocity, which is the important thing.

 

[ergospherical]

Classically there will always be some instant where you stop touching the mass. That can be defined as the instant where the mass is ejected, giving an infinite mass ejection rate.

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.

 

[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.