I Rocket Fuel Ejection: Intuitive & Math Explained

[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.

 

[jbriggs444]

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.

If you throw the mass off continuously there is not much controversy to be found. No one in their right mind would throw out the reaction mass at more than a single relative velocity. You use up the energy in an incremental bit of fuel and you use 100% of that incremental energy to expel 100% of the incremental expended fuel at the maximum velocity that energy can allow.

You would not, for instance, use your liquid hydrogen and liquid oxygen to power a fuel cell, use the energy from the fuel cell to power an ion rocket and then dribble the waste water out the back at zero relative velocity. That would be an example of throwing your reaction mass out at two different exhaust velocities. It would be a daft thing to do. You get much better performance by expelling the waste water out the back. So nobody does the daft thing.

If you are handling the fuel in discrete chunks, you would ideally want to do much the same thing. You would want to use up all of the energy in a chunk of fuel and expel all of the expended fuel in that chunk at a single velocity so that all of the expended fuel in that chunk moves off as a blob rather than as an expanding cloud. You could save energy that way (at what would probably be an intolerable expense of practicality).

The Tsiolkovsky rocket equation does not contemplate such an approach. Nor does merely stating "I am going to burn all my fuel at once" necessarily entail such an approach. But if you do burn your fuel all at once, you should pay attention to the freedom you have to choose how to distribute the exhaust velocity of the material you are expending in the impulsive burn.

 

[bob012345]

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.

It would help if I explained my formula and how I got it.

My rocket is a system of masses consisting of $$M_{total} = M_r + M_f$$ where $M_f = n m_f$, and n is the number of units we divide the reaction mass into thus for the first reaction mass, $$ v_{f1} = \frac {v_e m_f}{(M_{total} - m_f}$$ Likewise, $$ v_{f2} = \frac {v_e m_f}{(M_{total} - 2m_f} and , v_{f3} = \frac {v_e m_f}{(M_{total} - 3m_f}$$ and
$$v_f = v_f1 + v_f2 + v_f3 + ...v_fn$$ so $$v_f = \sum\limits_{k=1}^n \frac{v_e m_f}{M_{t} - k m_f}$$ which is $$v_f = \sum\limits_{k=1}^n \frac{v_e m_f}{M_r + M_f- k m_f}$$$$v_f = \sum\limits_{k=1}^n \frac{v_e }{\frac{nM_r}{M_f}+ n - k }$$

using $M_f = n m_f$ this can be written as

$$v_f = \sum\limits_{k=1}^n \frac{v_e }{\frac{nM_r}{M_f}+ n - k }$$

$$\large\frac{v_f}{v_e} = \sum\limits_{k=1}^n \frac{1 }{n(1 + \frac{M_r}{M_f}) - k }$$

 

[Filip Larsen]

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.

I think we are trying to model the same thing, but what I try to point out is that the relative speed between the ejected mass and the remaining mass at each ejection needs to be $v_e$ if it has to model a discrete version of the rocket equation.

Apparently I am good at botching up my calculations when I try to squeeze them in between morning coffee and leaving for work, but I hope we can agree than if a rocket at rest of initial mass $m$ ejects the mass $pm$ (i.e. $p(m_r + m_f) = m_f$) in one chunk then conservation of momentum gives $(1-p)m v_1 = pm v_p$ where the speed $v_1$ of the remaining mass and $v_p$ of the ejected fuel are opposite (and relative to the inital rest frame). If we then require $v_e = v_1 + v_p$ in order to let this process "simulate" a rocket with ejection speed $v_e$ it follows that $v_1 = p v_e = \frac{m_f} {m_r + m_f} v_e$. Compare this with your equation and my first attempt (based on $v_p = v_e$) which for one ejected chunk (i.e. $n = 1$) gives $v_f = v_1 = \frac{m_f}{m_r} v_e$.

My argument for requiring $v_e = v_1 + v_p$ is that the original question asks if (and if so why) it is better to eject in smaller chunks (i.e. continuously in the limit) instead of in one go. For a comparison between the two situations to make sense it is sensible to require the relative speed of any ejected mass at the time of ejection to be fixed at $v_e$. This is also equivalent to requiring the total amount of work in the two situations to be the same.

 

[bob012345]

I just lost all my edits for some unknown reason and have to recompose post #72 above.

 

[bob012345]

I just lost all my edits for some unknown reason and have to recompose post #72 above.

Ok, so my final equation is;

$$\large\frac{v_f}{v_e} = \sum\limits_{k=1}^n \frac{1 }{n(1 + \frac{M_r}{M_f}) - k }$$

I think this is very general. It only uses the total fuel and dry masses and the number of chunks $n$ the fuel is broken into. The ejection velocity is $v_e$ in every instance in the inertial frame where the rocket is at rest. It works for any $n$ from 1 to ∞ where it gives the same results as the rocket equation.

$\frac{M_r}{M_f}$	$n$	$\frac{v_f}{v_e}$	Rocket eq.
1	1	1	0.693147
1	2	0.83333	0.693147
1	5	0.745635	0.693147
1	10	0.71877	0.693147
1	100	0.69565	0.693147
1	$10^3$	0.693397	0.693147
1	$10^4$	0.693172	0.693147
1	$10^5$	0.693149	0.693147
1	$10^6$	0.693147	0.693147
0.1	1	10	2.397895
0.1	2	5.83333	2.397895
0.1	5	3.57460	2.397895
0.1	10	2.928968	2.397895
0.1	100	2.441754	2.397895
0.1	$10^3$	2.402449	2.397895
0.1	$10^4$	2.398350	2.397895
0.1	$10^5$	2.397940	2.397895
0.1	$10^6$	2.397899	2.397895
0.1	$10^8$	2.397895	2.397895

 

[Filip Larsen]

It works for any from 1 to ∞ where it gives the same results as the rocket equation.

We both agree on the limit as $n$ goes to infinity, namely the rocket equation, but we also have to decide, as the original question asks, if the final speed of the the rocket when ejecting all fuel at once is higher, lower or equal, compared to ejecting it continuously, and herein lies the difference between the models.

With a simple model (yours, and my first one) where after each ejection the rocket gets a speed change opposite the fuel such that the fuel always has speed $v_e$ relative to the instantaneous rest frame, then you always get $v_f > v_e$ for $n = 1$ (and $m_f > 0$) which means the equation goes towards the rocket equation from above and this also means this model supports that you get higher final speed if you eject all mass at once, rather than continuously, which is contrary to what is asked to show. As a numeric example, if we take your table entry for $n = 1$ and $m_r/m_f = 0.1$ we get $v_f = 10 v_e$, which is to say the relative speed between the rocket and the fuel is $11 v_e$. My question now is, if we end up with $11 v_e$ does this process really model the same rocket as in the continuous case?

In order to get closer to model the same process I then tried to put in the constrain that after each fuel ejection, the relative speed between the rocket and the ejected fuel chunk has to be $v_e$. This results in a similar equation but now the rocket equation limit is approached from below. This also means that this model affirms the original question. So which one is correct?

I am now not so sure that either of the two above models are a particularly good fit for a discrete model of a continuous rocket. Going further, one could perhaps try to require that the total work in the single-chunk, multi-chunk and continuous case is all the same and proportional to the fuel mass. My immediate problem with this approach is that it quickly becomes fairly messy, and its also not clear to me what the total work in the continuous case should be.

 

[Leo Liu]

It would help if I explained my formula and how I got it.

My rocket is a system of masses consisting of $$M_{total} = M_r + M_f$$ where $M_f = n m_f$, and n is the number of units we divide the reaction mass into thus for the first reaction mass, $$ v_{f1} = \frac {v_e m_f}{(M_{total} - m_f}$$ Likewise, $$ v_{f2} = \frac {v_e m_f}{(M_{total} - 2m_f} and , v_{f3} = \frac {v_e m_f}{(M_{total} - 3m_f}$$ and
$$v_f = v_f1 + v_f2 + v_f3 + ...v_fn$$ so $$v_f = \sum\limits_{k=1}^n \frac{v_e m_f}{M_{t} - k m_f}$$ which is $$v_f = \sum\limits_{k=1}^n \frac{v_e m_f}{M_r + M_f- k m_f}$$$$v_f = \sum\limits_{k=1}^n \frac{v_e }{\frac{nM_r}{M_f}+ n - k }$$

using $M_f = n m_f$ this can be written as

$$v_f = \sum\limits_{k=1}^n \frac{v_e }{\frac{nM_r}{M_f}+ n - k }$$

$$\large\frac{v_f}{v_e} = \sum\limits_{k=1}^n \frac{1 }{n(1 + \frac{M_r}{M_f}) - k }$$

https://www.physicsforums.com/threads/people-jumping-from-a-flatcar.993490/#post-6391280
This is actually similar to a homework question I asked a year ago. As #5 by etotheipi (ergospherical) pointed out, the continuous ejection of masses results in a different terminal velocity than the discrete rejection does.

 

[A.T.]

...is the equation goes towards the rocket equation from above ... now the rocket equation limit is approached from below...So which one is correct?

If they both approach the rocket equation in the continuous limit, they are both valid.

Which approach is more appropriate in the non-continuous case depends on additional information about the scenario that is not explicitly provided in the question. It can only be deduced from the premise, that the approach from below is to be assumed here.

 

[bob012345]

We both agree on the limit as $n$ goes to infinity, namely the rocket equation, but we also have to decide, as the original question asks, if the final speed of the the rocket when ejecting all fuel at once is higher, lower or equal, compared to ejecting it continuously, and herein lies the difference between the models.

With a simple model (yours, and my first one) where after each ejection the rocket gets a speed change opposite the fuel such that the fuel always has speed $v_e$ relative to the instantaneous rest frame, then you always get $v_f > v_e$ for $n = 1$ (and $m_f > 0$) which means the equation goes towards the rocket equation from above and this also means this model supports that you get higher final speed if you eject all mass at once, rather than continuously, which is contrary to what is asked to show. As a numeric example, if we take your table entry for $n = 1$ and $m_r/m_f = 0.1$ we get $v_f = 10 v_e$, which is to say the relative speed between the rocket and the fuel is $11 v_e$. My question now is, if we end up with $11 v_e$ does this process really model the same rocket as in the continuous case?

In order to get closer to model the same process I then tried to put in the constrain that after each fuel ejection, the relative speed between the rocket and the ejected fuel chunk has to be $v_e$. This results in a similar equation but now the rocket equation limit is approached from below. This also means that this model affirms the original question. So which one is correct?

I am now not so sure that either of the two above models are a particularly good fit for a discrete model of a continuous rocket. Going further, one could perhaps try to require that the total work in the single-chunk, multi-chunk and continuous case is all the same and proportional to the fuel mass. My immediate problem with this approach is that it quickly becomes fairly messy, and its also not clear to me what the total work in the continuous case should be.

Ahh, Now I see the difference! My equation sets $v_e$ as the exhaust velocity in the CM frame taken as the instantaneous rest frame and your equation forces $v_e$ to be the relative velocity compared to the rocket. My gut tells me your interpretation is correct. Even for $n = 1$ my formula give a relative separation of $2v_e$ instead of $v_e$. I need to rethink my derivation and see if I get the same result as you did. Thanks!

 

[bob012345]

Also, in an effort to either confuse or enlighten things, consider firing a gun in space. The muzzle velocity is fixed by the chemistry of the reaction. But there is a reaction velocity of the gun too as a result of the impulse. We would need to consider the energy conservation as well as momentum conservation to get the velocities. So far, I have only been considering momentum.

 

[Filip Larsen]

We would need to consider the energy conservation as well as momentum conservation to get the velocities.

Yes, I also think that by including "proper" energy conservation one should be able to model a discrete rocket that is more "similar" to the physical limits involved in a continuous rocket. However, exactly how to do this in an easy and elegant way is still a bit elusive ...

 

[vanhees71]

Trying to formulate the "sudden ejection model" leads to a mathematical indefiniteness, though. Just assuming
$$m(t)=m_f \Theta(t)+m_i \Theta(-t)$$
leads to
$$\dot{m}(t)=(m_f-m_i) \delta(t).$$
Now plug this into the standard rocket equation of motion (with $t$ as the indpendent variable) leads to
$$m \dot{v}=-(m_f-m_i) V \delta(t),$$
where $V$ is the relative velocity of the ejected fuel wrt. the rocket.

It seems clear that the solution must be something like
$$v(t)=v_f \Theta(t) + v_i \Theta(-t),$$
but it's impossible to uniquely determine $v_f$, because it's not clear, how to define $\Theta(t)$ at $t=0$ for this problem.

As soon as you regularize the $\delta$ distribution, making $m(t)$ smooth, you inevitably end up with the rocket equation, for which the change of velocity is uniquely given by the Tsiolkovsky equation. So I'd say it's pretty safe to say that this "sudden ejection model" is simply ill defined.

 

[hmmm27]

The most energy-efficient space-to-space journey leaves ejecta with 0 KE : it all goes into the ship. Irrelevant for Tsiolkovsky's equation with fixed thrust... except for the point where the speeds of ship and thrust are equal.

 

[bob012345]

Trying to formulate the "sudden ejection model" leads to a mathematical indefiniteness, though. Just assuming
$$m(t)=m_f \Theta(t)+m_i \Theta(-t)$$
leads to
$$\dot{m}(t)=(m_f-m_i) \delta(t).$$
Now plug this into the standard rocket equation of motion (with $t$ as the indpendent variable) leads to
$$m \dot{v}=-(m_f-m_i) V \delta(t),$$
where $V$ is the relative velocity of the ejected fuel wrt. the rocket.

It seems clear that the solution must be something like
$$v(t)=v_f \Theta(t) + v_i \Theta(-t),$$
but it's impossible to uniquely determine $v_f$, because it's not clear, how to define $\Theta(t)$ at $t=0$ for this problem.

As soon as you regularize the $\delta$ distribution, making $m(t)$ smooth, you inevitably end up with the rocket equation, for which the change of velocity is uniquely given by the Tsiolkovsky equation. So I'd say it's pretty safe to say that this "sudden ejection model" is simply ill defined.

I did not take the all at once statement on the OP to literally mean a process happening at infinitesimal time but rather all the reaction mass leaving together as opposed to a continuous process.

 

[jbriggs444]

The most energy-efficient space-to-space journey leaves ejecta with 0 KE : it all goes into the ship. Irrelevant for Tsiolkovsky's equation with fixed thrust... except for the point where the speeds of ship and thrust are equal.

You may have to formulate this optimization problem better. By my reckoning, there is no optimum and no restriction on exhaust velocity.

For any stationary point to stationary point journey you devise, I can devise one that takes less energy and uses a higher exhaust velocity. It'll just take longer.

 

[Filip Larsen]

Let me try again, now with energy conservation.

Using as before $m_f = p m = p (m_r + m_f)$, conservation of momentum gives $$(1-p) m v_1 = p m v_p , $$ and conservation of energy can be set as $$\frac{1}{2}(1-p)m v_1^2+\frac{1}{2}pm v_p^2 = \frac{1}{2}pm v_e^2 ,$$ where I have used $\frac{1}{2}v_e$ as a measure of the fuels specific (chemical) potential to do mechanical work.

Solving for for the two speeds I get $$v_1 = v_e \frac{p}{\sqrt{1-p}} , $$ and $$v_p = v_e\sqrt{1-p} .$$

Repeating the above but now with the total fuel mass $pm$ ejected in $n$ chunks of mass $pm/n$ each, I get $$s_n = \frac{v_f}{v_e} = \sum_{i=1}^n \frac{p}{\sqrt{(n-pi)p + (n-pi)^2}} .$$

Putting in some numbers (as shown below), it seems that this again converges on the continuous rocket in the limit, but again from above in all cases. So, if we compare the continuous rocket with a burning the same amount of fuel and use that energy to send the exhaust out as one single chunk it looks like it does pay off a little bit, especially for higher $p$. For example, for $p = 0.9$ we should get around 24% more speed by collecting all the exhaust and sending it out as one chunk. I am not sure I really can make sense of that statement.

$p$	$s_1$	$s_2$	$s_3$	$s_4$	$s_5$	$s_\infty = -ln(1-p)$
0.1​	0.105409​	0.105373​	0.105366​	0.105364​	0.105362​	0.105361​
0.2​	0.223607​	0.223260​	0.223196​	0.223173​	0.223162​	0.223144​
0.3​	0.358569​	0.357159​	0.356891​	0.356797​	0.356753​	0.356675​
0.4​	0.516398​	0.512282​	0.511479​	0.511194​	0.511062​	0.510826​
0.5​	0.707107​	0.696923​	0.694856​	0.694115​	0.693769​	0.693147​
0.6​	0.948683​	0.925515​	0.920530​	0.918707​	0.917847​	0.916291​
0.7​	1.278019​	1.226715​	1.214719​	1.210176​	1.207995​	1.203973​
0.8​	1.788854​	1.671098​	1.640113​	1.627637​	1.621431​	1.609438​
0.9​	2.846050​	2.525586​	2.425515​	2.380467​	2.356248​	2.302585​

 

[hmmm27]

You may have to formulate this optimization problem better. By my reckoning, there is no optimum and no restriction on exhaust velocity.

For any stationary point to stationary point journey you devise, I can devise one that takes less energy and uses a higher exhaust velocity. It'll just take longer.

True, "journey" does imply a deceleration at some point ; I misspoke - the thread is about whether a greater velocity can be achieved through a big boom, or an extended burn. The former wastes (at least) half the expended energy, not counting thermal.

 

[bob012345]

Let me try again, now with energy conservation.

Using as before $m_f = p m = p (m_r + m_f)$, conservation of momentum gives $$(1-p) m v_1 = p m v_p , $$ and conservation of energy can be set as $$\frac{1}{2}(1-p)m v_1^2+\frac{1}{2}pm v_p^2 = \frac{1}{2}pm v_e^2 ,$$ where I have used $\frac{1}{2}v_e$ as a measure of the fuels specific (chemical) potential to do mechanical work.

​

Please check your energy conservation equation; $$\frac{1}{2}(1-p)m v_1^2+\frac{1}{2}pm v_p^2 = \frac{1}{2}pm v_e^2 ,$$ What does $\frac{1}{2}pm v_e^2$ mean? I think $s_1$ for $p = 0.5$ where mass of the rocket and fuel are equal, should be exactly 0.5 shouldn't it?

 

[bob012345]

Meanwhile, I re-derived my formula keeping $v_e$ fixed as the relative velocity;

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

Where $v_f$ means final velocity of the rocket after the ejection and $m_f$ means a mass of fuel It's almost identical except for the additional 1 in the denominator. For a mass ratio of 1 and one chunk I get $$\large\frac{v_f }{v_e} = \large\frac{1}{2}$$

$\frac{m_r}{m_f}$	$n$	$\frac{v_f}{v_e}$	Rocket equation
1	1	0.5	0.693147
1	2	0.583333	0.693147
1	5	0.645635	0.693147
1	10	0.668771	0.693147
1	100	0.690653	0.693147
1	$10^3$	0.692897	0.693147
1	$10^4$	0.693122	0.693147
1	$10^5$	0.693145	0.693147
1	$10^6$	0.693147	0.693147

Before I was assuming $v_e$ was with respect to an instantaneous Center of Mass frame;
$$ v_{f1} = \frac {v_e m_f}{(M_{total} - m_f)}$$
Now, forcing $v_e$ to be the relative velocity I get
$$ v_{f1} = \frac {v_e m_f}{M_{total} }$$
If we let $m_f$ approach a differential, it is easy to see how both would give the Tsiolkovsky/Goddard rocket equation;

$$\large dv = \large\frac{v_e dm }{m}$$

$$\large v_f = -v_e \large\int_{m_i}^{m_f} \frac{1}{m} \,dm = v_e \ln(\frac{m_i}{m_f})$$

So I conclude the OP was correct that the continuous burn is better. I believe that is related to the Oberth effect;

https://en.wikipedia.org/wiki/Oberth_effect

 

[jbriggs444]

True, "journey" does imply a deceleration at some point ; I misspoke - the thread is about whether a greater velocity can be achieved through a big boom, or an extended burn. The former wastes (at least) half the expended energy, not counting thermal.

Hmmm. Does it? If I shoot a bullet from a gun, almost all of the energy in the propellant winds up in the bullet, not in the gun. (Idealizing away the waste energy in the muzzle gasses).