I Rocket Fuel Ejection: Intuitive & Math Explained

[Filip Larsen]

What does $\frac{1}{2}pm v_e^2$ mean?

The ejected fuel mass $pm = m_f$ has a effective chemical potential for work that is released and converted into kinetic energy of the fuel and the remaining mass. For $m_f \approx 0$ (i.e. in the limit when $m_f \ll m_r$) we know the fuel is ejected with speed $v_e$, hence I choose to model the available energy when $m_f$ is ejected as $\frac{1}{2}m_f v_e^2$. When $m_f \not\ll m_r$ this energy is then distributed between the ejected chunk and the remaining mass.

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?

No. For $p = 0.5$ and $n = 1$ the speed of ejected fuel and the remaining rocket (relative to the center of mass) should be equal, and they are as both are equal to $v_e / \sqrt{2}$. As you can also see in the table, $s_1$ is relatively much closer to the rocket equation speed at $s_\infty$ than $0.5 v_e$ is.

 

[jbriggs444]

No. For $p = 0.5$ and $n = 1$ the speed of ejected fuel and the remaining rocket (relative to the center of mass) should be equal, and they are as both are equal to $v_e / \sqrt{2}$. As you can also see in the table, $s_1$ is relatively much closer to the rocket equation speed at $s_\infty$ than $0.5 v_e$ is.

OK. I saw no assertion about p=0.5 in #87.

 

[bob012345]

The ejected fuel mass $pm = m_f$ has a effective chemical potential for work that is released and converted into kinetic energy of the fuel and the remaining mass. For $m_f \approx 0$ (i.e. in the limit when $m_f \ll m_r$) we know the fuel is ejected with speed $v_e$, hence I choose to model the available energy when $m_f$ is ejected as $\frac{1}{2}m_f v_e^2$. When $m_f \not\ll m_r$ this energy is then distributed between the ejected chunk and the remaining mass.No. For $p = 0.5$ and $n = 1$ the speed of ejected fuel and the remaining rocket (relative to the center of mass) should be equal, and they are as both are equal to $v_e / \sqrt{2}$. As you can also see in the table, $s_1$ is relatively much closer to the rocket equation speed at $s_\infty$ than $0.5 v_e$ is.

Assuming the CM stays fixed for the $p = 0.5, n = 1$ case, if both velocities are equal and opposite, would not the relative velocity then be twice $v_e / \sqrt{2}$ or $\sqrt{2} v_e$which violates the assumption that $v_e$ is the relative velocity?

 

[bob012345]

There are two historical scenarios this thread brings to mind.

First, The Baltimore Gun Club in Jules Verne's From the Earth to the Moon where the "rocket" was launched by a rather large cannon. This has obvious problems with massive acceleration. It is interesting that Verne did not think of rockets since they existed. The second is project Orion where the proposed interstellar spaceship was powered by a series of discreet nuclear explosions. This has obvious problems with politics and international agreements.

https://www.ans.org/news/article-1294/nuclear-pulse-propulsion-gateway-to-the-stars/

 

[hutchphd]

The second is project Daedalus

You are missing, of course, the best exemplar: Project Orion with the likes of Theodore Taylor and Freeman Dyson working apace. Great films.

 

[bob012345]

You are missing, of course, the best exemplar: Project Orion with the likes of Theodore Taylor and Freeman Dyson working apace. Great films.

That's what I meant, Orion with pulsed nuclear bombs. The link discussed both Orion and Daedalus.

 

[bob012345]

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).

True, but compare the same bullet fired from a pistol and a rifle. I would expect the rifle fired bullet to end up with a higher velocity. In effect, it has a longer "burn".

 

[Filip Larsen]

Assuming the CM stays fixed for the $p = 0.5, n = 1$ case, if both velocities are equal and opposite, would not the relative velocity then be twice $v_e / \sqrt{2}$ or $\sqrt{2} v_e$which violates the assumption that $v_e$ is the relative velocity?

You and I have now discussed three different models on top of conservation of momentum that leads to different expressions for single-chunk speeds. The last one you are quoting from uses energy conservation such that at across each ejection the total kinetic energy after is equal to the available chemical potential for work in the fuel to be ejected. This means after each ejections the speed does not necessarily sum up to anything special, but the corresponding kinetic energy does sum up to what it should.

In the quoted case with $p = 0.5$ and $n = 1$ both speeds are $v_e / \sqrt{2}$ so both the ejected fuel and the remaining rocket, each of mass $\frac{1}{2} m$, has the kinetic energy $\frac{1}{2} m (v_e / \sqrt{2})^2 = \frac{1}{4} m v_e^2$ which means the total kinetic energy is $\frac{1}{2} m v_e^2$ as required.

 

[Filip Larsen]

OK. I saw no assertion about p=0.5 in #87.

You lost me with the reference to post #87 so I am not sure what you are trying to point out. Can you clarify?

 

[jbriggs444]

True, but compare the same bullet fired from a pistol and a rifle. I would expect the rifle fired bullet to end up with a higher velocity. In effect, it has a longer "burn".

Not if one idealizes away the energy in the muzzle gasses. The point of the rifle is to capture more of that.

 

[bob012345]

Not if one idealizes away the energy in the muzzle gasses. The point of the rifle is to capture more of that.

In that case yes so its probably not the best example in this debate. This problem could equally use idealized springs.

 

[bob012345]

You and I have now discussed three different models on top of conservation of momentum that leads to different expressions for single-chunk speeds. The last one you are quoting from uses energy conservation such that at across each ejection the total kinetic energy after is equal to the available chemical potential for work in the fuel to be ejected. This means after each ejections the speed does not necessarily sum up to anything special, but the corresponding kinetic energy does sum up to what it should.

In the quoted case with $p = 0.5$ and $n = 1$ both speeds are $v_e / \sqrt{2}$ so both the ejected fuel and the remaining rocket, each of mass $\frac{1}{2} m$, has the kinetic energy $\frac{1}{2} m (v_e / \sqrt{2})^2 = \frac{1}{4} m v_e^2$ which means the total kinetic energy is $\frac{1}{2} m v_e^2$ as required.

OK, that's fine but I just wasn't clear you were changing the assumption of fixed $v_e$ for that case. For this problem I am not convinced we need to concern ourselves with the energy. It is assumed to be whatever is necessary to eject a quantity of mass $m_f$ at the velocity $v_e$ relative to the rocket and is assumed the rocket engine is ideally efficient in converting chemical energy to kinetic energy. In my model it was assumed conserved at each ejection too. It should be linear in mass ejected relative to the rocket engine.

 

[jbriggs444]

In my model it was assumed conserved at each ejection too. It should be linear in mass ejected relative to the rocket engine.

The kinetic energy should be linear in the amount of mass expelled, yes. But that does not entail that the exhaust velocity of a non-infinitesimal exhaust packet will be constant independent of packet size. The exhaust velocity relative to the CM frame will vary with the mass ratio of the exhaust packet to the remaining gross vehicle weight. [Apologies if I am preaching to the choir]

 

[bob012345]

The kinetic energy should be linear in the amount of mass expelled, yes. But that does not entail that the exhaust velocity of a non-infinitesimal exhaust packet will be constant independent of packet size. The exhaust velocity relative to the CM frame will vary with the mass ratio of the exhaust packet to the remaining gross vehicle weight. [Apologies if I am preaching to the choir]

Of course but $v_e$ is not relative to the CM frame but to the rocket. For a massive rocket and small mass of ejected fuel they are almost equal. That's what took me a while to get but I did thanks to @Filip Larsen.

 

[jbriggs444]

Of course but $v_e$ is not relative to the CM frame but to the rocket. That's what took me a while to get.

OK. For a finite exhaust pellet, this $v_e$ is still not completely determined by the fuel energy density, but by the fuel energy density in conjunction with the mass ratio of the packet to the payload.

 

[bob012345]

OK. For a finite exhaust pellet, this $v_e$ is still not completely determined by the fuel energy density, but by the fuel energy density in conjunction with the mass ratio of the packet to the payload.

Right, which is why in my corrected model the bump in rocket speed is
$$ v_f = \frac {m_f v_e}{M}$$ where $M$ is the mass of rocket plus remaining fuel.

This assumes the energy is supplied in proportion to the ejected mass so I don't have to worry about it.

 

[bob012345]

I think it would be instructive to find real rocket engine data and extrapolate from that.

 

[Filip Larsen]

For this problem I am not convinced we need to concern ourselves with the energy.

I set out with an aim to find a way to compare a single-chunk ejection with continuous ejection via a model that makes sense in both situation in order to answer the original question about which of the two is yielding most rocket speed. In order for this to make sense, there must be some "physical similarity" between the two situations, otherwise we are just comparing apples and oranges. Constraining conservation of momentum by just requiring a fixed relative speed of $v_e$ is simple but also means the energy used is not proportional to the amount of spent fuel, potentially giving either situation an "unfair advantage" in the comparison.

So a natural next step is to constrain the energy to be "equal" in both cases and see what happens. And I do feel that including energy gives a more correct or interesting model for the comparison, even if it also do seem indicate that single-chunk ejection with such energy constrain is more speed-efficient than a continuous rocket, that is, the opposite conclusion the original question posed.

But that said, all in all, I still think the question is not well-defined enough for a conclusive answer, but that does of course not exclude us from having interesting discussions about it.

 

[bob012345]

I don't why you think the energy used is not proportional to the ejected fuel? I think it is exactly that. The energy used ideally is exactly that which makes the exhaust velocity $v_e$ relative to the rocket and is linear with ejected mass whether it is one lump or small lumps.

Instead of rocket engines just think of mass ejectors that eject mass with a fixed energy ratio per unit mass at a fixed relative velocity $v_e$.

I agree this is an idealized problem because there is no practical way to do this with real rockets. But in that spirit we can imagine such a one shot rocket.

I envision a spring loaded device that uses one large ideal spring vs. $n$ small ideal springs. Each gives the rocket a relative boost of $v_e$.

 

[jbriggs444]

I don't why you think the energy used is not proportional to the ejected fuel? I think it is exactly that. The energy used ideally is exactly that which makes the exhaust velocity $v_e$ relative to the rocket and is linear with ejected mass whether it is one lump or small lumps.

Instead of rocket engines just think of mass ejectors that eject mass with a fixed energy ratio per unit mass at a fixed relative velocity $v_e$.

I agree this is an idealized problem because there is no practical way to do this with real rockets. But in that spirit we can imagine such a one shot rocket.

I envision a spring loaded device that uses one large ideal spring vs. $n$ small ideal springs. Each gives the rocket a relative boost of $v_e$.

You are ignoring the recoil of the rocket. If you push off from a finite sized packet, the rocket recoil involves non-zero energy. By contrast, the energy from an infinitesimal packet winds up 100% in the packet, 0% in the rocket.

This is just a [time-reversed] inelastic collision. The energy in such a collision as measured in the center of mass frame is divvied out in inverse proportion to the mass ratio. The smaller ejected piece gets the larger share of the energy. If you eject something infinitesimal, all of the energy goes there. If you eject something finite, not all of the energy goes that way.

You can change your reference frame to the post-burn payload frame and do the energy accounting there, but you still have the same behavior. Energy usage and delta v are invariants. They do not change depending on the chosen reference frame. Choice of reference frame can change where the kinetic energy winds up. But it will not change the total amount by which kinetic energy increments or the delta v that results.

 

[bob012345]

You are ignoring the recoil of the rocket. If you push off from a finite sized packet, the rocket recoil involves non-zero energy. By contrast, the energy from an infinitesimal packet winds up 100% in the packet, 0% in the rocket.

This is just a [time-reversed] inelastic collision. The energy in such a collision as measured in the center of mass frame is divvied out in inverse proportion to the mass ratio. The smaller ejected piece gets the larger share of the energy. If you eject something infinitesimal, all of the energy goes there. If you eject something finite, not all of the energy goes that way.

You can change your reference frame to the post-burn payload frame and do the energy accounting there, but you still have the same behavior. Energy usage and delta v are invariants. They do not change depending on the chosen reference frame. Choice of reference frame can change where the kinetic energy winds up. But it will not change the total amount by which kinetic energy increments or the delta v that results.

I am not ignoring the recoil of the rocket! At least not knowingly. It has been in all my models. That it can be ignored in the infinitesimal limit is exactly why each of these models approach the rocket equation in that limit.

 

[Filip Larsen]

I don't why you think the energy used is not proportional to the ejected fuel?

For the model with $v_e$ as fixed relative speed when ejecting the fuel mass $m_f$ with $m_r$ remaining rocket mass, I get the total kinetic energy to be $\frac{1}{2}m_f v_e ^2 \frac{m-m_f}{m}$. Notice the trailing factor making this non-linear in $m_f$, especially for $m_f \gg m_r$ where the total kinetic energy goes towards zero even though nearly all of the initial mass has been ejected as fuel.

 

[bob012345]

For the model with $v_e$ as fixed relative speed when ejecting the fuel mass $m_f$ with $m_r$ remaining rocket mass, I get the total kinetic energy to be $\frac{1}{2}m_f v_e ^2 \frac{m-m_f}{m}$. Notice the trailing factor making this non-linear in $m_f$, especially for $m_f \gg m_r$ where the total kinetic energy goes towards zero even though nearly all of the initial mass has been ejected as fuel.

Is this the sum of both rocket and ejected mass? In the instantaneous CM frame? You have to be specific as to where you are referencing the kinetic energy from. Also, did you mean to write;

$$\frac{1}{2}m_f v_e ^2 \large \frac{m_r-m_f}{m_r}$$

This goes highly negative for $m_f \gg m_r$!

 

[Filip Larsen]

The total kinetic energy of what with respect to what reference frame?

I was referring to the single-chunk equations so there is really only one interesting frame. Before ejection the total mass sits at rest, after ejection the two masses move in opposite directions with some total kinetic energy that, as mentioned, does not vary linearly with the mass of the ejected fuel for the model in question. Please calculate the energy yourself and you will see.

 

[bob012345]

I was referring to the single-chunk equations so there is really only one interesting frame. Before ejection the total mass sits at rest, after ejection the two masses move in opposite directions with some total kinetic energy that, as mentioned, does not vary linearly with the mass of the ejected fuel for the model in question. Please calculate the energy yourself and you will see.

I'll happily do that. Whether we totally agree or not I find the discussion interesting.

First though, what about the negative kinetic energy I pointed out above and are we still forcing $v_e$ as the relative velocity in all mass ratio's? Thanks.

 

[Filip Larsen]

did you mean to write;

$$\frac{1}{2}m_f v_e ^2 \large \frac{m_r-m_f}{m_r}$$

No, you read it wrong.

I wrote the kinetic energy out like I did in terms of $m_f$ and the total (constant) mass $m = m_f + m_r$ to clearly show that it varies quadraticly, and not linearly, in $m_f$ and with the first part (i.e. $\frac{1}{2} m_f v_e^2$) separate because that is the equivalent energy released by the the fuel in the continuous case.

 

[bob012345]

No, you read it wrong.

I wrote the kinetic energy out like I did in terms of $m_f$ and the total (constant) mass $m = m_f + m_r$ to clearly show that it varies quadraticly, and not linearly, in $m_f$ and with the first part (i.e. $\frac{1}{2} m_f v_e^2$) separate because that is the equivalent energy released by the the fuel in the continuous case

Thanks!

I think I see what's happening here. There are different reference frames to consider and they are not the same nor is the kinetic energy required to be equivalent.

First, in the rocket's frame $R$, an observer sees the mass ejecta at $v_e$ as specified by the rocket engine and combustion chemistry or just by whatever is ejecting the mass. That is always $m_f$ moving at $v_e$ and its kinetic energy in that frame is always $K_r = \frac{1}{2}m_f \large v_e ^2$ and linearly proportional to $m_f$. The rocket velocity is by definition zero so that is the total kinetic energy. Also, one could look in the reference frame of the recoiling fuel mass where the total energy is $K_f = \frac{1}{2}m_r \large v_e ^2$ since $v_f$ is zero in that frame.

Next, we look at the velocities in the instantaneous center of mass reference frame. That is where you calculated;
$$K_{cm} = \frac{1}{2}m_f \large v_e ^2 \frac{m-m_f}{m}$$ or
$$K_{cm} = \frac{1}{2}\large v_e ^2 \frac{m_rm_f}{m_r + m_f}$$ I agree that is the total kinetic energy in the center of mass reference frame.

for $m_r \gg m_f$ we are essentially in the rocket frame $R$
$$K_{cm} ≈ \frac{1}{2}\large m_f \large v_e ^2$$
and for $m_r = m_f$
$$K_{cm} = \frac{1}{2}\frac{1}{2}\large m_f \large v_e ^2$$

The point is that in general $K_r ≠ K_f ≠ K_{cm}$ but all are valid.It is in the rocket frame however where the fuel is burned or just ejected that we measure the chemical or stored energy released. A real rocket is all about constant thrust or power which is mass flow times velocity or $\large m_f' ⋅ v_e$ where the energy per unit mass is ≈constant and $v_e$ is fixed by reaction chemistry.

The original OP question was simply;

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

I think as long as we fix the relative velocity of the fuel mass wrt the rocket, as rockets do, we are keeping with the spirit of the question and we both saw that the rocket does better with a slow burn than throwing off a single lump.

 

[Filip Larsen]

There are different reference frames to consider

I am not sure why you want to bring in more than one frame of reference, but I hope we agree that for the process of ejecting a single chunk of mass there is only need for one, namely the frame where the center of mass is at rest.

There is actually a ejection mechanism that sensible will cover both cases (constrainted ejection speed vs constrained ejection energy) if we imagine the rocket being equipped with a linear accelerator that can accelerate a chunk of mass put on it.

In constrained ejection speed mode the accelerator will accelerate any chunk of mass put on it until the chunk has exactly the set speed $v_e$ relative to the accelerator (and rest of the rocket) itself. The energy to do so comes from some other internal or external energy source that can provide up to at least some maximum of energy. Note that we don't care how long the accelerator take to do its work, conservation of momentum works regardless of the time taken. The important part (for this to be a "discrete" rocket) is only that the ejected mass is ejected as a single chunk, that is, the particles of the ejected mass is all at rest relative to each other after ejection. In this mode, the conclusion then is, that you will get more total speed of the rocket if you eject the same mass in smaller chunks than if you do it in one big chunk.

In constrained energy mode the accelerator instead works by slurping in an amount of fuel (and oxidizer), combusting it to get energy, but collecting all the combustion products into a small (mass-less) tank, and then use exactly the released energy to accelerate the combustion products out the back of the rocket. In this mode, however silly it sounds, the conclusion is that the rocket will get less final speed if the same chunk is broken up into smaller chunks.

Since the original question does not specify details that can discern between the two models and since the two models give opposite conclusions I will stil join the consensus in this thread that the question is ill-posed.

One may argue that since the constrained speed model is nice and simple and since it seems to affirm the original question then this is the type of rocket that whoever posed the original question must have had in mind. But I would rather answer the original question with "yes, there are models of ejections mechanisms where this will be true, but there are also other models where it is not true. Which one are you thinking off?"

 

[bob012345]

I am not sure why you want to bring in more than one frame of reference, but I hope we agree that for the process of ejecting a single chunk of mass there is only need for one, namely the frame where the center of mass is at rest.In constrained ejection speed mode the accelerator will accelerate any chunk of mass put on it until the chunk has exactly the set speed $v_e$ relative to the accelerator (and rest of the rocket) itself. The energy to do so comes from some other internal or external energy source that can provide up to at least some maximum of energy. Note that we don't care how long the accelerator take to do its work, conservation of momentum works regardless of the time taken. The important part (for this to be a "discrete" rocket) is only that the ejected mass is ejected as a single chunk, that is, the particles of the ejected mass is all at rest relative to each other after ejection. In this mode, the conclusion then is, that you will get more total speed of the rocket if you eject the same mass in smaller chunks than if you do it in one big chunk.

In constrained energy mode the accelerator instead works by slurping in an amount of fuel (and oxidizer), combusting it to get energy, but collecting all the combustion products into a small (mass-less) tank, and then use exactly the released energy to accelerate the combustion products out the back of the rocket. In this mode, however silly it sounds, the conclusion is that the rocket will get less final speed if the same chunk is broken up into smaller chunks.

Rockets combust propellent and expel exhaust gases from within the rocket's frame of reference not the instantaneous CM frame although in the usual case where $m_r >> m_f$ discussed above, the rockets frame and the CM frame are essentially the same. As I showed above in that case the total kinetic energy in the CM frame is almost the same as in the rocket frame.

Thus the natural frame to think about energy is the rocket frame. Only in that frame are both cases of constrained $v_e$ and constrained energy simultaneously true. Constraining the kinetic energy in the CM frame forces the $v_e$ to be different for each case the mass ratio is different because it is dependent on both masses. Focusing on the CM is the original mistake I made in which you corrected me where I took $v_e$ relative to the CM and not the rocket. It is a good approximation in the continuous case but not the discreet case.

I think an answer to the OP regarding why is simply momentum conservation. Since delta $v_r$ is inversely proportional to total mass, the summation of single, discreet or continuous dictates that behavour.

 

[jbriggs444]

Rockets combust propellent and expel exhaust gases from within the rocket's frame of reference not the instantaneous CM frame

Things happen in all reference frames, not just in some.

although in the usual case where $m_r >> m_f$ discussed above, the rockets frame and the CM frame are essentially the same.

The invariants of the problem such as the total increment to kinetic energy as a result of burning a prescribed quantity of fuel and ejecting a particular sized packet of exhaust are invariant. They do not depend on the choice of reference frame. The change in payload velocity as a result of a particular burn is also an invariant.

The choice of reference frame within which to do the analysis is completely arbitrary. All choices will yield identical results for all invariants of the problem.

The CM frame is a nice choice because of the symmetry it offers and the fact the the pre-burn momentum and kinetic energy are both zero.