100% energy efficient, constant power rocket

In summary, the conversation discusses a model for a rocket that aims to maximize energy efficiency by matching exhaust speed closely to rocket speed. The model involves a constant power supply and the ejection of propellant at a specific rate and speed. It also considers the equation of motion and compares it to the Tsiolkovsky rocket equation. The efficiency of the rocket is also discussed in terms of energy and mass, with the conclusion that it may not be mass efficient but could potentially be power efficient in certain scenarios.
  • #1
SteveR001
I gather rocket energy efficiency is maximal when exhaust speed closely matches the rocket speed (but oppositely directed), so to an outside observer such a rocket would leave a stationary propellant trail. I’ve found little on this model and am trying to learn more.

Say we have a rocket of mass M that has a constant power W available for propulsive use (from solar panels perhaps) and it ejects a mass of propellant (m) at a rate (r) per second at a speed (u) relative to the rocket (zero to an outside observer) such that power (W) used remains constant. We could also restrict u to be a maximum of say 3000 m/s and cut off the engine at or before this point; because that is the only situation I’m interested in. The rocket will be given an initial speed u, otherwise it would never accelerate. The mass/s of exhaust ejected will diminish with speed (V) and would be calculated.

As far as I can tell from numerical simulation, the power, acceleration and rocket momentum remain constant, while the overall the performance and flight characteristics appear related to initial speed (u), plus of course the initial mass and power available. Has this model been analysed anywhere? What is its equation of motion and how does it compare to the Tsiolkovsky rocket equation? Is this rocket energy and mass efficient?
 
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  • #2
SteveR001 said:
I gather rocket energy efficiency is maximal when exhaust speed closely matches the rocket speed (but oppositely directed), so to an outside observer such a rocket would leave a stationary propellant trail. I’ve found little on this model and am trying to learn more.

Say we have a rocket of mass M that has a constant power W available for propulsive use (from solar panels perhaps) and it ejects a mass of propellant (m) at a rate (r) per second at a speed (u) relative to the rocket (zero to an outside observer) such that power (W) used remains constant. We could also restrict u to be a maximum of say 3000 m/s and cut off the engine at or before this point; because that is the only situation I’m interested in. The rocket will be given an initial speed u, otherwise it would never accelerate. The mass/s of exhaust ejected will diminish with speed (V) and would be calculated.

As far as I can tell from numerical simulation, the power, acceleration and rocket momentum remain constant, while the overall the performance and flight characteristics appear related to initial speed (u), plus of course the initial mass and power available. Has this model been analysed anywhere? What is its equation of motion and how does it compare to the Tsiolkovsky rocket equation? Is this rocket energy and mass efficient?

Sorry, I realized after posting that I have used the same variable (u) twice, once for exhaust speed relative to the rocket (which will actually be -V anyway) and again for the initial rocket speed. The latter variable name should have been v. so u = -V and v = initial rocket speed, which will be some small arbitrary speed. Actually the magnitude of the starting speed (v) is one area of my interest.
 
  • #3
SteveR001 said:
Is this rocket energy and mass efficient?
It would certainly not be mass efficient, just conceptually. I assume by energy and mass efficient you mean ##\Delta v/E## and ##\Delta v/m##
 
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  • #4
SteveR001 said:
I gather rocket energy efficiency is maximal when exhaust speed closely matches the rocket speed (but oppositely directed), so to an outside observer such a rocket would leave a stationary propellant trail.
This can be modeled by choosing a frame of reference that matches the exhaust velocity. The spent fuels kinetic energy goes from non-zero to zero as it's expelled, so all of the energy goes into accelerating the rocket. This only momentary, since a moment later, the exhaust will have a small velocity in the same direction as the rocket.

When considering the rocket efficiency, the total change in energy of both the fuel as well as the rocket need to be considered. The total change in energy will be (inertial) frame independent. The greater the exhaust speed with respect to the rocket, the more efficient.

As a bit of trivia, say a rocket starts off with zero velocity relative to some inertial frame. Assuming constant exhaust velocity with respect to the rocket, at the point where the mass of rocket and fuel is reduced to 1/e ~= 1/2.718 ~= 0.368 of it's original mass, the rocket velocity will exceed the exhaust velocity and the exhaust will also be moving in the same direction as the rocket as observed from that initial frame of reference.
 
  • #5
Dale said:
It would certainly not be mass efficient, just conceptually. I assume by energy and mass efficient you mean ##\Delta v/E## and ##\Delta v/m##

Yes, though come to think of it I should have said 'power efficient' rather than energy efficient, as I'm imagining a fixed power supply scenario. For example in a situation that might occur in future where propellent is in copious supply (plasma engines already expel a wide range of materials) but where the power supply is fixed and quite low (e.g. from solar panels). I suppose what I'm after is a 'constant power, variable exhaust velocity rocket equation', as opposed to a usual fixed exhaust speed, to explore various things, such as what would such a rocket look like. I wondered if maintaining exhaustVelocity = -rocketVelocity (constantly adjusted during flight to keep the power fixed) would maximise the use of the available power. And whether ejecting a greater mass at slower speed would be more power efficient than ejecting a lesser mass at higher speed, delivering the same momentum to an albeit initially heavier rocket. I suppose at the end of the day it would come down to how much time you have on your hands to get a given payload up to a given deltaV.

I think i may have something here https://forum.nasaspaceflight.com/index.php?topic=37511.0 but am not yet sure what the variable names represent.
 
  • #6
rcgldr said:
When considering the rocket efficiency, the total change in energy of both the fuel as well as the rocket need to be considered. The total change in energy will be (inertial) frame independent. The greater the exhaust speed with respect to the rocket, the more efficient.

And that is indeed the operating principle of practical electrical rocket engines I have read about - to maximise propellent velocity - which confirms that you are right. But on the other hand I can't square that realisation with a variable exhaust speed rocket that arrives at a destination empty of fuel with its propellent stationary wrt to your inertial observer, along its entire length. In the first case a great deal of energy has ended up in the exhaust and in the second case all the energy appears to end up in the payload. And yet the first case scenario is the more energy efficient.
 
  • #7
SteveR001 said:
I gather rocket energy efficiency is maximal when exhaust speed closely matches the rocket speed (but oppositely directed), so to an outside observer such a rocket would leave a stationary propellant trail...

Say we have a rocket of mass M that has a constant power W available for propulsive use (from solar panels perhaps) and it ejects a mass of propellant (m) at a rate (r) per second at a speed (u) relative to the rocket (zero to an outside observer) such that power (W) used remains constant. We could also restrict u to be a maximum of say 3000 m/s and cut off the engine at or before this point; because that is the only situation I’m interested in. The rocket will be given an initial speed u, otherwise it would never accelerate. The mass/s of exhaust ejected will diminish with speed (V) and would be calculated...

Is this rocket energy and mass efficient?
rcgldr said:
When considering the rocket efficiency...

The greater the exhaust speed with respect to the rocket, the more efficient...
SteveR001 said:
Yes, though come to think of it I should have said 'power efficient' rather than energy efficient, as I'm imagining a fixed power supply scenario. For example in a situation that might occur in future where propellent is in copious supply (plasma engines already expel a wide range of materials) but where the power supply is fixed and quite low (e.g. from solar panels). I suppose what I'm after is a 'constant power, variable exhaust velocity rocket equation', as opposed to a usual fixed exhaust speed, to explore various things, such as what would such a rocket look like. I wondered if maintaining exhaustVelocity = -rocketVelocity (constantly adjusted during flight to keep the power fixed) would maximise the use of the available power. And whether ejecting a greater mass at slower speed would be more power efficient than ejecting a lesser mass at higher speed, delivering the same momentum to an albeit initially heavier rocket. I suppose at the end of the day it would come down to how much time you have on your hands to get a given payload up to a given deltaV.
I think more needs to be put into the definition of "efficiency" here because the naked use of the word means power (or energy) out divided by power (or energy) in, expressed as a percentage. Also known as thermodynamic efficiency. But is that really useful for a rocket?

For a car, "fuel efficiency" is in miles per gallon because the point of the definition is to get from one place to another on the least amount of fuel(few people ever look at the thermodynamic efficiency of a car). Is that what you are looking for here? Or are you looking to do it fastest? Cheapest?

One common definition of "efficiency" for rocket engines is specific impulse, which is force times time (total change in momentum) per unit of fuel. This is based on the idea that what you really want out of a rocket engine is to apply as large of a force as possible for as long as possible and with as little fuel as possible.

An ion engine might have an efficiency of 80%, but who cares: what is really great is that it has a specific impulse of several times (up to about 10 times) that of a conventional rocket.

But, if reaction mass can be made smaller, maybe that's not what you really care about? You specifically mentioned electric power, so it may be instructive to browse the specs of some ion engines to see how lower thrust generally means lower power for the same impulse. Since the energy is free (presumably, solar), you make "better" use of your reaction mass by colleting more energy from the sun to apply to it over a longer period of time. But is that more "efficient" or less?
https://en.wikipedia.org/wiki/Ion_thruster
 
  • #8
russ_watters said:
I think more needs to be put into the definition of "efficiency" here because the naked use of the word means power (or energy) out divided by power (or energy) in, expressed as a percentage. Also known as thermodynamic efficiency. But is that really useful for a rocket?

For a car, "fuel efficiency" is in miles per gallon because the point of the definition is to get from one place to another on the least amount of fuel(few people ever look at the thermodynamic efficiency of a car). Is that what you are looking for here? Or are you looking to do it fastest? Cheapest?

One common definition of "efficiency" for rocket engines is specific impulse, which is force times time (total change in momentum) per unit of fuel. This is based on the idea that what you really want out of a rocket engine is to apply as large of a force as possible for as long as possible and with as little fuel as possible.

An ion engine might have an efficiency of 80%, but who cares: what is really great is that it has a specific impulse of several times (up to about 10 times) that of a conventional rocket.

But, if reaction mass can be made smaller, maybe that's not what you really care about? You specifically mentioned electric power, so it may be instructive to browse the specs of some ion engines to see how lower thrust generally means lower power for the same impulse. Since the energy is free (presumably, solar), you make "better" use of your reaction mass by colleting more energy from the sun to apply to it over a longer period of time. But is that more "efficient" or less?
https://en.wikipedia.org/wiki/Ion_thruster

The motivation is to maximise the use of a fixed power supply (other details below), though my original question asked what you now ask me to specify, i.e. to define more precisely in what way the rocket should be efficient.

Here's the rocket specification: A rocket engine is provided with a constant power supply of W watts and its function is to accelerate a rocket payload of mass M to a speed V. The time to do it is unlimited and the initial propellent mass is also unlimited. The rocket must eject mass as the same speed as the rocket wrt the rocket however (zero speed relative to an inertial observer), which in turn means the rocket must be supplied with some initial momentum. This momentum would be some initial value that is also to be determined. What is the equivalent of the rocket equation for this rocket? It could be an unrealistic rocket but doesn't seem so to me at present, though it could well be impractical. From this I can compare it to the usual rocket equations and gauge efficiency of this versus say the standard rocket equations that you mention, with a constant and maximal exhaust velocity.

What I have discovered so far from numerical simulation: The momentum of the rocket remains constant during the flight, i.e. what was provided to it initially. Its acceleration remains constant. At the risk of using the word efficient, all the energy seems to end up in the rocket and none in the exhaust, so something interesting efficiency-wise may be occurring here. Is this not spookily interesting in it own right even if the rocket isn't efficient in any way? Its acceleration and time to reach a deltaV is dependent on initial conditions (power, initial mass and initial momentum). Propellent exhaust speed wrt to the rocket increases with time (with a constant acceleration) as does the rocket speed. The momentum supplied to the rocket decreases with time.

Since the acceleration is constant, we can easily work out the time to reach the target delatV. Likely all the final values and its performance characteristics can be determined easily in closed form, but I don't have a rocket equation yet.
 
  • #9
If you are throwing unburnt fuel out the back as reaction mass then you are not being as efficient as you should be.

If you constrain yourself to throwing reaction mass out the back at a zero ground-relative velocity then, for a rocket starting from rest relative to the ground, you will be throwing large quantities of reaction mass out the back.
 
  • #10
jbriggs444 said:
If you are throwing unburnt fuel out the back as reaction mass then you are not being as efficient as you should be.

My propellant isn’t standard rocket fuel, it’s just any mass that is in large supply. It could be moon dust for all I care (perhaps plasma engines will be able to eject that one day!).

Anyway I think I have answered my own question to my satisfaction now (though could be wrong), and the answer turned out to be simple:

Noting that the momentum of this rocket is conserved at all times: Mf * Vf = M0 * V0

[M = the total mass of the rocket at any time, V = Velocity of the rocket, subscripts f = final, 0 = initial)

Therefore Vf = (M0/Mf)*V0

So for example to get the rocket up to 30,000 m/s with a starting velocity of 10m/s, (M0/Mf) would have to be 3000! Not propellant efficient, but that wasn’t a limiting condition and is only one example where the starting velocity is low.

Similarly noting that my power supply is constant and that all the input energy ends up in the rocket (I am not confident about this assumption) and that energy = power * time, therefore the time to reach a particular mass ratio can be formulated. And from that an equation of motion, or a rocket equation, could be created I think. I’ve not created it yet though as am a bit unsure about my assumptions. A bit of a non-vigorous, non-standard approach, but oh well, it works for me for now.

A bit concerning that no one else seems impressed with the energy efficiency of this rocket, which indicates I am wrong about something somewhere, but thanks for your inputs.

[I ascertained that momentum remains constant by numerical simulation, but that could also be determined by noting that the exhaust trail, as viewed by an inertial observer, will be stationary. Ditto for energy all ending up in the rocket]
 
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  • #11
SteveR001 said:
30,000 m/s with a starting of 1m/s, (M0/Mf) would have to be 3000!
30,000 rather.
 
  • #12
jbriggs444 said:
30,000 rather.
Yes! I had corrected my post in the meantime and have added a few other details. Cheers.
 
  • #13
SteveR001 said:
no one else seems impressed with the energy efficiency of this rocket
This "rocket" design is realized currently by essentially all trains, cars and trucks. One uses the Earth as the reaction mass.
 
  • #14
jbriggs444 said:
This "rocket" design is realized currently by essentially all trains, cars and trucks. One uses the Earth as the reaction mass.

Which is handy, because if we tried to move around by ejecting high velocity exhaust gases, the electricity consumption would make the lights go out :) Where as my 'car rocket' would need to be pushed by hand to say 1 m/s and would weigh 30 times its final mass, but at least the lights would stay on.
 
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  • #15
The car-rocket I use masses about 3x1021 as much as the payload. I do not bother with a push start.
 
  • #16
SteveR001 said:
The motivation is to maximise the use of a fixed power supply (other details below)...

Here's the rocket specification: A rocket engine is provided with a constant power supply of W watts and its function is to accelerate a rocket payload of mass M to a speed V. The time to do it is unlimited...

[separate post]
A bit concerning that no one else seems impressed with the energy efficiency of this rocket, which indicates I am wrong about something somewhere, but thanks for your inputs.
Since the time and therefore energy available is infinite, why do we even care about efficiency?

Or from the other direction: I don't see the point of a rocket that doesn't have to go anywhere. Are you sure that's really the constraint you are after?
 
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  • #17
SteveR001 said:
A bit concerning that no one else seems impressed with the energy efficiency of this rocket, which indicates I am wrong about something somewhere, but thanks for your inputs.

[I ascertained that momentum remains constant by numerical simulation, but that could also be determined by noting that the exhaust trail, as viewed by an inertial observer, will be stationary. Ditto for energy all ending up in the rocket]

I see two problems here:

The first problem is that it shouldn't matter what the initial speed is. The kinetic energy is maximized if and only the delta-v is maximized (for motion in a straight line) and this should be independent of the reference frame chosen.

The second problem with this, is that the solution is only optimal if you have just the right famount of reaction mass and energy.
If you have more energy it seems obvious that you can only use that by accelerating the reaction mass to a higher speed, and if you have more reaction mass, it can't be right to not use some of this but there is simply not enough energy to accelerate it all to the required velocity.

It may be possible to resolve this by choosing the reference frame in such a way, that thrusting with an exhaust speed equal to your own speed exactly uses up both the available mass and energy.
 
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  • #18
willem2 said:
I see two problems here:

The first problem is that it shouldn't matter what the initial speed is. The kinetic energy is maximized if and only the delta-v is maximized (for motion in a straight line) and this should be independent of the reference frame chosen.

The second problem with this, is that the solution is only optimal if you have just the right famount of reaction mass and energy.
If you have more energy it seems obvious that you can only use that by accelerating the reaction mass to a higher speed, and if you have more reaction mass, it can't be right to not use some of this but there is simply not enough energy to accelerate it all to the required velocity.

It may be possible to resolve this by choosing the reference frame in such a way, that thrusting with an exhaust speed equal to your own speed exactly uses up both the available mass and energy.

Problems, great!

Unfortunately I only have time to address the first one, let me know if it's still there after I reply to that one.

Okay, let’s try an example. As the inertial frame I have been using leaves the propellant stationary after the burn, I’ll do the calculation as a single impulsive burn in that frame and then switch to another to see if everything all ties up, okay? The first frame is where a rocket has some small speed and the second frame I pick will be one where the rocket is initially stationary which I think is your preference. I won't use my formula on that one, I'll switch frames manually. It will probably need two diagrams to follow the below.

First frame: Initial mass (M0) =100 kg, speed (V0)= 1 m/s, final mass(Mf)= 10 kg. So these are the inputs to my little formula Vf = (M0/Mf)*V0, which comes to 10 m/s. So the momentum of the rocket before the burn was 100 kg * 1 m/s and momentum after = 10 kg * 10 m/s. We have used 100-10=90 kg propellant and in this frame the propellant has zero momentum and energy. The energy of the rocket in this frame was 50 J and is now 500 J. We used 450 J of energy.

Second frame: In this frame the rocket was initially stationary and after the burn is moving at 9 m/s, and is 10 kg afterwards as before. The rocket momentum before the burn was zero and after the burn is = 10 kg * 9 m/s. The propellant (90 kg as in the first case) was stationary before the burn and is now moving at 1 m/s, so its momentum = 90 kg * 1 m/s. Momentum is conserved. The energy of the rocket before the burn was 0 and after the burn is 0.5 * 10*9*9=405 J. The energy is the exhaust = 0.5 * 90 *1 * 1=45 J. Total energy = 450 J. Same as in the first case.

I'm not sure if I can get my head around creating a frame independent version of my formula at present, but switching frames manually seems to work out.
 
  • #19
To accelerate a 10kg mass to increase the velocity by 9 m/s by expelling mass, the energy spent decreases as the amount of ejected mass increases. The total energy output can be less that 450 J if the ejected mass is greater than 90 kg. Using the second frame of reference with an initial velocity of 0.

EM = Ejected Mass = 180 kg @ 0.5 m/s
Initial momentum = 0
Final momentum = 10 kg x 9 m/s - 180 kg x 0.5 m/s = 0
Initial energy = 0
Final energy = 1/2 x 10 kg x (9 m/s)^2 + 1/2 x 180 kg x (0.5 m/s)^2 = 405 J + 22.5 J = 427.5 J

EM = Ejected Mass = 360 kg @ 0.25 m/s
Initial momentum = 0
Final momentum = 10 kg x 9 m/s - 360 kg x 0.25 m/s = 0
Initial energy = 0
Final energy = 1/2 x 10 kg x (9 m/s)^2 + 1/2 x 360 kg x (0.25 m/s)^2 = 405 J + 11.25 J = 416.25 J

EM = Ejected Mass = 900 kg @ 0.1 m/s
Initial momentum = 0
Final momentum = 10 kg x 9 m/s - 900 kg x 0.1 m/s = 0
Initial energy = 0
Final energy = 1/2 x 10 kg x (9 m/s)^2 + 1/2 x 900 kg x (0.1 m/s)^2 = 405 J + 4.5 J = 409.5 J

Using the first frame of reference for the 900kg case:

Initial energy = 1/2 x 910 x 1^2 = 455 J
Final energy = 1/2 x 10 x 10^2 + 1/2 x 900 x .9^2 = 500 + 364.5 = 864.5 J
Increase in energy = 864.5 J - 455 J = 409.5 J

This is the same reason that gliders and other efficient aircraft use long wing spans: to accelerate a greater mass with a lesser change in velocity.

As for required power, the required power decreases as time increases. Say the constant power available is 450 watts applied for 1 second. Then the energy output is 450 J, and the example using 90 kg for the ejected mass in the prior posts would apply. Consider what happens if the time is doubled to 2 seconds. Now the energy output is 900 J, allowing for a much smaller mass to be used. In this case the ejected mass is 90/11 kg ~= 8.182 kg, and the velocity of the ejected mass is -11 m/s.

initial momentum = 0
final momentum = 10 kg x 9 m/s - 90/11 kg x 11 m/s = 0
initial energy = 0
final energy= 1/2 x 10 kg x (9 m/s)^2 + 1/2 x 90/11 kg x (-11 m/s)^2 = 405 J + 495 J = 900 J
 
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  • #20
willem2 said:
I see two problems here:

The first problem is that it shouldn't matter what the initial speed is. The kinetic energy is maximized if and only the delta-v is maximized (for motion in a straight line) and this should be independent of the reference frame chosen.

The second problem with this, is that the solution is only optimal if you have just the right famount of reaction mass and energy.
If you have more energy it seems obvious that you can only use that by accelerating the reaction mass to a higher speed, and if you have more reaction mass, it can't be right to not use some of this but there is simply not enough energy to accelerate it all to the required velocity.

It may be possible to resolve this by choosing the reference frame in such a way, that thrusting with an exhaust speed equal to your own speed exactly uses up both the available mass and energy.

I liked these insights so much that I have come back again.

I have given further thought to your point about reference frame dependence in my solution and I think one answer would be to introduce a second variable to hold the starting frame velocity, which is subtracted from the rocket velocity in the equation. Or indicate that the rocket velocity is relative to a frame the rocket was in prior to it being given its initial momentum, i.e. its starting frame, which is also the frame in which the exhaust velocity is zero.

willem2 said:
only optimal if you have just the right amount of reaction mass and energy...is simply not enough energy to accelerate it all to the required velocity
The propellent mass ejected per unit time isn't constant in my solution, but the power is constant. The reaction mass/s goes as 2*ConstPower/RocketVelocity^2, so the propellent can always be ejected with full power. The reaction mass/s gets smaller as it and the rocket get faster and lighter. A sample plot from a run using my rocket equation that I have created but won't post until it is tested:

deletelater.png
 
  • #21
SteveR001 said:
It will probably need two diagrams to follow the below.

Below is what I had in mind:
ImpulseEg.png
 

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  • #22
rcgldr said:
As for required power, the required power decreases as time increases. Say the constant power available is 450 watts applied for 1 second. Then the energy output is 450 J, and the example using 90 kg for the ejected mass in the prior posts would apply. Consider what happens if the time is doubled to 2 seconds. Now the energy output is 900 J, allowing for a much smaller mass to be used. In this case the ejected mass is 90/11 kg ~= 8.182 kg, and the velocity of the ejected mass is -11 m/s.

Thanks for your inputs!

I agree with what you say and we are getting close to answering my questions I think. And that is ultimately on whether there is anything special about an exhaust trail that contains zero momentum and energy along its entire path, wrt the rocket starting rest frame, and how such a rocket would compare to the usual [seemingly high energy] rocket equations. And secondly whether some benefit could be had from carrying and subsequently ejecting a greater mass of propellant at slower speed than usual when propellant is in large supply.

Your cases support the view that ejecting a large mass can be more energy efficient, and one way of looking at that is that minimal energy remains in the exhaust, no? This is the process I am interested in too, but in a rocket. A single impulsive burn using a mass the size of a planet is efficient, but might involve high accelerations howver. Admittedly I used one in my example above for another reason. My interest is to see if we can obtain the same advantage that we appear to agree on in a rocket (no momentum or energy left in the exhaust) and to see what cost this would incur in mass or other practical difficulties. I envisage power and acceleration would likely be low in any practical example of this rocket, if it is feasible at all, and that the engine would operate over a long period, and be far from impulsive.

In your penultimate paragraph an impulse is adjusted to match the ejected mass to the available energy, my algorithm similarly outputs a continuously changing impulse to match the constant power input over a period of time in order to leave a stationary exhaust trail over that period. In the rocket starting rest frame. I insert that last sentence because I am haunted by the comment made by willem2 regarding dependence on single frames of reference and I may be doing just that. At 0.1 second intervals my rocket mass with those parameters would go as : 100 , 53 , 36 , 27 , 22 , 18 , 16 , 14 , 12 , 11 , 10 , 9 , 8 , 8 , 7 , 7 , 6 , 6 , 6 , 6 , 5 kg and so on, for example. The final value (5kg) is actually less than the target payload of 10 kg, so some adjustment to the input parameters would be necessary to meet the constant power, initial momentum and mass ratio demands in that example. I haven’t got to stage of fixing parameters to a payload yet and there would likely be some restrictions on what is possible.

I’m not actually sure if anyone can help here aside from highlighting any obvious errors in my reasoning, as you have been doing, but I think I have all the tools I need now to investigate those myself if necessary, though obviously would prefer to have an answer rather spending time digging. It is a bit ominous that this proposal seems far from ‘textbook' however.
 
  • #23
SteveR001 said:
Your cases support the view that ejecting a large mass can be more energy efficient, and one way of looking at that is that minimal energy remains in the exhaust, no?
My examples are easiest to understand if using the "second frame" as the frame of reference, where the rockets initial speed is zero. The 10 kg rocket always ends up at 9 m/s with 405 J of energy, so the only variable is the amount of energy that the ejected mass ends up with, and this energy decreases as the ejected mass increases. The impulse is always the same, +90 kg m/s for the 10 kg rocket, - 90 kg m/s for the ejected mass. Assuming the time period is the same, then the Newton third law pair of forces have the same magnitude regardless of the amount of ejected mass. If the time is 1 second, then the magnitude of each of the forces is 90 Newtons.
 
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  • #24
rcgldr said:
My examples are easiest to understand if using the "second frame" as the frame of reference, where the rockets initial speed is zero. The 10 kg rocket always ends up at 9 m/s with 405 J of energy, so the only variable is the amount of energy that the ejected mass ends up with, and this energy decreases as the ejected mass increases. The impulse is always the same, +90 kg m/s for the 10 kg rocket, - 90 kg m/s for the ejected mass. Assuming the time period is the same, then the Newton third law pair of forces have the same magnitude regardless of the amount of ejected mass. If the time is 1 second, then the magnitude of each of the forces is 90 Newtons.

Yes, you are right. Nothing was making sense to me in the end, wrong frame. Back to the drawing board for me tomorrow. Cheers!
 
  • #25
rcgldr said:
My examples are easiest to understand if using the "second frame" as the frame of reference, where the rockets initial speed is zero. The 10 kg rocket always ends up at 9 m/s with 405 J of energy, so the only variable is the amount of energy that the ejected mass ends up with, and this energy decreases as the ejected mass increases. The impulse is always the same, +90 kg m/s for the 10 kg rocket, - 90 kg m/s for the ejected mass. Assuming the time period is the same, then the Newton third law pair of forces have the same magnitude regardless of the amount of ejected mass. If the time is 1 second, then the magnitude of each of the forces is 90 Newtons.

Hmm, another day and I’m back to thinking my ‘global frame’ may not be wrong after all! Could I trouble you further. First for ease of expression below can I define two frames, one called the ‘global frame’, which is the frame where the rocket started with zero velocity and a second frame called the ‘rocket frame’ where the rocket always has zero momentum. My preference is the first frame and your preference is the second, I think.

I see that you are a programmer, and if you were to program a rocket numerically would you not employ a global frame? Even if the calculations are done in the rocket frame, the result would have to be converted to a global frame at some stage? So a global frame shouldn’t really be too out of the ordinary or the cause of anything unphysical occurring? It’s just for ease?

The rocket frame is great for simple intuitive calculation for an impulse as you have shown, and your burn in the rocket frame shows a low exhaust speed in that frame, and in one example it was zero is the global frame too (-1 m/s in the rocket frame) which was promising. But the problem to my mind comes with subsequent burns in a rocket frame that is moving and accelerating in the global frame, because this is the one I am interested in. And the reason for that perhaps unusual perspective is because my question revolves around knowing if a zero exhaust speed in the global frame, along its entire path, is 100% energy efficient. A low exhaust speed in the rocket frame may not be a low speed in the global frame on subsequent iterations.

Are you able to do multiple burns in the rocket frame while keeping the exhaust speed at zero in the global frame? Or is that unphysical?
 
  • #26
My preference is for an inertial frame where the initial velocity of the rocket is zero. In this frame the rocket moves one way, the ejected mass the other way, but the center of mass of the rocket + mass never moves. In any inertial frame, the center of mass of the rocket + mass moves at constant velocity, I choose a frame where the constant velocity is zero. The "rocket" frame is an accelerating frame, which makes it more complicated to calculate momentum and energy changes.
 

Related to 100% energy efficient, constant power rocket

1. How does a 100% energy efficient, constant power rocket work?

A 100% energy efficient, constant power rocket works by using a combination of chemical reactions and physics principles to propel itself through the air. The rocket uses a fuel source, such as liquid hydrogen and oxygen, which is ignited to create a controlled explosion. This explosion produces hot gases that are forced out through a nozzle, creating thrust and propelling the rocket forward.

2. Is it possible to achieve 100% energy efficiency in a rocket?

While it is theoretically possible to achieve 100% energy efficiency in a rocket, it is not currently possible with our current technology. This is because some energy is always lost due to factors such as friction, heat transfer, and air resistance. However, scientists and engineers are constantly working to improve the efficiency of rockets.

3. How is constant power maintained in a rocket?

In order to maintain constant power in a rocket, the fuel supply must be regulated and controlled. This is usually done through a sophisticated system of valves and pumps that deliver the correct amount of fuel and oxidizer to the combustion chamber. Additionally, the shape and design of the rocket's nozzle play a crucial role in maintaining a constant flow of hot gases and therefore, a constant power output.

4. What are the advantages of a 100% energy efficient, constant power rocket?

The main advantage of a 100% energy efficient, constant power rocket is that it can achieve higher speeds and greater distances compared to less efficient rockets. This is because it is able to convert a larger percentage of its fuel into thrust, allowing it to overcome the resistance of Earth's atmosphere and travel further into space. Additionally, a more efficient rocket can carry heavier payloads, making it more practical for space exploration and commercial use.

5. What are some challenges in developing a 100% energy efficient, constant power rocket?

Developing a 100% energy efficient, constant power rocket is a complex and challenging task. One of the biggest challenges is finding ways to minimize energy loss due to factors such as heat transfer and air resistance. Additionally, developing the technology and materials to withstand the extreme temperatures and pressures involved in rocket propulsion is a major hurdle. There is also a significant cost and time commitment involved in designing, testing, and launching a rocket, which adds to the challenge of developing a highly efficient one.

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