Rocket Ship Question: How Much Gas for 2000 MPH?

[Submission1]

We are having a debate here at work and we need an answer from some smart folks (that's you).

If you have a rocket ship in space that takes 1 gallon of gas to reach 1000 mph then, negating friction, would it take less, more or the same amount to go from 1000 mph to 2000 mph.

Thanks,
Sub.

 

[mgb_phys]

Kinetic energy, the energy needed to move something depends on the speed squared.
So to travel at 2000 mph 4x as much energy as 1000mph. Since you started at 1000mph you will need to add 3x as much extra energy to get to 2000mph than you did to get to 1000mph.


You also have to take into account that the mass of the rocket is changing as you use up the fuel so it takes less energy to accelarate the rocket since it weighs less - if you doing this for real.

 

[Submission1]

Cool. Thanks.

 

[chronon]

Kinetic energy, the energy needed to move something depends on the speed squared.
So to travel at 2000 mph 4x as much energy as 1000mph. Since you started at 1000mph you will need to add 3x as much extra energy to get to 2000mph than you did to get to 1000mph.

But there's nothing special about the starting frame of reference. So it shouldn't take any more fuel to add an extra 1000mph starting in the 1000mph frame as to get to the 1000mph frame starting from the zero frame.

 

[mgb_phys]

Does that apply if it's accelerating ?

 

[D H]

This topic ws recently discussed in this thread.

The governing equation is the Tsiolokovsky rocket equation. Applying this equation to a rocket that burns all of its fuel,

[tex]\Delta v = v_e\ln\left(\frac {m_\text{rocket}+m_\text{fuel}}{m_\text{rocket}}\right)[/tex]

Suppose you find you have to load some rocket with a quantity of fuel [itex]m_{\text{fuel}|1000\text{mph}}[/itex] to make the rocket attain a velocity of 1000 mph after burnout. Atfer churning the crank on the rocket equation, the amount of fuel it would take to bring the rocket to 2000 mph is

[tex]m_{\text{fuel}|2000\text{mph}} =
m_{\text{fuel}|1000\text{mph}}\left(2+ \frac{m_{\text{fuel}|1000\text{mph}}}{m_\text{rocket}}\right)[/tex]

In other words, one must more than double the quantity of fuel to double a rocket's final velocity.

 

[mgb_phys]

This topic ws recently discussed in this thread.

Thanks - I'm away for a week and look what I miss.

 

[chronon]

In other words, one must more than double the quantity of fuel to double a rocket's final velocity.

Indeed. But more of that fuel will be burnt in getting it from 0 to 1000mph than from 1000mph to 2000mph.

 

[D H]

A little more detail regarding previous post:

The reason it takes more than double the fuel to go from 0 to 1000 mph than from 0 to 2000 mph (or whatever) is because the rocket carries its fuel.

Assume we have some rocket and want to make two test flights with it. The first test flight involves flying the rocket from rest to a final speed of 1000 mph, and the second, rest to 2000 mph. Both tests end with the rocket devoid of all fuel. The amount during the second test of fuel left in the rocket at the point the rocket reaches 1000 mph is exactly the same as the amount of fuel initially placed in the rocket for the first test. In this sense, it takes the same amount of fuel to go from 1000 to 2000 mph as it does from 0 to 1000 mph.

However, the rocket has to first achieve that 1000 mph during the second test. It is the first 1000 mph that costs more. At the start of the second test, the rocket comprises the dry mass of the rocket, the quantity fuel (call this f1) needed to bring the dry mass of the rocket from 1000 to 2000 mph plus the quantity of fuel needed to bring the dry mass of the rocket plus f1 from 0 to 1000 mph.

 

[D H]

Indeed. But more of that fuel will be burnt in getting it from 0 to 1000mph than from 1000mph to 2000mph.

Indeed. We just cross-posted.

 

[rcgldr]

Mentioned in the other thread is that there is work also peformed on the spent fuel as it is ejected behind the rockets engine. If the kinetic energy of both spent fuel, and the rocket (plus it's remaining unspent fuel) are summed, and if the rate of fuel consumption is constant, then thrust will be constant, and the sum of the kinetic energies of spent fuel, rocket (and remaining fuel) will increase linearly with time. This means that the power involved is constant (with a constant thrust).

Also from the other thread, it's easier to grasp this if you consider the rocket to be held in place so that all of the work done by the engine is to acclerate the spent fuel. The terminal velocity of the fuel is constant, and the rate of fuel consumption (mass ejected) is constant, so the kinetic energy of the fuel increases linearly with time.

 

[Dale]

From this thread and the other I have a kind of vague impression that rockets are better understood in terms of conservation of momentum than conservation of energy. I think that is because conservation of momentum requires you to think of the exhaust whereas conservation of energy does not.

 

[TVP45]

We are having a debate here at work and we need an answer from some smart folks (that's you).

If you have a rocket ship in space that takes 1 gallon of gas to reach 1000 mph then, negating friction, would it take less, more or the same amount to go from 1000 mph to 2000 mph.

Thanks,
Sub.

You always have to ask, 1000 mph relative to what? A rocket ship just traveling along at constant velocity will never know it. You always need a reference against which you measure your velocity. And, it is in that reference frame where it takes 3 times as much energy to go from 1000 mph to 2000 mph as it does from 0 to 1000 mph.

 

[D H]

From this thread and the other I have a kind of vague impression that rockets are better understood in terms of conservation of momentum than conservation of energy. I think that is because conservation of momentum requires you to think of the exhaust whereas conservation of energy does not.

That is how I prefer to look at things. A conservation of energy viewpoint is tough to do properly. The rocket fuel remaining in the rocket gains kinetic energy as the rocket accelerates. Moreover, some of the released chemical potential energy is wasted in the form of hot exhaust. Failing to account for either leads to erroneous results.

Conservation of linear momentum is a much simpler proposition. Adding conservation of angular momentum makes things a bit hairier. Adding the fact that in a real rocket the center of mass and inertia tensor change as the rocket burns fuel makes things a lot hairer. I wouldn't dream of attacking a non-point mass rocket with fuel located away from the center of mass with a conservation of energy perspective -- unless I was doing so at a micro level and using a CFD model.

The driving reason for investigating a problem from the point of view of any of the conservation principles is that doing so makes the answer fall out. I don't have to use a full-blown CFD model to get a very, very good model of what firing a roket engine does to the state of a vehicle. A full-blown CFD model that takes advantage of all of the conservation laws is needed to characterize behavior. However, CFD models cannot "see the forest for the trees". CFD models examine the spots on a beetle that sits on a tree in a huge forest. Once the behavior has been properly characterized, a conservation of momentum model does wonders.

 

[alvaros]

So the answer is ... ??
A 1000 kg rocket push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph ( respect to the Frame of Reference it was before starting the engine )
Again
A 999 kg rocket ( the same rocket ) push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph.

( I use very high energy fuel, so we don't have to discuss about carrying the fuel or using Tsiolokovsky formula )

 

[TVP45]

A kind of new thought on this thread...
This kind of question pops up a lot and there always seems to be (in the mind of the questioner) some kind of disconnect between common-sense examples and the equations.

Is the work-energy theorem the best way to teach people? I didn't get it for many years after that freshman class.

Is there another way?

 

[alvaros]

A kind of new thought on this thread...
This kind of question pops up a lot and there always seems to be (in the mind of the questioner) some kind of disconnect between common-sense examples and the equations.

Is the work-energy theorem the best way to teach people? I didn't get it for many years after that freshman class.

Is there another way?

Im very sorry, but I don't understand your english.
I keep asking: the answer is ...

 

[TVP45]

Im very sorry, but I don't understand your english.
I keep asking: the answer is ...

Alvaros,
I wasn't answering your reply. I was just wondering (wandering?) in general about this whole frame of reference question and how it ought to be presented in introductory physics.
What exactly is your example?

 

[alvaros]

What exactly is your example?

I think my example is clear enough:

A 1000 kg rocket push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph ( respect to the Frame of Reference it was before starting the engine )
Again
A 999 kg rocket ( the same rocket ) push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph.

It seems that the second kg of fuel gives more kinetic energy to the rocket than the first Kg ( ?? )

I was just wondering (wandering?) in general about this whole frame of reference question and how it ought to be presented in introductory physics.

I made a thread "what is a IFR" but everybody seems to have it very clear. I think the main mistakes are representing forces with an arrow -> and talking about IFR as some abstract.
In this problem the forces are between the rocket and the fuel ejected:
fuel <-> rocket
And the IFR is the center of mass of the system ( rocket + fuel ). The IFR always refers to something material ( with mass ).

 

[Dale]

It seems that the second kg of fuel gives more kinetic energy to the rocket than the first Kg.

Yes. Don't forget the KE of the exhaust. It is better to use conservation of momentum principles because they force you to consider the exhaust.

I am trying out some new (for me) applications of old ideas here here, so feel free to point out any mistakes in my logic:

Since the exhaust velocity is constant (assuming the mass of the rocket is large relative to the mass of the fuel) each kg fuel spent gives the same Δp and therefore the same Δv to the rocket.

Each successive kg with its Δv results in a larger ΔKE for the rocket since KE is proportional to v^2.

"But conservation of energy you protest!" The PE of the fuel does not only go into KE of the rocket, but also into the rather large KE of the exhaust.

Because the exhaust is going in the opposite direction of the rocket each successive kg of exhaust gains less KE, exactly compensating the more KE gained by the rocket so that the increased KE of the rocket-exhaust system is always equal to the PE in the spent fuel (neglecting heat).

Bottom line: for rockets always use conservation of momentum so that you cannot neglect the exhaust.

 

[TVP45]

Yes. Don't forget the KE of the exhaust. It is better to use conservation of momentum principles because they force you to consider the exhaust.

I am trying out some new (for me) applications of old ideas here here, so feel free to point out any mistakes in my logic:

Since the exhaust velocity is constant (assuming the mass of the rocket is large relative to the mass of the fuel) each kg fuel spent gives the same Δp and therefore the same Δv to the rocket.

Each successive kg with its Δv results in a larger ΔKE for the rocket since KE is proportional to v^2.

"But conservation of energy you protest!" The PE of the fuel does not only go into KE of the rocket, but also into the rather large KE of the exhaust.

Because the exhaust is going in the opposite direction of the rocket each successive kg of exhaust gains less KE, exactly compensating the more KE gained by the rocket so that the increased KE of the rocket-exhaust system is always equal to the PE in the spent fuel (neglecting heat).

Bottom line: for rockets always use conservation of momentum so that you cannot neglect the exhaust.

DaleSpam,
This is the sort of example I was asking about. If I were a high school teacher (thank G-d I'm not; that's a hard job), how would I present this example so that students can grasp it? I'm not being in any way critical, any more than I was of Alvaros; I am simply curious about how this can be developed conceptually.

 

[Dale]

If I were a teacher in a high-school level physics course I would not even mention conservation of energy in this problem. I would stick entirely with conservation of momentum. You can solve for everything of interest that way, and the conservation of momentum principle does what a good conservation principle should: it simplifies things.

As a teacher I would not introduce conservation of energy into the problem since it does nothing to simplify the problem. If a student asked I would tell them that energy is conserved, but it doesn't make the problem any easier. I think students would understand that.

 

[TVP45]

At first blush, I agree. Momentum is much easier to deal with than energy. Two questions occur to me:
(1) Can a student grasp the concept of momentum without being sidetracked by preconceptions about energy?
(2) Is the calculation of change of momentum rather than absolute momentum sensible?
Thanks for the reply.

 

[alvaros]

DaleSpam:

Because the exhaust is going in the opposite direction of the rocket each successive kg of exhaust gains less KE, exactly compensating the more KE gained by the rocket so that the increased KE of the rocket-exhaust system is always equal to the PE in the spent fuel (neglecting heat).

Wrong, the change in kinetic energy on the fuel is 1000 times the change in kinetic energy on the rocket.

Momentum is an abstract concept, its much easier the 3rd Newtons law: the burning fuel pushes the rocket and the rocket pushes the fuel fuel <-> rocket
( you see the double arrow <-> , all forces ( ? ) are double arrow )
( It seems, as I read here, that the conservation of momentum is more "universal" than 3rd Newtons law, but, in this case, both laws are the same and true )

But, at the end, there isn't any clear answer to the problem.

 

[D H]

Invoking Newton's third law on this problem is in a sense more ad-hoc and more abstract than using conservation of momentum with regard to this problem. The concept of force (Newton's second law) is very abstract; anything change to an object's momentum involves some force. Applying conservation of momentum is no less abstract than applying Newton's third law and yields a deeper answer.

Put the spaceship in deep space, far from any massive object, so that there are no measurable external forces acting on the vehicle. Now, what exactly is the force that is making the spaceship accelerate? By assumption, there are no external forces. You can say posit some ad-hoc force [itex]F[/itex] that results from burning the fuel and get an answer via Newton's second law. Note well: Since the rocket's mass is changing, the simpler form [itex]F=ma[/itex] is not valid here. We have to use [itex]F=dp/dt[/itex] instead.

[tex]F
= \frac{dp_r} {dt}
= \frac{d}{dt} ( m_r \, v_r )
= \dot m_r v_r}
+ m_r \, a_r
[/tex]

Solving for the rocket's acceleration,

[tex]a_r =
\frac {F} {m_r}
- \frac {\dot m_r} {m_r} v_r}
[/tex]

This is not very satisying. What exactly is this force? It's purely ad-hoc for one thing. Moreover, it will turn out that the force is frame-dependent.

Here is how things turn out from conservation of momentum point of view. For this derivation, I will use some math that mathematicians don't particular like but physicists use willy-nilly -- things [itex]\Delta v[/itex]. Things can be done more formally using continuum physics (classical treatment of gases), but that is a lot messier.

Nomenclature:
[tex]\vec v_r(t)[/tex] Rocket velocity at time [itex]t[/itex], inertial observer
[tex]\vec v_e(t)[/tex] Rocket exhaust velocity at time [itex]t[/itex], relative to vehicle velocity
[tex]m_r(t)[/tex] Rocket mass at time [itex]t[/itex]
[tex]\dot m_f(t)[/tex] Rate at which rocket consumes fuel at[itex]t[/itex]

Over a small time interval [itex]\Delta t[/itex], the rocket will eject a small mass of fuel [itex]\dot m_f(t)\,\Delta t[/itex]. At the start of the interval, the rocket has mass [itex]m_r(t)[/itex], velocity [itex]\vec v_r(t)[/itex], and momentum [itex]m_r(t) \, \vec v_r(t)[/itex]. At the end of the interval, the rocket has mass, velocity, and linear momentum

[tex]m_r(t+\Delta t) = m_r(t)-\dot m_f(t)\,\Delta t[/tex]
[tex]\vec v_r(t+\Delta t) = \vec v_r(t)+\Delta \vec v_r(t)[/tex]
[tex]\vec p_r(t+\Delta t) = (m_r(t) -\dot m_f(t)\,\Delta t)\, (\vec v_r(t)+\Delta \vec v_r(t))[/tex]

The bit of ejected fuel carries some momentum from the vehicle. The mass, inertial observer velocity, and momentum of the exhaust are

[tex]\Delta m_e(t) = \dot m_f(t)\,\Delta t[/tex]
[tex]\vec v_{e_{\text{inertial}}}) = \vec v_r(t)+\vec v_e(t)[/tex]
[itex]\Delta \vec p_e(t+\Delta t) =
\dot m_f(t)\,\Delta t\, (\vec v_r(t)+\vec v_e(t))[/tex]

The momentum of the rocket+exhaust at the end of the time interval is thus
[tex]\vec p_{r+e}(t+\Delta t) =
(m_r(t) -\dot m_f(t)\,\Delta t)\, (\vec v_r(t)+\Delta \vec v_r(t)) +
\dot m_f(t)\,\Delta t\, (\vec v_r(t)+\vec v_e(t))
[/tex]

Dropping the second-order term [itex]\Delta t \Delta \vec v_r(t)[/itex] and simplifying,

[tex]\vec p_{r+e}(t+\Delta t) =
m_r(t)\,\vec v_r(t)+
m_r(t) \, \Delta \vec v_r(t) +
\dot m_f(t)\,\Delta t\, \vec v_e(t)
[/tex]

Assuming no external forces act on the rocket and the ejected fuel during this time interval, the rocket and the ejected fuel form a closed system. Momentum is conserved in a closed system, so [itex]\vec p_{r+e}(t+\Delta t)=\vec p_r(t)[/tex]:

[tex]
m_r(t)\,\vec v_r(t)+
m_r(t) \, \Delta \vec v_r(t) +
\dot m_f(t)\,\Delta t\, \vec v_e(t) = m_r(t)\,\vec v_r(t)
[/tex]

or

[tex]
m_r(t) \, \Delta \vec v_r(t) +
\dot m_f(t)\,\Delta t\, \vec v_e(t) = 0
[/tex]

Dividing by [itex]\Delta t[/itex] and taking the limit [itex]\Delta t \to 0[/itex],

[tex]
\frac {d\vec v_r(t)}{dt} = - \, \frac {\dot m_f(t)} {m_r(t)} \, \vec v_e(t)
[/tex]

This is the equation for the acceleration of the rocket at time [itex]t[/itex].

By conservation of mass, the time derivative of the rocket's mass is just the additive inverse of the fuel consumption rate: [itex]\dot m_r(t) = -\dot m_f(t)[/itex]. If the relative exhaust velocity is a constant vector, both the left and right hand sides of the above acceleration equation are integrable. Integrating from some initial time [itex]t_0[/itex] to some final time [itex]t_1[/itex] yields

[tex]
\vec v_r(t_1) - \vec v_r(t_0) = \ln\left(\frac{m_r(t_1)}{m_r(t_0)}\right) \vec v_e
[/tex]

This is the Tsiolokovsky rocket equation. Since the final mass is smaller than the initial mass, the logarithm will be negative. The change in velocity is directed against the exhaust direction.

 

[Dale]

Wrong, the change in kinetic energy on the fuel is 1000 times the change in kinetic energy on the rocket.

Are you sure? Then where does the extra KE come from when you burn the second kg of fuel?

 

[D H]

OK, I finished now. Is this (post #25) what you guys are looking for?

 

[DaveC426913]

Ye, I still don;t see a concensus. Let's simplify the situation momentarily and ignore the change in mass due to fuel. Does it not take the same energy to go from 0 to 1000 as 1000 to 2000?

 

[TVP45]

Ye, I still don;t see a concensus. Let's simplify the situation momentarily and ignore the change in mass due to fuel. Does it not take the same energy to go from 0 to 1000 as 1000 to 2000?

In what reference frame? We have to agree how we are seeing this in order to get the same answer.

 

[D H]

All other things being equal, of course it takes the same energy to go from 0 to 1000 as 1000 to 2000. It takes a certain amount of fuel to go from 0 to 1000 MPH in a reference frame in which the rocket is initially at rest. Exact same circumstances, except now the observer sees the rocket with an initial velocity of 1000 MPH opposite the thruster direction. The final velocity will be 2000 MPH. The only thing that matters is the change in velocity.

More important is the amount of fuel required to attain a delta V of 1000 MPH versus a delta V 2000 MPH. A naive argument from a conservation of energy viewpoint says it should take four times as much fuel because [itex]E=1/2mv^2[/itex]. Applying the rocket equation says something quite different. I'll look at this two ways.

First, suppose we initially load the rocket with a fixed amount of fuel [itex]m_f[/itex] capable of bringing the rocket to a very high speed. Denoting the mass of the dry rocket plus initial fuel load as [itex]m_0[/itex] and the mass after the rocket has achieved some change in velocity [itex]\Delta v[/itex] as [itex]m_0 + \Delta m_f(\Delta v)[/itex], the rocket equation becomes

[tex]
\Delta v = \ln\left(1-\frac{\Delta m_f(\Delta v)}{m_0}\right) \vec v_e
[/tex]

The fuel required to double the [itex]\Delta v[/itex] is given by

[tex]
\ln\left(1-\frac{\Delta m_f(2\Delta v)}{m_0}\right) =
2 \ln\left(1-\frac{\Delta m_f(\Delta v)}{m_0}\right)
[/tex]

or

[tex]
1-\frac{\Delta m_f(2\Delta v)}{m_0} =
\left(1-\frac{\Delta m_f(\Delta v)}{m_0}\right)^2
[/tex]

Simplifying,

[tex]
\Delta m_f(2\Delta v) =
2 \Delta m_f(\Delta v) - \frac{\Delta m_f(\Delta v)^2}{m_0}
[/tex]

It takes less than twice the fuel to double the rocket's velocity, assuming the rocket is initially loaded with at least enough fuel to reach that doubled velocity.The above caveat begs the question, What if the rocket isn't initially loaded with the requisite amount of fuel to reach the doubled delta V? The second case investigates this problem. Suppose now the rocket is only loaded with enough fuel to achieve some final delta V. How much more fuel needs to be added to double this final delta V? In this case, it is the final mass that is constant. Denoting the mass of the dry rocket [itex]m_r[/itex] and the initial fuel loaded needed to achieve some final delta V as [itex]m_f(\Delta v)[/itex], the rocket equation becomes

[tex]
\Delta v = \ln\left(\frac{m_r}{m_r+m_f(\Delta v)}\right) \vec v_e
= -\, \ln\left(1+\frac{m_f(\Delta v)}{m_r}\right) \vec v_e
[/tex]

The fuel required to double the final [itex]\Delta v[/itex] is given by

[tex]
\ln\left(1+\frac{m_f(2\Delta v)}{m_r}\right) =
2 \ln\left(1+\frac{m_f(\Delta v)}{m_r}\right)
[/tex]

or

[tex]
1+\frac{\Delta m_f(2\Delta v)}{m_r} =
\left(1+\frac{\Delta m_f(\Delta v)}{m_r}\right)^2
[/tex]

Simplifying,

[tex]
m_f(2\Delta v) =
2 \Delta m_f(\Delta v) + \frac{\Delta m_f(\Delta v)^2}{m_r}
[/tex]

It takes more than twice the fuel to double the rocket's final velocity, assuming the rocket consumes all of the available fuel.

 

[D H]

Bottom line: The question "how much fuel is needed to go from 0 to 1000 MPH versus going from 1000 to 2000 MPH" is an ill-phrased question. If you want an unambiguous answer, ask an unambiguous question.

----------

Note well: The above analysis ignored a lot of details, all of which are very important in real rocket science. The rocket was implicitly assumed to be a point mass and was explicitly assumed to be traveling through empty space. Real rockets have thrusters located away from the center of mass, the center of mass moves as the rocket consumes fuel, the atmosphere adds drag, and perform less efficiently (lower specific impulse) in the atmosphere than in space.

 

[Dale]

Hi DH,

Thank you for the very well-worked and informative post. I think it is clear that conservation of momentum is (1) more fundamental than conservation of energy and (2) more general than f=ma. I really see no reason to bring energy considerations into the rocket equations, particularly for teaching purposes.

However, for the purpose of this thread and to demonstrate that energy is conserved I thought it would be useful to expand on your post and include KE terms.

At the beginning of the interval, the rocket has KE
[tex]KE_r(t) = 1/2 (m_r(t))(\vec v_r(t))^2[/tex]

At the end of the interval the KE of the rocket is
[tex]KE_r(t+\Delta t) = 1/2 (m_r(t)-\dot m_f(t)\,\Delta t)(\vec v_r(t)+\Delta \vec v_r(t))^2[/tex]

The KE of the exhaust is
[tex]KE_e(t+\Delta t) = 1/2 (\dot m_f(t)\,\Delta t)(\vec v_r(t)+\vec v_e(t))^2[/tex]

So the change in KE for the rocket and exhaust system is
[tex]\Delta KE(t+\Delta t)) = (KE_e(t+\Delta t)+KE_r(t+\Delta t))-KE_r(t)[/tex]

By solving the conservation of momentum expression we obtain
[tex]\Delta \vec v_r(t) = - \, \frac {\Delta t \,\vec v_e(t)^2 \,\dot m_f(t)\,}{m_r(t)-\Delta t \,\dot m_f(t)\,} [/tex]

Substituting this into the KE expression and simplifying we obtain
[tex]\Delta KE(t+\Delta t) = \frac {\Delta t \, m_r(t) \,\vec v_e(t)^2 \,\dot m_f(t)\,}{2\, m_r(t) - 2 \,\Delta t \,\dot m_f(t)\,}[/tex]

Dividing by [itex]\Delta t[/itex] and taking the limit [itex]\Delta t \to 0[/itex],
[tex]\frac {dKE(t)}{dt} = \frac {1}{2}\,\vec v_e(t)^2 \,\dot m_f(t)[/tex]

So, in the end the expression for the change in KE of the whole system (rocket and exhaust) is not a function of the rocket's velocity. It is, in fact, exactly equal to the KE of the exhaust in the rocket's comoving frame, which should have been expected from the beginning although it surprised me. This change in KE is then equal to the useable chemical energy in the fuel (chemical energy minus engine ineffeciencies such as waste heat), so energy is conserved.

Note, however, that although energy is conserved, energy conservation cannot be used by itself to solve this problem. Without conservation of momentum there is no way to determine [tex]\Delta \vec v_r(t)[/tex]. In other words, we know that some of the chemical energy goes into kinetic energy of the rocket and rest goes into kinetic energy of the exhaust, but without conservation of momentum there is no way to determine what fraction goes to each. So, since the problem can be solved by conservation of momentum alone, and since the problem cannot be solved by conservation of energy alone, the problem should be approached strictly with a conservation of momentum approach. Energy conservation can be demonstrated, but not used.

 

[rcgldr]

So, in the end the expression for the change in KE of the whole system (rocket and exhaust) is not a function of the rocket's velocity. It is, in fact, exactly equal to the KE of the exhaust in the rocket's comoving frame, which should have been expected from the beginning although it surprised me.

In my previous post, I mentioned the example of a rocket "held" in place, so that the only work done was to the exhaust (spent fuel). In this case it's clear that the KE is related to the exhaust, and extending this to the case where the rocket is moving and other frames of reference are used, the total KE will be the same since the source of the work being done is the same.

 

[Dale]

You are exactly right. I read your post about the stationary rocket, but the connection with the moving rocket didn't hit me until I did the math myself. For some reason I thought the moving-rocket case was fundamentally different from the stationary-rocket case. That is why I said that I was surprised by my own result but I should have expected it.

 

[alvaros]

I expect you are not teachers !
Could anyone give a brief and clear explanation about what happens on the rocket ?
( if possible with numbers )