Rocket Ship Question: How Much Gas for 2000 MPH?

[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.

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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
KE_r(t) = 1/2 (m_r(t))(\vec v_r(t))^2

At the end of the interval the KE of the rocket is
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

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

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

By solving the conservation of momentum expression we obtain
\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)\,}

Substituting this into the KE expression and simplifying we obtain
\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)\,}

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

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 \Delta \vec v_r(t). 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 )

 

[D H]

Could anyone give a brief and clear explanation about what happens on the rocket ?

That is exactly what we just did. In post #25 I derived the rocket acceleration and the Tsiolokovsky equation. In post #30 I analyzed what it is involved in doubling a rocket's velocity. Dale did a nice job adding kinetic energy concerns to the analysis in post #32.

 

[DaveC426913]

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.

You can state your assumptions in the answer - but I would think that the spirit of the OP's question includes the following conditions at a minimum:

1] Ignore mass changes due to fuel consumption

2] Calculate speed wherein the rocket started off at rest wrt to the observer: s=0,t=0. (It seems to me, the OP actually explicitly stated this by saying "0 mph")

 

[TVP45]

You can state your assumptions in the answer - but I would think that the spirit of the OP's question includes the following conditions at a minimum:

1] Ignore mass changes due to fuel consumption

2] Calculate speed wherein the rocket started off at rest wrt to the observer: s=0,t=0. (It seems to me, the OP actually explicitly stated this by saying "0 mph")

If I understand your conditions, you are saying the observer remains in a non-accelerated frame (where the rocket used to be at 0 mph relative to that frame) and observes the rocket. Further, the rocket uses a propulsion system that does not decrease mass. And, we don't have any messy nonconservative stuff going on. Is this correct?

If so, there is clearly 3 times as much fuel required to go from 1000 mph to 2000 mph as was required to go from 0 mph to 1000 mph, all as measured from the non-accelerated observer's point of view.


This, I think, is the example often used by high school Physics teachers, and often explained incorrectly. It might be better to have the student calculate free-fall in a vacuum on a very large planet.

 

[D H]

The answer to the question "How much fuel does it a rocket take to go from 0 to 1000 MPH versus 1000 to 2000 MPH" is the same, more, or less, depending on what exactly is being asked. In this post I will ask the question in three ways.


1. All other things being equal, how much fuel does it a rocket take to go from 0 to 1000 MPH versus 1000 to 2000 MPH?
Answer: The same amount of fuel. "All other things being equal" means the same rocket loaded with the same quantity of fuel; the only difference is the initial velocity. The only things that matter are the change in velocity and the "mass ratio", the ratio of the total vehicle mass at the end to the total vehicle mass at the start. The delta-V is 1000 MPH in both cases and the mass ratios are the same thanks to the assumption of "all other things being equal.


2. A rocket goes from 0 to 2000 MPH. How much fuel did it take to go from 0 to 1000 MPH versus 1000 to 2000 MPH?
Answer: It took more more fuel to go from 0 to 1000 MPH than from 1000 to 2000 MPH. It takes a certain amount of fuel to go from 1000 to 2000 MPH. During the first part of the flight (0 to 1000 MPH) that extra fuel to achieve the final 1000 MPH of delta-V is just dead weight. The mass ratios for each leg (0 to 1000 MPH versus 1000 to 2000 MPH) of the journey must be equal to achieve the same 1000 MPH delta-V. The mass ratio for the second leg of the journey is
m_r/(m_r+\Delta m_f(leg 2))
and for the first leg, it is
(m_r+\Delta m_f(leg 2))/(m_r+\Delta m_f(leg 2)+\Delta m_f(leg1))
Equating these two ratios and simplifying,
\Delta m_f(leg1) = \Delta m_f(leg2)*(1+\Delta m_f(leg2)/m_r)


3. A rocket goes from 0 to 1000 MPH and then runs out of fuel. Assuming the fuel tanks can hold the extra fuel, how much more fuel needs to be added to make the rocket go from 0 to 2000 MPH?
Answer: The quantity of fuel needs to be doubled, and then some. In other words, for this version of the question it takes more fuel to go from 1000 to 2000 MPH than 0 to 1000 MPH. In this case, the rocket has the same final mass at the end of each mission. It is the initial fuel load that determines whether the rocket goes from 0 to 1000 MPH or 0 to 2000 MPH. Here the mass ratios are related by
2\ln(m_r/(m_r+m_f(1000 MPH))) = ln(m_r/(m_r+m_f(2000 MPH))).
Expanding this relation and simplifying,
m_f(2000 MPH) = m_f(1000 MPH)*(2+m_f(1000 MPH)/m_r)[/tex].<br /> The extra fuel needed to go the extra 1000 MPH is<br /> m_f(2000 MPH)-m_f(1000 MPH) = m_f(1000 MPH)*(1+m_f(1000 MPH)/m_r)

 

[Dale]

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

Here is the "Ritalin summary": A rocket ejects exhaust off the back, momentum is conserved, so the rocket moves forward. It really is that simple conceptually.

Here is a concrete example you could use for teaching after deriving the equation given by DH:

A rocket is initially at rest next to a space station in deep space. The rocket consists of 1000 kg of fuel and 100 kg of vehicle/payload. The exhaust velocity is 5 km/s.

After the rocket burns its fuel, what is its final velocity? (12 km/s)

What is the final momentum of the vehicle/payload? (1200 kg km/s)

What is the final total momentum of the exhaust gas? (-1200 kg km/s)

What is the average velocity of the exhaust gas? (-1.2 km/s)

Why is the average velocity of the exhaust gas not equal to the exhaust velocity? (The exhaust velocity is measured relative to the rocket, so as the rocket changes its velocity relative to the space station so does the exhaust it ejects. For example, at the end, when the rocket is moving at 12 km/s the exhaust gases are moving at 7 km/s relative to the space station, not -5 km/s)

You should stop there in a real class, but if one of your students asks about conservation of energy then you can derive my equations and show them that the final KE of the vehicle/payload is 7.2 GJ, the work done by the fuel is 12.5 GJ, and by conservation of energy the KE of the exhaust is 5.3 GJ.

 

[Dale]

there is clearly 3 times as much fuel required to go from 1000 mph to 2000 mph as was required to go from 0 mph to 1000 mph, all as measured from the non-accelerated observer's point of view.

No, this is wrong. See the derivations above by me and DH. You are trying to use conservation of energy but you are neglecting the KE of the exhaust.

The same amount of rocket fuel is required to move a given mass from 0 to 1000 mph as is required to move that mass from 1000 to 2000 mph. Use conservation of momentum!

 

[DaveC426913]

Maybe we should take a poll...
Highest votes = correct answer.

Physics by consensus...

 

[D H]

You can also use conservation of energy to derive an upper bound on the exhaust velocity. Chemical potential energy is often expressed in units of kilojoules per mole. Dividing by the molar mass yields another form of expressing the chemical potential energy, typically megajoules/kilogram. Well, joules/kilogram is just another way of saying (meters/second) squared. The square root of the chemical potential energy is the maximum exhaust velocity. An exhaust velocity higher than this limit would violate conservation of energy. In practice, the exhaust velocity will be less than this upper limit because some of the energy of combustion is wasted in the form of hot exhaust gas.

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I haven't mentioned this yet, but propulsion engineers often use something called specific impulse to characterize the efficiency of a rocket instead of the exhaust velocity. Specific impulse has units of time, and is directly proportional to the exhaust velocity via Isp = v_e / g_0, where g_0 = 9.80665 m/s^2, one Earth standard gravity.

 

[TVP45]

No, this is wrong. See the derivations above by me and DH. You are trying to use conservation of energy but you are neglecting the KE of the exhaust.

The same amount of rocket fuel is required to move a given mass from 0 to 1000 mph as is required to move that mass from 1000 to 2000 mph. Use conservation of momentum!

No. I was replying to a previous post and I said "If I understand your conditions...Further, the rocket uses a propulsion system that does not decrease mass" I understand your and DH's derivations (and I quite agree, for the conditions you cite, but I was trying to reach a clearer understanding of a previous question.

I believe there is a general confusion that occurs for three primary reasons:
(1) Most questions do not clearly state the observer's reference frame;
(2) Many questions assume rocket fuel has no exhaust;
(3) There is a facile skipping back and forth between dF/dt = 0 and dF/dx = 0 when talking about constant force.

I have seen this confusion dozens (hundreds?) of times and I think there is not a lucid explanation presented in most general physics courses.

Can anybody raise Arnold Arons from the dead to help? Did he ever look at this in any detail?

 

[Dale]

"If I understand your conditions...Further, the rocket uses a propulsion system that does not decrease mass" I understand your and DH's derivations (and I quite agree, for the conditions you cite, but I was trying to reach a clearer understanding of a previous question.

OK, point taken and caveat accepted.

However, any propulsion system that will work in space will have to work via conservation of momentum. So it will still be best to analyze it wrt conservation of momentum rather than energy. I think that this is the primary cause for confusion, trying to apply a conservation law when it does nothing to simplify the problem.

 

[TVP45]

OK, point taken and caveat accepted.

However, any propulsion system that will work in space will have to work via conservation of momentum. So it will still be best to analyze it wrt conservation of momentum rather than energy. I think that this is the primary cause for confusion, trying to apply a conservation law when it does nothing to simplify the problem.

Yes, you're right.

I think the reason so many teachers ask questions about rocket ships is that they think that's a good way to avoid friction, drag, and gravity, and of course they ignore the fact that rocket engines are fuel hogs. They might be better off just putting a streetcar on a frictionless track, with a 3rd rail, and low enough speed that drag doesn't screw it up and use that as an example.

 

[rcgldr]

The thrust from the rocket is due to the change in speed (acceleration) of the fuel. From a non-accelerating observer with the same initial velocity of the rocket, as fuel is used, the speed of both the rocket and it's remaining fuel increase. The amount of thrust generated is the same regardless of the velocity of the rocket. To a non-accelerating observer, power varies with the perceived velocity, since thrust remains constant (as long as the rate of fuel consumption is constant). When trying to calculate the work done, and change in KE, both the KE of the exhaust and the rocket (and it's remaining fuel) need to be taken into account.

Gravity provides a similar situation, the application of force is independent of the velocity of the objects involved, and only depends on the total mass and distance between the objects.

In the case of a car, the point of application is between car and pavement, and as the speed differential between car and pavement increases, more power is required to maintain the same force at the point of application. Since the engine of a car has a maximum power output, then in the ideal situation where a CVT (continuously variable transmission) is used to keep the engine running at maximum power, force at the driven tires will decrease linearly with speed.

 

[Dale]

They might be better off just putting a streetcar on a frictionless track, with a 3rd rail, and low enough speed that drag doesn't screw it up and use that as an example.

In the case of a car, the point of application is between car and pavement, and as the speed differential between car and pavement increases, more power is required to maintain the same force at the point of application.

I agree with both of you. Use rockets to teach conservation of momentum, use cars or trains (neglecting friction) to teach conservation of energy.

 

[TVP45]

I agree with both of you. Use rockets to teach conservation of momentum, use cars or trains (neglecting friction) to teach conservation of energy.

Thanks to all who responded to my comment. I now understand the apparent paradox and know how to reconcile it. Now all I need is a good linguist to help me explain it.