# Kinetic energy gained by a rocket

1. May 22, 2009

### phyzmatix

1. The problem statement, all variables and given/known data

Find the gain in kinetic energy when a rocket emits a small amount of matter. (Well, there's more, but the rest of the question is built around this initial problem)...

3. The attempt at a solution

My logic revolves around the total kinetic energy of the rocket after ejecting matter, which I took to be the kinetic energy before the emission added to the gain in kinetic energy. Furthermore, the mass of a rocket emitting a small amount of matter is reduced by dM = -dm and its velocity increases wth dv to (v+dv). If we let u = velocity at which matter is ejected, then after combining all of the above into a single equation we have:

$$K_{f}=K_{i}+K_{gain}$$
$$\frac{1}{2}(M-dm)(v+dv)^2=\frac{1}{2}Mv^2+\frac{1}{2}dm(v-u)^2$$

But this doesn't seem to be leading in the right direction and I personally believe there is a flaw in my reasoning.

So how am I supposed to set up this problem?

Thanks!
phyz

2. May 22, 2009

### TVP45

There are two conservation laws that seem to apply to such problems - energy and momentum. There can be great differences in ease of use and in understanding.

I would suggest making a little two panel sketch, "Before" and "After" and trying to identify everything in both. See if you can account for all mass, all velocities, all momenta, and all energy.

Then try telling yourself a "story" about what has happened.

A long, roundabout way to do this, but you're asking for a way to understand.

3. May 22, 2009

### phyzmatix

There is an example in my textbook based on conservation of momentum (a different type of question asking for the rocket's velocity that I basically adapted that example to get to what I put in the OP, obviously unsuccessfully so). At first I thought I could use the same approach to calculate the rocket's velocity and somehow get from there to the gain in kinetic energy, but I haven't had any luck there either.

Since we don't know anything about the potential energy (I assume for a rocket it's of a chemical nature), I'm not sure how I can incorporate it into an equation. How will it help?

4. May 22, 2009

### Cyosis

There is no external force on the system so could you write down the conservation of momentum equation and solve for v?

5. May 22, 2009

### D H

Staff Emeritus
Is that truly the problem, or is it your interpretation of the problem? There are two issues with using kinetic energy to analyze rocket behavior. First, kinetic energy isn't conserved. The goal of a rocket is to use potential energy (e.g., the chemical potential energy of the fuel to change the rocket's velocity. Moreover, some of that potential energy is wasted in the form of hot exhaust. The momentum of the rocket+exhaust system, on the other hand, is conserved.

Secondly, kinetic energy (and hence the change in kinetic energy) is a frame-dependent quantity. A rocket that is slowing down is losing kinetic energy while a rocket that is speeding up is gaining kinetic energy. For a given rocket, you can always find a frame in which the kinetic energy is decreasing. You can find another frame in which the kinetic energy is increasing. You can find yet another frame in which the instantaneous change in kinetic energy is zero. This final frame is the most useful frame for analyzing rocket behavior.

Bottom line: Analyzing rocket behavior from the point of view of energy is a bit challenging. It can be done, but it is not the easiest way to solve the problem.

6. May 22, 2009

### chiro

I think also that you have to take into account the potential energy of the rockets fuel. So essentially there is a conversion from potential energy into kinetic energy and you have to accordingly work with the transformation in this context.

7. May 22, 2009

### TVP45

You might consider, when he has solved this, giving him your long solution to this problem. That is the best explanation I've ever seen.

8. May 23, 2009

### phyzmatix

First of all, thank you for your input!

In answer to your question, that is the opening sentence. The complete question is:

Find the gain in kinetic energy when a rocket emits a small amount of matter. Hence calculate the total energy which must be supplied from chemical or other sources to accelerate the rocket to a given velocity. Show that this is equal to the energy required if an equal amount of matter is ejected while the rocket is held fixed on a test-bed.

$$\frac{1}{2}M_0u^2(1-e^{-v/u})$$

which I assume is the answer to the first part...

However, getting from the question to the answer has proven to be beyond my meagre abilities so far. From the input received so far I gather that I should rather focus on a conservation of momentum approach? But how?

9. May 24, 2009

### RaStevey

Phyz, we meet again

I am struggling with the same problem as phyz and also tried all different ways of reasoning, but none seem to make sense.

Finally I tried to reverse engineer the given answer and found:

$$\frac{1}{2}M_{0}u^{2}-\frac{1}{2}Mu^{2}$$

by taking into account that

$$M=M_{0}e^{-v/u}$$

which makes sense, since the change in kinetic energy is the initial kinetic energy minus the final kinetic energy, but why are both the components dependent on the initial velocity? Does that imply that the velocity doesn't change? And if the velocity does stay the same, doesn't that imply that there will be a loss of kinetic energy since the mass decreases and velocity stays the same and kinetic energy is dependent on the mass and velocity only.

If I were to calculate the change in kinetic energy I would calculate it as follows:

$$\frac{1}{2}M_{0}u^{2}-\frac{1}{2}Mv^{2}$$

where v is the final velocity, but this doesn't seem to yield the correct answer???

10. May 25, 2009

### D H

Staff Emeritus
Could you write the question exactly as written in your text?

11. May 25, 2009

### RaStevey

the whole question is exactly as phyz stated in his last post, it is problem 8.6 from the textbook "Classical Mechanics" by Kibble and Berkshire (5th edition) if you are perhaps fimiliar with that, but like i said, phyz did state the complete question in his last post.

12. May 25, 2009

### phyzmatix

So we do! Good to know I'm not the only one struggling!

Hi DH! Post #8.

13. May 25, 2009

### D H

Staff Emeritus
Part of the problem here is nomenclature. From looking at the book,
• The initial velocity is zero
• The rocket's final velocity is v
• The exhaust velocity, relative to the rocket, is u.

A bigger part of the problem is that you aren't looking at the whole picture. The question is asking for the "total energy which must be supplied ... to accelerate the rocket to a given velocity". Hint: Some of that energy is not being applied to the vehicle. The exhaust cloud trailing behind the vehicle has kinetic energy, and that kinetic energy came from fuel combustion.

(For other homework helpers, the problem is viewable at google books; see http://books.google.com/books?id=0a8dk0eDxgEC&pg=PA193&lpg=PA193)

14. May 25, 2009

### RaStevey

Ok, could you elaborate a bit more on your explanation please, I don't really get what your quoted post implies, which obviously shows that i don't understand the problem correctly...

15. May 25, 2009

### D H

Staff Emeritus
Look at the problem from the point of view of conservation of energy, which means you must find an isolated system. The fuel+rocket (potential energy + kinetic energy) does not form an isolated system. A lot of energy is going out the back end of the rocket.

Consider instead the fuel+rocket+exhaust system. This is an isolated system: conservation of energy applies. Suppose you calculate the change in the total kinetic energy of the rocket and the exhaust cloud after the rocket has achieved some velocity v. Thanks to conservation of energy, this must be equal to the amount of potential energy used to achieve that velocity. (Not quite, actually. Some of the chemical energy is converted to thermal energy. You can ignore that for this problem.)

16. May 26, 2009

### RaStevey

Ok, in the previous posts I mistakenly took "u" to be the initial velocity, since I'm use to a certain convention where "u" is the initial velocity, but in this case it is the exhaust velocity, sorry

Well, I think I'm beginning to understand a bit what's going on here.

The gain in kinetic energy due to the rocket emitting a small amount of matter only and not from added source, is given by:

(I still don't completely understand why this is the equation and why it is only determined by the exhaust velocity and not other velocity components)

Now you can calculate the total kinetic energy gained by the rocket as follows (this total kinetic energy gained is due to the rocket emitting small amounts of matter plus extra supplied kinetic energy from other sources.)

Consider the problem consisting of three different components:

1.) The exhaust cloud
2.) The fuel inside the rocket (this is the part of the total mass of the rocket which will be decreasing)
3.) The shell of the rocket (without fuel) (this is the part of the total mass of the rocket which will stay the same)

Each of these components have a mass and a velocity.

1.) For the shell of the rocket:

Initial mass: $$m_{r}$$
Initial velocity: $$v_{0}$$
Final mass: $$m_{r}$$
Final velocity: $$v_{0}+\Delta v$$

2.) For the fuel inside the rocket:

Initial mass: $$m_{f}$$
Initial velocity: $$v_{0}$$
Final mass: $$m_{f}-\Delta m$$
Final velocity: $$v_{0}+\Delta v$$

3.) For the exhaust cloud:

Initial mass: 0
Final mass: $$\Delta m$$
Final velocity: $$v_{0}+\Delta v-u$$

It can be seen that the initial and final velocity of the fuel and the rocket is the same, since they are moving together and the final velocity of the exhaust cloud is again an inertial frame, thus it has a component due to moving with the rocket ( $$v_{0}+\Delta v$$ ) minus the component at which it is emitting from the rocket, since it is emitted in the opposite direction to which the rocket is moving, to accelerate the rocket ( $$u$$ )

The initial kinetic energy is thus:

$$T_{0}=\frac{1}{2}m_{r}v^{2}_{0}+\frac{1}{2}m_{f}v^{2}_{0}$$

which is just due to the shell of the rocket and the fuel, since the exhaust cloud does not exist yet.

The final kinetic energy is thus:

$$T_{F}=\frac{1}{2}m_{r}\left(v_{0}+\Delta v\right)^{2}+\frac{1}{2}\left(m_{f}-\Delta m\right)\left(v_{0}+\Delta v\right)^{2}+\frac{1}{2}\Delta m\left(v_{0}+\Delta v-u\right)^{2}$$

$$T_{F}=\frac{1}{2}m_{r}\left(v^{2}_{0}+2v_{0}\Delta v+\Delta v^{2}\right)+\frac{1}{2}\left(m_{f}-\Delta m\right)\left(v^{2}_{0}+2v_{0}\Delta v+\Delta v^{2}\right)+\frac{1}{2}\Delta m\left(v^{2}_{0}+2v_{0}\Delta v-2v_{0}u+\Delta v^{2}-2\Delta vu+u^{2}\right)$$

$$T_{F}=\frac{1}{2}m_{r}v^{2}_{0}+m_{r}v_{0}\Delta v+\frac{1}{2}m_{r}\Delta v^{2}+\frac{1}{2}m_{f}v^{2}_{0}+m_{f}v_{0}\Delta v+\frac{1}{2}m_{f}\Delta v^{2}-\frac{1}{2}\Delta mv^{2}_{0}-\Delta mv_{0}\Delta v-\frac{1}{2}\Delta m\Delta v^{2}$$
$$+\frac{1}{2}\Delta mv^{2}_{0}+\Delta mv_{0}\Delta v-\Delta mv_{0}u+\frac{1}{2}\Delta m\Delta v^{2}-\Delta m\Delta vu+\frac{1}{2}\Delta mu^{2}$$

$$T_{F}=\frac{1}{2}m_{r}v^{2}_{0}+m_{r}v_{0}\Delta v+\frac{1}{2}m_{r}\Delta v^{2}+\frac{1}{2}m_{f}v^{2}_{0}+m_{f}v_{0}\Delta v+\frac{1}{2}m_{f}\Delta v^{2}-\Delta mv_{0}u-\Delta m\Delta vu+\frac{1}{2}\Delta mu^{2}$$

Thus the gain in kinetic energy is equal to the final kinetic energy minus the initial kinetic energy:

$$T_{F}-T_{0}=m_{r}v_{0}\Delta v+\frac{1}{2}m_{r}\Delta v^{2}+m_{f}v_{0}\Delta v+\frac{1}{2}m_{f}\Delta v^{2}-\Delta mv_{0}u-\Delta m\Delta vu+\frac{1}{2}\Delta mu^{2}$$

Now I know you can group terms together to reduce it more, but I didn't do it, because I don't really know which terms to group together to get the best result, because if I'm right you have subtract $$\frac{1}{2}M_0u^2(1-e^{-v/u})$$ from the above answer to get the energy that needs to be supplied by other sources, right?

Plus if the rocket is "held fixed on a test bed" does it imply that it is stationary and have a constant velocity of 0 or that it has a velocity, but can't accelerate?

17. May 26, 2009

### D H

Staff Emeritus
That is the answer to part 2, not to part 1.

Suppose the rocket is moving at some velocity v and spits out a small amount of mass with a velocity v-u relative to the rocket. This is going to make the rocket's velocity change by a small amount. You can find the change in velocity via conservation of momentum. Using your nomenclature, conservation of momentum says

$$(M-\Delta m)(v+\Deltav) + \Delta m(v-u) = Mv$$

or

$$M \Delta v - u \Delta m = 0$$

To compute the corresponding change in total kinetic energy, compute the difference between the kinetic energy of the system before and after the rocket spits this small chunk of mass. For the after state, you need to account for the energy in the rocket and in the exhausted piece of mass.

Be careful of your signs. You should get a whole lot of cancellation and you should come up with an incredibly simply expression.

18. May 26, 2009

### RaStevey

That is what i thought i did in my previous post

If i did make any mistakes, can you please show me on them, because i can't seem to find them

19. May 26, 2009

### RaStevey

I think I have it

Ok, here goes:

Firstly you should approach the problem from a point of view of infinitesimal changes.

So from the problem and the example on page 179 of the textbook (Classical Mechanics, 5th edition, by Kibble and Berkshire) you have that the rocket has a mass of M and it's moving at a velocity of v.

The rocket then emits a small amount of matter of mass dm at an exhaust velocity of u relative to the rocket. The effective velocity of the emitted matter will thus be v-u

Thus the mass of the rocket now becomes M-dm and the velocity of the rocket will now become v+dv, due to conservation of momentum.

The total kinetic energy before emission is thus:

$$T_{I}=\frac{1}{2}Mv^{2}$$

The total kinetic energy after emission can also be calculated as follows:

$$T_{F}=\frac{1}{2}dm\left(v-u\right)^{2}+\frac{1}{2}\left(M-dm\right)\left(v+dv\right)^{2}$$

Where $$\frac{1}{2}dm\left(v-u\right)^{2}$$ is the kinetic energy contained in the exhaust cloud and $$\frac{1}{2}\left(M-dm\right)\left(v+dv\right)^{2}$$ is the kinetic energy still contained in the rocket.

This can now be simplified as follows by neglecting all the second-order infinitesimals as well:

$$T_{F}=\frac{1}{2}dm\left(v^{2}-2vu+u^{2}\right)+\frac{1}{2}\left(M-dm\right)\left[v^{2}+2vdv+\left(dv\right)^{2}\right]$$
$$T_{F}=\frac{1}{2}dmv^{2}-dmvu+\frac{1}{2}dmu^{2}+\frac{1}{2}\left(M-dm\right)\left(v^{2}+2vdv\right)$$
$$T_{F}=\frac{1}{2}dmv^{2}-dmvu+\frac{1}{2}dmu^{2}+\frac{1}{2}Mv^{2}+Mvdv-\frac{1}{2}dmv^{2}-\frac{1}{2}dmvdv$$
$$T_{F}=-dmvu+\frac{1}{2}dmu^{2}+\frac{1}{2}Mv^{2}+Mvdv$$

The gained kinetic energy is thus:

$$T_{F}-T_{I}=-dmvu+\frac{1}{2}dmu^{2}+Mvdv$$

But as seen in the example in the textbook, the following relationship was deduced from the conservation of momentum equation:

$$Mdv=udm$$

Substituting this relationship for dv into the equation for the gained kinetic energy, gives the following result:

$$T_{F}-T_{I}=-dmvu+\frac{1}{2}dmu^{2}+Mv\left(u\frac{dm}{M}\right)$$
$$T_{F}-T_{I}=-dmvu+\frac{1}{2}dmu^{2}+dmvu$$
$$T_{F}-T_{I}=\frac{1}{2}dmu^{2}$$

Now it's known that dm is equal to the total initial mass of the rockket minus the total final mass of the rocket, thus:

$$dm=M_{0}-M$$

And also in the example in the textbook it was found that:

$$M=M_{0}e^{-v/u}$$

given that the rocket has an initial velocity of 0 and an initial mass of $$M_{0}$$, which is true for this problem, thus:

$$T_{F}-T_{I}=\frac{1}{2}\left(M_{0}-M\right)u^{2}$$
$$T_{F}-T_{I}=\frac{1}{2}\left(M_{0}-M_{0}e^{-v/u}\right)u^{2}$$
$$T_{F}-T_{I}=\frac{1}{2}M_{0}u^{2}\left(1-e^{-v/u}\right)$$

Please tell me if this is right and where I have gone wrong if it is not.

Now I just want to know how you determine the gain in kinetic energy only due to the rocket emitting the matter (part 1 of the problem) and how would you prove it's the same for a rocket fixed on a test-bed (part 3)

20. May 26, 2009

### D H

Staff Emeritus
You have a sign error but you have the correct result. You sign error results from your notation that dm is positive; you canceled this sign error by (incorrectly) saying dm=M0-M.

Fix your sign error and parts 1 and 2 are done. For part 3, assume all of the energy goes into sending exhaust out of the back end of the rocket (i.e., the rocket itself doesn't move).