# Rocket Ship Question

1. Nov 6, 2007

### 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.

2. Nov 6, 2007

### 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.

Last edited: Nov 6, 2007
3. Nov 6, 2007

### Submission1

Cool. Thanks.

4. Nov 6, 2007

### chronon

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.

5. Nov 6, 2007

### mgb_phys

Does that apply if it's accelerating ?

6. Nov 6, 2007

### D H

Staff Emeritus
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,

$$\Delta v = v_e\ln\left(\frac {m_\text{rocket}+m_\text{fuel}}{m_\text{rocket}}\right)$$

Suppose you find you have to load some rocket with a quantity of fuel $m_{\text{fuel}|1000\text{mph}}$ 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

$$m_{\text{fuel}|2000\text{mph}} = m_{\text{fuel}|1000\text{mph}}\left(2+ \frac{m_{\text{fuel}|1000\text{mph}}}{m_\text{rocket}}\right)$$

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

7. Nov 6, 2007

### mgb_phys

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

8. Nov 6, 2007

### chronon

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

9. Nov 6, 2007

### D H

Staff Emeritus
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.

Last edited: Nov 6, 2007
10. Nov 6, 2007

### D H

Staff Emeritus
Indeed. We just cross-posted.

11. Nov 6, 2007

### 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.

12. Nov 6, 2007

### Staff: Mentor

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.

13. Nov 6, 2007

### TVP45

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.

14. Nov 6, 2007

### D H

Staff Emeritus
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.

15. Nov 8, 2007

### 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 dont have to discuss about carrying the fuel or using Tsiolokovsky formula )

16. Nov 8, 2007

### 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?

17. Nov 9, 2007

### alvaros

Im very sorry, but I dont understand your english.

18. Nov 9, 2007

### TVP45

Alvaros,

19. Nov 10, 2007

### alvaros

I think my example is clear enough:

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

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 ).

20. Nov 10, 2007

### Staff: Mentor

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.

Last edited: Nov 10, 2007
21. Nov 10, 2007

### TVP45

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.

22. Nov 10, 2007

### Staff: Mentor

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.

23. Nov 11, 2007

### 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?

24. Nov 11, 2007

### alvaros

DaleSpam:
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 isnt any clear answer to the problem.

25. Nov 11, 2007

### D H

Staff Emeritus
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 $F$ 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 $F=ma$ is not valid here. We have to use $F=dp/dt$ instead.

$$F = \frac{dp_r} {dt} = \frac{d}{dt} ( m_r \, v_r ) = \dot m_r v_r} + m_r \, a_r$$

Solving for the rocket's acceleration,

$$a_r = \frac {F} {m_r} - \frac {\dot m_r} {m_r} v_r}$$

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 $\Delta v$. Things can be done more formally using continuum physics (classical treatment of gases), but that is a lot messier.

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

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

$$m_r(t+\Delta t) = m_r(t)-\dot m_f(t)\,\Delta t$$
$$\vec v_r(t+\Delta t) = \vec v_r(t)+\Delta \vec v_r(t)$$
$$\vec p_r(t+\Delta t) = (m_r(t) -\dot m_f(t)\,\Delta t)\, (\vec v_r(t)+\Delta \vec v_r(t))$$

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

$$\Delta m_e(t) = \dot m_f(t)\,\Delta t$$
$$\vec v_{e_{\text{inertial}}}) = \vec v_r(t)+\vec v_e(t)$$
$\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 $$\vec p_{r+e}(t+\Delta t) = (m_r(t) -\dot m_f(t)\,\Delta t)\, (\vec v_r(t)+\Delta \vec v_r(t)) + \dot m_f(t)\,\Delta t\, (\vec v_r(t)+\vec v_e(t))$$ Dropping the second-order term [itex]\Delta t \Delta \vec v_r(t)$ and simplifying,

$$\vec p_{r+e}(t+\Delta t) = m_r(t)\,\vec v_r(t)+ m_r(t) \, \Delta \vec v_r(t) + \dot m_f(t)\,\Delta t\, \vec v_e(t)$$

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 $\vec p_{r+e}(t+\Delta t)=\vec p_r(t)[/tex]: $$m_r(t)\,\vec v_r(t)+ m_r(t) \, \Delta \vec v_r(t) + \dot m_f(t)\,\Delta t\, \vec v_e(t) = m_r(t)\,\vec v_r(t)$$ or $$m_r(t) \, \Delta \vec v_r(t) + \dot m_f(t)\,\Delta t\, \vec v_e(t) = 0$$ Dividing by [itex]\Delta t$ and taking the limit $\Delta t \to 0$,

$$\frac {d\vec v_r(t)}{dt} = - \, \frac {\dot m_f(t)} {m_r(t)} \, \vec v_e(t)$$

This is the equation for the acceleration of the rocket at time $t$.

By conservation of mass, the time derivative of the rocket's mass is just the additive inverse of the fuel consumption rate: $\dot m_r(t) = -\dot m_f(t)$. 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 $t_0$ to some final time $t_1$ yields

$$\vec v_r(t_1) - \vec v_r(t_0) = \ln\left(\frac{m_r(t_1)}{m_r(t_0)}\right) \vec v_e$$

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.

Last edited: Nov 11, 2007