# Homework Help: A rocket question

1. Jun 1, 2005

### Volcano

Hello,

I want to learn something but below question is better to learn for me. If, any help.

Q: A 6090kg space probe, moving nose-first toward Jupiter at 105 m/s relative to tthe sun, fires its rocket engine, ejecting 80kg of exhaust at a speed of 253m/s relative to the space probe. What is the final velocity of the probe?

for my opininon, this question must solve with,

Vf = Vi + Vrel * ln(Mi/Mf) ........................ (*)

but beside,

6090 * 105 = (6090-80)Vf + 80(Vf - 253) ................(1)

above where,

Vf = 105 + $${\Delta}$$ V

6090 *105 = (6090-80)V' + 80(105 - 253) ...............(2)

may solve this for a approx. solution. But which one will be more closer to (*) equation? I think it is (1) eq. but ...

Any idea?

Thanks

Last edited: Jun 1, 2005
2. Jun 1, 2005

### Dr.Brain

I think you are using the correct equation,But the basic equation is :

$F_ext = M \frac {dv}{dt} + V_r \frac {dM}{dt}$

Your equation is just the derived form of above equation.I think all the values are given in the question and its easy just put them up in the equation above , firther check your calculations and you will definitely get a correct answer.

Last edited by a moderator: Jun 2, 2005
3. Jun 1, 2005

### Volcano

Thank you Dr.Brain,

do you mean (1) equation? BTW, i am not familiar with Latex, but i suppose your eq. is kind of $$\frac{dm}{M} = \frac{dV}{Vrel}$$ eq. but with external force. If is, (1) must be true, is it?

4. Jun 1, 2005

### Dr.Brain

Yes your equations is the derived form of my equation where $F_e_x_t$
is absent . And I think you are going right about the question.Please tell about any difficulty you are facing in concept.Do the calculations yourself.

5. Jun 1, 2005

### Volcano

Thanks, yes calculations are myself. Honestly, above (2) equation is from a book which containing solves of the "Fundamentals of Physics". A Prof. and a Doc. has solved and printed in a book. But i confused with this solve and think to ask others. The question is, problem 48E from "4. Edition Extended".

6. Jun 1, 2005

### Staff: Mentor

Yes, this is the standard "rocket equation".
This looks like you are trying to apply conservation of momentum. But realize that the 80 kg of exhaust does not have a single speed, so this equation will not apply.

7. Jun 2, 2005

### Volcano

Hi Doc Al,

Thank you,
İ can not completely understand this. As i know exhaust speed is a characteristic for a rocket. But surely rocket speed is changing because of rocket thrust, so has an accelerate. But in this question we may approve, we are only dealing with the velocity difference.

BTW, in my opinion, for a sensitive solution, we must use (*) equation. But in my solutions book, this problem is solved with (2) equation. Anyhow, (*) equation is a more develop expression of (1) eq. Once again, what is your preferred equation for this problem.

Note: Please forgive my intervals between my replies. I can only use internet for approx. an hour in evenings.

Again thank you for replies.

Regards

8. Jun 2, 2005

### Staff: Mentor

My point was that if the exhaust were ejected at some finite rate (dM/dt), then an initial small piece of ejected fuel would have a different speed (with respect to an inertial frame fixed to the sun) than the final ejected piece since the probe does accelerate. Equations 1 and 2 both assume that the entire 80 kg of fuel is ejected as a unit with a single speed (relative to that inertial frame).

All three equations give pretty much the same answer for this problem. I agree that the rocket equation (*) is the better model for a real rocket thrust. Equation 1 underestimates the final speed of the probe; Equation 2 overestimates it. (The difference is in the speed assigned to the ejected fuel.) Take your pick.

9. Jun 3, 2005

### Volcano

Thank you Doc Al,

Yes, this is the point. Agree with this. Honestly, i was dealing with the same thing. But could not read an exact phrase about this. And also, the (2) equation is the original method to solve in the book (by a professor). So, for me hard to say some differente things about it. But was confused.

And a last help, after this problem i have researced this a lot and found lots of solve methods and finding rocket equation. But, for me, the best approach is; if we have

U : Exhaust velocity (relative to the sun)
Vrel : Exhaust velocity (relative to the rocket)
Vi : First velocity of rocket (relative to the sun)
Vf : Final velocity of rocket (relative to the sun)

first assume $$\Delta V = V_f - V_i$$ then,

$$U = (V_i + \Delta V) - V_{rel}$$ ....... (A)

or

$$U = V_i - V_{rel}$$ ............(B)

Surely either (A) or (B) needs differantial further. I suppose the right one is (A) eq. is it?

Thank you

10. Jun 4, 2005

### Volcano

Hi,

Maybe you wonder, what this guy want to get exactly with this questions?

If we want to solve a smilar problem like; say the last stage of a rocket traveling in space without fuel. The last stage is made up of two parts which are clamped together. Say, the clamp is released, a compressed spring causes to separate them. Now, if we deal with their final speeds,

Mr : Mass of rocket case
Mp : Mass of payload capsule
Vr': Final velocity of rocket case
U : Relative velocity of payload capsule (to rocket)
V : Initial velocity of both

$$V{^'}_r = V + \Delta V$$
$$(M_r + M_p) V = M_r V{^'}_r + M_p (V{^'}_r - U)$$

And if we expel fuel backward beside a rocket part, we must use rocket equation,

$$M \frac {dv}{dt} = - v_{ex} \frac {dM}{dt}$$

because rocket velocity is changing at acceleration as mentioned previously.

But for a model, must choice a relation equation. In my opinion, as above solution, the velocity of exhaust for an observer is,

[last velocity of rocket] - [relative velocity] ..........(i)

this is the same as the solution above. And same as the problem in my first post, as (1). But in some sources, it is,

[initial velocity of rocket] - [relative velocity] .........(ii)

and this is more look like the (2) solution in my first post. Surely i only need the simple, shared result, the (*) eq. And surely if there is external force, will need a little addition as Dr. Brain indicate. But i want to understand, not only using formulas to solving problems. Meanwhile some sources which find rocket equation;

As (i):

http://ffden-2.phys.uaf.edu/211.fall2000.web.projects/I. Brewster/physics.html
and also "Fundamentals of Physics".

As (ii):

and "Serway Physics" too.

both coming near same results but first approaches different. I can approve both, but first approachs complately different for macro solutions(Delta beside differantial). For example, if i solve the above question like (ii), will this be correct?

Note: Sorry friends who deal my problem so far. But i started to doubt many thing. So, please approve my insistent behaviour. I know, not easy to deal and help others. I also bored with dealing the same thing. Hope now my question(problem) is more clear. Hope my English is enough to explain.

Regards

Last edited by a moderator: May 2, 2017
11. Jun 9, 2005

### Volcano

Hi,

After my latest two posts, i have complately explained my question. The latest two posts are the reason of my question.

Everyday looking this thread with a lot of hope. Thank you everybody who interested in my question.

A latest help possible?

12. Jun 17, 2005

### Volcano

Hi again,

Before please accept my apologizes about insist on this question. This is the last time to ask this question.

If somebody think that, can not understand my latest question, feel free to ask. Or, somebody know a better place to ask or suggest any other thing, i am waiting for this. I could not understand why my latest questions is in the fix.

13. Jun 17, 2005

### OlderDan

I am hoping that I understand your question, and that this reply will be clear to you. As I understand it, the issue is a comparison between two different situations.

Case 1: A rocket ejects fuel gradually as it accelerates from some initial velocity to some final velocity

Case 2: A rocket is suddenly separated into two pieces that are thrust apart by an explosion or a spring.

Both of these cases conserve linear momentum. I am going to call u a positive number, and the direction the rocket moves the positive direction. In the first case, because the fuel is being ejected gradually with velocity -u relative to the rocket, the velocity of the fuel relative to some far off reference point like the sun is not all the same. For example, if the rocket starts from rest and eventually attains a velocity of 2u, the fuel ejected in the beginning will have a velocity of -u relative to the sun. When the rocket is already moving with velocity u relative to the sun, the ejected fuel will not be moving at all relative to the sun. When the rocket reaches a velocity of 2u relative to the sun, the ejected fuel will have velocity u relative to the sun, and will now be moving in the same direction as the rocket. The rocket leaves behind a trail of fuel that is all travelling at different velocities and the line of fuel keeps spreading out farther and farther as time goes on. In the example I have used, if the fuel was ejected at a constant rate, dm/dt, until the rocket achieved velocity 2u the center of mass of the ejected fuel would be stationary relative to the sun with the two ends of the trail moving opposite one another with speed u.

Because different parts of the fuel are ejected at different velocities relative to the sun, we have to look at what is happening at every instant in time and focus attention on a very small amount of fuel, all of which has the same velocity relative to the sun. That leads to the "rocket equation" where the the final velocity of the rocket is calculated as a sum (integral) of little changes in velocity (dv) that result from ejecting little bits of fuel of mass (dm) with velocity v - u relative to the sun.

Case 2 also involves conservation of linear momentum, but this case is different because the big piece of the rocket that is ejected has only one velocity relative to the sun. Conservation of momentum in this case involves only two objects, each of which has a mass, and an initial velocity and a final velocity. Before separation both objects have the same velocity, so calculating the initial momentum is easy. After separation the object left behind has only one velocity relative to the sun, and the object that continues has only one velocity relative to the sun.

Notice that if the ejected mass has velocity -u relative to the continuing mass, and that if the rocket starts from rest, the two objects will always be moving in opposite directions relative to the sun. A simple calculation shows that in this case the continuing portion of the rocket has to be moving with velocity less than u. Its speed will approach u as the ratio of its mass to the total mass approaches zero. Compare this to the case of a rocket that, according to the rocket equation, achieves velocity u by ejecting about 63% of its initial mass, and can achieve a velocity of 2u by ejecting less than 87% of its mass.

When you take into consideration the enormous amount of force that would be required to separate two large objects with relative speed u compared to the force required to eject a small amount of fuel at relative speed u, you will understand that there is a huge difference between the final speeds that can be attained by a rocket as compared to a single sudden explosion that ejects the same total amount of mass.

I hope this helps.

14. Jun 18, 2005

### Volcano

Thank you OlderDan,

I will read your post again. But obviously your exclamations and examples are the best i have seen so far. Again big thanks.

In fact so far, i approved the two cases which you explained above, after the helps of other friends. In the last stage i keep my mind on the reason of two different approachs to finding rocket equation. As i pointed my previos post, at least

http://ffden-2.phys.uaf.edu/211.fall2000.web.projects/I. Brewster/physics.html

and at least,

starting from different initiates.

According to first url, (vector diagram can also be seen)

$$U=V+dV-Vgas$$

but for second url,

$$U=V-Vgas$$

honestly i modified to imitate the symbols and for easily check.

Both has different approachs but the last equations are identical. I can not say: "Which one is correct one?". But can not completely understand this.

Last edited by a moderator: May 2, 2017
15. Jun 18, 2005

### OlderDan

I cannot read the equations of the second article. Apparently they are using something my browser does not support for the equations. However, I think I understand your question. It does not matter which way you write it. Remeber, these equations are all being written in the limit as differential quantities approach zero. dV is of no consequence compared to V and the gas velocity.

In my opinion, the first article should not have included the dV, but they did it so that the resulting dmdV product would cancel another dmdV product later on. They were careless in the use of dm and dM as well. But it does not matter. If you wind up with a dmdV product in an equation that otherwise involves only single differentials (no products) you would simply drop it.

Do you remember in calculus learning how to find derivatives using the delta process? When you did that and took the limit as delta x goes to zero, all the terms you still had with delta x in them were thrown away. The same thing is happening here.

Last edited by a moderator: May 2, 2017
16. Jun 20, 2005

### Volcano

Thank you OlderDan,

I don't know how will i thank to you and Dr. Brain, Doc Al too. These were great helps for me. Unfortunatly, i could not discuss this in my country and surroundings. Because of my language problems, had to read the same lines again and again for to be sure what i understand(if i could).

As i understood you did not prefered $$U = V + dV - Vgas$$ approach for initial. But if we have a problem, (a quote from above)
this problem looking like rocket+fuel question except there is solid parts which has single speed. So, call this rocket+rocket.

Is above equation ok for this problem?

If, is OK, for my opinion we only replaced a rocket part with the place of fuel. From now on we have to use rocket equation. But, above equation i used $$V{^'}_r = V + \Delta V$$. And this is including $$\Delta V$$ and if we accept this approach this delta will be $$dV$$ later(if use fuel beside rocket part). So, we will have $$dm dV$$ and will have to drop it later as you pointed before. But already this model is much suitable. Isn't it? Why we are giving up using $$\Delta V$$ (however will need to drop later)?

17. Jun 20, 2005

### OlderDan

I think you know this, but just to be sure, be aware that the U in the last scenario
is not the same thing as the U in the equation

$$U = V + dV - V_{gas}$$

In this equation for the rocket that is gradually ejecting fuel, U is the velocity of the gas as seen by the distant observer. In the latest scenario U is the relative velocity. It corresponds to the $$V_{gas}$$ in the equation just above.

In the last post I explained why it does not matter whether you used

$$U = V + dV - V_{gas}$$
or

$$U = V - V_{gas}$$

when you are considering differential changes in mass and velocity, but that I preferred the second form. I had not looked carefully enough at your equation of the rocket case-payload separation. This equation

$$(M_r + M_p) V = M_r V{^'}_r + M_p (V{^'}_r - U)$$

which, after substitution becomes

$$(M_r + M_p) V = M_r (V + \Delta V) + M_p (V + \Delta V - U)$$

needs to be interpreted carefully. First, we need to agree that it is the rocket case that is being left behind, and the payload capsule that is being thrust forward. In that case, $$\Delta V$$, as you have defined it, is a negative quantity. The rocket case is slowed down by the separation. The final velocity of the payload is greater than $$V{^'}_r$$ not less than $$V{^'}_r$$. The equation is OK, as long as U is also taken to be negative. If U is taken to be positive, which is implied by the way you have stated it, the equation should be

$$(M_r + M_p) V = M_r (V + \Delta V) + M_p (V + \Delta V + U) = M_r V{^'}_r + M_p (V{^'}_r + U)$$

I would prefer to change the meaning of $$\Delta V$$ to make it analogous to the dV of the rocket gradually ejecting fuel. I would have written

Mr : Mass of rocket case
Mp : Mass of payload capsule
Vp' : Final velocity of payload capsule
U : Relative velocity of payload capsule (to rocket)
V : Initial velocity of both

$$V{^'}_p = V + \Delta V$$
$$(M_r + M_p) V = M_p V{^'}_p + M_p (V{^'}_p - U)$$

Now $$\Delta V$$ is a positive quantity that represents the increase in velocity of the payload, and U is a positive quantity representing the relative velocity. After substitution this becomes

$$(M_r + M_p) V = M_p (V + \Delta V) + M_r (V + \Delta V - U)$$

After looking at this more carefully, I have to change my earlier opinion about the two differential forms of the equation. The preferred way to write that equation should be analogous to this last equation. That would be

$$U = V + dV - V_{gas}$$

where U in these two equations mean completely different things. The analogy is between the final velocity of the expended rocket case

$$V + \Delta V - U$$

and final velocity of the ejected fuel

$$V + dV - V_{gas}$$

where U is the positive relative velocity in the first equation and $$V_{gas}$$ is the positive relative velocity in the second case. It would be wrong to drop the $$\Delta V$$ in the first equation, so it is preferred to keep the $$dV$$ in the second equation.

18. Jun 21, 2005

### Volcano

I have read your latest post yet. And as i mentioned before, have to read again because of my language dififulty. But i suppose understand what you wrote. Maybe will ask one more time with my all shame. But not so far.

As i understand, your last post has everything i need. And now completely agree with you.

BTW, have you ever write your own physics book? It must. You are a real teacher.

Best Regards