A system with varying mass - a rocket

[Eus]

In Fundamental of Physics 6th ed. by Halliday, Resnick, Walker pg. 181-183, the section is about systems with varying mass: a rocket.
It is written:

Assume that we are at rest relative to an inertial reference frame, watching a rocket accelerate through deep space with no gravitational or atmospheric drag forces acting on it.
For this one-dimensional motion, let M b the mass of the rocket and v its velocity at an arbitrary time t (see Fig. a)
{~~~~~~~~~~~~~}--System boundary
{ |--------\ ... }
{ |... M ... > ---> v }
{ |--------/ ... }
{~~~~~~~~~~~~~}
------------------------------------------- x
(a)

{~~~~~~~~~~~~~~~~~~~~}--System boundary
{ -dM ... |--------\ ... }
{ <<<< --> U |. M+dM . > ---> v }
{ ....|--------/ ... }
{~~~~~~~~~~~~~~~~~~~~}
------------------------------------------- x
(b)
Figure b shows how things stand a time interval dt later.
The rocket now has velocity v+dv and mass M+dM, where the change in mass dM is a negative quantity.
The exhaust products released by the rocket during interval dt have mass -dM and velocity U relative to our inertial reference frame.

Is it okay if I change M+dM with M-dM because I know that the rocket loses its mass and -dM with dM because the exhaust product has mass in positive quantity?
So that figure (b) becomes
{~~~~~~~~~~~~~~~~~~~~}--System boundary
{. dM ... |--------\ ... }
{ <<<< --> U |. M-dM . > ---> v }
{ ... |--------/ ... }
{~~~~~~~~~~~~~~~~~~~~}
------------------------------------------- x
And dM is not a negative quantity anymore because I've taken account of it in the figure (the rocket loses its mass and the exhaust products gain mass).

Next it is written:

Our system consists of the rocket and the exhaust products released during interval dt.
The system is closed and isolated, so the linear momentum of the system must be conserved during dt; that is,
Pi = Pf ... (9-37)
where the subscripts i and f indicate the values at the beginning and end of time interval dt. We can rewrite Eq. 9-37 as
M * v = -dM * U + (M + dM) * (v + dv) ... (9-38)
where the first term on the right is the linear momentum of the exhaust products released during interval dt and the second term is the linear momentum of the rocket at the end of interval dt.

If I'm allowed to change M+dM with M-dM and -dM with dM then Eq. 9-38 becomes
M * v = dM * U + (M - dM) * (v + dv) ... (Eus-1)

Then it is written:

We can simplify Eq. 9-38 by using the relative speed vrel between the rocket and the exhaust products, which is related to the velocities relative to the frame with
(velocity of rocket relative to frame) = (velocity of rocket relative to products) + (velocity of products relative to frame).
In symbols, this means
(v + dv) = vrel + U, or
U = v + dv - vrel ... (9-39)
Substituting this result for U into Eq. 9-38 yields, with a little algebra,
-dM * vrel = M * dv ... (9-40)
Dividing each side by dt gives us
-(dM/dt) * vrel = M * (dv/dt) ... (9-41)
We replace dM/dt (the rate at which the rocket loses mass) by -R, where R is the (positive) mass rate of fuel consumption, and we recognize that dv/dt is the acceleration of the rocket. With these changes, Eq. 9-41 becomes
R * vrel = M * a ... (first rocket equation) ... (9-42)
Equation 9-42 holds at any instant, with the mass M, the fuel consumption rate R, and the acceleration a evaluated at that instant.
The left side of Eq. 9-42 has the dimensions of a force (kg m/s^2 = N) and depends only on design characteristics of the rocket engine, namely, the rate R at which it consumes fuel mass and the speed vrel with which that mass is ejected relative to the rocket. We call this term R * vrel the thrust of the rocket engine and represent it with T. Newton's second law emerges clearly if we write Eq. 9-42 as T = M * a, in which a is the acceleration of the rocket at the time that its mass is M.

If I do Eq. Eus-1 with little algebra it becomes
dM * vrel = M * dv ... (Eus-2)
Dividing each side by dt gives us
(dM/dt) * vrel = M * (dv/dt)
I replace dM/dt by R not -R because I've taken account of this decrement in rocket mass since the beginning.
Am I right?
Therefore, I got the first rocket equation without any problem that is R * vrel = M * a.

Now I run into a problem when finding the velocity of the rocket
It is written:

How will the velocity of a rocket change as it consumes its fuel? From Eq. 9-40 we have
dv = -vrel * (dM/M)
Integrating leads to
vi S vf dv = -vrel * [Mi S Mf (1/M) dM]
Note: I use S as integral symbol.
in which Mi is the initial mass of the rocket and Mf its final mass. Evaluating the integrals then gives
vf - vi = vrel * ln( Mi / Mf ) ... (second rocket equation) ... (9-43)
for the increase in the speed of the rocket during the change in mass from Mi to Mf.
(The symbol "ln" in Eq. 9-43 means the natural logarithm.)

My problem is that when I use Eq. Eus-2 as below
dv = vrel * (dM/M)
Integrating each side leads to
vi S vf dv = vrel * [Mi S Mf (1/M) dM]
And evaluating the integrals then gives
vf - vi = vrel * ln( Mf / Mi ) ... (Eus-3)
Yup! This is the problem. I cannot replace vrel by -vrel because I don't have any reason to do that.
Because Mf is always less than Mi, Eq. Eus-3's right hand side always has a negative value and that means vf is less than vi (the rocket is slowing down)
Mmmm... do you think I've done a big mistake by changing M+dM by M-dM and -dM by dM? Why is it wrong?
If I'm not wrong then is there something that I've taken account into my calculation when I calculate the second rocket equation? What is it?

Thank you very much for you help! ^_^
I really appreciate it! ^^v

 

[vincentchan]

a very long typing, how long did you take to type all this?
I am not going to type this long... my typing is slow

I think you have already noticed your mistake is on the sign only... you changed the M+dM to M-dM' (i added a prime in your dM to make thing looks nicer)
Here is your problem:
when you doing the integral since dM' = -dM, the limit of the integral changed:
$$\int_{b}^a dM' = \int_{b}^a (-dM) = -\int_{b}^a dM=\int_{a}^b dM$$

the negative sign in Eus-3 is due to the integral... vrel is fine...
if you carry a prime notation all the way in your calculatioin... you will see this problem had never been taken care of... even if you changed the sign a couple times

 

[Eus]

Hi Ho! ^_^

Mmm... about an hour ^^;;

Ummm... I know that this is all about the sign but what make me wonder actually is not the sign but the author's reason.
Why did he use M+dM when he knew that the rocket loses its mass?
If I were the author I would use M-dM because I know that the rocket loses its mass.
Here -dM is really a -dM not -dM' because the rocket loses its mass.

 

[vincentchan]

I used the prime notation because want to distinguish your dM from the auther's dM. sure the dM is the rocket mass... I can't say your dM=-dM... that doesn't make sense mathematically and conceptaully... One small question, do u understand my reasoning above?

if the auther use M-dM... while he do the integral in eq 9-43, he have to make the same argument------the limit of the integral must flip------ that will make this complicate calculation more complicated...

let me know do you understand my reasoning...

 

[Eus]

Mmmm... yup! I know exactly that dM is not equal to -dM
And I know your reason why you used dM'.
Wait a minute, I think I see the light.

Could you please check my story below?
A rocket initially has mass M and velocity v relative to an inertial reference frame.
During dt, it ejects exhaust products with mass dExhaust.
Therefore, at the end of dt, the rocket has velocity v+dv and mass M-dExhaust.
Whereas, the exhaust products have velocity u and mass dExhaust.
Then using the conservation of momentum,
M * v = dExhaust * u + (M-dExhaust) * (v+dv)
M * v = dExhaust * (v+dv-vrel) + (M-dExhaust) * (v+dv)
M * v = dExhaust * (v+dv) - vrel * dExhaust + M * v + M * dv - dExhaust * (v+dv)
0 = -vrel * dExhaust + M * dv
vrel * dExhaust = M * dv ... (1)
vrel * (1/M) * dExhaust = dv
Okay! So, dExhaust = -dM because ... ?
Could you please tell me why is dExhaust equal to -dM?
If I continue, I'll find the second rocket equation without any problem.
But I'll screw up the first rocket equation.

If dExhaust = -dM then ...
Back to equation (1)
vrel * dExhaust = M * dv
Dividing each side by dt gives us
vrel * (dExhaust/dt) = M * (dv/dt)
Because dExhaust = -dM,
vrel * (-dM/dt) = M * (dv/dt)
-vrel * R = M * a
Argh, I screw up the first rocket equation because R is positive.

Do you like Final Fantasy 7? ^^;;;

Ah, my problem arose when I saw M+dM in my textbook.
I don't have any problem when the author wrote v+dv because I know the rocket will gain additional speed by ejecting something.
Hehehe... why didn't I think that the rocket will slow down? Because in TV I see the rocket will gain speed ^^;;;
But I don't think that the rocket will gain additional mass.
It ejects something so it loses its mass.
Therefore, I have a problem with M+dM.

 

[jdstokes]

Hi Ho! ^_^

Mmm... about an hour ^^;;

Ummm... I know that this is all about the sign but what make me wonder actually is not the sign but the author's reason.
Why did he use M+dM when he knew that the rocket loses its mass?
If I were the author I would use M-dM because I know that the rocket loses its mass.
Here -dM is really a -dM not -dM' because the rocket loses its mass.

The equations work perfectly well if you assume dM > 0. In fact, I prefer the derivation your way. We assume that initially the rocket is moving with velocity $v$ and the fuel is ejected with velocity $U$, so
$\begin{align*} P_1 & = P_2 \\ Mv & = (M - \mathrm{d}M)(v + \mathrm{d}v) + U\,\mathrm{d}M. \end{align*}$

Expanding this out gives

$Mv = Mv + M\,\mathrm{d}v - v\,\mathrm{d}M - \mathrm{d}M\,\mathrm{d}v + U\,\mathrm{d}M$

The vector term $- \mathrm{d}M\,\mathrm{d}v$ is negligible, so

$Mv & = Mv + M\,\mathrm{d}v - v\,\mathrm{d}M - \mathrm{d}M\,\mathrm{d}v + U\,\mathrm{d}M.$

Substituting $U = v_\mathrm{ex} + v$ and solving for $M\,\mathrm{d}v$, before dividing both sides by $\mathrm{d}t$ gives


$M\frac{\mathrm{d}v}{\mathrm{d}t} = - v_\mathrm{ex}\frac{\mathrm{d}M}{\mathrm{d}t}.$

Separating the variables and integrating gives

$v_\mathrm{f} - v_\mathrm{i}= v_\mathrm{ex} \ln\frac{M_\mathrm{i}}{M_\mathrm{f}}.$

 

[vincentchan]

sorry, I don't play Final Fantasy...

Could you please tell me why is dExhaust equal to -dM?

it is easy... because the textbook define the mass of rocket is M+dM... the rocket is lossing mass... so dM is negative... but you define the rocket's mass as M - dExhaust... the dExhaust is positive... therefore

dExhaust = -dM

dExhaust equal to a negative (negative number) => dExhaust is positive... simple logic...

 

[Eus]

Hi Ho! ^_^

Although I'm still confused, thank you very much for all of your helps! ^^v
I really appreciate it! ^^