I The rocket equation, one more time

[Rick16]

TL;DR

Trying to understand the setup of the rocket equation

I already posted a similar thread a while ago, but this time I want to focus exclusively on one single point that is still not clear to me.

I just came across this problem again in Modern Classical Mechanics by Helliwell and Sahakian. Their setup is exactly identical to the one that Taylor uses in Classical Mechanics: a rocket has mass m and velocity v at time t. At time $t+\Delta t$ it has (according to the textbooks) velocity $v + \Delta v$ and mass $m+\Delta m$. Why not $m - \Delta m$? This is all the more strange since just 2 pages further on Helliwell and Sahakian give the example of an open railroad boxcar moving in the rain, which accumulates a mass $\Delta m_r$ due to the rain falling in, and which loses mass $\Delta m_l$ due to water leaking out. Then they write "... at time $t+\Delta t$ ... boxcar of mass $M+\Delta m_r - \Delta m_l$, indicating that the boxcar has gained mass $\Delta m_r$ due to the falling rain, while losing mass $\Delta m_l$ due to the leak."

So in the case of the boxcar they write $m - \Delta m$ to indicate a decreasing mass, whereas in the case of the rocket they write $m + \Delta m$ to indicate the exact same situation. How can one make sense of this?

 

[PeroK]

$\Delta m$ is generally defined as the change in $m$. If the mass reduces, then $\Delta m$ is negative.

It's a common point of confusion to think that negative quantities must be subtracted. In general, we write $m + \Delta m$ regardless of whether $\Delta m$ is negative or positive.

 

[Rick16]

$\Delta m$ is generally defined as the change in $m$. If the mass reduces, then $\Delta m$ is negative.

It's a common point of confusion to think that negative quantities must be subtracted. In general, we write $m + \Delta m$ regardless of whether $\Delta m$ is negative or positive.

Ok, but why do they then write $m - \Delta m_l$ for the boxcar? I guess they do it to distinguish it from the increasing term $\Delta m_r$. Should I always write $m + \Delta m$, except when I have both an increasing and a decreasing term? But I don't see how this makes sense. If I indicate a decreasing mass by $-\Delta m$ in one case, it would seem that I always have do it like this. Isn't there a contradiction between the rocket case $m + \Delta m$ and the boxcar case $m +\Delta m_r - \Delta m_l$?

 

[kuruman]

Furthermore, what matters in variable mass problems is the rate of change $\Delta m/\Delta t$. Since $\Delta t$ is always positive, when the system loses mass (as in the case of a rocket) the ratio is negative; if it gains mass (as in the case of a moving container accumulating rainwater) the ratio is positive.

 

[Rick16]

Furthermore, what matters in variable mass problems is the rate of change $\Delta m/\Delta t$. Since $\Delta t$ is always positive, when the system loses mass (as in the case of a rocket) the ratio is negative; if it gains mass (as in the case of a moving container accumulating rainwater) the ratio is positive.

I am sorry, but I don't understand. In the rocket case $\Delta m$ is negative, and they write $+\Delta m$. In the boxcar case $\Delta m_r$ is positive and $\Delta m_l$ is negative, and they write $+\Delta m_r$ and $-\Delta m_l$. In one case a negative quantity is added, and in the other case a positive quantity is added and a negative quantity is subtracted. I don't see how I can reconcile the two cases.

 

[kuruman]

I am sorry, but I don't understand. In the rocket case $\Delta m$ is negative, and they write $+\Delta m$.

Not quite. If you look in the figure, they write $\Delta m =-|\Delta m|.$ This means that $\Delta m$ is negative. That is also explicitly mentioned in the text

I think you are confusing a symbol representing an algebraic quantity with its value. When you write $x+2=0$, you can see that $x$ is equal to $-2$, a negative number. Should you then write $-x+2=0$ because $x$ is negative? Of course not.

That said, I can see how the derivation in your book can be confusing when you have systems that exchange mass and momentum. You may wish to look at this article that tries to address this confusion by following an approach that departs a bit from the standard textbooks.

 

[PeroK]

I am sorry, but I don't understand. In the rocket case $\Delta m$ is negative, and they write $+\Delta m$. In the boxcar case $\Delta m_r$ is positive and $\Delta m_l$ is negative, and they write $+\Delta m_r$ and $-\Delta m_l$. In one case a negative quantity is added, and in the other case a positive quantity is added and a negative quantity is subtracted. I don't see how I can reconcile the two cases.

They can do what they like. As long as it adds up.

 

[Dale]

Should I always write

You can choose whichever convention you like. You just have to be careful to be clear about the convention you use and be careful to identify the convention that is used by an author.

Whatever convention you choose, it is important to be consistent within a single problem. But you are free to use different conventions in different problems. It looks like your book author does that.

 

[PeroK]

You can choose whichever convention you like. You just have to be careful to be clear about the convention you use and be careful to identify the convention that is used by an author.

Whatever convention you choose, it is important to be consistent within a single problem. But you are free to use different conventions in different problems. It looks like your book author does that.

As an example, imagine we have a financial spreadsheet. Let's say column A is the starting balance, column B is a deposit and column C is a withdrawal, then we have a choice.

1) We could enter the withdrawal as a negative number in column C and use the equation
$D = A + B + C$

2) We could enter the withdrawal as a positive number in column C and use the equation to: $D = A + B - C$

The choice is yours. Neither is more right or wrong than the other.

 

[Rick16]

The choice is yours. Neither is more right or wrong than the other.

This sounds very simple, but it does not work for me. If, for the rocket, instead of writing

$$(m+\Delta m)(v+\Delta v)-\Delta m(v-v_{gas})=mv$$

I write

$$(m-\Delta m)(v+\Delta v)+\Delta m(v-v_{gas})=mv$$

I find that the rocket slows down as its mass decreases. I guess there is something wrong with the second equation, but I don't see where the mistake is. Can somebody show me how to set up the equation using $m-\Delta m$?

 

[PeroK]

I can look at this later. In the meantime, what are the possible values of $v$ and $v_{gas}$?

 

[PeroK]

PS try $v =0$ and see what happens.

 

[Rick16]

I can look at this later. In the meantime, what are the possible values of $v$ and $v_{gas}$?

That is another tricky point. $v$ can be less than or greater than $v_{gas}$. One would think that one should set up 2 equations, one for each case. But this is not how it is done in textbooks.

 

[Rick16]

PS try $v =0$ and see what happens.

If $v=0$, nothing changes, because all the terms containing $v$ cancel each other anyway.

 

[PeroK]

I suspect that $v_{gas}$ is the problem.

That is another tricky point. $v$ can be less than or greater than $v_{gas}$. One would think that one should set up 2 equations, one for each case. But this is not how it is done in textbooks.

If you use velocities rather than speeds, then you can use a single equation.

Moreover, the velocity of the gas is relative to the rocket. Its speed is the modulus of the term you are using.

 

[Rick16]

If the rocket moves along the x-axis to the right, then $\Delta m(v-v_{gas})$ can move either to the right or to the left, depending on whether $v$ is less than or greater than $v_{gas}$. I would think that this should be reflected in the equation, but this is not my major concern here. My primary concern is now the sign of the deltas.

 

[PeroK]

I write

$$(m-\Delta m)(v+\Delta v)+\Delta m(v-v_{gas})=mv$$

I find that the rocket slows down as its mass decreases. I guess there is something wrong with the second equation, but I don't see where the mistake is. Can somebody show me how to set up the equation using $m-\Delta m$?

Let's look at this equation in the rest frame of the rocket ($v = 0$). We have:
$$(m-\Delta m)(\Delta v)+\Delta m(0-v_{gas})=0$$Which becomes:
$$\Delta v = \frac{\Delta m v_{gas}}{m - \Delta m} \approx \frac{\Delta m v_{gas}}{m}$$That look right. Assuming $v_{gas}$ is the speed of the gas expellant.

If we take an arbitrary rocket speed $v$, then we get the same final equation. As all the terms in $v$ cancel out.

So, I don't see the problem.

 

[PeroK]

PS if you want to take $v_{gas}$ as the velocity of the gas relative to the rocket, then you need $v + v_{gas}$ in that equation. In which case, $v_{gas}$ would be a negative quantity.

Or, to be more precise, it would be opposite the required acceleration of the rocket. If you wanted to use the expellant to slow the rocket down, then the gas would be fired in the same direction as the rocket's velocity, and $v_{gas}$ would be positive.

 

[Rick16]

The problem is, when I solve this equation, I get $v=v_0 +v_{gas}ln \frac {m}{m_0}$. Since $m$ is less than $m_0$, the logarithm is negative and consequently $v$ is less than $v_0$. When I use the approach from the book with $m+\Delta m$ the result is $v=v_0 +v_{gas}ln \frac {m_0}{m}$, as it should be.

 

[PeroK]

The problem is, when I solve this equation, I get $v=v_0 +v_{gas}ln \frac {m}{m_0}$. Since $m$ is less than $m_0$, the logarithm is negative and consequently $v$ is less than $v_0$. When I use the approach from the book with $m+\Delta m$ the result is $v=v_0 +v_{gas}ln \frac {m_0}{m}$, as it should be.

Okay, the problem must be with your solution to the differential equation.

 

[PeroK]

I suspect the problem is that, because you've taken $\Delta m$ to be positive when the rocket is losing mass, you have:
$$\frac{dm}{dt} \approx - \frac{\Delta m}{\Delta t}$$Where $m$ is the mass of the rocket.

 

[Rick16]

I suspect the problem is that, because you've taken $\Delta m$ to be positive when the rocket is losing mass, you have:
$$\frac{dm}{dt} \approx - \frac{\Delta m}{\Delta t}$$Where $m$ is the mass of the rocket.

I am afraid this does not convince me. $dm$ is just the limiting case of $\Delta m$. Where would the change of sign come from?

 

[A.T.]

you've taken $\Delta m$ to be positive when the rocket is losing mass,

$dm$ is just the limiting case of $\Delta m$.

If the above convention was indded used, then $dm$ is the limiting case of $-\Delta m$.

 

[Rick16]

If the above convention was indded used, then $dm$ is the limiting case of $-\Delta m$.

Could you elaborate a little on this?

 

[PeroK]

I am afraid this does not convince me. $dm$ is just the limiting case of $\Delta m$. Where would the change of sign come from?

You have no choice with the derivative: it's positive when a quantity is increasing. You have chosen the opposite convention for $\Delta m$. Quite explicity in your calculations $\frac{\Delta m}{\Delta t}$ is positive for a rocket that is losing mass.

Whereas, $\frac{dm}{dt}$ is negative for a rocket that is losing mass.

 

[Dale]

I write

When you write something then it is up to you to be clear in all of your sign conventions and the meaning of each variable.

If you do that and set up and solve the equations consistent with your conventions, then you will get valid answers.

 

[PeroK]

When you write something then it is up to you to be clear in all of your sign conventions and the meaning of each variable.

If you do that and set up and solve the equations consistent with your conventions, then you will get valid answers.

The problem that has arisen here is that if $\frac{dm}{dt}$ is positive for a rocket losing mass, then all the theorems of the differential and integral calculus need to be adjusted to reflect this change of convention!

Although, it really means that the choice of sign convention does not extend beyond finite quantities. And, in particular, it does not extend to the derivative, which has a single well-defined definition that implies a certain sign convention. Namely, positive is increasing and negative is decreasing. There is no other option.

 

[Rick16]

Although, it really means that the choice of sign convention does not extend beyond finite quantities. And, in particularm does not extend to the derivative, which has a single well-defined definition that implies a certain sign convention.

I think this comment will set me on the right track. But I will have to think about it. In any case, thanks a lot.

 

[Dale]

The problem that has arisen here is that if dmdt is positive for a rocket losing mass, then all the theorems of the differential and integral calculus need to be adjusted to reflect this change of convention!

Sure, that is part of “solve the equations consistent with your conventions”. I was just making a general statement. There is nothing wrong with picking a different convention, you just have to be clear at first and then be consistent from that point on.

 

[Rick16]

So, my equation with the deltas was correct. The mistake happened when I switched from $\Delta$ to $d$. That is something new. In all the situations that I have encountered so far, it didn't matter whether I wrote $\Delta$ or $d$. I just automatically changed from one to the other. Also, textbooks often treat the two as if they are interchangeable. It didn't even occur to me that I had to think about this point. I think I have learned something really important today.

 

[A.T.]

In all the situations that I have encountered so far, it didn't matter whether I wrote $\Delta$ or $d$.

In those situations $\Delta$ was apparently defined conveniently, to match the sign of the derivative.

For $\Delta y$ you always have a choice of which $y$ value is subtracted from which other $y$ value, and you can even force a positive sign by defining $\Delta y$ as the absolute difference.

For $\frac{dy}{dx}$ the sign this is determined by how $y$ changes increasing $x$.