How to develop the rocket equation?

[Rick16]

I am referring specifically to chapter 3.2 in John Taylor's textbook Classical Mechanics where he develops the equation of motion for a rocket, but not in enough detail for me.

Taylor first writes down the change in the momentum of the rocket as dP = P(t + dt) - P(t) = (m + dm)(v + dv) - mv = mdv + vdm, where he neglected the small term dmdv. So far, so good.

Next he writes down the change in momentum of the expelled fuel. He defines the velocity v_ex of the expelled fuel, so that the velocity of the fuel relative to the ground or some other fixed point would be v - v_ex. Then he claims that the change in momentum of the expelled fuel would be -dm(v - v_ex) which is negative because the momentum of the fuel is opposite the momentum of the rocket.

Now my question: Shouldn't the change in momentum of the expelled fuel also depend on dv? And how could I develop the expression for dP of the fuel systematically, similar to how Taylor did it for the rocket? In a time Δt the mass of the expelled fuel would be Δm, and as Δt goes to zero, Δm would also go to zero. This would mean that it is not possible to determine a momentum of the expelled fuel at a specific instant t, and that only a change in momentum can be expressed. If this is true, then I cannot develop the expression for dP of the fuel as P(t + dt) - P(t). But how else can I do it? Taylor does not convince me because he does not mention what happens to dv. What exactly is going on here?

 

[kuruman]

Then he claims that the change in momentum of the expelled fuel would be -dm(v - vex) which is negative because the momentum of the fuel is opposite the momentum of the rocket.

I have Taylor in front of me and I don't see where he claims that. What I see is that -dm(v - vex) is the momentum (not the change) of dm in the equation for P(t + dt) just above equation (3.4).

This would mean that it is not possible to determine a momentum of the expelled fuel at a specific instant t, and that only a change in momentum can be expressed.

If you know dP and are looking for P(t), you can add a whole bunch of dP's starting at time t' = 0 when the momentum is zero and ending at time t' = t. Then the sum of all these changes (a.k.a. the integral over time) would be the P(t) that you are looking for.

 

[Rick16]

Thanks a lot. You solved my confusion. I wrongly assumed that the expression for the momentum of the expelled fuel was a change in momentum because of the term dm.

Could I just ask one more thing? I did not express myself right when I wrote

that it is not possible to determine a momentum of the expelled fuel at a specific instant t.

What I meant is: is it possible for the expelled fuel to have an instantaneous momentum? I would think not because the expelled fuel is not a concrete object and has no actual mass at anyone point in time.

 

[kuruman]

What I meant is: is it possible for the expelled fuel to have an instantaneous momentum? I would think not because the expelled fuel is not a concrete object and has no actual mass at anyone point in time.

The expelled fuel takes the form of gaseous combustion byproducts which have a distribution of momenta. In the absence of external forces and if the rocket starts from rest, the momentum of the center of mass of the gases will be equal in magnitude and opposite in direction to the momentum of the rocket at any time. That's required by momentum conservation.

 

[DrStupid]

Taylor first writes down the change in the momentum of the rocket as dP = P(t + dt) - P(t) = (m + dm)(v + dv) - mv = mdv + vdm, where he neglected the small term dmdv.

Why so complicate? Momentum is P = m·v and the total derivate is dP = m·dv + v·dm.

Shouldn't the change in momentum of the expelled fuel also depend on dv?

The already ejected reaction mass doesn't interact with the rocket anymore. Therefore the exhausted mass can always assumed to be zero - just as if the rocket engine has just been started. With m=0 you get dP = v·dm.

Of course you can also consider the momentum of the ejected mass. But in that case v is not the velocity of the currently leaving reaction mass but the velocity of the center of mass of the total exhaust. That would result in an overly complicated calculation without any benefit.

 

[kuruman]

Why so complicate? Momentum is P = m·v and the total derivate is dP = m·dv + v·dm.

Sure, that's always the case. However, here P is the total momentum of rocket plus ejected mass. One is concerned with finding expressions for each term and, to do that, one normally considers the rocket and the ejected mass separately. Taylor goes on to derive equation 3.6, $$m \dot v=-\dot m v_{ex}$$where $m$ = mass of rocket, $v$ = rocket velocity and $v_{ex}$ = velocity of exhausted gas relative to the rocket. How would you derive this equation without considering the ejected mass? Did I misunderstand what you are saying?

 

[DrStupid]

However, here P is the total momentum of rocket plus ejected mass.

That would result in dP=0 (due to conservation of momentum). P is the momentum of the rocket including the reaction mass that has not yet been ejected.

How would you derive this equation without considering the ejected mass?

Change of the momentum of the rocket: $\dot P_R = m_R \cdot \dot v_R + v_R \cdot \dot m_R$

Change of the momentum of the exhaust: $\dot P_E = m_E \cdot \dot v_E + v_E \cdot \dot m_E$

Conservation of momentum: $\dot P_R + \dot P_E = 0$

Conservation of mass: $\dot m_R + \dot m_E = 0$

Relative velocity or the currently expelled reaction mass: $v_{ex} : = v_R - v_E$

Already ejected reaction mass is not considered: $m_E = 0$

Everything together results in equation 3.6.

 

[Rick16]

Thank you for the answers. I am still thinking about this, it is still not 100% clear to me. The last approach using conservation of momentum and conservation of mass is very systematic, but I cannot convince myself to consider m_E being zero, and at the same time dm_E/dt being non-zero. Even if I try to keep the two concepts apart, considering m_E the already ejected reaction mass and dm_E/dt the change of the mass of the rocket, there still remains the problem that mathematically, one expression is the derivative of the other one, and how can the derivative of zero be non-zero?

 

[stevendaryl]

Thank you for the answers. I am still thinking about this, it is still not 100% clear to me. The last approach using conservation of momentum and conservation of mass is very systematic, but I cannot convince myself to consider m_E being zero, and at the same time dm_E/dt being non-zero. Even if I try to keep the two concepts apart, considering m_E the already ejected reaction mass and dm_E/dt the change of the mass of the rocket, there still remains the problem that mathematically, one expression is the derivative of the other one, and how can the derivative of zero be non-zero?

The problem with talking about $m_E$ is that the exhaust is not all traveling at the same speed. So the exhaust is not really a single object. But it's also true that the exhaust that the rocket expelled an hour ago is pretty much irrelevant to the motion of the rocket right now. The only thing that is relevant is how much exhaust is being expelled right now.

A way to think about it that gives you the right answer in the limit is to think of the fuel as composed of small little pellets, each with mass $m_E$. Every $\delta t$ seconds, the rocket throws a pellet out the back, at speed $v_E$ relative to the rocket. Then you can just use conservation of momentum applied to one single pellet:

Before:

The rocket (plus its remaining fuel) has mass $m_R$. It is traveling at speed $v_R$.

After:

The rocket (plus its remaining fuel) has mass $m_R - m_E$. It is traveling at speed $v_R + \delta v_R$.

The fuel pellet has mass $m_E$. It is traveling at speed $v_R - v_E$.

By conservation of momentum, $m_R v_R = (m_R - m_E) (v_R + \delta v_R) + m_E (v_R - v_E)$

So $\delta v_R = \frac{m_E v_E}{m_R - m_E}$

Now, divide through by $\delta t$ and write $m_E = \frac{- \delta m_R}{\delta t}$

$\frac{\delta v_R}{\delta t} = \frac{\frac{-\delta m_R}{\delta t} v_E}{m_R + \frac{\delta m_R}{\delta t} \delta t}$

If you take the limit as $\delta t \rightarrow 0$ while letting $\delta m_R/\delta t$ remain constant, this becomes:

$\frac{dv_R}{dt} = - \frac{v_E}{m_R} \frac{d m_R}{d t}$

 

[stevendaryl]

If you take the limit as $\delta t \rightarrow 0$ while letting $\delta m_R/\delta t$ remain constant, this becomes:

$\frac{dv_R}{dt} = - \frac{v_E}{m_R} \frac{d m_R}{d t}$

The important thing is that $m_E$ should not be taken to be the mass of ALL the exhaust since the beginning of time, but the exhaust during some small period of time $\delta t$, small enough that $m_E \ll m_R$.

 

[Rick16]

I posted this question some time ago, and I recently discovered a very good treatment of it in Shankar, Fundamentals of Physics, Volume I, page 135, which I want to share here in case someone else stumbles upon this thread looking for an answer.

The basic approach to the rocket equation is to write down the momenta of each component of the system (rocket and exhaust fumes) at time $t$ and at time $t+dt$.

At time $t$:
The rocket has momentum $P_{rocket}(t)=mv$.
The exhaust fumes have momentum $P_{ex}(t)=0$.

At time $t+dt$:
With its engines running, the rocket loses mass and gains velocity, so it should have momentum$P_{rocket}(t+dt)=(m-dm)(v+dv)$.
The exhaust fumes have momentum $P_{ex}(t+dt)=-v_{ex}dm$. Here $-v_{ex}$ is the velocity of the exhaust fumes relative to the rocket. Relative to some fixed origin (the same fixed origin that is used for the rocket velocity) the exhaust fumes have velocity (at $t+dt$) $v+dv-v_{ex}$. So the momentum of the exhaust fumes relative to the same origin as the one used for the rocket is $P_{ex}(t+dt)=(v+dv-v_{ex})dm$.

Then $dP=P(t+dt)-P(t)=(m-dm)(v+dv)+(v+dv-v_{ex})dm-mv$.

Unfortunately, this equation is not correct. The problem is with the $dm$. Taylor writes about this (Taylor, Classical Mechanics, page 85): "... at $t+dt$, the rocket's mass is $(m+dm)$, where $dm$ is negative..." This is the most confusing point for me, to write $+~dm$, when $dm$ should be negative.

Here is Shankar's method (slightly modified): You do not immediately write down the differential $dm$, instead you use $\delta m$ for the mass difference. The resulting expression for $dP$ is exactly as above, except that $\delta m$ is used instead of $dm$: $dP=(m-\delta m)(v+dv)+(v+dv-v_{ex})\delta m-mv$.

Now the crucial step to determine $dm$: $dm=m(t+dt)-m(t)=m-\delta m-m=-\delta m$. So $\delta m=-dm$. Using this result in the above equation: $dP=(m+dm)(v+dv)-(v+dv-v_{ex})dm-mv$ = $mv+mdv+vdm+dmdv-vdm-dvdm+v_{ex}dm-mv=mdv+v_{ex}dm$. Then the rocket equation is:$$mdv+v_{ex}dm=Fdt$$
Basically, this approach is the same as Taylor's (and those of others), but there is one additional step that I find crucial for my understanding.

 

[erobz]

It's funny how these topics seem to come in waves. Here is another way you might not see.

Group effort derived in:

https://www.physicsforums.com/threa...d-about-newtons-2nd-law.1050482/#post-6861856

$$\sum F_{ext} = \frac{d}{dt} \left( p_{\rm{rocket}} + p_{\rm{fuel}} + p_{\rm{ejecta}} \right) \tag{1}$$

Mass of the rocket is $M_r$, a constant

Let $M_f$ be the mass of the fuel carried along with the rocket.

It follows that the mass of fuel lost from the rocket is gained by the ejecta, thus:

$$ \frac{dM_e}{dt} = -\frac{dM_f}{dt} \tag{2}$$

The velocity of the ejecta in the rest frame $O$ is given by $v_{e/O} = v_r - v_{e/r}$

In the inertial frame the velocity of the ejecta $v_{e/O}$ is constant after being ejected at time $t$, however the momentum of the ejecta is accumulating over time such that:

$$ p_{\rm{ejecta}} = \int \frac{dM_e}{dt}( v_r - v_{e/r}) ~dt $$

Substituting for each contribution to the total momentum $(1)$ of the system (rocket + fuel + ejecta):

$$\sum F_{ext} = \frac{d}{dt} \left( M_r v_r + M_f v_r + \int \frac{dM_e}{dt}( v_r - v_{e/r}) ~dt \right)$$

Applying the derivative on the RHS:

$$\sum F_{ext} = \overbrace{\left( M_r\frac{dv_r}{dt} \right)}^{\rm {rocket}} +\overbrace{\left(M_f\frac{dv_r}{dt} + \frac{dM_f}{dt}v_r\right)}^{\rm{fuel}}+ \overbrace{ \frac{dM_e}{dt}( v_r - v_{e/r})}^{\rm{ejecta}} $$

Group terms and simplify substituting $(2)$:

$$\sum F_{ext} = \left( M_r + M_f \right) \frac{dv_r}{dt}+ \cancel{ \left( \frac{dM_f}{dt}v_r + \frac{dM_e}{dt}v_r \right)}^0 -\frac{dM_e}{dt} v_{e/r}$$

We are left with "The Rocket Equation":

$$ \sum F_{ext} = \left( M_r + M_f \right) \frac{dv_r}{dt} + \frac{dM_f}{dt} v_{e/r} \tag{Rocket Eq.}$$

 

[PeroK]

Here is Shankar's method (slightly modified): You do not immediately write down the differential $dm$, instead you use $\delta m$ for the mass difference.

Technically, the rocket must be propelled by a finite number of reactions involving expellant of a finite mass. The transformation to a continuous differential equation is something of an approximation, where we model the expellant as a flow of continuous matter.

 

[Rick16]

Technically, the rocket must be propelled by a finite number of reactions involving expellant of a finite mass. The transformation to a continuous differential equation is something of an approximation, where we model the expellant as a flow of continuous matter.

Do you mean this as a general statement, because in nature everything is quantized and continuity exists only in mathematics? I have actually been wondering about this point, about the interface between physics and mathematics in this specific case. Taylor repeatedly stresses (in his book Classical Mechanics) that in physics differentials are just shorthand for very small increments, but when he sets up the expression for the rocket momentum (page 85), he seems to distinguish between the physical quantity $-dm$ and the mathematical object $dm$. I wonder: Is it possible that Taylor writes $m+dm$ (instead of $m-dm$) for mathematical reasons?

 

[jbriggs444]

Do you mean this as a general statement, because in nature everything is quantized and continuity exists only in mathematics? I have actually been wondering about this point, about the interface between physics and mathematics in this specific case. Taylor repeatedly stresses (in his book Classical Mechanics) that in physics differentials are just shorthand for very small increments, but when he sets up the expression for the rocket momentum (page 85), he seems to distinguish between the physical quantity $-dm$ and the mathematical object $dm$. I wonder: Is it possible that Taylor writes $m+dm$ (instead of $m-dm$) for mathematical reasons?

Possibly you are not aware of how the notation works. The fact that a differential is negative does not require that one put a minus sign in front of it.

The form of an integral from high to low is exactly the same as an integral from low to high.$$\int_1^2 x^2 dx = - \int_2^1 x^2 dx$$The dx in the first is positive. The dx in the second is negative. No minus sign is required in the formula being integrated.

 

[PeroK]

Do you mean this as a general statement, because in nature everything is quantized and continuity exists only in mathematics? I have actually been wondering about this point, about the interface between physics and mathematics in this specific case. Taylor repeatedly stresses (in his book Classical Mechanics) that in physics differentials are just shorthand for very small increments, but when he sets up the expression for the rocket momentum (page 85), he seems to distinguish between the physical quantity $-dm$ and the mathematical object $dm$. I wonder: Is it possible that Taylor writes $m+dm$ (instead of $m-dm$) for mathematical reasons?

To add to what @jbriggs444 says, note that in the formal definition of a limit:
$$\lim_{h \to 0} f(x_0 + h)$$that $h$ can be positive or negative.

 

[Rick16]

Possibly you are not aware of how the notation works. The fact that a differential is negative does not require that one put a minus sign in front of it.

The form of an integral from high to low is exactly the same as an integral from low to high.$$\int_1^2 x^2 dx = - \int_2^1 x^2 dx$$The dx in the first is positive. The dx in the second is negative. No minus sign is required in the formula being integrated.

Indeed, I was not aware of this. I suspected it, but I did not know it. Now I know. Thank you very much.

 

[Rick16]

To add to what @jbriggs444 says, note that in the formal definition of a limit:
$$\lim_{h \to 0} f(x_0 + h)$$that $h$ can be positive or negative.

Things are becoming clearer, but one question remains: If a negative differential does not require a minus sign in front of it, does that mean that I am not allowed to put a minus sign in front of it? I cannot see anything wrong with writing $lim_{h \to 0} f(x_0-h)$. But if Taylor had written $m-dm$, he would not have got the correct solution. So, do I always must write the differential without a minus sign?

 

[PeroK]

In this respect, a differential is no different from a real number. A real number $x$ can be positive or negative. We don't write $-x$ when we want a negative number. Instead, we can simply say: Let $x < 0$.

This confuses some students. Take for example the symmetry of the cosine function, which can be expressed as$$\cos( x) = \cos(-x)$$Some students assume that $x$ here is positive and $-x$ is negative. But, that's not correct. If we let $x = -\frac \pi 4$, then that equation becomes specifically:$$\cos(-\frac \pi 4) = \cos(\frac \pi 4)$$

Similarly $dx$ or $\Delta x$ are infinitesimal and finite changes in $x$ that may be positive or negative. If the mass of a rocket is decreasing, we can still simply use $dx$ or $\Delta x$. A negative sign is neither necessary, nor correct.

 

[jbriggs444]

Things are becoming clearer, but one question remains: If a negative differential does not require a minus sign in front of it, does that mean that I am not allowed to put a minus sign in front of it? I cannot see anything wrong with writing $\lim_{h \to 0}f(x_0 - h)$. But if Taylor had written $m - dm$, he would not have got the correct solution. So, do I always must write the differential without a minus sign?

[Your post above was somewhat malformed with your own text buried in [QUOTE] tags. It was challenging to quote the result. But I think I've managed it]

I agree that $\lim_{h \to 0}f(x_0 - h)$ and $\lim_{h \to 0}f(x_0 + h)$ are equivalent.

You can chase that equivalence back to the epsilon/delta definition of a limit. There is an absolute value buried in there. Or, in slightly more general topological terms, a metric distance. In topology, the metric distance between two distinct things is always positive, regardless of the order in which those things might appear on a number line. The absolute value function $|x-y|$ is a familiar first example of the general notion of a metric distance.

Despite the equivalence, it would invite pointless confusion to use the version with the minus sign.

I do not want to say that it is never appropriate to put a minus sign in front of a differential. But I cannot think of an instance where it is a good idea.

It is worth noting that the "dx" that appears in an integral is as much an artifact of notation as a real numeric quantity. It is just a notationally appealing way to identify the variable that is being integrated over and ought not be taken literally as an actual mathematical operation. For instance$$\int (x^2 + y^2) dx$$integrates over x to produce a result which is a function which depends on the free variable $y$ while$$\int (x^2 + y^2) dy$$integrates over y to produce a result which is a function which depends on the free variable $x$.

 

[Rick16]

Thank you both very much for your detailed explanations. This point has troubled me for quite a while and your answers have clarified a lot for me.

 

[PeroK]

There's also a somewhat subtle point here. If $x$ stands for a specific quantity, then clearly you cannot interchange $x$ and $-x$. Sometimes, but not always, we use $x_0$ or $a$ in these cases. For example, for the roots ofa quadratic equation we can write either:$$x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{2a}$$or$$x_1, x_2 = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$In any case, here we are using $x$ as a specific value.

But, if we use $x$ as an arbitrary real number, then $-x$ is also an arbitrary real number. In other words, the following two sets are equivalent:$$\{x: x\in \mathbb R \} = \{-x: x \in \mathbb R\}$$Here we are using $x$ as a variable.

 

[Rick16]

Sorry, that I keep coming back to this. I looked at Taylor's solution again, and there is still a point that bothers me. At the bottom of page 85 in Classical Mechanics Taylor writes:

"At time $t$ the (rocket's) momentum is $P(t) = mv$. At short time later at $t+dt$, the rocket's mass is $(m+dm)$, where $dm$ is negative, and its momentum is $(m+dm)(v+dv)$."

So far, so good. $dm$ is a variable that can represent both positive and negative entities, so he writes $+dm$, even though the rocket's mass decreases. But then he continues:

"The fuel ejected in the time $dt$ has mass $(-dm)$ and velocity $v-v_{ex}$ relative to the ground. Thus the total momentum (rocket plus the fuel just ejected) at $t+dt$ is $P(t+dt)=(m+dm)(v+dv)-dm(v-v_{ex})$"

Why is it justified to use $(-dm)$ here? And why would the mass of the ejected fuel be negative in the first place? Because $dm$ is negative? Then why does he choose to mark it as negative for the fuel and not for the rocket?

This is no longer a question about how to find a solution. The method that I outlined under #11 is perfectly clear to me, and there are other ways to set this up. But I would also like to understand Taylor's reasoning, because as long as I don't understand it, I feel like I am missing something fundamental about how to set up equations of this type.

 

[Frabjous]

Sorry, that I keep coming back to this. I looked at Taylor's solution again, and there is still a point that bothers me. At the bottom of page 85 in Classical Mechanics Taylor writes:

"At time $t$ the (rocket's) momentum is $P(t) = mv$. At short time later at $t+dt$, the rocket's mass is $(m+dm)$, where $dm$ is negative, and its momentum is $(m+dm)(v+dv)$."

So far, so good. $dm$ is a variable that can represent both positive and negative entities, so he writes $+dm$, even though the rocket's mass decreases. But then he continues:

"The fuel ejected in the time $dt$ has mass $(-dm)$ and velocity $v-v_{ex}$ relative to the ground. Thus the total momentum (rocket plus the fuel just ejected) at $t+dt$ is $P(t+dt)=(m+dm)(v+dv)-dm(v-v_{ex})$"

Why is it justified to use $(-dm)$ here? And why would the mass of the ejected fuel be negative in the first place? Because $dm$ is negative? Then why does he choose to mark it as negative for the fuel and not for the rocket?

This is no longer a question about how to find a solution. The method that I outlined under #11 is perfectly clear to me, and there are other ways to set this up. But I would also like to understand Taylor's reasoning, because as long as I don't understand it, I feel like I am missing something fundamental about how to set up equations of this type.

Mass is conserved so you need dm+(-dm)=0 which is independent of the specific value of dm.

 

[Rick16]

Mass is conserved so you need dm+(-dm)=0.

Okay, that is one step forward. But why does he mark $dm$ as negative for the fuel and not for the rocket, when it is actually the rocket that loses the mass? Does it have something to do with the mathematics, i.e. that in an expression like $x+dx$ the differentials are always written as positive, and if they represent negative values, this fact has to be compensated for by marking them as negative elsewhere?

 

[PeroK]

Okay, that is one step forward. But why does he mark $dm$ as negative for the fuel and not for the rocket, when it is actually the rocket that loses the mass?

If $dm$ is negative, then $-dm$ is positive. You are confusing the negative sign with negative quantities.

If we followed you we would have to write, for the most general quadratic equation:
$$\pm ax^2 \pm bx \pm c = 0$$Where, according to you, $a,b,c$ are positive unless they have a negative sign in front of them. Not to mention, that you might want to restrict $x$ to non-negative values and use $-x$ when your variable is negative.

Mathematics doesn't work like that. A minus sign does not denote a negative quantity. It denotes the negative of a quantity. The constant $a$ may be positive or negative and the constant $-a$ may then be negative or positive.

 

[Rick16]

I will try to reformulate my question.

I need to arrive at the expression: $$P(t+dt)=(m+dm)(v+dv)-dm(v-v_{ex})$$

$dm$ appears twice in this expression, once with a plus sign and once with a minus sign in front of it. The question for me is: How do I decide where to put the minus sign?

How do I decide whether to write $P(d+dt)=(m+dm)(v+dv)-dm(v-v_{ex})$ or $P(d+dt)=(m-dm)(v+dv)+dm(v-v_{ex})$?

 

[Frabjous]

The way to write a differential is x+dx. So when you do the positive on the exhaust the variable is referring to the exhaust mass not the rocket mass.

 

[Rick16]

The way to write a differential is x+dx. So when you do the positive on the exhaust the variable is referring to the exhaust mass not the rocket mass.

But both masses vary. The rocket mass varies, and the exhaust mass varies. Why choose one over the other?

I have been wondering from the start if the reason to mark $dm$ positive for the rocket is due to the way how expressions like $x+dx$ are written. Is this the underlying reason, that it is generally not possible to write something like $x-dx$?

 

[jbriggs444]

But both masses vary. The rocket mass varies, and the exhaust mass varies. Why choose one over the other?

Since the exhaust does not have a single velocity, it will be more convenient to use the rocket mass.

 

[Frabjous]

I have been wondering from the start if the reason to mark dm positive for the rocket is due to the way how expressions like x+dx are written. Is this the underlying reason, that it is generally not possible to write something like x−dx?

Using the standard definition of calculus, x+dx is the way to write things. If you are willing to use another definition of calculus (i.e., a slightly different set of rules), you could write x-dx.

 

[Rick16]

In order to finally wrap this up, I want to try and present a detailed argument why the change in the rocket's mass has to be written $m+dm$ (and not $m-dm$).

First the physics: At some moment in time $t$ the rocket has mass $m$:
$$m(t)=m~~~~~(1)$$
In a certain time interval the rocket's mass changes by an amount $\Delta m$. Both $m$ and $\Delta m$ are positive, as masses usually are. Since the rocket's mass decreases when its engines are running, its mass at a later time is $m-\Delta m$:
$$m(t+dt)=m-\Delta m~~~~~(2)$$
Then the mathematics: The differential $dm$ of a variable $m(t)$ is defined as: $dm=m(t+dt)-m(t)$. Plugging in the expressions from eq. (1) and (2) gives:
$$dm=m(t+dt)-m(t)=m-\Delta m -m=-\Delta m$$
Now using this result in eq. (2): $m(t+dt)=m+dm$.

If the rocket's mass were increasing for some reason, I would get: $m(t+dt)=m+\Delta m$, $dm=m(t+dt)-m(t)=m+\Delta m-m=\Delta m$ and $m(t+dt)=m+dm$ -- the same result as for decreasing $m$.

So no matter what happens to the rocket's mass, whether it increases or decreases, I always get the expression $m+dm$ (and never $m-dm$). This is a direct consequence of the definition of the differential. I suppose this is obvious for more experienced people, but I had to go through this detailed procedure to convince myself.

There is one last point that bothers me: In eq. (2) I have $dt$ and $\Delta m$ in the same equation. Is this legitimate, or should I have either two differentials or two deltas? I have tried to use $m(t+\Delta t)=m-\Delta m$, but with this I ran into further difficulties.

 

[erobz]

Do you feel any better about the following where we let the final velocity of the rocket be $v$?

$$ \begin{aligned} \sum F ~dt &=( p + dp )- p \\ \quad \\
& = ( M_r + M _f ) v + dm (v-u) - \left( M_r + M_f + dm \right)\left( v - dv \right) \\ \quad \\
&= \cancel{( M_r + M _f ) v} + \cancel{dm v} - dm u - \cancel{( M_r + M _f ) v} + ( M_r + M _f ) dv -\cancel{dm v} + \cancel{dm~dv}^0 \\ \quad \\
&= -dm u + ( M_r + M _f ) dv \end{aligned}$$

$$ \sum F = -\frac{dm}{dt}u + ( M_r + M _f )\frac{dv}{dt} $$

And with $\frac{dm}{dt} = -\frac{dM_f}{dt}$ we have:

$$ \sum F = \frac{dM_f}{dt} u +( M_r + M _f )\frac{dv}{dt} $$

$$ \text{"The Rocket Equation"}$$

 

[sophiecentaur]

I would think not because the expelled fuel is not a concrete object and has no actual mass at anyone point in time.

I'm late to the party, perhaps but there must be mass involved for momentum to be transferred. "Concrete" is not an appropriate word here. Anything with mass fits the equations. The amount of mass to consider is the mass (dm) ejected in time dt. The "dt" is implied and the rate of ejection must be the same as the rate of mass reduction.

 

[Rick16]

I'm late to the party, perhaps but there must be mass involved for momentum to be transferred. "Concrete" is not an appropriate word here. Anything with mass fits the equations. The amount of mass to consider is the mass (dm) ejected in time dt. The "dt" is implied and the rate of ejection must be the same as the rate of mass reduction.

You quote what I wrote quite a while ago. I have since understood that the situation is exactly like you explain. My concern now is only to get the signs right.