I How to develop the rocket equation?

[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.

 

[Rick16]

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:

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

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

Now you confuse me with the $v-dv$. I mean, it seems to make sense. If the final velocity is $v$, then the initial velocity should be $v-dv$. But to me it also seemed to make sense to write $m$ for the initial mass and $m-dm$ for the final mass. If $m-dm$ is wrong, why is $v-dv$ correct?

You have $v(t+dt)=v$ and $v(t)=v-dv$. If you switch to $v(t)=v$, you get $v(t+dt)=v+dv$. This looks perfectly okay.

But if I do the same with the mass, I run into a problem. If I use $m(t+dt)=m$ and $m(t)=m+dm$, and I switch to $m(t)=m$, then I get $m(t+dt)=m-dm$. And this is not what I want to get.

 

[Rick16]

I want to come back to what I wrote in #32. It was inspired by Shankar’s solution in Fundamentals of Physics. But Shankar does not use $\Delta m$, instead he uses $\delta$. I had wondered why he did that. Perhaps I am beginning to see the reason now. When I write $m(t+dt)=m-\Delta m$, this $\Delta m$ suggests that it can be made randomly small and will eventually end up as $dm$. If instead I write $m(t+dt)=m-\delta$, things look better. $\delta$ is meant to represent a fixed quantity, not a changing quantity. I have to think some more about this, perhaps this is enough to convince myself.

All this certainly looks terribly pedantic, but I think you must be pedantic in physics, or else you end up in strange places.

 

[PeroK]

All this certainly looks terribly pedantic, but I think you must be pedantic in physics, or else you end up in strange places.

It's not pedantic to insist that $\Delta m$ is postive. It's simply wrong.

 

[Rick16]

It's not pedantic to insist that $\Delta m$ is postive. It's simply wrong.

$\Delta m$ is a mass, it is a certain amount of mass. How can it not be positive?

 

[weirdoguy]

For me Delta means difference between final and initial value (that is "final minus initial"). If final mass is less than initial then Delta m will be negative.

 

[Rick16]

For me Delta means difference between final and initial value (that is "final minus initial"). If final mass is less than initial then Delta m will be negative.

It means the same for me, but when I write $m-\Delta m$ and the result is less than the initial value, then $\Delta m$ is positive. If $\Delta m$ were negative, then I would subtract a negative value, which would mean that the result would be bigger than the initial value.

 

[vanhees71]

I think, this overcomplicates the issue more and more rather than to clarify it. It's simply a question of definition which sign you give your quantities. The physics is independent of this. To derive the rocket equation you simply write down the momentum balance of the rocket (=body of the rocket + fuel contained with mass $m(t)$ in it) and exhaust (fuel being exhausted from the rocket) within an infinitesimal time $\delta t$. If you assume that $\dd m_e>0$ is the mass of the exhausted fueld during this time the balance reads and the velocity of the exhausted fuel relative to the rocket is $v_e>0$, then the momentum balance reads
$$m(t) v(t)=m(t+\mathrm{d} t) v(t+\mathrm{d} t) +\mathrm{d} m_e (v-v_e),$$
i.e., expanding up to order $\mathrm{d} t$
[EDIT: corrected in view of #47]
$$m(t) v(t) = m(t) v(t) + [\dot{m} \mathrm{d} t] v(t) + m(t) \dot{v}(t) \mathrm{d} t+ \mathrm{d} m_e(v-v_e)+m g \dd t.$$
From mass conservation you have
$$\mathrm{d}m_e=-\mathrm{d} t \dot{m}$$
and thus finally
$$m \dot{v} + \dot{m} v_e=-m g.$$
So you need $m(t)$ and $v_e(t)$ and have an equation of motion for $v$.

 

[erobz]

Now you confuse me with the $v-dv$. I mean, it seems to make sense. If the final velocity is $v$, then the initial velocity should be $v-dv$. But to me it also seemed to make sense to write $m$ for the initial mass and $m-dm$ for the final mass. If $m-dm$ is wrong, why is $v-dv$ correct?

You have $v(t+dt)=v$ and $v(t)=v-dv$. If you switch to $v(t)=v$, you get $v(t+dt)=v+dv$. This looks perfectly okay.

But if I do the same with the mass, I run into a problem. If I use $m(t+dt)=m$ and $m(t)=m+dm$, and I switch to $m(t)=m$, then I get $m(t+dt)=m-dm$. And this is not what I want to get.

If you don't use $v-dv$ it doesn't seem to work out. That was my only motivation. I understand your frustration with the $\Delta$'s though.

Thats why I like #12. It doesn't use them. As a trade off, you have the integration that is immediately nullified by the derivative.

 

[Rick16]

If you don't use $v-dv$ it doesn't seem to work out. That was my only motivation. I understand your frustration with the $\Delta$'s though.

Thats why I like #12. It doesn't use them. As a trade off, you have the integration that is immediately nullified by the derivative.

Finally somebody understands my frustration. Thank you!

 

[erobz]

I'm going to give it one last try though. I think it solves the problem you (we) are having?

$$ \begin{aligned} \sum F~dt &= ( p + dp) - p \\ \quad \\

&=\overbrace{ ( M_r + M + dM )( v + dv ) + dm(v-u)}^{p+ dp} - \overbrace{( M_r + M)v }^{p} \\ \quad \\

&= \cancel{(M_r + M)v }+ (M_r + M)dv + dM~v + \cancel{dM~dv}^0 + dm~v - dm~u -\cancel{( M_r + M)v} \\ \quad \\

&= (M_r + M)dv +[ dM~v + dm~v ]- dm~u \end{aligned}$$

The change in mass of the rocket is opposite the change in mass of the ejecta, hence the bracketed term:

$dM~v + dm~v \equiv 0$

Subbing that in and dividing through by $dt$:

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

Changing from mass of the ejecta to mass of the fuel:

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

we are left with the Rocket Equation...No subtracting of negative differential masses or velocities that I can see.

 

[Rick16]

I'm going to give it one last try though. I think it solves the problem you (we) are having?

$$ \begin{aligned} \sum F~dt &= ( p + dp) - p \\ \quad \\

&=\overbrace{ ( M_r + M + dM )( v + dv ) + dm(v-u)}^{p+ dp} - \overbrace{( M_r + M)v }^{p} \\ \quad \\

&= \cancel{(M_r + M)v }+ (M_r + M)dv + dM~v + \cancel{dM~dv}^0 + dm~v - dm~u -\cancel{( M_r + M)v} \\ \quad \\

&= (M_r + M)dv +[ dM~v + dm~v ]- dm~u \end{aligned}$$

The change in mass of the rocket is opposite the change in mass of the ejecta, hence the bracketed term:

$dM~v + dm~v \equiv 0$

Subbing that in and dividing through by $dt$:

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

Changing from mass of the ejecta to mass of the fuel:

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

we are left with the Rocket Equation...No subtracting of negative differential masses or velocities that I can see.

This is pretty great. At first sight your solution seems to be the same as Taylor’s, with the only difference that instead of writing $dm$ for the change in the rocket mass and $-dm$ for the change in the mass of the ejecta, you write $dM$ for the rocket and $dm$ for the ejecta, where $dM=-dm$. But your method has one major advantage: you completely eliminate the need to decide where to put the minus sign. The mass of the rocket changes, the mass of the ejecta changes, one of them decreases, the other one increases. The big question for me was how to decide which one to mark as negative in my equation. With your method you don’t have to take this decision, all you need to know is that one is the negative of the other one. You did not directly solve my dilemma, instead you made it disappear. Thanks a lot.

 

[Rick16]

I think, this overcomplicates the issue more and more rather than to clarify it. It's simply a question of definition which sign you give your quantities. The physics is independent of this. To derive the rocket equation you simply write down the momentum balance of the rocket (=body of the rocket + fuel contained with mass $m(t)$ in it) and exhaust (fuel being exhausted from the rocket) within an infinitesimal time $\delta t$. If you assume that $\dd m_e>0$ is the mass of the exhausted fueld during this time the balance reads and the velocity of the exhausted fuel relative to the rocket is $v_e>0$, then the momentum balance reads
$$m(t) v(t)=m(t+\mathrm{d} t) v(t+\mathrm{d} t) +\mathrm{d} m_e (v-v_e),$$
i.e., expanding up to order $\mathrm{d} t$
$$m(t) v(t) = m(t) v(t) + [m(t)+\dot{m} \mathrm{d} t] v(t) + m(t) \dot{v}(t) + \mathrm{d} m_e(v-v_e)+m g \dd t.$$
From mass conservation you have
$$\mathrm{d}m_e=-\mathrm{d} t \dot{m}$$
and thus finally
$$m \dot{v} + \dot{m} v_e=-m g.$$
So you need $m(t)$ and $v_e(t)$ and have an equation of motion for $v$.

Thank you very much. This is basically the same approach as in #45. If I understand correctly, $dm$ in #45 corresponds to your $dm_e$ and $dM$ in #45 corresponds to your $\dot{m}dt$. What I find confusing in your equation is the term $m(t)$ in $[m(t)+\dot{m}dt]v(t)$. Shouldn’t this just be $\dot{m}dt v(t)$? In any case, the main idea is not having to worry about where to put a minus sign, and this is what I needed.

 

[vanhees71]

You are right, this is a typo. I've corrected it in the posting.