How to develop the rocket equation

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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 vex of the expelled fuel, so that the velocity of the fuel relative to the ground or some other fixed point would be v - vex. 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.

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?
 

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  • #2
kuruman
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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.
 
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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 any one point in time.
 
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kuruman
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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 any one 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.
 
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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.
 
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kuruman
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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?
 
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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: [itex]\dot P_R = m_R \cdot \dot v_R + v_R \cdot \dot m_R[/itex]

Change of the momentum of the exhaust: [itex]\dot P_E = m_E \cdot \dot v_E + v_E \cdot \dot m_E[/itex]

Conservation of momentum: [itex]\dot P_R + \dot P_E = 0[/itex]

Conservation of mass: [itex]\dot m_R + \dot m_E = 0[/itex]

Relative velocity or the currently expelled reaction mass: [itex]v_{ex} : = v_R - v_E[/itex]

Already ejected reaction mass is not considered: [itex]m_E = 0[/itex]

Everything together results in equation 3.6.
 
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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 mE being zero, and at the same time dmE/dt being non-zero. Even if I try to keep the two concepts apart, considering mE the already ejected reaction mass and dmE/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?
 
  • #9
stevendaryl
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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 mE being zero, and at the same time dmE/dt being non-zero. Even if I try to keep the two concepts apart, considering mE the already ejected reaction mass and dmE/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}##
 
  • #10
stevendaryl
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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##.
 

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