Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Recently I decided to brush on the those parts of classical mechanics that always gave me trouble. One of those parts was the rocket equation so I figured that I would try to derive the appropriate equations from scratch. Evidently I ran into trouble somewhere, so I am hoping that someone can point out my errors and steer me in the right direction. Anyway here goes ...

Suppose a rocket expels exhaust behind it with constant velocity (relative to the rocket) $\mathbf{u}$. Fix an inertial reference frame and consider the rocket + exhaust system. If $m_R$ denotes the instantaneous mass of the rocket, then the momentum of the rocket can be written as $m_R\mathbf{v}$, where $\mathbf{v}$ is the instantaneous velocity of the rocket. The momentum of the exhaust ought to be given by the integral:

$$\int (\mathbf{v}+\mathbf{u})\frac{\mathrm{d}m_E}{\mathrm{d}t} \; \mathrm{d}t$$

where $m_E$ denotes the instantaneous mass of the exhaust. Combining these expressions for the momentum, we arrive at the total momentum for the system:

$$\mathbf{p} = m_R\mathbf{v} + \int (\mathbf{v}+\mathbf{u})\frac{\mathrm{d}m_E}{\mathrm{d}t} \; \mathrm{d}t$$

Now according to the 2nd Law of Motion, the force acting on the rocket + exhaust system is obtained from $\mathbf{p}$ by taking time derivatives. In other words:

$$\mathbf{F} = \dot{\mathbf{p}} = \frac{\mathrm{d}m_R}{\mathrm{d}t}\mathbf{v}+m_R\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}+(\mathbf{v}+\mathbf{u})\frac{\mathrm{d}m_E}{\mathrm{d}t} = m_R\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} - \frac{\mathrm{d}m_R}{\mathrm{d}t}\mathbf{u}$$

This expression superficially matches the ones in my college mechanics books, but there is one big difference. In this expression $\mathbf{F}$ is the total force acting on the rocket + exhaust system, while in my texts this $\mathbf{F}$ is just the force acting on the rocket alone.

So based on this I have made some fundamental error somewhere in my argument. As I asked before, if anyone could help me identify this error and correct it, I would be much obliged.

share|improve this question
    
"the force acting on the rocket alone" -- I'd express as $\mathbf{ F }_R := m_R \, \mathrm{d/d}t( \mathbf{ v } ) = \mathbf{ F }_R^{external} + \mathbf{ u } \, \mathrm{d/d}t( m_R )$; while "the total force acting on the rocket + exhaust system" -- I'd express as $\mathbf{ F }^{external}$, where it is perhaps usually assumed that it is irrelevant which forces act on the exhaust matter once it has been expelled, so perhaps $\mathbf{ F }^{external} := \mathbf{ F }_R^{external}$. I conclude that your texts are sloppily worded ... –  user12262 Oct 9 '13 at 21:58

1 Answer 1

I don't see an error on your part. The force on the rocket must equal the time rate of change of the rocket momentum only so either the texts are misleading or plain wrong.

There is no external force acting on the rocket / propellent system so the center of mass of the rocket / propellent system, as a whole, does not change.

Thus:

$m_R \cdot d\mathbf{v} = \mathbf{u} \cdot dm_R$

or

$dv = -u \dfrac{dm_R}{m_R} $

which leads to

$\Delta v = u [\ln(m_{R,i}) - \ln(m_{R,f})] = u \ln\dfrac{m_{R,i}}{m_{R,f}}$

share|improve this answer
    
Right so this leads to the correct equation for the velocity in the absence of external forces. But when I try my approach in say, the presence of a gravitational field, I get incorrect results. For example, if I want to find the velocity in this case, I get $(m_R+m_E)g = m_R\dot{v}-u\dot{m}_R$ and I can solve for $v$ from here. However, Kleppner and Kolenkow say that I should have $m_Rg = m_R\dot{v} - u\dot{m}_R$ and that I should solve for $v$ from there. Hence my confusion. –  James Miller Oct 9 '13 at 22:21
    
Ah I think I identified the source of my error. In the case where there are no external forces acting on the system, the derivation goes through just fine. But when I consider external forces acting on the system, the expression I have for the momentum of the exhaust is incorrect. An additional term is needed to correct for whatever momentum is gained/lost due to the external force acting on the exhaust. Do I have this right? –  James Miller Oct 9 '13 at 22:39
    
Yes, an external force on a system will change its momentum according to the formula you stated in you question. However the propulsion due to expelled exhaust gasses will apply an internal force on the rocket/propellant system, so no change of momentum of the system. –  fibonatic Jan 8 at 0:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.