4
$\begingroup$

I have read many different rocket equation derivations but I do not understand them. Here is my derivation:

At time $t$, assume we have an accelerating rocket with mass $m$ and velocity $v$. At time $t+\delta t$, the rocket would have ejected post-reaction propellant with mass $\delta m$ at exhaust velocity $v_e$. The rocket would have mass $m-\delta m$ and velocity $v+\delta v$. By the conservation of momentum from an observer's reference frame, we have $$mv = (m-dm)(v+dv)+\delta m(v-v_e) \qquad\qquad \text{right is positive}$$ Note: Most of the derivations I have read use $(v+v_e)$ instead, but isn't the exhaust velocity in the opposite direction of the rocket?

Expanding and simplifying leaves us with $$m\delta v-v_e\delta m-\delta m\delta v=0$$ We can ignore higher order terms. $$\delta v = v_e \frac{\delta m}{m}$$ Integrating gives $$\Delta v = v_e ln{\frac{m_f}{m_0}}$$ I do not understand where I went wrong. I have read posts saying that $m -\delta m $ should be $m+\delta m$ but then the other $\delta m$ term would be affected too.

$\endgroup$
2
  • $\begingroup$ If yes the use q =dm/dt $\endgroup$
    – Priyan
    Commented Feb 10, 2020 at 6:33
  • 1
    $\begingroup$ Apply principle of impulse. $\endgroup$
    – Priyan
    Commented Feb 10, 2020 at 6:41

2 Answers 2

2
$\begingroup$

There is a subtle conceptual error in your derivation regarding the differantials; So let's review the problem: At the start the mass in our reference frame consists solely of the rocket and is described by a function $m(t)$. Now after the time $\mathrm{d}t$ has elapsed, a propellant particle with mass $\mathrm{d}m_e$ got ejected. So the total mass in our reference frame is $m(t + \mathrm{d}t) + \mathrm{d}m_e$. We may expand the function $m(t + \mathrm{d}t)$ and apply the principle of mass conservation: $$m(t) = m(t + \mathrm{d}t) + \mathrm{d}m_e = m(t) + \mathrm{d}m + \mathrm{d}m_e.$$ From here it follows that the infinitesimal small mass $\mathrm{d}m$ has the opposite sign of $\mathrm{d}m_e$, otherwise the above equation couldn't be true. With this knowledge, you can correct the momentum equation (remember the principle of momentum tells us that the change of momentum equals the resulting force $R$): \begin{align} R \, \mathrm{d}t & = I(t + \mathrm{d}t) - I(t) = (m(t) + \mathrm{d}m)(v(t) + \mathrm{d}v) - m(t)v(t) + \mathrm{d}m_e v_e \\ & = m(t) \, \mathrm{d}v + \mathrm{d}m \,v(t) + \mathrm{d}m_e v_e + \mathrm{d}m\,\mathrm{d}v \\ & = m(t) \, \mathrm{d}v + \mathrm{d}m \,v(t) - \mathrm{d}m v_e + \mathrm{d}m\,\mathrm{d}v. \end{align} Now dividing by $\mathrm{d}t$ and doing some algebraic manipulation gives you $$m \frac{\mathrm{d}v}{\mathrm{d}t} = (v_e - v) \frac{\mathrm{d}m}{\mathrm{d}t} + R$$ Solving this differential equation for $R = 0$ and $v_e - v = \Delta v$ should result in a correct expression for $v(t)$. I hope I could clarify your issues (including why $\mathrm{d}m =\mathrm{d}m_{\text{rocket}} = -\mathrm{d}m_e$), the point is you should always be careful when it comes to expansion of functions.

$\endgroup$
2
  • $\begingroup$ I do not understand. This is what I get: $$m(t+\delta t) = m(t) -\delta m_e$$ $$m(t) = m(t+\delta t) +\delta m_e = m(t) + \delta m + \delta m_e$$ therefore $$\delta m = - \delta m_e$$ I understand this so far. From here, by the conservation of momentum, we have $$mv = (m-\delta m_e)(v+\delta v) + \delta m_e(v-v_e)$$ I get the same result. At time $t + \delta t$, the mass of the rocket is the mass of the rocket at time $t$ minus $\delta m_e$, the mass of the ejected fuel $\delta m_e$ has a velocity $v - v_e$. What am I doing wrong? $\endgroup$
    – user572780
    Commented Feb 10, 2020 at 14:00
  • $\begingroup$ You have to substitute $-\delta m$ for $\delta m_e$ in your equation for the momentum, since we need an equation that depends only on $v$ and $m$. Regarding the velocities: $v_e$ is the velocity of the expelled fuel regarding a non-moving reference frame while $v_e - v$ is the particle velocity relative to the moving rocket and for completeness $v - v_e$ would be the velocity of the rocket relative the the fuel particles. $\endgroup$
    – Tobi7
    Commented Feb 10, 2020 at 19:23
1
$\begingroup$

Using $m+\delta m$ in the initial momentum will not change your result, if you change the right side of the equation accordingly. The result you got the wrong sign is because when you integrate, you need to integrate the mass of the rocket, so for the rocket, so your $\delta m_{exhaust}=-\delta m_{rocket}$. See here.

$\endgroup$
4
  • 2
    $\begingroup$ feedback would be more polite, instead of being a trigger happy $\endgroup$
    – user65081
    Commented Feb 10, 2020 at 6:19
  • $\begingroup$ Hi Wolphram, it was not me who downvoted you. I do not understand what you mean. At what step should I change $\delta m_{exhaust} = -\delta m_{rocket}$?And why? $\endgroup$
    – user572780
    Commented Feb 10, 2020 at 10:23
  • 1
    $\begingroup$ You can do at any time, the confusion is because you named it $\delta m$, but it is actually $d m_{exhaust}$, the rest of the m's in the equation are $m_{rocket}$. You can replace $d m_{exhaust}=-d m_{rocket}$ at any time $\endgroup$
    – user65081
    Commented Feb 10, 2020 at 14:03
  • $\begingroup$ Ah, I see. Makes more sense to me now. Thank you. $\endgroup$
    – user572780
    Commented Feb 10, 2020 at 14:09

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