Rocket in a Constant Gravitational Field - Burning Fuel Rapidly

Question

According to Kleppner and Kolenkow in "An Introduction to Mechanics" it's better to burn the fuel of a rocket quickly since this will net a greater final velocity. I'm a bit confused with this.

Using the rocket equation we have:

$\vec{F}_{ext}=\frac{d\vec{P}}{dt} \leftrightarrow M\vec{g}=M\frac{d\vec{v}}{dt}-\vec{u}\frac{dM}{dt}$

After integrating with respect to time and setting $\vec{v}_0=0$ and $t_0=0$ we obtain:

$v_f=ulog(\frac{M_o}{M_f})-gt_f$

It appears that when reducing $t_f$ we indeed end up with a greater velocity. My confusion is that I'm integrating the expression as something like this (with the same initial conditions):

$v(t)=ulog(\frac{M_o}{M(t)})-gt$

Which means we are only selecting a time at which we measure the velocity? I think I'm not understanding the integration step properly perhaps. I would very much appreciate some help with this.

Answer 1

The reason you can't compare $v_f(t)$ between fast-burning and slow burning rockets, is that those scenarios do not share the same burn duration, so a comparison at time $t$ is meaningless. One rocket will have burned more fuel fraction and have better speed, the other will have more reserve fuel. You need to take the time out of the picture.

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According to Kleppner and Kolenkow in "An Introduction to Mechanics" it's better to burn the fuel of a rocket quickly since this will net a greater final velocity. (Emphasis mine)

You must consider the duration of the manoeuvre, end particularly the end result. That requires you to introduce a fuel burn rate $\dot{m}$.

Your total burn time is $t_f=\frac{M_f-M_0}{\dot{m}}$

So your final speed is (using your own eq.):

$$v_f=u\log(\frac{M_0}{M(T_f)})-g T_f$$ $$v_f=u\log(\frac{M_0}{M_f})-\frac{g}{\dot{m}}(M_f-M_0)$$

The left member $u\log(\frac{M_0}{M_f})$is a constant, it's the optimal result that is typically found using impulse propulsion. The right member is negative (it's called the gravity drag), so to maximize $v_f$ you must maximize $\dot{m}$. This is why burning more quickly gives better efficiency, by reducing gravity drag.