# Rocket in a Constant Gravitational Field - Burning Fuel Rapidly

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.

• The time $t_{rm f}$ is the time at which all the fuel is used up and the final mass is $M_{rm f}$. That time is dictated by the rate at which you burn the fuel. Burn it at a faster rate and you reach $M_{rm f}$ in a shorter time. – Farcher Jul 21 '17 at 10:24

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.

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.

• Oh, I get it now, we are just looking at the total duration of propulsion and the velocity at the end of that period. Seeing the fuel burn rate explicitly in the equation helps a lot. Just to confirm could we simply state reducing gravity drag as "reducing the mass as quickly as possible since this means reduced impulse due to weight"? – phytab Jul 21 '17 at 12:36
• Hmm, better not mix impulse and gravity drag, because the former assumes instant burn, the latter assumes a non-zero burn duration. What is reduced by the drag is the $\Delta V$, the total change in speed, which is the only thing that really matters in rocket dynamics. So you could really say reducing drag by reducing the mass as quickly as possible since it reduces the $\Delta V$ lost to weight – MrBrushy Jul 21 '17 at 13:33