I'm trying to understand why rockets have multistages releasing their fuel tanks. Say a rocket $R$ has two fuel tanks $A$ and $B$, which respectively have masses $m_a$ and $m_b$, and the mass of the fuel in the respective tanks is $m_{fa}$ and $m_{fb}$. Let $M$ be the total initial mass, and suppose the rocket starts from rest. We use the equation $v-v_0=v_{ex}\ln(m_0/m)$, where $v_{ex}$ is the (constant) speed of the expent fuel.

First suppose the rocket spends its fuel all at once. Then

$$v_f=v_{ex}\ln\left(\frac{M}{M-(m_{fa}+m_{fb})}\right).$$

Now suppose the rocket first spends its fuel in tank $A$: $v_1=v_{ex}\ln(\frac{M}{M-m_{fa}}).$ Now the rocket detaches tank $A$, so the mass is now $M-(m_{fa}+m_a).$ Consequently, after spending the fuel in tank $B$, we have

\begin{align}

v_f&=v_1+v_{ex}\ln\left(\frac{M-(m_{fa}+m_a)}{M-(m_{fa}+m_a+m_{fb})}\right)\\&=v_{ex}\left[\ln\left(\frac{M}{M-m_{fa}}\right)+\ln\left(\frac{M-(m_{fa}+m_a)}{M-(m_{fa}+m_a+m_{fb})}\right)\right].\end{align}

Now I'm not seeing how to show $v_f$ is greater in the second case, so I'm thinking I made some mistake in the argument. Can anyone clear things up?

## Best Answer

Put your math aside for a minute, and take a lesson from

Robert H. Goddard, in one of my all-time favorite papers.Basically your rocket consists of a payload H, and the rest of the rocket consisting of fuel mass P, plus non-fuel mass (i.e. tank) K. The secret is, as you shed P through combustion, you must also shed K. Otherwise as P gets smaller and smaller it is trying to accelerate dead weight.