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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?

Kyle Kanos
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    Intuitively it's pretty simple: if the rocket keeps boosting the mass of the first stage fuel tank, it becomes part of the payload mass, so the overall velocity is reduced because the effective mass ratio is reduced (if both stage exhaust velocities are the same, the calculation reduces to a single rocket stage in that case!). I think you may see it in your math easier, if you did the calculation with mass ratios, rather than absolute masses. – CuriousOne Sep 30 '14 at 23:45
  • Related: http://physics.stackexchange.com/questions/88145/ – Kyle Kanos Oct 01 '14 at 02:07

2 Answers2

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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.

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Mike Dunlavey
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The answer is right there in your own math.

You derived that the delta V that results from using the rocket as a single stage rocket is $$\Delta V_{\text{single stage}} = v_{ex}\ln\Bigl(\frac{M}{M-(m_{fa}+m_{fb})}\Bigr)$$ while the delta V from using the rocket as a two stage rocket is $$\Delta V_{\text{two stage}} = v_{ex} \Biggl( \ln\Bigl(\frac{M}{M-m_{fa}}\Bigr) +\ln\Bigl(\frac{M-(m_{fa}+m_a)}{M-(m_{fa}+m_a+m_{fb})}\Bigr)\Biggr)$$

So the problem at hand is to determine the conditions needed to make $\Delta V_{\text{two stage}} > \Delta V_{\text{single stage}}$.

First, let's rewrite the two stage result using the fact that $\ln a + \ln b = \ln(ab)$: $$\Delta V_{\text{two stage}} = v_{ex} \ln\Bigl(\frac{M}{M-m_{fa}}\frac{M-(m_{fa}+m_a)}{M-(m_{fa}+m_a+m_{fb})}\Bigr)$$

Since the natural logarithm is monotonically increasing function, the condition needed is that $$\frac{M}{M-m_{fa}}\frac{M-(m_{fa}+m_a)}{M-(m_{fa}+m_a+m_{fb})} > \frac{M}{M-(m_{fa}+m_{fb})}$$ Clearing the denominators yields $$M(M-(m_{fa}+m_a))\;(M-(m_{fa}+m_{fb})) > M(M-m_{fa})\;(M-(m_{fa}+m_a+m_{fb}))$$ This nicely reduces to $$m_a m_{fb} > 0$$ In other words, given a rocket designed as a two stage rocket, it's always better to use that rocket as a two stage rocket than as a single stage rocket.

This does not quite mean that it's better to build a two stage rocket than a single stage rocket. If it did, the same logic would dictate that a hundred stage rocket would be better than a ten stage rocket, and so on. There are penalties in building that hundred stage rocket (and in a two stage rocket). A multi-stage rocket has extra engines, extra fuel tanks, and extra structure, one set per extra stage. That's extra mass, and that extra mass means the lower stages also need extra structure to support that extra mass in the upper stages. Each additional stage also introduces extra places where things can go wrong. At some point, dividing a vehicle into ever more stages becomes counterproductive.

In most cases, that break even point is a two or three stage launch vehicle, possibly carrying a payload that is itself a rocket. (A single stage rocket that can carry a reasonable sized payload to orbit is beyond the capability of currently available engineering, physics, and chemistry.) The most complex vehicle ever built was the rocket that brought humans to and back from the Moon, which comprised eight stages in total:

  1. S-IC first stage, 2,300,000 kg gross mass, 131,000 kg empty mass.
  2. S-II second stage, 480,000 kg gross mass, 36,000 kg empty mass.
  3. S-IVB third stage, 120,800 kg gross mass, 10,400 kg empty mass.
  4. Launch escape system (thankfully never used for its intended purpose), 4,200 kg gross mass.
  5. Service module, 24,500 kg gross mass, 6,100 kg empty mass.
  6. Lunar descent stage, 10,300 kg gross mass, 2,100 kg empty mass.
  7. Lunar ascent stage, 4,547 kg gross mass, 2,213 empty mass.
  8. Command module, 5,560 kg gross mass, including 120 kg fuel.
David Hammen
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    Actually, as most Kerbal Space Program players know, you don't need extra engines even if you have stages. You just have to attach the tanks to the side instead of stacking them. In reality, there's another benefit: if you have multiple engines, you can optimize them for different ambient pressures and trust levels. – MSalters Oct 01 '14 at 08:20
  • To my knowledge, the Space Shuttle is the only spacecraft that actually uses an external tank. Most other launch systems (liquid propellant) have dedicated engines for each stage. This makes sense, because you can optimize the different engines for different environments. The first stage operates in the densest possible atmosphere. The second stage operates in thin air, and so on. So, it actually pays off, to carry and drop additional engines. – Dohn Joe Oct 01 '14 at 08:28
  • The Delta IV Heavy uses strap-on boosters. The central engines are throttled back to 50% shortly after liftoff, and remain at 50% until burnout of the strap-ons. Not quite a fuel tank approach, but getting there. The Falcon Heavy will also use strap-on boosters, but these cross-feed fuel to the central engines. The Falcon Heavy strap-ons are doing double duty as an external tank and as strap-on boosters. One other difference between these vehicles: The Delta IV Heavy has flown three times, while the Falcon Heavy is still in development. – David Hammen Oct 01 '14 at 09:24