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Looking at the numbers that are given, for example, at the SpaceX launches, it seems that the velocities at first stage separation are a bit above 2 km/s. The second stage does all the rest of the work, accelerating to the final velocity. So, the first stage delivers just a bit more of 25% of total delta-v on the system.

By comparison, the Saturn V reached around 2.7 km/s with its first stage, quite a lot more.

Why was it designed that way for the Falcon 9? Of course, the Tsiolkovsky equation governs this (plus the effect of drag and gravity), but I assume some cost optimization must have gone into it as well.

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    $\begingroup$ Obligatory (assuming no Wertz/Larson at hand): projectrho.com/public_html/rocket/multistage.php $\endgroup$
    – Deer Hunter
    Commented Feb 24, 2015 at 23:39
  • $\begingroup$ The Saturn V upper stages were hydrogen fueled; at first blush one would assume you wanted as much of the delta-v as possible delivered on hydrogen/LOX instead of kerosene/LOX because of its ISP advantage. However, hydrogen stages are vastly larger in volume than equivalent delta-v of kerosene/LOX, so one possible explanation could be that the S-II stage was as big as was practical for a hydrogen stage and the rest had to be done in kerosene. $\endgroup$
    – Russell Borogove
    Commented Feb 25, 2015 at 0:03
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    $\begingroup$ The Saturn V is a poor comparison because it’s a 3-stage rocket and had a much higher ideal delta-V, since it’s mission was the moon, not LEO. Rocket design is exceedingly complex, and for an actual vehicle determining the ideal delta-V split depends on the specifics of almost all aspects of the design: propellant, staging, engines, aerodynamics, etc. $\endgroup$
    – Adam Wuerl
    Commented Feb 25, 2015 at 6:15
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    $\begingroup$ I would imagine that it might also be easier to make the first stage reusable if it reaches a lower maximum velocity $\endgroup$
    – neelsg
    Commented Feb 25, 2015 at 7:37
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    $\begingroup$ This is one of the most important fundamental questions I've ever seen on this site. $\endgroup$
    – Deer Hunter
    Commented Feb 25, 2015 at 19:02

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Short answer:

You can get pretty close with the rocket equation and optimizing for a rocket of given mass to lift a maximum payload to a specified velocity.

Long answer:

I finally tried some math (though in a slightly different way than the link above).

First, I made the problem dimensionless

  • dividing all masses by the total mass of the rocket without the payload. The mass of the rocket is then equal to one.
  • introducing a factor $\phi$ which equals the total mass of the first stage divided by the total mass of the rocket. $\phi=0$ would mean a zero stage of no mass (rather silly) and $\phi=1$ would mean a second stage of zero mass (equally silly).
    • Calling the dry mass fraction of the first stage $e_1$ and the dry mass fraction of the second stage $e_2$
  • Assuming a mass $m_3$ which is the payload, which comes on top
  • Dividing all velocities by the target velocity $v_t$ (in this case, needed to reach LEO).
  • Dividing effective exhaust velocities, $I_{sp} g$, by the required velocity, to get $v_{e,1} = I_{sp,1} g / v_t $ as the effective dimensionless velocity for the first stage and $v_{e,2} = I_{sp,2} g / v_t$

Plugging this into the rocket equation gives $$\Delta v_1 = v_{e,1} \ln\left(\frac{1+m_3}{\phi * e_1 + 1 - \phi + m_3}\right) $$ $$\Delta v_2 = v_{e,2} \ln \left( \frac{1-\phi+m_3}{(1-\phi) e_2 + m_3} \right) $$ with $\Delta v = \Delta v_1 + \Delta v_2$. Plugging in the parameters and solving the equation $$\Delta v_1 + \Delta v_2 = 1$$ (iteratively) for $m_3$ will give you the payload for a given rocket mass, which should be maximized.

So I plugged in some values for a Falcon 9: $$v_t = 9.5 km/s$$ $$I_{sp,1} = 290s$$ $$v_{e,1}=0.305$$ $$v_{e,2}=0.358$$ $$I_{sp,2}=340 s$$ $$e_1=18 t/385 t = 0.0468$$ $$e_2= 4.9t/90t=0.0544$$

After varying $phi$ and also calculating the velocity of the second stage, I got

    phi     m_3      delta v 2
    0.2    0.0160      0.895
    0.4    0.0241      0.858
    0.6    0.0308      0.753
    0.7    0.0331      0.683
    0.8    0.0339      0.592
    0.9    0.0313      0.456

So, there is a relatively broad optimum where the first stage has around 80% of the mass of the total rocket and the second stage is responsible for around 60% of the total $\Delta v$.

Plug this into the original assumption of a $\Delta v$ of around 9.5 km/s, and you get to the (approximate) 6 km/s for the second stage.

Why doesn't the first stage become faster? Probably because it is the one which has to fight gravity and drag. 180 s at 10 m/s^2 could cost you around 1.8 km/s if you were firing straight up (which nobody is doing, I know...)

Compare this to the actual mass fraction of the first stage of a Falcon 9, which is around 0.809... I probably came out a bit closer than I had a right to expect.

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