When I graph the rocket equation, substituting arbitrary values for v(exhaust) and m1, so m0 because m1 - m0, the graph implies that increasing propellant mass past a certain point does not increase the delta V by very much.

Likewise, if I plot the inverse of the function, the graph implies that for increasing delta V values, extremely high propellant masses are required, and there is a vertical asymptote. Does this not imply that a rocket has a maximum delta V?

Also, what is the significance of the graph of the derivative? of the integral?

  • $\begingroup$ Why do you think NASA jumps through so many hoops to save a bit of delta V?? $\endgroup$ Commented Jan 29, 2016 at 23:28

3 Answers 3


The implication of the rocket equation is that linear increases in ∆v require exponential increases in mass ratio for a single stage.

There's not strictly a maximum delta-v -- if you redo your plot on a log scale, you'll see that it doesn't go vertical.

Getting very high mass ratios (much above 10:1) is difficult to do on a single stage, so there is a practical ∆v limit there.

Multiple stage rockets can reach arbitrarily high total mass ratios (if you have the budget), so linear increases in ∆v simply require exponential increases in overall mass.

The derivative of that plot is a rate of change of mass with respect to rate of change in velocity. For a rocket at a particular point in flight, it tells how much propellant mass needs to be expended in order to increase its speed by a given amount; for a rocket on the drawing board, it tells how much more propellant needs to be carried in order to increase its capability as defined by ∆v.

I don't know if there are any useful interpretations of the integral.

  • $\begingroup$ Of course if you consider launch from Earth, where the rocket needs to fight a constant Earth acceleration and atmospheric drag, there IS a maximum delta-V; scaling the amount of fuel up you increase mass, decreasing TWR. Scaling the thrust up you increase size of the rocket = atmospheric drag. We can safely say a coal-and-steam powered rocket will never reach orbit. $\endgroup$
    – SF.
    Commented Jan 29, 2016 at 14:21
  • $\begingroup$ As discussed on the miniature Saturn V thread, if you assume that the engines are scaled up with the rocket, the cross-sectional drag gets smaller relative to the thrust force. Assuming that the payload mass is fixed, you can scale the engines at the same rate as the fuel and continue to increase the mass ratio. $\endgroup$ Commented Jan 29, 2016 at 16:46
  • $\begingroup$ Yes, but you will be limited by practical factors. The amount of coal available, the atmospheric air availability, structural integrity etc. You just can't scale up indefinitely. $\endgroup$
    – SF.
    Commented Jan 29, 2016 at 17:01

Your question is about the behavior of the Tsiolkovsky rocket equation itself, in the limit of very small final mass (dry mass). Roughly: "is there any limit to delta-v in theory?"

Using MathJax:

$$ \Delta v=v_e \ln\frac{m_0}{m_f}. $$ If you just look at the velocity ratio and the mass ratio: $$ \frac{\Delta v}{v_e}=\ln\frac{m_0}{m_f}=-\ln\frac{m_f}{m_0}, $$ you can see that one is just the log of the other. Log keeps increasing forever, but more and more slowly. It never reaches an asymptote. It shoots up fast on the linear plot, but if you change the x-axis from linear to log, the velocity ratio just continues to increase.

click to enlarge: Tsiolkovsky rocket equation lin log

If the final mass is one billionth of the start mass, you can get a delta-v that is over 20 times the exhaust velocity. Theoretically.

So I think that the answer to your question, just from the mathematical point of view, based on the rocket equation is no. There's no limit here. If your rocket is made of pure fuel, with a magic massless nozzle, it can go very very fast!

But for sure, as the other answers suggest, a physical rocket we can build and launch does have a very real (and so far very expensive) physical limit.

Here is some Python code for the plot:

import numpy as np
import matplotlib.pyplot as plt

mf_over_m0 = np.logspace(-0.5, -10, 20)

delta_v_over_ve = -np.log(mf_over_m0)

# that's it - the rest is just plotting @@!

fig = plt.figure(figsize=[16, 8])

xlab,   ylab   = "mf / m0",             "delta v / v exhaust"
title1, title2 = "x-axis linear scale", "x-axis log scale"

ax1 = plt.subplot(121, xlabel=xlab, ylabel=ylab, title=title1)
ax2 = plt.subplot(122, xlabel=xlab, ylabel=ylab, title=title2)

# from here: http://stackoverflow.com/a/14971193/3904031
for ax in [ax1, ax2]:
    for item in ([ax.title, ax.xaxis.label, ax.yaxis.label] +
                 ax.get_xticklabels() + ax.get_yticklabels()):

for ax in [ax1, ax2]:
    ax.plot(mf_over_m0, delta_v_over_ve, 'ok')
    ax.plot(mf_over_m0, delta_v_over_ve, '-b')

plt.savefig("rocket equation question")

No, that is not quite right. Let's first state and describe the Tsiolkovsky Rocket Equation:

$\displaystyle \Delta v = V_e \times \ln(\frac{m_i}{m_f})$

$\Delta v$ is delta v, the change in velocity in km/s

$V_e$ is the effective exhaust velocity in km/s (it's another way of measuring specific impulse)

$\ln()$ is just the natural logarithm, or log base e (e is an interesting mathematical constant approximately equal to 2.7). Almost all scientific and graphing calcs has an ln() function.

$m_i$ and $m_f$ are initial and final masses in kg

The ratio of masses is just that, a ratio. So if we start with 100 kg and end with 10, or start with 1,000 kg and end with 100, the change in velocity will stay the same. This of course assumed that Ve didn't change either.

So the equation is basically the natural log of a ratio, r, then multiplied by a constant. To analyze any asymptote behavior all you really need to understand is the behavior of the natural log function.

Go to fooplot.com and type in "ln(x)" without quotes. You will see a black curve. In this graph, our ratio r is now called x, and y represents the delta v.

Yes there is a vertical asymptote on the left, coinciding with the y-axis itself. This would imply some really weird stuff about an unbounded negative limit of delta v (so you could go infinitely fast with a negative speed!)...except that our ratio cannot be less than 1...because initial mass is always greater than final mass. That's just how chemical rockets work. They burn mass and thus lose mass. You always start with more fuel than you end with. In mathematical terms, this means that the valid domain starts at x > 1. x <= 1 is just not physically possible.

Now you should notice that the curve gets shallower as it goes to the right. You may be wondering if the curve has a horizontal asymptote? If so, this would indeed set a max on delta v as you suspect. But the answer is, no, there's no horizontal asymptote in the graph. The curve always gets higher and higher as it goes to the right...just at shallower and shallower angles.

Another way of saying that, is that logarithms behave with exponential decay. However, "exponential" is used very loosely these days. I don't recommend that terminology! Just to confuse you more, a horizontal asymptote is also called exponential decay! (of another kind).

BTW you can multiply "ln(x)" by whatever Ve you want and it will not change the graph's behavior. Fair warning though, if you type in a typical number that real-world rockets use, such as 3,000 (m/s), you will have to zoom way, way out in order to see the curve. That's why I like working in km/s so you only multiply by 3.

Bottom line about the equation: there is no max delta v. If you're willing to exponentially increase the mass ratio, you can linearly increase the delta v to any high value you want. This does not mean, however, that it's physically possible. There are of course technological limits on how much thrust, heat, and other stuff we can controllably achieve in our rocket.

You asked about the derivative and integral too. You also mentioned the inverse of the Tsiolkovsky Rocket Equation in your second paragraph. I will try to explain what those mean in a future edit.


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