# What does ln mean in the Delta V equation?

I am working on a project for school and can't figure out what ln means in the Delta-V equation, as in Δv = VE * ln(ML / ME). I know that Δv is Delta V, VE is exhaust velocity, and ML and ME is the difference between a rocket that has not undergone its burn, and when it has. Does anyone know what ln means?

• It is explained here en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation
– Uwe
Feb 22, 2022 at 21:06
• The essence is: the end speed of the rocket, after burning all the fuel, will be the same as the fuel exhaust speed, if its fuel ratio is 1-1/e. e is 2.7173 . Feb 23, 2022 at 9:02
• @peterh (for small values of e :-) ) Feb 23, 2022 at 9:06
• @Uwe It's also explained by looking up ln on Wikipedia: the first entry in the "Math, science and technology" section is for "Natural Logarithm". I'm not trying to be excessively critical here, but Wikipedia's own search facility is woefully underused: far too many people search for something by Googling for e.g. "ln wikipedia" and get results which are essentially useless. Feb 23, 2022 at 14:05
• @MarkMorganLloyd My experience is that for the vast majority of sites google delivers vastly superior search result than the site's own search function. A lamentable state, for sure, but using google as the default search everywhere is a reasonable, time-tested stragegy. Feb 24, 2022 at 15:22

ln is a math function, the "natural log"

Most scientific calculators have a key for it. • Even for calculators that don't have a specific ln key, it can be calculated as long as you have a log key. ln is simply a logarithm in base $e$, and you can compute $ln(X) = log_e (X) = log_{10} (X)/ log_{10} (e)$. This works with any other base, although base 10 will usually be the default of the log key. Feb 23, 2022 at 18:09
• @NuclearHoagie if you are my age you can look it up in a table. Feb 23, 2022 at 19:11
• @OrganicMarble What? No slide rule? Feb 25, 2022 at 15:49
• @Tonny that's awesome. I actually have a slide rule but never really used it, pocket calculators were just appearing when I was in high school. Feb 25, 2022 at 15:51
• @OrganicMarble Same for me, but in high-school I had a physics teacher that constantly astounded the kids by being able to do all the calculations in his head faster than we could use the calculator. It was months before I realized that he only demonstrated that when he was sitting behind his desk and he was always fiddling with the ruler lying on his desk when doing the calculations. So, 1st chance I had I took a very good look at that ruler. My initial thought was that it was a ruler with build-in mini-calculator. But it was a slide rule. I had to get one myself after that. Feb 25, 2022 at 16:00

It's the logarithmus naturalis. That wikipedia page was the first result when I googled "ln". If you encounter situations like this in the future and you're concerned that two letters aren't enough to get results, you can add the word "math" to the search to give the search engine more context. And as Mark Morgan Lloyd mentioned in a comment, sometimes wikipedia's search function can give better results than Google's. There's also a website called wolframalpha.com that is a good place to get information about math. Additionally, searching "rocket equation" yields a wikipedia page that explain what all the terms, including ln, are.

• I am upvoting this just because you correctly said "when I googled" instead of "when you google". Thanks for that. It's worth noting that search engines generally produce bespoke results for its users based on context.
– Wyck
Feb 25, 2022 at 19:17

As others have explained, ln is the natural logarithm.

What's noteworthy about logarithms in general is that they grow very slowly when their argument grows. In fact, each multiplication of the argument — say, a doubling — only adds a constant value to the logarithm. Let's look at an example:

With a loaded:empty mass ratio of 10, ln(ML/ME) is ~2.3: You can expect to accelerate to ~2.3 times the exhaust velocity. Now assume that you more than double your fuel: $$ln (20) \approx 3.0$$. You double again: ~3.7. Again: ~4.4. Each doubling adds only a constant speed increase of 0.7 of the exhaust speed. The reason is that early in the flight, you need fuel to accelerate the fuel you'll need only later. And earlier, you need fuel to accelerate the fuel that you'll need to accelerate the fuel. And so on. This "stacking" is typical for exponential processes. ("The fuel needed grows exponentially with the desired speed" is the flip side of "the resulting speed grows logarithmically with the fuel": Logarithm and exponential function are inverse to each other).

You are now at 8 times the original fuel and you are not even twice as fast. This completely ignores the structural changes to the rocket needed to carry eight times the original fuel: The empty rocket is now also many times heavier than the original model, which means you'll actually need substantially more fuel than eight times the original amount in order to achieve the needed ML/ME ratio of 80, for a speed that's not even double the original.

"And this, dear friends, is why fast is hard." ;-)

• Also known as "the tyranny of the rocket equation". Feb 24, 2022 at 16:34
• Good answer! I started writing my somewhat similar answer before yours was posted. Feb 24, 2022 at 18:50
• @WaterMolecule Ah, I wondered :-) Yes, it takes time to write something properly. Feb 24, 2022 at 18:52

If you are unfamiliar with the notation $$\ln$$, then you might also be unfamiliar with the concept of a logarithm. There are lots of places to learn about it, but let's talk about it in the context of the rocket equation.

What's important to understand about this function for the rocket equation is that it grows very slowly with the ratio of initial to final masses: $$\Delta v = v_e \ln (\frac{m_0}{m_f})$$. Let's say the original delta-v of our space craft is $$\Delta v_\mathrm{old}$$. Our spacecraft weighs 120 tons unfuelled and 2400 tons with fuel, giving a mass ratio $$\frac{m_0}{m_f}=20$$. Our exhaust velocity is 3 km/s, so $$\Delta v_\mathrm{old} \approx 9~\mathrm{km/s}$$. (I've picked numbers roughly similar to the Saturn V first stage).

We work really hard to make the spacecraft lighter without reducing the amount of fuel it can carry. We replace heavy steel structural supports with titanium or beryllium. We remove insulation from the hull. And with all that work we reduce the unfuelled weight of the spacecraft from 120 tons to 40 tons. This is a major accomplishment! The engineers are very proud of themselves.

Our mass ratio tripled! We might expect that our delta-v will triple too! But we plug it into the rocket equation and find that the delta-v doesn't triple. It doesn't double either. When the mass ratio triples (really multiplied by $$2.718...$$), the $$\ln$$ just gets one bigger. So if you can make your mass ratio 3 times better, the delta-v just gets one more $$v_e$$ added to it.

$$\Delta v_\mathrm{new} = v_e \ln (\frac{3m_0}{m_f}) \approx v_e + v_e\ln (\frac{m_0}{m_f}) = v_e + \Delta v_\mathrm{old}$$

So, we don't go from 9 km/s to 27 km/s or even 18 km/s. $$\Delta v_\mathrm{new}$$ is just about 12 km/s.

Our engineers are sad, but are sure they can do better. We replace all of the metal with carbon nanotubes and borophene. We don't let the astronauts take their game consoles. We reduce the dry mass to 20 tons, 7 times less than the 140 tons it used to be. Surely now, with these science-fiction level improvements to the materials, we will be able to get a bigger delta-v.

We increase the mass ratio by a factor of 7 relative to our original craft, but the delta-v doesn't increase by a factor 7. It doesn't even double!

$$\Delta v_\mathrm{new} = v_e \ln (\frac{7m_0}{m_f}) \approx 2v_e + v_e\ln (\frac{m_0}{m_f}) = 2v_e + \Delta v_\mathrm{old}$$

We've reached the realm of science fiction, but still we only get 15 km/s rather than 9 km/s. Another way to understand this is that you have to carry fuel to carry fuel. This is why getting off this planet is so hard.

• Interesting that we used complementary approaches to illustrate the same circumstance: I increased the fuel, you decreased the dry weight. Somehow I find your example more intuitive: It seems more obvious that reducing the dry weight by a few tons doesn't make such a big difference if you still carry thousands of tons of fuel. In fact, I find the gain rather surprisingly high ;-) -- it is only achieved during the last phase, when the tank is almost empty and it matters a lot that your payload is light. Feb 24, 2022 at 18:58
• @Peter-ReinstateMonica I might sometimes describe myself as a materials scientist (depending on which funding agency I'm applying to), so the materials perspective was the first I thought of. However, I wish that my brain had produced this gem: '"The fuel needed grows exponentially with the desired speed" is the flip side of "the resulting speed grows logarithmically with the fuel": Logarithm and exponential function are inverse to each other).' Feb 24, 2022 at 21:01