What does `ln` mean in the Delta V equation?

Question

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?

Answer 1

ln is a math function, the "natural log"

Most scientific calculators have a key for it.

See https://en.wikipedia.org/wiki/Natural_logarithm

Answer 2

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.

Answer 3

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." ;-)

Answer 4

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.