# An Actually Intuitive Explanation of the Oberth Effect

*This is a linkpost for* [*An Actually Intuitive Explanation of the Oberth Effect*](https://outsidetheasylum.blog/an-actually-intuitive-explanation-of-the-oberth-effect/)*.*

Like anyone with a passing interest in ~~Kerbal Space Program~~ physics and spaceflight, I eventually came across the [Oberth Effect](https://en.wikipedia.org/wiki/Oberth_effect). It's a very important effect, crucial to designing efficient trajectories for any rocket ship. And yet, I couldn't understand it.

Wikipedia's explanation focuses on how kinetic energy is proportional to the square of the speed, and therefore more energy is gained from a change in speed at a higher speed. I'm sure this is true, but it's not particularly helpful; [simply memorizing formulae is not what leads to understanding of a phenomenon](https://edisciplinas.usp.br/pluginfile.php/7610891/mod_resource/content/2/Excerpt%20from%20Surely%20Youre%20Joking%2C%20Mr.%20Feynman.pdf). You have to know what the numbers *mean*, how they correspond to the actual atoms moving around in the real universe.

This explanation was particularly galling as it seemed to violate relativity; how could a rocket's behavior change depending on its speed? What does that even mean; its speed relative to what? Whether a rocket is traveling at 1 m/s or 10000000 m/s relative to the Earth, the people on board the rocket should observe the exact same behavior when they fire their engine, right?

So I turned to the internet; [Stack Overflow](https://space.stackexchange.com/questions/4229/oberth-effect-for-earth-vehicles), [Quora](https://www.quora.com/What-is-an-intuitive-explanation-for-the-Oberth-effect), [Reddit](https://www.reddit.com/r/KerbalAcademy/comments/1jhne0/clearing_up_misconceptions_about_the_oberth_effect/), [random physicists' blogs](https://www.askamathematician.com/2013/01/q-how-does-the-oberth-effect-work-and-where-does-the-extra-energy-come-from-why-is-it-better-for-a-rocket-to-fire-at-the-lowest-point-in-its-orbit/). But they all had the same problem. Every single resource I could find would "explain" the effect with a bunch of math, either focusing on the quadratic nature of kinetic energy, or some even more confusing derivation in terms of work.

A few at least *tried* to link the math up to the real world. Accelerating the rocket stores kinetic energy in the propellant, and this energy is then "reclaimed" when it's burned, leading to more energy coming out of the propellant at higher speeds. But this seemed unphysical; kinetic energy is not a property of the propellant itself, it depends on the reference frame of the observer! So this explanation still didn't provide me with an intuition for why it worked this way, and still seemed to violate relativity.

It took me *years* to find someone who could explain it to me in better terms.

### **Asymmetric gravitational effects**

Say your spacecraft starts 1 AU away from a planet, on an inertial trajectory that will bring it close to the planet but not hit it. It takes a year to reach periapsis going faster and faster the whole way. Then it takes another year to reach 1 AU again, slowing down the whole time.

Two things to note here: The coordinate acceleration experienced by the spacecraft (relative to the planet) is higher the closer it gets, because that's where gravity is strongest. Way out at 1AU, the gravitational field is very weak, and there's barely any effect on the ship. Secondly, note that the trajectory is symmetric, because orbital mechanics is time-reversible. That's how we know that if it takes 1 year to fall in it will also take 1 year to get back out, and you'll be traveling at the same speed as you were at the beginning.

Now imagine that you burn prograde at periapsis. Now you'll be traveling faster as you leave than you were as you came in. This means that *gravity has less time to act on you on the way out than it did on the way in*. Of course the gravitational field extends all the way out to 1 AU, but if we take just a subregion of it, like the region within which the acceleration is at least 1 m/s^2^, you'll spend less time subject to that level of acceleration.

So the Oberth effect is just a consequence of you maximizing the amount of time gravity works on you in the desired direction, and minimizing it in the other direction. (And of course you'd get the inverse effect if you burned retrograde; a more efficient way to slow down.)

This has nothing to do with propellant. Maybe instead of thrusters, there's a giant automatic baseball bat machine that will smack your rocket ship as you pass by, adding 1 m/s to whatever your current speed is. It would be more efficient to put the machine on the surface of a planet[^k4vjlnl1a7c] than to put it out in space.

### **It really is gravity, not speed**

The confusing explanations of the Oberth effect focus on speed being the relevant factor, but it's really not. Being *inside a gravitational well* is what matters.

Think about the baseball bat machine placed on a planet. Why does the inbound part of the trajectory matter? Couldn't we just imagine the spaceship getting launched directly from that planet, and it would have to behave the same way? Yeah, it would. And indeed, the equations for escape speed show this is the case.

Let's say you have a gun that shoots a bullet at 1 m/s. You shoot it in deep space, and wait until the bullet has traveled 1AU. It will still be going 1 m/s, obviously. If you increase the muzzle speed to 2 m/s, the bullet will be going at 2 m/s. But what it you fire it on the surface of a planet that has an escape speed of 1 m/s? If the muzzle speed is 1 m/s, then the bullet will be asymptotically slowing down towards 0; we can say that it's speed "at infinity" is 0. But if the bullet speed at launch is 2 m/s, its [speed at infinity](https://en.wikipedia.org/wiki/Hyperbolic_trajectory#Velocity) will be sqrt((initial speed^2^) \- (escape speed^2^)), or about 1.7 m/s.

Look at that! Increasing the efficiency of your space gun has more of an impact if you're standing on the surface of a planet than if you're in deep space. Making the gun shoot faster has compounding returns, because the faster the bullet leaves the gravity well of the planet, the less of its energy gravity is able to "steal" as it leaves.

### **Bringing propellant back into it**

Of course 1.7 m/s is still less than 2 m/s, so to maximize projectile speed it's still best to put the gun in deep space, away from the planet. Why is this? With the traditional Oberth effect using the giant baseball bat machine, it was more efficient to put it on the planet. But with our space gun, it's better to put it out in deep space. What's the difference?

The difference is that the starting position of the whole system is not the same between the two scenarios, so it's not a symmetric comparison. With the baseball bat machine, the ship starts in outer space, and then goes back to outer space. With the space gun, the bullet starts on the surface of the planet. It costs energy to bring your space gun up into space!

In other words, the spaceship got to cheat. It started already having some potential energy for being outside of a gravity well, whereas the bullet didn't start with that energy.[^ry0vzjc157] Taking *everything* into account, it actually is more efficient to just launch the bullet from the surface at a higher speed, rather than spend a bunch of extra energy lugging the whole gun into space too.

Could the gun in space cheat too? Sure; let it fall into the gravity well and then fire the bullet, and you'll get the same positive effect on the bullet's total speed. Oh look, we've reinvented a rocket.

So this is another framing of the Oberth effect: If the rocket thrusts in deep space, it only gains the chemical energy of the propellant. All that propellant's potential energy from being far above the surface of the planet is wasted, with the propellant flying off to infinity, never going anywhere near the planet. If the rocket instead falls into the gravity well, the propellant's potential energy turns into kinetic energy. Then when the rocket fires its engine at periapsis, that propellant has to escape the gravity well, meaning it will be traveling slower at infinity that it would have been from an engine firing in deep space. That extra kinetic energy has to go somewhere, and it goes into the rocket. The rocket gets to make use of the potential energy of the propellant in addition to its chemical energy.[^iqvvago67vg]

(The seeming asymmetry is due to the direction of travel. The propellant is itself "thrusting" retrograde by ejecting a rocket behind it, and the Oberth effect causes it to decelerate more efficiently than it would have in deep space. The rocket is thrusting prograde, and gains that energy. The Oberth effect depends on the direction of travel relative to the gravity source at the time of the burn.[^d0u2dgk52qi])

### **Ok, maybe it's about speed after all**

Hmm, my observation that the Oberth effect depends on the direction of travel and not just "being in a gravity well" reminds me of the original claim that the Oberth effect is just a general consequence of being moving. And the typical descriptions of the Oberth effect claim that it applies to speed anywhere, even with no planets in sight.

I think this is just a mathematical curiosity; a natural consequence of the fact that kinetic energy scales with the square of the speed. The K=1/2mv^2^ formula itself is already thing thing that seems to violate relativity, and the Oberth framing just makes this more obvious. Given that kinetic energy is proportional to the square of the speed, it *must* be the case that a faster speed results in a larger gain to kinetic energy.

The solution to the seeming relativity-violation in the formula for kinetic energy is that energy is only conserved between two measurements *from the same reference frame*. If you measure the energy of the universe, then accelerate, then measure it again, you will indeed get a higher number, and that's fine, because you've changed reference frames. The actual numerical value of energy is arbitrary, and all that matters are changes and relative quantities. (Just like how we can either define an object on a planet to have 0 potential energy and it to have positive energy at infinity, or for it to have 0 energy at infinity and negative energy on the planet's surface. Both are equally valid as long as you use them consistently.) So sure, from a stationary perspective an object seems to gain more energy when it accelerates at higher speeds, while from a different stationary perspective where it's going a lower speed it would seem to gain less energy, but that's fine, since those two reference frames are not comparable. (They'd also measure a passing planet as having a different kinetic energy.)

So this aspect of the Oberth effect is unphysical and of no practical consequence when there's no gravity well nearby. The Oberth effect only affects your rocket ship trajectory planning when there's a gravity well that lets you convert kinetic energy into speed at a better-than-normal rate[^3zoax2py6u], with the rate depending on the reference frame of the planet. But when you're in empty space, the Oberth effect just cancels itself out.

[^k4vjlnl1a7c]: With no atmosphere. 

[^ry0vzjc157]: It feels kind of weird to define objects floating in space to have positive potential energy just because there happens to be a planet somewhere else in the universe, which is why we traditionally define the gravitational potential as negative and let the objects be at zero. But in this context I think the positive energy framing is clearer, so that's what I'm using. 

[^iqvvago67vg]: One question this raises is why this effect doesn't compound with the first framing of "spends less time being decelerated by gravity". It almost seems like a 1 m/s burn should have more of an impact on the rocket than a 1 m/s strike by a baseball bat, since both situations have the differential time spent in the gravity well, but in the former case there's also the additional transfer of kinetic energy from the propellant to the rocket. But of course this kinetic energy transfer is non-physical, and since the baseball bat experiences a backwards force from hitting the rocket, the bat case is really just equivalent to an engine burn where the propellant is fired at below escape speed and falls back to the planet. Thinking about this gets me confused about reference frame invariance again. 

[^d0u2dgk52qi]: One interesting questions is at what angle of thrust does the effect on the propellant go from negative to positive? I didn't do the math to check, but I'm pretty sure it's just the angle at which the speed of the propellant in the planet's reference frame is the exact same as the rocket's speed. 

[^3zoax2py6u]: I don't get exactly how this works.