# An Actually Intuitive Explanation of the Oberth Effect

post by Isaac King (KingSupernova) · 2024-01-10T20:23:17.216Z · LW · GW · 33 comments## Contents

Asymmetric gravitational effects It really is gravity, not speed Bringing propellant back into it Ok, maybe it's about speed after all None 33 comments

*This is a linkpost for **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. 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. 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, Quora, Reddit, random physicists' blogs. 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^{[1]} 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 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.^{[2]} 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.^{[3]}

(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.^{[4]})

**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^{[5]}, 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.

^{^}With no atmosphere.

^{^}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.

^{^}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.^{^}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.

^{^}I don't get exactly how this works.

## 33 comments

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## comment by habryka (habryka4) · 2024-01-10T21:07:49.366Z · LW(p) · GW(p)

I would have appreciated an explanation of what the basic Oberth Effect is. Even just the first two sentences from Wikipedia would have made me a bunch more interested in reading the rest of the post:

In astronautics, a

powered flyby, orOberth maneuver, is a maneuver in which a spacecraft falls into a gravitational well and then uses its engines to further accelerate as it is falling, thereby achieving additional speed.^{[1]}The resulting maneuver is a more efficient way to gain kinetic energy than applying the same impulse outside of a gravitational well. The gain in efficiency is explained by theOberth effect, wherein the use of a reaction engine at higher speeds generates a greater change in mechanical energy than its use at lower speeds.

## comment by Charlie Steiner · 2024-01-11T05:28:24.484Z · LW(p) · GW(p)

Remember that work is force · *distance*, not force · time. When reasoning using energy conservation, ignore the time and just think about the distance something happens over.

Force · time is momentum, so when reasoning using momentum, then think about time again.

## comment by Algon · 2024-01-10T21:06:16.760Z · LW(p) · GW(p)

I think that you're still confused about the Oberth Effect, and don't intuitively understand it. But good news! I didn't either, and you've made me confused in a *new* way, which is valuable. Since this post has pretty low karma, I'll give it a weak upvote. But it needs to either: be renamed to reflect that it is a dialect with an uncertain ending or be polished into a correct intuitive explanation so clear that with a bit of thought, readers will realize why the other frames are wrong.

Also, typo:

Replies from: KingSupernovaIf you increase the muzzle speed to

2 km/s, the bullet will be going at2 m/s.

## ↑ comment by Isaac King (KingSupernova) · 2024-01-10T21:15:42.332Z · LW(p) · GW(p)

I'm still confused about some things, but the primary framing of "less time spent subject to high gravitational deceleration" seems like the important insight that all other explanations I found were missing.

## comment by simon · 2024-01-11T09:37:36.193Z · LW(p) · GW(p)

I'm not sure my perspective is significantly different than yours, but:

Using conservation of energy: imagine we have a given amount of mechanical (i.e. kinetic+potential) energy produced by expelling exhaust in the rocket's reference frame. The total mechanical energy change will be the same in any reference frame. But in another reference frame we have:

- the faster the rocket is going, the more kinetic energy the exhaust loses (or less it gains, depending on relative speeds) when it is dumped the other way, which means more energy for the rocket.
- the further down a gravity well you dump the exhaust, the less potential energy it has, which means more energy for the rocket.

Both are important from this perspective, but related since kinetic+potential energy is constant when not thrusting, so it's moving faster when it's down in the gravity well. Yeah, it also works with it using a gun or whatever instead of exhaust, but it's more intuitive IMO to imagine it with exhaust.

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.

I am not quite sure I understand the question, but when the thrust is at 90 degrees to the trajectory, the rocket's speed is unaffected by the thrusting, and it comes out of the gravity well at the same speed as it came in. That would apply equally if there were no gravity well.

Replies from: KingSupernova, dmitry-vaintrob## ↑ comment by Isaac King (KingSupernova) · 2024-01-11T19:49:39.154Z · LW(p) · GW(p)

when the thrust is at 90 degrees to the trajectory, the rocket's speed is unaffected by the thrusting, and it comes out of the gravity well at the same speed as it came in.

That's not accurate; when you add two vectors at 90 degrees, the resulting vector has a higher magnitude than either. The rocket will be accelerated to a faster speed.

Replies from: simon## ↑ comment by simon · 2024-01-12T18:52:20.292Z · LW(p) · GW(p)

In the limit where the perpendicular side vector is infinitesimally small, it does not increase the length of the main vector it is added to.

If you keep thrusting over time, as long as you keep the thrust continuously at 90 degrees as the direction changes, the speed will still not change. I implicitly meant, but did not explicit say, that the thrust is continuously perpendicular in this way. (Whereas, if you keep the direction of thrust fixed when the direction of motion changes so it's no longer at 90 degrees, or add a whole bunch of impulse at one time like shooting a bullet out at 90 degrees, then it will start to add speed).

Replies from: KingSupernova## ↑ comment by Isaac King (KingSupernova) · 2024-01-12T20:13:55.980Z · LW(p) · GW(p)

How is that relevant? In the limit where the retrograde thrust is infinitesimally small, it also does not increase the length of the main vector it is added to. Negligibly small thrust results in negligibly small change in velocity, regardless of its direction.

Replies from: simon## ↑ comment by simon · 2024-01-12T21:11:30.473Z · LW(p) · GW(p)

In the limit where the retrograde thrust is infinitesimally small, it also does not increase the length of the main vector it is added to.

I implicitly meant, but again did not say explicitly, that the ratio of the contribution to the length of the vector from adding an infinitesimal sideways vector, as compared to the length of that infinitesimal vector, goes to zero of as the length of the sideways addition goes to zero (because it scales as the square of the sideways vector).

So adding a large number of tiny instantaneously sideways vectors, in the limit that the size of each goes to zero and holding to the total amount of thrust added constant, in that limit results in a non-zero change in direction but zero change in speed.

Whereas, if you add a large number of tiny instantaneous aligned vectors, the ratio of the contribution to the length of the vector to the length of each added tiny vector is 1, and if you add up a whole bunch of such additions, it changes the length and not the direction, regardless of how large or small each addition is.

Replies from: KingSupernova## ↑ comment by Isaac King (KingSupernova) · 2024-01-13T01:22:59.356Z · LW(p) · GW(p)

I don't understand how that can be true? Vector addition is associative; it can't be the case that adding many small vectors behaves differently from adding a single large vector equal to the small vectors' sum. Throwing one rock off the side of the ship followed by another rock has to do the same thing to the ship's trajectory as throwing both rocks at the same time.

Replies from: simon## ↑ comment by simon · 2024-01-13T02:30:56.385Z · LW(p) · GW(p)

Yes, it's associative. But if you thrust at 90 degrees to the rocket's direction of motion, you aren't thrusting in a constant direction, but in a changing direction as the trajectory changes. This set of vectors in different directions will add up to a different combined vector than a single vector of the same total length pointing at 90 degrees to the direction of motion that the rocket had at the start of the thrusting.

Replies from: KingSupernova## ↑ comment by Isaac King (KingSupernova) · 2024-01-13T04:03:43.334Z · LW(p) · GW(p)

...Are you just trying to point out that thrusting in opposite directions will cancel out? That seems obvious, and irrelevant. My post and all the subsequent discussion are assuming burns of epsilon duration.

Replies from: simon## ↑ comment by simon · 2024-01-13T06:03:58.162Z · LW(p) · GW(p)

...Are you just trying to point out that thrusting in opposite directions will cancel out?

No.

I'm pointing out that continuous thrust that's (continuously during the burn) perpendicular to the trajectory doesn't change the speed.

This also means that (going to your epsilon duration case) if the burn is small enough not to change the direction very much, the burn that doesn't change the speed will be close to perpendicular to the trajectory (and in the low mass change (high exhaust velocity) limit it will be close to halfway between the perpendiculars to the trajectory before and after the burn, even if it does change the direction a lot). That's independent of the exhaust velocity, as long as that velocity is high, and when it's high it will also tend not to match the ship's speed since it's much faster, which maybe calls into question your statement in the post, quoted above, which I'll requote:

Replies from: KingSupernovaOne 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.

## ↑ comment by Isaac King (KingSupernova) · 2024-01-14T23:24:38.281Z · LW(p) · GW(p)

Ok now I'm confused about something. How can it be the case that an instantaneous perpendicular burn adds to the craft's speed, but a constant burn just makes it go in a circle with no change in speed?

Replies from: simon## ↑ comment by simon · 2024-01-15T20:00:04.531Z · LW(p) · GW(p)

The trajectory is changing during the continuous burn, so the average direction of the continuous burn is between perpendicular to where the trajectory was at the start of the burn and where it was at the end. The instantaneous burn, by contrast, is assumed to be perpendicular to where the trajectory was at the start only. If you instead made it in between perpendicular to where it was at the start and where it was at the end, as in the continuous burn, you could make it also not add to the craft's speed.

Going back to the original discussion, yes this means that an instantaneous burn that doesn't change the speed is pointing slightly forward relative to where the rocket was going at the start of the burn, pushing the rocket slightly backward. But, this holds true even if you have a very tiny exhaust mass sent out at a very high velocity, where it obviously isn't going at the same speed as the rocket in the planet's reference frame.

Replies from: KingSupernova## ↑ comment by Isaac King (KingSupernova) · 2024-01-17T21:56:54.078Z · LW(p) · GW(p)

I don't understand what "at the start" is supposed to mean for an event that lasts zero time.

Replies from: simon## ↑ comment by simon · 2024-01-18T01:24:01.291Z · LW(p) · GW(p)

In the case where it's instantaneous, "at the start" would effectively mean right before (e.g. a one-sided limit).

## ↑ comment by Dmitry Vaintrob (dmitry-vaintrob) · 2024-01-11T19:23:40.203Z · LW(p) · GW(p)

Right - looking at energy change of the exhaust explains the initial question in the post: why energy is preserved when a rocket accelerates, despite apparently expending the same amount of fuel for every unit of acceleration (assuming small fuel mass compared to rocket). Note that this doesn't depend on a gravity well - this question is well posed, and well answered (by looking at the rocket + exhaust system) in classical physics without gravity. The Oberth phenomenon is related but different I think

Replies from: simon## comment by polytope · 2024-01-11T21:07:11.138Z · LW(p) · GW(p)

Here's my intuition-driving example/derivation.

Fix a reference frame and suppose you are on a frictionless surface standing next to a heavy box equal to your own mass, and you and the box always start at rest relative to one another. In every example, you will push the box leftward, adding 1 m/s leftward velocity to the box, and adding 1 m/s rightward velocity to yourself.

Let's suppose we didn't know what "kinetic energy" is, but let's suppose such a concept exists, and that whatever it is, an object of your mass has 0 units of it when at rest, and it is a continuous monotonic function of the absolute value of that object's velocity. Let's also take as an assumption that when you perform such a push like the above, you are always adding precisely 1 unit of this thing called "kinetic energy" to you and the box combined.

Okay so suppose the box and you are at rest and you perform this push, and start moving at 1m/s left and right, respectively. You and the box started with 0 units of kinetic energy, and you added 1 unit total. Since you and the box have the same absolute value of velocity, your energies are equal, so you each must have gotten 1/2 of a unit. Great, therefore we derive 1 m/s is 1/2 unit of kinetic energy.

Now suppose you and the box *start out* at a velocity of 1m/s rightward, so you have 1/2 unit of energy each, for a total of 1 unit. You perform the same push, bringing the total kinetic energy to 2 units. The box ends up at 0 m/s, so it has 0 units of energy now. You end up going 2m/s rightward, with all the energy. Great, therefore we derive 2 m/s is 2 units of kinetic energy.

Now suppose you and the box *start out* at a velocity of 2m/s rightward, so you have 2 units of energy each, for a total of 4 units. You perform the same push, bringing the total kinetic energy to 5 units. The box ends up at 1 m/s, so it has 1/2 unit of energy now, since we derived earlier that 1m/s is 1/2 unit of energy. You end up going 3m/s rightward. So you must have the other 4.5 units of energy. Therefore we derive 3 m/s is 4.5 units of kinetic energy.

We can continue this indefinitely, without running into any inconsistencies or contradictions. This "kinetic energy" thing so far seems to be a self-consistent concept given these assumptions! In general, we derive that an object of our mass moving at velocity v is has a kinetic energy of 1/2 v^2 units.

And I hope this makes it clearer why kinetic energy has to behave quadratically. A quadratic function f is precisely the kind of function such the quantity f(x+c) + f(x-c) - 2f(x) is constant with respect to x. It's the only function that satisfies the property that a fixed "amount of push" of the propellant you are carrying away from you always adds the same total energy into the combined system of you + propellant.

And it also gives some intuition for why *you* end up with more energy when you fire the propellant while moving faster. When you pushed the box while initially at 0 m/s, your kinetic energy went from 0 units to 0.5 units (+0.5), but when you pushed the box while initially at 1 m/s, your kinetic energy went from 0.5 units to 2 units (+1.5), and when you pushed the box while initially at 2 m/s, your kinetic energy went from 2 units to 4.5 units (+2.5) and in all cases you only added 1 unit of energy yourself. Where does the extra energy come from? From slowing down the box's rightward motion, and/or from not speeding up the box to go leftward from rest.

## comment by Thomas Kwa (thomas-kwa) · 2024-01-10T23:47:12.598Z · LW(p) · GW(p)

This sounds overcomplicated to me compared to an explanation I once heard. Suppose you are orbiting a planet, you have a gun with a single bullet, and you want to maximize your total energy after firing the bullet backwards. At what point in the trajectory should you shoot the bullet? Since energy is conserved, you want to **minimize the bullet's energy**. Because KE = mv^2, you have maximal ability to decrease (or increase) the bullet's energy when its KE is already highest, and because KE+U is constant, this is at the lowest point in your trajectory.

I don't remember where I first heard this.

Replies from: daniel-kokotajlo## ↑ comment by Daniel Kokotajlo (daniel-kokotajlo) · 2024-01-11T00:00:46.787Z · LW(p) · GW(p)

Is this also how swinging on swings works? How does swinging on swings work exactly? Huh.

Replies from: KingSupernova, hastings-greer## ↑ comment by Isaac King (KingSupernova) · 2024-01-11T02:07:29.061Z · LW(p) · GW(p)

I don't think so. The difference in the gravitational field between the bottom point of the swing arc and the top is negligible. The swing isn't an isolated system, so you're able to transmit force to the bar as you move around.

There's a common explanation you'll find online of how swings work by you changing the height of your center of mass, which is wrong, since it would imply that a swing with rigid bars wouldn't work. But they do.

The actual explanation seems to be something to do with changing your angular momentum at specific points by rotating your body.

## ↑ comment by Hastings (hastings-greer) · 2024-01-15T20:36:27.156Z · LW(p) · GW(p)

There are two completely distinct ways to swing on a swing- You can rotate your body relative to the seat-chain body at the same frequency as your swinging but out of phase, or move your center of mass up and down the chain at twice the frequency. The power of the former is ~ torque applied to chain \* angular velocity, the power output of the latter is radial velocity of your body \* (angular velocity ^2 \* chain length).

To get to any height, you have to switch from one to the other once the angular velocity ^2 term dominates- this is why learning to swing is so unintuitive.

## comment by Samuel Hapák (hleumas) · 2024-01-11T23:37:11.304Z · LW(p) · GW(p)

It feels to me that you lack a good intuition for how kinetic energy, momentum conservation,, Newton laws, and Galileo's relativity all play together and this is causing you confusion.

Relativity says that there is no objective notion of "being still". We can't objectively distinguish between being still and moving at constant velocity (same speed, same direction).

2nd Newton law: Force is equal to mass times acceleration: .

3nd Newton law: All forces in the nature exhibit the property that if object A acts on B with a force F, then B acts on A with exactly opposite force.

1st Newton law is boring, it just says that if F = 0, then a = 0.

Force and Energy are tied together through the concept of work which says that change in objects energy = work received by that object. And formula for work is , notice the dot product here! Force perpendicular to movement doesn't make any impact on energy.

Putting all of this together, it's useful for you to do following exercises:

- Try to derive the formula for kinetic energy.
- Try to derive the formula for potential energy in the gravitational field of Earth near the surface (neglecting change of gravitational force with height).
- Try to derive the law of conservation of momentum.

Once you do the three above, I have a tricky paradox problem for you to solve:

Imagine two cars driving on the road with the speed . Suddenly the first car accelerates to double the speed, thus travelling . This naturally required consumption of energy in the form of fuel. The change in kinetic energy was: and this should somehow correspond to the amount of the fuel consumed.

However, from the point of the view of the second driver, the car was starting still and accelerated to the velocity , thus the change in energy is simply .

But, this is paradox. It's not possible that the car would spend 3 times less fuel from the point of view of the other car than from the point of view of the observer standing still on the ground.

Can you explain this paradox?

Replies from: KingSupernova## ↑ comment by Isaac King (KingSupernova) · 2024-01-12T10:06:30.499Z · LW(p) · GW(p)

Unfortunately I already came across that paradox a day or two ago on Stack Exchange. It's a good one though!

Yeah, my numerical skill is poor, so I try to understand things via visualization and analogies. It's more reliable in some cases, less in others.

## comment by Joseph Miller (Josephm) · 2024-01-12T22:18:44.296Z · LW(p) · GW(p)

This is a great post.

## comment by Shankar Sivarajan (shankar-sivarajan) · 2024-01-11T18:45:01.606Z · LW(p) · GW(p)

"Rocket fuel provides a fixed *impulse *('delta-v') and not a fixed change in kinetic energy."

For most people, intuition conflates the two, and understanding that they're different adequately explains the effect.

Of course there's a perfectly correct explanation for it all reference frames, but the frame of the planet is a good choice.

What you're doing sounds to me like saying "I don't think the concept of centrifugal force is intuitive. I shall explain, say, how roller coaster loop-the-loops work without using it."

## comment by Boris Kashirin (boris-kashirin) · 2024-01-11T01:58:01.450Z · LW(p) · GW(p)

Important fact about rocket is that it provides some fixed amount of change of velocity (delta-v). I think your observation of "how long strong gravity slows me" combined with thinking in terms of delta-v budget and where it is best spent brought intuitive understanding of Oberth effect to me. Analysing linear trajectory instead of elliptic one also helps.

## comment by qvalq (qv^!q) · 2024-01-11T19:43:03.874Z · LW(p) · GW(p)

kinetic energy scales with the square of the speed

Why is this?

Replies from: hleumas## ↑ comment by Samuel Hapák (hleumas) · 2024-01-11T23:17:17.519Z · LW(p) · GW(p)

That's an excellent question!

Change in energy must equal work done on the moving body .

Now, work is force over distance

But we also know that force is mass times acceleration, thus .

If you have a body moving under the influence of constant force over time , starting from the speed 0, it will have a speed at the end. It's easy to see that the average speed it travels will be though and thus the distance travelled will be .

Now, the kinetic energy equals work and thus .

Replies from: qv^!q## ↑ comment by qvalq (qv^!q) · 2024-01-17T15:25:09.296Z · LW(p) · GW(p)

I've learned the maths before.

I think maybe I have no idea what kinetic energy is.

Replies from: hleumas## ↑ comment by Samuel Hapák (hleumas) · 2024-01-18T11:03:40.240Z · LW(p) · GW(p)

Well, the correct question is “What is energy”. And the answer is that energy is some number that we can compute for any physical system and it doesn’t change no matter what as long as the system is reasonably isolated from its surroundings. Kinetic energy is just a portion of this quantity we can compute for something that is moving.

It’s not very intuitive honestly. The best explanation for what energy is I ever read is this one from Feynman: