# Why is the Oberth Effect still applicable when reaching interplanetary space?

## Recommended Posts

This is embarrassing, because I've been playing the game for years, but I realized something that I don't fully understand. I sort of get the Oberth Effect.  In my mind it boils down to “maneuvers are more efficient when you perform them close to a gravity well”. But once you reach interplanetary space, you've removed the "gravity well" part.  The sun is pretty far away regardless of what point you consider (around Kerbin's orbit).  Whether you end up at 12 or 13 billion meters above the sun, it's all roughly the same. So I imagined there wouldn't be much difference between leaving Kerbin, then burning towards Jool. Versus leaving Kerbin while also burning towards Jool.

I experimented with the maneuver node and found out it's not true at all.  There is a huge difference between the two.  In both cases, leaving Kerbin (from a Low circular orbit) costs around 940 m/s. But then pushing a bit more until you hit Jool costs about 1000 m/s. Whereas heading towards Jool (from simply sitting in interplanetary space at Kerbin's altitude from the sun) costs an enormous 2500 m/s. So my assumed scenario actually ends up costing 3440 m/s compared to just 1940 m/s in the optimal scenario.

In this instance I would have imagined the gravitational pull from Kerbin would have counteracted the Oberth effect. Kerbin is pulling against us, preventing us from leaving. Whereas the lone satellite in interplanetary space is just chilling.   But the satellite ends up at a disadvantage compared to the spacecraft that has to claw it's way out of a gravity well.  Does someone have a more thorough explanation for how the oberth effect is not only applying here, but ends up producing a huge difference?

Edited by PTNLemay

##### Share on other sites
12 minutes ago, PTNLemay said:

“maneuvers are more efficient when you perform them close to a gravity well”

Closer is "Prograde and retrograde maneuvers are more efficient the faster you're going."

Kinetic energy is mass times speed squared. So if you go from 1 to 2 speed units, you go from 1 to 4 kinetic energy units, gaining 3 kinetic energy units. If you go from 10 to 11 speed units you go from 100 to 121 kinetic energy units, for a gain of 21 kinetic energy units.

The fact that you tend to go faster lower in a gravity well is ancillary but by no means required or even part of it. The Oberth Effect for you is garbage on the launchpad, for instance, and that's closer to the COM than in LKO.

Edited by 5thHorseman

##### Share on other sites

That reminds me of another physics rule I learned, but never quite fully understood.

Say you have a gun that can impart a very specific amount of energy onto a bullet.   Say the bullet is 1 kg, and we impart 10000 joules of energy into it.  The bullet starts at rest.  By running the kinetic energy equation, we get afinal velocity of around 141 m/s.

But now we are on a train that is already going at 500 m/s (let's say it's on the moon or something with no air resistance).  You have that same gun, imparting that same amount of kinetic energy.  If we fire it while this train is moving, I would imagine the bullet would leave the barrel of the gun travelling around 1141 m/s (relative to the side of the road).  The speed of the train, plus the speed we added to it when it was at rest.  But if we put those numbers in the kinetic energy equation from earlier, using final speed energy minus initial speed energy, we get an energy difference of over 80000 joules.  Where did the extra energy come from?  The bullet's gunpowder only had so much.  I think this is the amount of energy that the bullet would release if it impacted the side of the road, but only the energy it gained from the gun, not the energy it gained from the train.

I see what the equations tell us, but it doesn't sit right in my head.

Edited by PTNLemay

##### Share on other sites
7 minutes ago, PTNLemay said:

But now we are on a train that is already going at 500 m/s. ...  Where did the extra energy come from?

Depends what kind of train it was. Probably diesel?

##### Share on other sites

But it couldn't have come from the train.  The train's speed is 500 m/s, putting that into the equation we get 125000 joules imparted onto the bullet.  That's the energy we would read if we just chucked the bullet off the train without firing it from the gun.

Now say we do fire the gun (while riding the train), which was designed to impart a specific 10000 joules onto the bullet.  When the bullet hits the side of the road, it releases over 200 kilojoules, more than the combined energy imparted by the train and the bullet.

Edited by PTNLemay

##### Share on other sites

The gun-on-a-train analogy is an interesting one.
(With the 500m/s fast train carrying a gun shooting a bullet at 141m/s,  I suppose you meant to say the bullet was 641m/s relative to the land outside.)

Given the limited energy in the gunpowder, the gun is able transfer a much larger amount of energy from the moving train to the bullet, when we measure energy in the reference frame of the land outside.

Given the way you set up the problem, probably doing the accounting will make you certain this is the correct explanation.  Suppose the bullet mass is m and the train mass is M.
The train slows down by a tiny amount (m/M)×141m/s.  Working out the changes in energy in detail, I figure an amount of energy m×141m/s×500m/s was transferred from the train to the bullet, in addition to the m(141m/s)²/2  from the gunpowder.

I don't have an intuitive explanation at the moment.  "from diesel" doesn't do it for me either.  Maybe there is another analogy where the same physics seems intuitive, and we can broaden the correct intuition from there.

##### Share on other sites

Scientists long ago realized that if you trust your instincts and ignore the data in front of your face you won't get anywhere.

##### Share on other sites

Of course I know it's me who's making an error in reasoning somewhere and the physics must be sound.  I've just never been able to fully visualize what's happening in these sort of problems.  And it bugs me.  Or like the Oberth thing where I thought I understood, but didn't.

And it's funny, whatever that discrepancy is in the original question (between what I thought and how it really is), I think it scales bigger and bigger with the further the planet you go to.  So when I went to Duna I didn't notice it as much, and I so rarely go to Jool that I never really realized I was being so inefficient with my interplanetary maneuvers to it.

##### Share on other sites

I think the point is that when you're already on escape trajectory, you're no longer being held up by the gravity, quite the opposite. Just look at the maneuver preview when you set up a node - it needs less and less d-v the higher your new Ap is going to be, and it stays true even when the rest of the trajectory is far away in interplanetary space.

Also, gravity assists. They are being used to change the trajectory, but if the engines are fired(?) at the Pe, at the highest speed, they can get even more from that maneuver. That's the oberth effect.

##### Share on other sites
8 hours ago, PTNLemay said:

So I imagined there wouldn't be much difference between leaving Kerbin, then burning towards Jool. Versus leaving Kerbin while also burning towards Jool.

I experimented with the maneuver node and found out it's not true at all.  There is a huge difference between the two.  In both cases, leaving Kerbin (from a Low circular orbit) costs around 940 m/s. But then pushing a bit more until you hit Jool costs about 1000 m/s. Whereas heading towards Jool (from simply sitting in interplanetary space at Kerbin's altitude from the sun) costs an enormous 2500 m/s. So my assumed scenario actually ends up costing 3440 m/s compared to just 1940 m/s in the optimal scenario.

Well, what you have "discovered" there is exactly the Oberth effect! If you do your transfer burn to Jool close to Kerbin then you need less dV than when you do it far away from Kerbin (at about the same distance to the sun).

And in contrast to what @5thHorseman wrote, for practical maneuver planning the gravity well is important. Kinetic energy depends on your reference frame! So while the "energy is proportional to the square of velocity" explanation might be more "physical", I found another way to look at it more useful:
When you are in LKO you already have some significant speed (that gets added to the orbital velocity of Kerbin around the Sun if you use that as a reference frame). If you only accelerate to Kerbins SOI so that you cross the SOI boundary with no speed relative to Kerbin, then you "lost" all that speed because Kerbin's gravity slowed you down. If you accelerate faster in LKO than just to get to the SOI boundary, then you spend less time in Kerbin's gravity well and thus get slowed down less. So if you you have 100 m/s more than needed to get to the SOI boundary in LKO then you'll cross the SOI boundary with more than 100 m/s. And the more you accelerate in LKO, the more of that orbital velocity (of the LKO) you can take with you across the SOI boundary. (Same goes for capturing in orbit: you want to slow down so that you don't have enough speed to make it to the SOI boundary, so you want to spend more time being slowed down by the CB that you want to orbit, so do you capture burn at PE.)
Now, this way of looking at the Oberth effect is harder to calculate - I actually would go back to the "energy"  view if I had to calculate it - but I find it more useful when planning maneuvers.

About the "bullet shot from a train" question:

7 hours ago, OHara said:

The train slows down by a tiny amount (m/M)×141m/s.  Working out the changes in energy in detail, I figure an amount of energy m×141m/s×500m/s was transferred from the train to the bullet, in addition to the m(141m/s)²/2  from the gunpowder.

Yupp, that's it. Independent of the mass of the train and its initial velocity, the train gets slowed down so much that it looses as much kinetic energy as the bullet gains when you shift the reference frame.

Edited by AHHans
fixed typo

##### Share on other sites
17 hours ago, PTNLemay said:

I experimented with the maneuver node and found out it's not true at all.  There is a huge difference between the two.  In both cases, leaving Kerbin (from a Low circular orbit) costs around 940 m/s. But then pushing a bit more until you hit Jool costs about 1000 m/s. Whereas heading towards Jool (from simply sitting in interplanetary space at Kerbin's altitude from the sun) costs an enormous 2500 m/s. So my assumed scenario actually ends up costing 3440 m/s compared to just 1940 m/s in the optimal scenario.

In this instance I would have imagined the gravitational pull from Kerbin would have counteracted the Oberth effect. Kerbin is pulling against us, preventing us from leaving. Whereas the lone satellite in interplanetary space is just chilling.   But the satellite ends up at a disadvantage compared to the spacecraft that has to claw it's way out of a gravity well.  Does someone have a more thorough explanation for how the oberth effect is not only applying here, but ends up producing a huge difference?

This isn't really an apples to apples comparison.  What's happening here is not just that you have the Oberth effect in play, when you're comparing to a vessel that's  in Kerbin's orbit but not circling Kerbin - you also have the vessel's velocity around Kerbin itself.

Think about it this way - if you suddenly deleted Kerbin when you're making your prograde maneuver from LKO, assuming you're in a 0degree inclination orbit, you would not be at 9285m/s around the Sun.  You'd be at 9285m/s + your Kerbin orbit velocity - so ~ 2300m/s = 11585m/s - enough by itself to do the Jool transfer.

When you make that comparison, you can see that it is still more expensive to depart for Jool from Kerbin, when you're in its gravity well - you just have a lot of kinetic energy in your orbit around Kerbin that you're leveraging.

##### Share on other sites
On 2/11/2020 at 9:55 PM, PTNLemay said:

That reminds me of another physics rule I learned, but never quite fully understood.

Say you have a gun that can impart a very specific amount of energy onto a bullet.   Say the bullet is 1 kg, and we impart 10000 joules of energy into it.  The bullet starts at rest.  By running the kinetic energy equation, we get afinal velocity of around 141 m/s.

But now we are on a train that is already going at 500 m/s (let's say it's on the moon or something with no air resistance).  You have that same gun, imparting that same amount of kinetic energy.  If we fire it while this train is moving, I would imagine the bullet would leave the barrel of the gun travelling around 1141 m/s (relative to the side of the road).  The speed of the train, plus the speed we added to it when it was at rest.  But if we put those numbers in the kinetic energy equation from earlier, using final speed energy minus initial speed energy, we get an energy difference of over 80000 joules.  Where did the extra energy come from?  The bullet's gunpowder only had so much.  I think this is the amount of energy that the bullet would release if it impacted the side of the road, but only the energy it gained from the gun, not the energy it gained from the train.

I see what the equations tell us, but it doesn't sit right in my head.

This is, as @OHara, said, an interesting analogy.

What follows in the spoiler is the extremely maths-heavy version, which I add for the benefit of others who do not know the equations.  Since you do, I'll do my best to give you a more intuitive approach.

Spoiler

Let's work through what it says in terms of numbers and see what we can figure out.

We have a bullet, a train, 10,000 joules of energy on the bullet, a bullet velocity of 141.4214 m/s, and a train velocity of 500 m/s.  The kinetic energy equation is (1/2)mv2, so if we run those numbers, then we get a 1 kg bullet--I will continue to call it a bullet and not a cannonball.

I'm not certain about how you get 1141 m/s when you fire the gun from the moving train.  I would expect 641.4124 m/s.  In terms of reference frames, the bullet is still travelling 141.4214 m/s faster than the 500 m/s train.  Things get more complicated at relativistic speeds, but this isn't anywhere near that amount of velocity.

If the bullet is now travelling at 641.4214 m/s, and to an observer on the surface of the moon (or otherwise at rest), it is, then its kinetic energy is now 205,711 joules.  Before we fired the gun, the bullet was travelling at 500 m/s, which gives it a kinetic energy of 125,000 joules, for a difference of 80,711 joules, as you said.  That's 70,711 joules more than you get when the bullet starts from rest.  Where did it get the extra energy?

What if we change the speed and direction?  Let's say that the train is travelling at 141.4214 m/s in the opposite direction and we fire the gun out the back.  The bullet still leaves the barrel at 141.4214 m/s relative to the train, but relative to the Moon (and observers at rest), it has zero velocity, zero energy, and falls to the dust of the lunar surface.  Where did the energy go?

Clearly, we're missing something.

The equation for kinetic energy is fairly straightforward to derive.  We begin with work-energy equivalence and go from there:

W = ΔKE = ∫F * ds

Where:

W = work
ΔKE = total change in kinetic energy
F = force
ds = instantaneous change in displacement (the difference between distance and displacement is the same as that between speed and velocity; one has a magnitude only, and the other has a magnitude and direction)

The ∫ symbol is the integral sign.

Force, by Newton's Second Law, is the time-derivative of momentum:

ΔKE = ∫ (d(mv) / dt) * ds

Where m is mass, v is velocity, the quantity mv is momentum, and dt is the instantaneous change in time.

We can consider mass to be constant in this example:

ΔKE = m ∫ (dv / dt) * ds

Now we can rearrange the differential terms:

ΔKE = m ∫ (ds / dt) * dv

Note that the differential (ds / dt) is equivalent to velocity (velocity is the time-derivative of displacement, which is what ds / dt represents):

ΔKE = m ∫ v * dv

Integrate from initial to final velocity:

ΔKE = m (1/2) v2 ]fi

The odd thing that I did with the square bracket indicates that we can solve this equation by entering the final velocity, vf, and subtracting from it the initial velocity, vi, which yields this:

ΔKE = m (1/2) vf2 - m (1/2) vi2

Generally, the assumption is that the object begins at rest, so the initial velocity (and thus energy) is zero and the second term disappears.  That's not the case here.  Even so, although we can see that the extra energy is now accounted, it still doesn't tell us from where that energy came in the first place.

We can still look at this in new ways.  For example, we know that energy is conserved.  However, that energy can change forms, such as from potential to kinetic and vice versa.  This exchange between forms is what causes spacecraft in elliptical orbits to change speed as they move from apoapsis to periapsis and back again.  In the case of the train and the bullet, clearly there is an exchange of energy taking place.  However, although the total energy in the system must be conserved, there is no law requiring the exchange of energy to be equally distributed.

On the other hand, conservation of momentum does require an equal exchange.  We can look at this problem in terms of momentum, then, and see whether we get new insights.

Specifically, conservation of momentum says that the momentum of the train-bullet combination must be the same after the bullet is fired as it was before the bullet was fired.  In other words, no matter what the initial momentum was, if we take the train momentum and bullet momentum as separate quantities, where the train is initially mass M and velocity V, the bullet is initially mass m and velocity V, and the changes in velocity are ΔV and ΔU respectively, then we get this:

MV + mV = M (V + ΔV) + m (V + ΔU)
MV + mV = MV + MΔV + mV + mΔU
MV + mV = MV + mV + mΔU + MΔV
0 = mΔU + MΔV
MΔV = -mΔU

Note that I used ΔU for the change in the bullet's velocity because its change is different from the train's.  I could call the bullet's initial velocity U and even have that be different from the train, but that would describe a collision.  It's important physically, but it is not relevant when the scenario calls for the train to carry the bullet until it is fired.

We can combine what we know into one large equation:

Energy before the gun is fired:

KEi = (1/2) (M + m) V2

Energy afterwards:

KEf = (1/2) M (V - ΔV)2 + (1/2) m (V + ΔU)2

And the difference between them:

ΔKE = (1/2)M(V - ΔV)2 + (1/2)m(V + ΔU)2 - (1/2)(M + m)V2
ΔKE = (1/2) (MV2 - 2MVΔV + M[ΔV]2) + (1/2) (mV2 + 2mVΔU + m[ΔU]2) - (1/2) (MV2 + mV2)
ΔKE = (1/2) (MV2 - 2MVΔV + M[ΔV]2 + mV2 + 2mVΔU + m[ΔU]2 - MV2 - mV2)
ΔKE = (1/2) (2MVΔV + M[ΔV]2 + 2mVΔU + m[ΔU]2)
ΔKE = MVΔV + (1/2)M[ΔV]2 + mVΔU + (1/2)m[ΔU]2

Now, a caveat:  what is ΔU?  From conservation of momentum above, we know that MΔV = -mΔU.  This gives an equivalence of ΔU = -(M/m)ΔV.  Substitute it here:

ΔKE = MVΔV + (1/2)M[ΔV]2 - mV[(M/m)ΔV] + (1/2)m[ΔU]2
ΔKE = MVΔV + (1/2)M[ΔV]2 - MVΔV + (1/2)m[ΔU]2
ΔKE = (1/2)M[ΔV]2 + (1/2)m[ΔU]2

This tells us that the total change in kinetic energy is equal to the change in kinetic energy of the train added to the change in kinetic energy of the bullet, which we would expect.  But attend very closely to what else it says:  there is no longer any reference to initial velocity, V.  The energy change is the same whether the train is at rest or moving at speed.

However, let's look at what the actual kinetic energy is before the bullet is fired.  It is already travelling at 500 m/s; I won't detail the equation.  The kinetic energy of a 500 m/s, 1 kg bullet is 125 kilojoules.  If we add 10 kilojoules to that, then we get 135 kilojoules--but that results in a total velocity of 519.6 m/s.  We need to account for the train again.

This time, we need to know the train's mass.  You did not give it, so I will assume a 1,000 tonne train.  This may be inaccurate with respect to reality; I am unfamiliar with lunar rail freight systems.

A 1,000,000 kg train travelling on the airless lunar surface over a straight, level track at a speed of 500 m/s has a kinetic energy of 125 gigajoules and a momentum of 5 x 108 kg*m/s (There's no unit for momentum).  The bullet's change in momentum is 141.4214 kg*m/s (from its 500 m/s initial state), so the train's change is the negative of that.  This yields a total of 499.999858579 x 108 kg*m/s, which, when divided out, gives a final velocity of 499.999858579 m/s.

ΔKE = (1/2)M[Vf]2 - (1/2)M[Vi]2
ΔKE = (1/2)(1,000,000)[499.9998585792 - 5002]
ΔKE = (1/2)(1,000,000)(249,999.858579 - 250000)
ΔKE = (1/2)(1,000,000)(-.141421)
ΔKE = -70,711

Add the ten thousand joules that you imparted to the bullet initially, and you get 80,711 joules.  Your extra energy came from the train--but more particularly, from the train's own motion.

The most common cause of confusion with the Oberth Effect is that people look at the relationship between the spacecraft and the nearest planet.  This is completely wrong.  Orbital energy is certainly a factor if you happen to be near a gravitational body, but only insofar as orbital energy gives an easy conversion between potential and kinetic energy.  The effect, itself, has to do with the way the kinetic energy is used, not how it is obtained, and so the really important relationship is between the rocket and its propellant, not the rocket and its orbit.  Let's take your train analogy, but instead of a bullet, let's use a tennis ball.  When the train is at rest, I can throw the ball some distance, x.  This is a representation of the chemical energy in the propellant.

When the train is in motion, I can throw the ball farther, x + y--provided I throw the ball forwards.  Prograde motion is important here.  If I stand next to the tracks and throw the ball at the oncoming train, then assuming no other velocity losses, the ball rebounds off the front of the train and goes a distance x + y (from the point of impact), as well.  Why does it do this?  The collision between the ball and the train is perfectly elastic, which is to say that the kinetic energy in it is conserved.  An inelastic collision is one where the ball enters the train and is lodged there--this is why I didn't continue to use the bullet analogy.  A physical representation of an inelastic collision in rocketry would be when the engine combustion chamber ruptures.  That is probably undesirable.

The important point to see here is that it doesn't matter whether I'm standing on the train or standing next to it; the train imparts energy to the ball at the instant the ball contacts it, or more specifically, the ball has the gained energy at the instant it breaks contact with the train (or my hand, which, when I'm standing on the train, is part of the train).  I certainly did nothing to make the ball go the extra +y distance.  Since my arm (or the chemical energy in the rocket propellant) can only impart a fixed amount of energy, and that energy only can carry the ball x distance, the energy to go farther must have come from the train.

In exact like kind, the extra energy from the Oberth Effect must come from something other than the chemical energy in the rocket propellant.  Unfortunately, our choices are very limited:  we have the rocket itself, which receives the energy and therefore cannot supply it, and we have the rocket propellant again, examined outside of the context of its chemical energy.

What, physically, happens when we burn propellant in a rocket engine?  When we burn it, the propellant exhaust products bounce about inside the combustion chamber, either off of the walls or off of other molecules of exhaust or propellant, until they are forced out of the nozzle.  By Newton's Third Law, the action of forcing the exhaust out the nozzle forces the rocket, with equal momentum, to move in the opposite direction.  In a physical sense, the exhaust bounces off of the forward wall of the combustion chamber (or, more accurately, the propellant between the forward wall and the nozzle) and out.  Efficiency can be increased with the engine bell, which directs exhaust moving at an angle to a more retrograde direction, again forcing the rocket to move in the opposite direction.  I'm certain that this is all elementary to you and I apologise for that, but consider, since the reaction in Newton's Third Law is equal, isn't it equally correct to say that the rocket bounces off of the exhaust?

Normally, one would not think of bouncing off of an exhaust plume so much as ploughing through it, because the exhaust, as a gas, is of course a fluid.  But the fluid is confined and it is made of matter; even though it is not dense, it is there, and so the rocket bounces off of it.  It is also true that this propellant exhaust begins by initially moving at speed with the rocket.  That much should be obvious; the propellant, until it exits the nozzle, is carried by the rocket.  The fact that the propellant must accelerate itself as well as the rocket is what leads to the tyranny of the rocket equation, but in this case, it is a somewhat benevolent tyranny because that acceleration results in higher energies that we can usefully extract.

Particularly, because the propellant does not stick to the engine or the rest of the rocket, the collision between it and the rocket is perfectly elastic.  Since the propellant, being carried by the rocket, has greater kinetic energy when it and the rocket collide while in motion than it would have if the rocket were at rest, more energy can be exchanged between them in that collision.

There is another common application of this that many people miss.  Have you considered why we burn the propellant in the first place?  Newton's Law only really requires that the momentum exchange be equal; there's nothing keeping you from, for example, venting the propellant out the back without burning it.  If you take a good slingshot and a ready supply of stones, then you can even have a rocket with real rocks in it.

When the propellant is burned, the chemical bonds are broken, and this liberates a lot of energy.  We see much of that energy as heat and pressure, but heat is also represented as the kinetic energy of molecular motion, and pressure is energy applied per unit area.  Pressure, in particular, can be increased by increasing the temperature of a confined gas or by forcing more molecules into a confined volume; rocket propellant mixtures can be selected to do either or both of these.  Combine combustion chamber pressure with a nozzle opening that has a cross-sectional area, and the kinetic energy bound up in the pressure can be liberated.  Consider that the increased temperature of the molecules is physically equivalent to the molecules moving faster, and you can begin to see that, when directed out the nozzle as exhaust, the burned propellant with its faster-moving molecules imparts more kinetic energy to the rocket than would cold, un-burned propellant that is vented more slowly.  Does this mean that the Oberth Effect works inside rocket engines to increase their efficiency by extracting more useful energy from faster-moving combustion products than can be had simply from venting the propellant?  Technically, yes.

This also means that, technically, burning the propellant is not necessary; as I mentioned above, the important part is how you use the energy, not how you obtain it.  You can obtain the kinetic energy simply by increasing the propellant temperature and pressure to levels normally seen in combustion, and it will work equally well.  That is exactly how nuclear thermal rockets work.  It isn't even necessary to heat or pressurise the propellant:  if we can increase the kinetic energy by increasing the velocity of the molecules directly, then that will also work, and that is how ion engines do it.

Nothing about the Oberth Effect is any more complicated than understanding that if I bounce a ball off of a moving train, then it will go farther than bouncing a ball off of a train at rest.  If I bounce a ball off of a train moving even faster, then the ball will go even farther.  The same is true whether I am on the train when I throw the ball or standing next to the train and bounce the ball off of the front.  The misunderstanding that people have is, in essence, the same misunderstanding that leads people to try rendezvous in orbit by thrusting towards the target, or to attempt to raise orbital altitude by thrusting radial out:  things in space work a little differently from what you have been led to expect from conditions on Earth, but the physics still work.

Edited by Zhetaan

##### Share on other sites

Thread moved to the Science subforum, since actual science is taking place here in that the game properly simulates Oberth stuff.

##### Share on other sites
On 2/11/2020 at 6:24 PM, PTNLemay said:

I sort of get the Oberth Effect.  In my mind it boils down to “maneuvers are more efficient when you perform them close to a gravity well”.

A better way to think of it is: "rockets are more efficient when they are already moving fast". It's not the gravity well bit that makes the Oberth Effect. It's the "moving fast" bit. And in a normal Keplerian orbit, you are moving fastest when you are deepest into the gravity well.

##### Share on other sites

Thanks for all the info, especially Zhetaan.  That's a lot of reading material, it'll take me a while to parse it.  It still warps my mind quite a bit, but I think I understand it a bit better now.

The bit about the train especially flummoxed me.  It really does sound like the bullet is getting a gravity assist when you describe the train slowing down a tiny bit.

##### Share on other sites
9 hours ago, PTNLemay said:

The bit about the train especially flummoxed me.  It really does sound like the bullet is getting a gravity assist when you describe the train slowing down a tiny bit.

In essence, that is correct.  The bullet gets an assist, albeit not a gravity assist.  Any time there is an elastic collision in physics, there is an energy and momentum exchange.  (Technically, inelastic collisions combine energy and momentum, but that is also a type of exchange.)  What makes it an assist is that you can arrange matters so that the exchange is favourable to the bullet for whatever purpose you have in mind.

The part that will really warp your mind is that a gravity assist is an elastic collision.  The reason for this is because the physical definition of a collision is the momentum exchange, and if, as in this case, momentum can be exchanged through forces that act at a distance, then it is not necessary for mechanical contact to occur (I say mechanical because physical contact does indeed occur).

This is not quite such a stretch for the imagination, when you think about it.  Matter, in the atomic sense, is mostly empty space (and amazingly so; I cannot properly relate the scale of it), and the reason that your fingers don't plunge through the keys of your keyboard is because of the mutual repulsion of their atoms' electron shells.  The atoms don't need to come into mechanical contact (and that is for the best) in order to interact with one another.

##### Share on other sites

Rockets are constant thrust devices. It doesn't how fast you're going, they push the same regardless.

Work is force times distance.

If you apply 10N to a stationary block and it doesn't move, you've done no work.

If you apply 10N to a trolley going 1m/s, you've done 10J of work.

If you apply 10N to a rocket going 10,000m/s, you've done 100kJ of work.

Rockets are therefore more efficient when they're going fast. The fastest part of an orbit is deep in a gravity well. Only velocity parallel to the direction of motion counts, so burning sideways is equivalent to a standing start.

Edited by RCgothic

##### Share on other sites

Damn! It's só simple...

Joule is the energy over a delta of space, Watt is the energy over a delta of time.

I don't know how in hell I managed to do not realise this for so many years...

##### Share on other sites
13 hours ago, PTNLemay said:

Thanks for all the info, especially Zhetaan.  That's a lot of reading material, it'll take me a while to parse it.  It still warps my mind quite a bit, but I think I understand it a bit better now.

The bit about the train especially flummoxed me.  It really does sound like the bullet is getting a gravity assist when you describe the train slowing down a tiny bit.

Fireing the gun create recoil who will slow the train down, an very tiny amount however. But on planes like the A-10 with its 30 mm gatling gun i think its noticeable.
The bullet shot from an gun on an airplane is also affected so it goes noticeable faster. If you have an gun turret like on an WW2 bomber things get weird as the bullet will move sideways at the speed of the plane if you fire 90 degree off the direction of travel.

##### Share on other sites

If anyone here played around with ArmA3's Apex expansion, specifically the AC-130 style gunship, it can be quite visible, in fact (especially when using the gatling). The bullets not only fly out and to the side, being slowed down by drag, they're also pulled down by gravity. Even the ArmA3 developers had trouble writing the code to compensate for this.  Before they managed that, it was not unusual, if the gunship was moving fast, to have the bullet impact point somewhere at the edge of your screen.

##### Share on other sites
8 hours ago, Dragon01 said:

Even the ArmA3 developers had trouble writing the code to compensate for this.

That sort of thing always makes me sad. I don't expect gameplay engineers to just know how to write this sort of thing. But then it really doesn't take much effort to find someone who does. Two things always make me really sad in games. Bad physics and bad networking. Both are always very easily fixable by somebody who knows what they are doing.

##### Share on other sites

Actually, physics were spot-on (ArmA series ballistics have been unmatched since the first games in the series). However, the gunsight could not cope with them.  Specifically, not only it needed to be stabilized to a point on the ground (because a gunship will typically circle the target), it also had to compute, once pointed at the target, bullet drop, windage (accounting for the gunship's motion), and account for target motion. For a long time, it fell short in the latter area, making the gunship very difficult to use.

##### Share on other sites

On 2/13/2020 at 3:59 AM, mikegarrison said:

A better way to think of it is: "rockets are more efficient when they are already moving fast". It's not the gravity well bit that makes the Oberth Effect. It's the "moving fast" bit. And in a normal Keplerian orbit, you are moving fastest when you are deepest into the gravity well.

Although true, it is hard to visualize the reasoning which is what OP struggled with especially if you are not used to think in energy potentials. Since you are only displayed the speed relative to the current body ins KSP, but overall speed counts, it is may as well not intuitiv to understand when you are actually fast. I mean around Kerbin you see ~ 2300 m/s, but once escaped Kerbin around the sun it shows > 9000 m/s !

My mental model is to not think only about the point closest to a body, but think how you are flying on an escape trajectory past a body: First you accelerate as you get closer because the gravity pulls in same direction as your velocity and behind the closest point the gravity pulls back and deaccelerates the rocket.

If you fire your engines at the closest point you dramatically change how long you are exposed to the deaccelerating force compared to the time when you where accelerating. So when burning prograde you escape the planet much faster, are less pulled back and therefore faster after escape. Vice versa if you burn retrograde to "catch a planet" you greatly extend the time.

Go ahead: Check how many days you are fighting Kerbin gravity if you barely escape it ( 9 days or more depending how close you make it) and compare it with direct transfer to Jool: around 1 day ! Do you now understand why the Oberth effect is so strong ?

Mathematically this is actually why energy is proportional to v²: Speed counts and the time you saved by speed.

Once you have this model in mind, you can do swing-by's: Clever choose your relative speed when passing a body to make asymmetry in gravity work for you. The effect is smaller than burning 1000m/s at the periapsis, but hey it is there !

##### Share on other sites

Oh and about the bullet on a train stuff: The additional energy if viewed from outside is indeed the recoil of the weapon. As the weapon is moving with faster train speed its energy change is higher as well as it changes with v² just like the bullet does.

##### Share on other sites
On 2/14/2020 at 4:29 AM, Dragon01 said:

Actually, physics were spot-on (ArmA series ballistics have been unmatched since the first games in the series). However, the gunsight could not cope with them.  Specifically, not only it needed to be stabilized to a point on the ground (because a gunship will typically circle the target), it also had to compute, once pointed at the target, bullet drop, windage (accounting for the gunship's motion), and account for target motion. For a long time, it fell short in the latter area, making the gunship very difficult to use.

Yeah, which just involves iteratively solving an integral equation for the integrated form of the diff eqs governing the bullet's motion. It's still the same physics, and if one struggles to produce good results in the sight tracking, they probably wouldn't be able to write good bullet simulation either.

That said, the engine and gameplay team are usually distinct. And it's probably the engine guys who wrote the simulation and gameplay who wrote the sight. Hence my comment. It's not that I expected gameplay guys to just know how to write it. I expected them to talk to the engine guys or somebody else who knows relevant physics. If that happened right away, it'd be done correctly right away. This is something that's pretty straight forward to implement if you've encountered it before and know methods used to solve these sorts of problems. But if you're just a good programmer who has been given this as a task without having relevant background in numerical methods, your first few attempts are going to be pretty far off, and probably far more computationally expensive than they need to be.

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.