Question about the Oberth effect

[Tatonf]

Hello !

I don't really understand how the Oberth effect work. If I understand correctly, the faster I go, the more delta-V I'll gain from it, which is why I should do my burns at periapsis.
What I don't understand, is why is it more efficient to do my burn at LKO for my interplanetary transfers, rather than in orbit around the Sun.
While in LKO my speed is about 2300 m/s, while an orbit around the Sun identic to Kerbin's one gives a speed of 9300 m/s.
So why the Oberth effect is better in LKO rather than in orbit around the Sun ? Edited November 17, 2015 by Tatonf

[LN400]

Just one thing. The speed you read in-game would be relative to the body you're orbiting. It's incorrect to compare 2 speeds that are relative to 2 different bodies and conclude your speed orbiting the sun is about 5 times greater than your orbital speed around Kerbin. Orbiting Kerbin, your speed around the sun is much greater than 2300 m/s. Orbiting the sun, your speed relative to Kerbin is nowhere near 9300 m/s if you're somewhere near Kerbin.

[LostOblivion]

You don't get more delta-v, but is able to use your delta-v more efficiently.

[davidpsummers]

[quote name='LN400']Just one thing. The speed you read in-game would be relative to the body you're orbiting. It's incorrect to compare 2 speeds that are relative to 2 different bodies and conclude your speed orbiting the sun is about 5 times greater than your orbital speed around Kerbin. Orbiting Kerbin, your speed around the sun is much greater than 2300 m/s. Orbiting the sun, your speed relative to Kerbin is nowhere near 9300 m/s if you're somewhere near Kerbin.[/QUOTE]

This is the thing I don't understand about the Oberth effect. It is simply that work (what you get out of a burn) depends on force (how big your rocket is) and the distance apply the force over (which is related to speed and burn time). All reference frames should be equal and two reference frames should give you the same answer. But a reference frame at rest with respect to Kerbin will give you a different optimal position than a reference frame at rest with respect to Kerbol.

[Pthigrivi]

These guys are correct. The 9300 m/s also counts the Kerbin's orbital velocity of 9284.5 m/s, where your reading of 2300 m/s is hiding that because it is counted relative to Kerbin.

In a related question though, I've been wondered in the past how much vessels with very low TWR suffer from their inability to take advantage of oberth. I often do at least two oberth passes for transfer burns, but surely the more focused your burn is on Pe the more efficient it is. Is there an easy way to figure out what this trade-off is? Im guessing the high ISP of low TWR engines more than compensates for this deficiency but I'd been curious by how much.

[GoSlash27]

Tatonf,
What gets you from place to place isn't actually "delta vee", it's "delta kinetic energy". We think of it as a change in [I]velocity[/I] because it simplifies the math. Since kinetic energy is proportional to the square of velocity, we add more kinetic energy with velocity at high speed than we do at low speed.
Best,
-Slashy

[Yasmy]

Take any orbit: elliptical, parabolic, hyperbolic. Doesn't matter.

With engines off, orbital energy stays constant. As you move farther away from whatever you are orbiting, you slow down, losing kinetic energy. That kinetic energy loss is exactly balanced by gain in gravitational potential energy due to climbing out of the gravity well. So orbital energy stays constant.

Thrusting, on the other hand, changes your velocity, but doesn't change the acceleration on your craft due to gravity. So thrusting does change your gravitational potential energy.

Initial kinetic energy = 1/2 m v^2.
Burn dv.
Final kinetic energy = 1/2 m (v+dv)^2 (Ignoring for the moment the fuel mass and non-prograde burns, just to keep it simple.)

So the change in energy is dE = 1/2 m (v+dv)^2 - 1/2 m v^2 = m v dv + 1/2 m dv^2 ~= m v dv.
This is the Oberth effect: The change in orbital energy due to burning dv is proportional to v.
Thus the faster you are moving, the bigger the change in energy. Therefore it is best to expend your fuel when you are moving faster, e.g., at periapsis.

Like LN400 said, you can't directly compare velocity on solar orbit with velocity on Kerbin orbit, because Kerbin is moving relative to the sun. Your sun-relative velocity at Kerbin departure is larger than your sun-relative velocity when you escape Kerbin by approximately Kerbin's escape velocity.

Edit: Ninjaed by Slashy's nice, succinct answer. Edited November 17, 2015 by Yasmy

[ajburges]

[quote name='Pthigrivi']These guys are correct. The 9300 m/s also counts the Kerbin's orbital velocity of 9284.5 m/s, where your reading of 2300 m/s is hiding that because it is counted relative to Kerbin.

In a related question though, I've been wondered in the past how much vessels with very low TWR suffer from their inability to take advantage of oberth. I often do at least two oberth passes for transfer burns, but surely the more focused your burn is on Pe the more efficient it is. Is there an easy way to figure out what this trade-off is? Im guessing the high ISP of low TWR engines more than compensates for this deficiency but I'd been curious by how much.[/QUOTE]

Long answer:

Trade-off is time vs efficiency of energy transfer. The efficiency of the energy transfer is determined by the sum of your losses (cosine in some form or another plus a little from Kepler orbits). You need a good model to calculate/approximate those losses to evaluate the trade-off.

[Tatonf]

Thanks for all the replies, but I still don't understand why I gain more "kinetic energy" from burning in LKO rather than in Sun's orbit.
You say that I can't compare my velocity from LKO and the velocity from my Sun orbit, but what does matter at the end ? My velocity relatively to the Sun ?
That would mean that, if I go orbiting the Sun at an altitude where my velocity is greater than 9284.5 + 3431.0 = 12 715.5 m/s, the Oberth effect I'll get from the Sun will be better than the one I get from LKO ?

[BenCushwa]

[quote name='Yasmy']Wise words.[/QUOTE]

This is an excellent answer.

One thing to consider when doing things like alignment burns (i.e. matching orbital inclinations) is that even though your engines are more efficient at periapsis, the amount of velocity change necessary is considerably lower at higher altitude.

For example: if you are in a 100km parking orbit around Kerbin and you change your orbital inclination to match Minmus, it will cost you about 200m/s of deltaV. You're moving faster, so you require more thrust to change direction. However, if you timed your burn such that you could adjust your orbital inclination near Minmus's orbit, it will cost you considerably less. That far away from Kerbin you're barely moving, so a change in orbital inclination requires very little thrust. Of course, the catch in this situation is that you must time your transfer burn to place you at the ascending or descending node of Minmus's orbit so it could take you far longer to maneuver for an intercept, but that's a general rule of orbital mechanics: if you want to minimize maneuvering costs, you'll usually need more time.

Also...very first forum post. Likely the first of many. B^)

[Aethon]

Ok. If you are in a car stopped at a stop sign and you press down the accelerator 2 cm, it has very little effect on your speed, but if you are on the highway going 120 and press the accelerator 2 cm it will have a much greater effect on your velocity.

[Red Iron Crown]

[quote name='Tatonf']Thanks for all the replies, but I still don't understand why I gain more "kinetic energy" from burning in LKO rather than in Sun's orbit.
You say that I can't compare my velocity from LKO and the velocity from my Sun orbit, but what does matter at the end ? My velocity relatively to the Sun ?
That would mean that, if I go orbiting the Sun at an altitude where my velocity is greater than 9284.5 + 3431.0 = 12 715.5 m/s, the Oberth effect I'll get from the Sun will be better than the one I get from LKO ?[/QUOTE]
Kerbin's velocity is added to the vessel's in the solar frame of reference. Relative to the sun, the vessel is moving faster at LKO than it is just outside Kerbin's SoI (despite what the speedometer says), so therefore it experiences more Oberth effect.

[quote name='Aethon']Ok. If you are in a car stopped at a stop sign and you press down the accelerator 2 cm, it has very little effect on your speed, but if you are on the highway going 120 and press the accelerator 2 cm it will have a much greater effect on your velocity.[/QUOTE]
I would stay away from car analogies, they don't work very well for explaining the Oberth Effect. That and your example is backwards, a 2cm throttle input will make a stopped car accelerate faster than the same input at high speed. (Cars don't use reaction drives, so Oberth doesn't apply anyway.) Edited November 17, 2015 by Red Iron Crown

[Lukaszenko]

[quote name='Aethon']Ok. If you are in a car stopped at a stop sign and you press down the accelerator 2 cm, it has very little effect on your speed, but if you are on the highway going 120 and press the accelerator 2 cm it will have a much greater effect on your velocity.[/QUOTE]

I don't think that's related...

[LN400]

Here's another one to ponder. Imagine you are the only object in the universe. No other object exists but you. In that universe, speed/velocity has no meaning.
Speed, and velocity, is always relative to something. What that something is, is your choice but your choice will affect the answer.
A car on a highly trafficed highway is travelling with all the other cars at say, 100 km/h, the speedometer tells you. The car's speed relative to the other cars going in the same direction is 0 km/h and the kinetic energy relative to the other cars is also zero. The cars coming the other direction are also going at 100 km/h and relative to them, your car is going 200 km/h and your kinetic energy will reflect that. Relative to the road/surface, you are going at 100 km/h and your kinetic energy, should you veer off the road and hit a garden gnome, will reflect that speed.

For speed/velocity to have any meaning at all, you need to choose a frame of reference. For the car, you might even have to juggle several frames of reference simultaneously. You need to mind the distance to the car in front of you, you need to mind the speed relative to the surface as you approach that curve with a mean garden gnome standing there watching you and you know it's better to bump into the car in front of you denting the bumper than meeting the oncoming car.

The frame you choose, should be chosen out of relevance to what you need to know, or accomplish. If your goal is to escape Kerbins's influence, then you don't need to worry about your speed around the sun. All calculations on kinetic energy would then be based on Kerbin as your reference. Kinetic energy relative to the sun won't enter the equation at all.

[GoSlash27]

[quote name='Tatonf']Thanks for all the replies, but I still don't understand why I gain more "kinetic energy" from burning in LKO rather than in Sun's orbit.
You say that I can't compare my velocity from LKO and the velocity from my Sun orbit, but what does matter at the end ? My velocity relatively to the Sun ?
That would mean that, if I go orbiting the Sun at an altitude where my velocity is greater than 9284.5 + 3431.0 = 12 715.5 m/s, the Oberth effect I'll get from the Sun will be better than the one I get from LKO ?[/QUOTE]

Tatonf,
For interplanetary trajectories, your velocity about the sun *does* matter. When orbiting Kerbin, your speed is added to (or subtracted from) Kerbin's orbit around the sun. If you burn after barely ejecting from Kerbin, you're starting with Kerbin's 9300 m/s velocity. If you burn while in LKO, you're starting with Kerbin's 9300 m/s plus another 2300 m/sec*.

An ill-advised car analogy :D
If you're in a car doing sixty and you drop a rock out of the window, the rock will be doing sixty. If, OTOH, you're sitting on the same car doing sixty and put the rock in a sling and twirl it, you can throw it much faster.

Best,
-Slashy

*technically the velocities are vector added when crossing SoI boundaries, not linear.

[Right]

In the fewest words I know how, I will venture an explanation...

Gravity is a force working over time. When you increase your velocity at the periapsis, you spend less time in the gravity well (losing speed) on your way out.

As per your question about the sun: its a malformed question. You're velocity hasn't actually changed that much when you leave kerbin SOI, your point of reference changed. Insert theory of relativity.

[Red Iron Crown]

[quote name='Right']Gravity is a force working over time. When you increase your velocity at the periapsis, you spend less time in the gravity well (losing speed) on your way out.[/QUOTE]
That's a poor way to explain it, because the Oberth Effect also works for retrograde burns which end up spending *more* time deeper in the gravity well.

Strictly speaking gravity doesn't cause the Oberth Effect, it's entirely a consequence of KE=0.5mv[sup]2[/sup] and reaction engines having the same thrust no matter the speed. Gravity just creates situations in which it can be harnessed.

***

I wrote a little [URL="http://forum.kerbalspaceprogram.com/threads/99758-Why-does-the-oberth-effect-work?p=1539004&viewfull=1#post1539004"]synopsis of the Oberth effect[/URL] a while back, perhaps some readers of this thread would benefit from it. Edited November 17, 2015 by Red Iron Crown

[Right]

[quote name='Red Iron Crown']That's a poor way to explain it, because the Oberth Effect also works for retrograde burns which end up spending *more* time deeper in the gravity well.[/QUOTE]

As far as I can tell, the explanation fits for retrograde burns just fine. When you slow down at the periapsis, you spend more time in the gravity well subsequently, which serves to slow you down more than otherwise.

I could be mistaken, I'll check out your link.

[Empiro]

[quote name='Tatonf']Thanks for all the replies, but I still don't understand why I gain more "kinetic energy" from burning in LKO rather than in Sun's orbit.
You say that I can't compare my velocity from LKO and the velocity from my Sun orbit, but what does matter at the end ? My velocity relatively to the Sun ?
That would mean that, if I go orbiting the Sun at an altitude where my velocity is greater than 9284.5 + 3431.0 = 12 715.5 m/s, the Oberth effect I'll get from the Sun will be better than the one I get from LKO ?[/QUOTE]

Indeed, if you drop to a lower periapsis around Kerbol, then your speed will be higher, and you'd get a bigger benefit out of the Oberth effect. The gain, however, would be offset by your lower total orbital energy. There are lots of different factors at play, which is why sometimes, it's good to drop your periapsis and then burn at periapsis, and sometimes, you want to do a direct burn.

The easiest way to test out the Oberth effect is to do the following: do one burn in LKO at 1000 m/s so that you're just barely escaping Kerbin in the prograde direction. Wait until you exist, then burn another 1000 m/s prograde. Look at where your Apoapsis is. Then, do a second test where you burn all 2000 in LKO so that you're ejected in Kerbin's prograde direction. Note where your Apoasis is. In the second case, your AP will be around Jool's orbit, while in the first case, it won't be close. You can also repeat the test in the retrograde direction.

[GoSlash27]

[quote name='Right']As far as I can tell, the explanation fits for retrograde burns just fine. When you slow down at the periapsis, you spend more time in the gravity well subsequently, which serves to slow you down more than otherwise.

I could be mistaken, I'll check out your link.[/QUOTE]

Right,
I'm with RIC on this one. the Oberth effect has nothing to do with gravity wells. It's simply the result of the [s]exponential[/s] [B]quadratic[/B] relationship between velocity and energy. It holds true even without gravity wells.

Best,
-Slashy Edited November 18, 2015 by GoSlash27

[pellinor]

[quote name='Tatonf']Thanks for all the replies, but I still don't understand why I gain more "kinetic energy" from burning in LKO rather than in Sun's orbit.
You say that I can't compare my velocity from LKO and the velocity from my Sun orbit, but what does matter at the end ? My velocity relatively to the Sun ?
That would mean that, if I go orbiting the Sun at an altitude where my velocity is greater than 9284.5 + 3431.0 = 12 715.5 m/s, the Oberth effect I'll get from the Sun will be better than the one I get from LKO ?[/QUOTE]

Maybe it helps to think in terms of gravity potential, which does not depend on the reference frame. The Oberth effect means that pro/retrograde burns are most effective when you are low in a gravity well. Now the sun generates a large gravity well spanning the whole system, and Kerbin causes a local dip in this potential (the SOI model in KSP is a close-enough approximation of this). So when you want to do a burn somewhere around Kerbin, you'd look for the local potential minimum, which is "as close to the planet as you can get away with".

Edit (overtaken by other posts): I agree that the Oberth effect does not need gravity wells, however they can avoid confusion about reference frames because you can formulate your kinetic energy in a handy way.

[quote name='GoSlash27']exponential relationship between velocity and energy[/QUOTE]
quadratic (just nitpicking, of course that does not invalidate your point) Edited November 18, 2015 by pellinor

[OhioBob]

[quote name='Tatonf']Thanks for all the replies, but I still don't understand why I gain more "kinetic energy" from burning in LKO rather than in Sun's orbit.
You say that I can't compare my velocity from LKO and the velocity from my Sun orbit, but what does matter at the end ? My velocity relatively to the Sun ?
That would mean that, if I go orbiting the Sun at an altitude where my velocity is greater than 9284.5 + 3431.0 = 12 715.5 m/s, the Oberth effect I'll get from the Sun will be better than the one I get from LKO ?[/QUOTE]

Let's say you want to get to Duna. If you perform this transfer from solar orbit near Kerbin, it takes a Δv of about 920 m/s. But to get into that solar orbit you must escape Kerbin orbit, which takes about 950 m/s. That's a total of 1870 m/s. Suppose now that instead of performing two burns, you just give yourself enough extra velocity during your escape from Kerbin that you have some velocity left over after escaping. You can do it so that you are already traveling 920 m/s relative to Kerbin after escaping so you don't have to perform the second burn. The extra velocity that you need to provide when close to Kerbin is much smaller than the amount you retain after escaping. To retain 920 m/s requires an ejection burn of only 1080 m/s. This is far less than the 1870 m/s that it took to perform the two-burn strategy.

[Aethon]

[quote name='Red Iron Crown']=I would stay away from car analogies, they don't work very well for explaining the Oberth Effect. That and your example is backwards, a 2cm throttle input will make a stopped car accelerate faster than the same input at high speed. (Cars don't use reaction drives, so Oberth doesn't apply anyway.)[/QUOTE]

I agree with the first sentence. It's simplistic to aid in understanding, and I respectfully disagree with the second sentence.

[GoSlash27]

[quote name='pellinor']quadratic (just nitpicking, of course that does not invalidate your point)[/QUOTE]

pellinor,
Yeah, you're right. Quadratic.

Best,
-Slashy

[Red Iron Crown]

[quote name='Aethon']I agree with the first sentence. It's simplistic to aid in understanding, and I respectfully disagree with the second sentence.[/QUOTE]
Most planetary transport methods use propulsion with a peak engine power, that power determines how quickly kinetic energy is added (since power is energy per unit time). So these vehicles are adding energy at a constant rate, but each m/s of additional speed costs more energy (because KE=0.5mv[sup]2[/sup]). This means that each additional m/s takes longer to add, i.e. acceleration decreases with increasing speed. Have a look at a typical speed vs time graph for an automobile:

[IMG]http://www.uselesspickles.com/files/jeep/drag_race_speed_stock_mag_ripp_prod_hemi.png[/IMG]

You can see that they all gain speed fastest at the beginning and at a gradually slower rate as speed increases. You can try this for yourself in a car, observe how long it takes to accelerate at full throttle from 40-60 and from 100-120, I guarantee that the latter will take longer.*

Rocket engines are different. They apply constant [i]force[/i] instead of power, so the rate at which they change the energy of the vehicle is not constant, it increases with speed. This is the critical observation that Oberth discovered, and it is so counterintuitive to the common, familiar behavior above that it is a tough concept to grasp (I certainly struggled with it). Have a look at a rocket stage's speed vs. time graph:

[img]http://www.mnealon.eosc.edu/RocketSciencePage5_files/image7071.png[/img]

[SIZE=1]* Observe all local traffic laws in your area, and do this in safe, controlled conditions. In fact, don't do this experiment, if anyone asks you I advised against it.[/SIZE] Edited November 18, 2015 by Red Iron Crown
Added second graph.