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I don't understand this and must ask a probably very stoopid question here:

The Oberth effect says that a rocket is much more efficient when (and in the orbital direction of) a payload when it already has the highest speed. This AFAIK is because force, or "work", is a function of the distance moved and not of the time spent, while instead the force from a rocket engine is a function of time. So the faster a rocket is, the larger is the distance over which a rocket engine produce force during its fuel=time window.

This is obviously anti-intuitive to everyday human experiences. Something which is already fast, such as a fast car or airplane, is less, not more, easy to accelerate further for the same amount of fuel consumed.

What causes the Oberth effect? Is it directly given from Newtonian mechanics? If so, why did Oberth "discover" it 300 years afterwards? Has it got something to do with microgravity or rocket reaction propulsion? If we imagine a maglev train in a vacuum tube on Earth (i.e. no friction, no air resistance, no diminishing engine efficiency), would it get more and more energy efficient the more it is accelerated, because of the Oberth effect?

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    $\begingroup$ The reason cars or airplanes are harder to accelerate further the faster they already are is because air resistance increases quadratically with speed. $\endgroup$ – Philipp Apr 10 '14 at 7:35
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Kinetic energy is 1/2 mv2. Here's my visualization:

1/2 mv2

When you add to that velocity v, you get a bigger square:

enter image description here

Here Vb is velocity imparted by your burn.

So for example, if you accelerate a kilogram at rest to 1 meter/sec, you get .5 joules of kinetic energy. If you accelerate a kilogram moving 10 meters/sec to 11 meters/second, you get 10.5 joules of kinetic energy. An Oberth benefit of 10 joules.

It is harder to accelerate stuff moving fast. If you have a mass driver on the moon, Accelerating something to 1 km/s would take 100 times as much energy as accelerating it to .1 km/s. Or accelerating something to 4 km/s would take 16 times as much energy as 1 km/s. But on earth the difficulty is even greater as you have atmospheric drag. Atmospheric drag scales roughly with velocity cubed.

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  • $\begingroup$ Is the following a correct explanation: The total energy is split between the rocket ship and the exhaust. The Oberth effect is an observation that your rocket ship ends up with more energy if the exhaust ends up with less energy. By "dumping" the exhaust when you're lower in the gravity well, it ends up in a lower orbit with less energy. Therefore, your rocket ship ends up with more energy. $\endgroup$ – LocalFluff Sep 17 '14 at 7:43
  • $\begingroup$ @LocalFluff: It happens to work out, but don't forget that the energy of the exhaust also has kinetic and potential energy. You can't just assume the exhaust has lowed energy because it's in a lower orbit. $\endgroup$ – MSalters Dec 4 '14 at 23:19
  • $\begingroup$ @MSalters This might be an intuition: A rocket which has spent its fuel at periapsis is lighter as it climbs up out of the gravity well. $\endgroup$ – LocalFluff Dec 4 '14 at 23:52
  • $\begingroup$ @LocalFluff This intuition sounds a bit too simple, I think; if e.g. the fuel is just dumped at periapse the rocket will also be lighter as it climbs up out of the gravity well, but it doesn't cause the Oberth effect. $\endgroup$ – matz Jun 17 at 18:30
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The basis of the Oberth effect is that kinetic energy goes up at the square of velocity but a rocket burn provides the same delta-v no matter what speed the rocket is moving at.

If you borrow some velocity from a gravity well and then burn your rocket you're going faster as you exit the gravity well and thus lose less velocity on the way back out.

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  • $\begingroup$ So it IS more economical to accelerate a fast train than a slow train? $\endgroup$ – LocalFluff Apr 9 '14 at 19:45
  • $\begingroup$ @LocalFluff This has nothing to do with fast and slow trains. It has to do with borrowed velocity--you can avoid paying back the whole loan. $\endgroup$ – Loren Pechtel Apr 9 '14 at 19:48
  • $\begingroup$ HopDavid provided a nice illustration of this. $\endgroup$ – Loren Pechtel Apr 10 '14 at 2:57
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Assume a vehicle in orbit about some gravitational body. The Oberth effect says that the way to get the biggest increase in energy from a maneuver that changes the vehicle's velocity by some $\Delta v$ is to perform that maneuver at periapsis (i.e., when the vehicle is going fastest with respect to the gravitating body). The total mechanical energy before the maneuver is $E_0 = \frac{v^2} 2 - \frac {\mu} r$, where $\mu=GM$ is the gravitational parameter for the planet/star. Assuming an impulsive maneuver, the total mechanical energy after the maneuver is $\frac{(v+\Delta v)^2} 2 - \frac {\mu} r$, or $E_0 + \frac{\Delta v^2} 2 + v \cdot \Delta v$.

Mechanical energy is a constant over the orbit, and $\Delta v^2$ will be the same wherever the maneuver is performed. The only thing that does change is that last term, $v \cdot \Delta v$. To maximize the change in energy we want two things to be true: Velocity $v$ needs to be at its maximum (i.e., periapsis) and the $\Delta v$ to be applied along the velocity vector.


Something which is already fast, such as a fast car or airplane, is less, not more, easy to accelerate further for the same amount of fuel consumed.

A car that's going fast is already in a high gear where you naturally get less torque. It's also using a good deal of power just to keep going the same speed. Let off the gas and the car will slow down. There's not much power left for acceleration. A plane that is already going fast (with respect to the airflow) similarly is using power just to maintain speed. Spacecraft operate in vacuum, and it doesn't matter how fast a vehicle is already going when a vehicle performs a maneuver. For a rocket, delta V is delta V. It is independent of the current velocity.


Update

One needs to be careful here. Imagine a rocket located in deep space, far removed from any gravitating bodies, about to test fire its thrusters. Suppose a number observers will watch this test, with different observers moving at different velocities relative to the rocket. Ignoring relativistic effects, all observers will agree on the rocket's change in velocity. The observers will not agree on how much the rocket's energy has changed. Some may even see the rocket as losing energy.

The observers will agree on the total change in energy of the rocket and its exhaust stream. Comparing $m\Delta v^2$ to this total change in energy yields a much more meaningful measure of efficiency than does comparing $m\Delta v^2$ to the change in the rocket's energy.

If we imagine a maglev train in a vacuum tube on Earth (i.e. no friction, no air resistance, no diminishing engine efficiency), would it get more and more energy efficient the more it is accelerated, because of the Oberth effect?

The Oberth effect isn't about efficiency so much as it is about effectiveness. An impulsive maneuver any along the rocket's orbit will instantaneously change velocity by the same amount, regardless of where the maneuver occurs. (For a non-impulsive burn, acceleration will be the same regardless of where the maneuver occurs.) The effectiveness of that change is the issue.

There are bound to be inefficiencies in a maglev system, even with a train running in vacuum. These inefficiencies will inevitably vary with velocity. In other words, acceleration from a given amount of power will be a function of velocity.

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  • $\begingroup$ I appreciate the explanations from all of you here. I understand the mathematical formulation of the effect, but it still seems counter intuitive that your current speed multiplies with your speed increase. But this is so for all motions, it's only our being used to resistances on Earth which makes it feel counter intuitive? $\endgroup$ – LocalFluff Apr 12 '14 at 14:18
  • $\begingroup$ Change of orbital inclination, on the other hand, is most efficient at apoapsis where the orbital speed is the lowest. Is that a mirror of the Oberth effect, or does that have a much different explanaition? $\endgroup$ – LocalFluff Apr 12 '14 at 14:24

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