Off Topic > General Discussion
Oberth Effect?
Northern Lurker:
Hi all
I have trouble about getting a hang on Oberth Effect. The effect of the Effect is simple enough; the faster the rocket is going, more delta-V it gets from a kg propellant.
I have been reading on it and have found three explanations: The first is that faster the propellant is going one way, more momentum it has. When it is expelled from nozzle in opposite direction it gives more momentum thrust. Second explanation is that the faster the rocket is going, more and more energy produced by combustion goes to kinetic energy of the rocket and less and less to kinetic energy of exhaust gases. Third explanation is that Work = Force x Distance. Force of the rocket engine produces is constant. Higher the velocity of the rocket, more distance it covers in unit of time. Constant force times longer distance equals more work.
Upon further reading I reached understanding that second and third explanations are actually the same thing, only described in two different ways.
So my questions are: Does momentum of propellants have part of Oberth Effect? Is the second and third explanations the same and are they correct? Does these explanations cover the Effect or am I missing something?
Lurky
PS. English is not my primary language and while I read a lot in it, I don't write or speak nearly enough in it. So any constructive criticism on my grammar is welcome.
JayUtah:
Go with the second explanation.
Imagine a rocket stage strapped down on a test stand. All the kinetic energy is going into the exhaust. From the observer's point of view, the rocket stage has zero velocity and the exhaust plume has much. The complete opposite is where the rocket is flying free in space at a tremendous velocity past a fixed observer. If the rocket is traveling at exactly the exhaust velocity as it passes the observer, the rocket will be reckoned to have all the kinetic energy since it's the only thing that has velocity relative to the observer. The exhaust that was emitted just as the rocket passed the observer would remain in the general vicinity of that observer, since the velocities cancel. It's like driving your car at 80 km/h and throwing a ball out of it backwards at 80 km/h (relative to the car). The ball would just drop to the pavement in that place. The magic of the Oberth Effect comes from correctly defining the observer's velocity state and leaving it alone.
If algebraic math is more your thing, imagine that kinetic energy is proportional to the square of the velocity. A given kilogram of fuel will give you a specific change in velocity. Forget for a moment that a rocket is a variable-mass vehicle. Just imagine two cases where 1 kg of fuel is expended from a given starting mass, the rocket being 1 kg less massive at the end of the burn. If the burn starts at, say, 10 m/s and ends at 12 m/s, the delta-v is 2 m/s. The mistake here would be to compute the change in kinetic energy as the square of the delta-v. You reckon it instead from the squares of the starting and ending values (accounting, of course, for the drop in mass, which we'll slightly ignore for the moment). So the change in kinetic energy is 144 - 100, or 44 units.
Now perform the same experiment starting at 20 m/s and ending at 22 m/s. The delta-v is the same, because we have the same mass rocket and burn the same mass of fuel as before. But the change in kinetic energy here is 484 - 400 = 84 units, almost twice as much of an increase as from the slower starting speed. This is because, as you know, a parabola -- the graph of a polynomial of order 2 -- gets steeper the farther away from the origin you get. So identical deltas along the x-axis result in greater changes in the y-direction the farther away from the origin they get.
If you're wondering where this extra energy comes from, just do the same calculation on the exhaust. If your exhaust velocity is 100 m/s, then the exhaust velocity relative to the observer in the first example is 100-10 m/s, or 90 m/s. In the second example it's 100-20 m/s, or 80 m/s. Relative to the observer, the exhaust has less kinetic energy in the second run. And relative to the observer, the rocket has more. And that's because of how the velocity reference frames are defined. You have to make the leap of regarding the exhaust velocity not as relative to the rocket, which is how it's canonically specified, but as relative to a fixed observer.
Northern Lurker:
A very big thank you, Jay. That cleared it up a lot.
Lurky
smartcooky:
--- Quote from: Northern Lurker on May 21, 2019, 02:46:09 AM ---A very big thank you, Jay. That cleared it up a lot.
Lurky
--- End quote ---
Don't you just LOVE the nitty-gritty details of rocket science!!?
JayUtah:
--- Quote from: Northern Lurker on May 21, 2019, 02:46:09 AM ---A very big thank you, Jay. That cleared it up a lot.
Lurky
--- End quote ---
You're quite welcome. I sympathize with the plight of students trying to understand the effect. It's one of several cases in science where you struggle to envision what's happening and you just have to trust the numbers.
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