When does the Oberth effect reach its apex?

[Skyler4856]

The title says it all. If the energy given into velocity through the Oberth effect grows exponentially it must reach an effeincy apex. What is this value?

[SunJumper]

The apex occurs when you lose more energy due to the atmosphere (or, y'know, the ground), than you can gain from thrusting. This value depends on the body.

[diomedea]

I fail to read where is said that energy grows exponentially (http://en.wikipedia.org/wiki/Oberth_effect), instead it grows linear with speed. For a given orbit, the effect is maximum where speed is the greatest, that is at periapsys. Of course a lower periapsys means higher speed, therefore higher energy due to the Oberth effect. But the limit for a low periapsys remains the radius of the principal body.

[DonLorenzo]

'....But the limit for a low periapsys remains the radius of the principal body.

This is fancy speak for 'don't crash'

[K^2]

Hm. That might actually be an interesting question. Suppose, you have a point mass. Well, a black hole. Obviously, you still don't want to dip bellow the event horizon, so the "don't crash," rule is still in effect. But could the optimal periapsis for boosting past a black hole actually be at some specific point above the event horizon? Once we include relativity, I'm not sure that the answer would be the trivial, "as low as possible."

[Skyler4856]

Hm. That might actually be an interesting question. Suppose, you have a point mass. Well, a black hole. Obviously, you still don't want to dip bellow the event horizon, so the "don't crash," rule is still in effect. But could the optimal periapsis for boosting past a black hole actually be at some specific point above the event horizon? Once we include relativity, I'm not sure that the answer would be the trivial, "as low as possible."

Yes this is exactly what I am asking. eventually more energy would be lost from the amplified gravity of flying as low as possible than is gained by the Oberth effect

[Moach]

without an atmosphere, the most 'help' you can get from the Oberth effect is found at the very last inches before you crash

i.e. low, fast → faster

atmospheres are lil' buggers known to get in the way... you can skim a little on the top, but it'll cost ya... when the drag losses exceed the gains from the effect itself, you're past the point of best efficiency

getting exact figures on this is quite tricky, and requires an ungodly amount of math that no reasonable individual should ever have to do by head.... of course, we're not all that reasonable, are we?

here's' a fun read: http://en.wikipedia.org/wiki/Oberth_effect

Edited December 1, 2013 by Moach

[Ralathon]

Yes this is exactly what I am asking. eventually more energy would be lost from the amplified gravity of flying as low as possible than is gained by the Oberth effect

I don't think you understand how orbits work. An orbit is all about conservation of energy, potential energy is turned into kinetic energy and vica versa. That's why you move fastest at periapsis and slowest at Apoapsis. The Oberth effect exists because kinetic energy is related to the square of velocity. If you increase your velocity by 10m/s while starting at 0m/s you gain 100*C energy, if you increase your velocity by 10m/s while starting at 100m/s you gain C*(110^2-100^2)=2100*C. So the same velocity change equates to a much bigger energy change when you go fast. Rockets don't care how fast they're going, 10dV is 10dV for them. So it is most efficient to burn when you go fast meaning you have to burn at periapsis.

There is no such thing as energy loss from the stronger gravity, the only thing adding or removing orbital energy from the system is the rocket itself.

[Skyler4856]

I don't think you understand how orbits work. An orbit is all about conservation of energy, potential energy is turned into kinetic energy and vica versa. That's why you move fastest at periapsis and slowest at Apoapsis. The Oberth effect exists because kinetic energy is related to the square of velocity. If you increase your velocity by 10m/s while starting at 0m/s you gain 100*C energy, if you increase your velocity by 10m/s while starting at 100m/s you gain C*(110^2-100^2)=2100*C. So the same velocity change equates to a much bigger energy change when you go fast. Rockets don't care how fast they're going, 10dV is 10dV for them. So it is most efficient to burn when you go fast meaning you have to burn at periapsis.

There is no such thing as energy loss from the stronger gravity, the only thing adding or removing orbital energy from the system is the rocket itself.

Thanks for clearing that up!

[peadar1987]

Does it not have something to do with exhaust speed as well? Because the kinetic energy gained from forcing burning fuel through a nozzle is split between the rocket and the exhaust, so in an ideal world, you'd have the fuel stopped dead in space (zero kinetic energy), and all that energy going to the rocket, as opposed to wasting kinetic energy by giving "backwards velocity" to the fuel? Or am I overthinking things?

[K^2]

You are right, but this isn't quite what we are talking about. If you could vary velocity of the exhaust, launching it at the speed equal to that of the rocket maximizes energy benefit in the chosen frame. (There is some question of frame choice^*, but they are easy to resolve for a 2-body problem.) But energy is not the only thing you have to worry about. Consider a rocket traveling at 6km/s in low Earth orbit. Energy-optimal exhaust velocity is going to be the same 6km/s in the other direction. If you expel 1kg of mass, you'll gain 18MJ of energy. Now, suppose, the same rocket is traveling near Geo-stationary orbit at just 3km/s. If it can achieve exhaust of 3km/s, it's going to have the same energy-efficiency as before. But now to get the same 18MJ difference, it has to expel 4kg of mass. Clearly, lower orbit is more efficient in terms of propellant.

In case of a realistic chemical rocket, you are dealing with an engine that produces constant exhaust velocity. Here, it is easy to convince yourself that once again, lower orbit allows you to gain most energy for a given quantity of fuel burned. Hence the Oberth effect.

^* As a toy problem, consider a rocket in nearly-empty space traveling at velocity v. As discussed above, you get most energy gain if you launch exhaust at velocity -v. Now, suppose there in alien on a nearby asteroid, and that asteroid is buzzing by fast enough to observe the rocket moving by at velocity 2v. So from perspective of the alien, rocket should be accelerating exhaust to 2v. Who's right? Well, it doesn't matter. Trying to optimize energy only makes sense in problems with gravity. Otherwise, you are optimizing delta-V, and that's a different problem.

[Seret]

On any body with an atmosphere the drag of your vehicle will come into play, so technically the optimum altitude for your periapse will be different for every vehicle, surely?

[LLlAMnYP]

K^2, there is something in your reasoning that just doesn't let me accept it. You are saying, that energy lost by propellant is necessarily transferred to the kinetic energy of the rocket? Imagine if we expel propellant at double velocity of the rocket relative to the parent body. The expelled propellant will have the same kinetic energy as before (it will just be travelling in the opposite direction), however the gain in the kinetic energy of the rocket will be considerably greater.

Someone mentioned black holes on the first page, by the way. Is the Oberth effect still just as valid when general (and/or special) relativity kicks in?

Edited December 3, 2013 by LLlAMnYP

[peadar1987]

K^2, there is something in your reasoning that just doesn't let me accept it. You are saying, that energy lost by propellant is necessarily transferred to the kinetic energy of the rocket? Imagine if we expel propellant at double velocity of the rocket relative to the parent body. The expelled propellant will have the same kinetic energy as before (it will just be travelling in the opposite direction), however the gain in the kinetic energy of the rocket will be considerably greater.

As I understand it, you'd have to liberate a lot more chemical energy from the fuel to do this though. The most efficient way to use the chemical energy locked up in the fuel is to stop the exhaust "dead" in space, but if efficiency isn't really your concern, you can get more thrust by using a far higher exhaust velocity.

[Ralathon]

K^2, there is something in your reasoning that just doesn't let me accept it. You are saying, that energy lost by propellant is necessarily transferred to the kinetic energy of the rocket? Imagine if we expel propellant at double velocity of the rocket relative to the parent body. The expelled propellant will have the same kinetic energy as before (it will just be travelling in the opposite direction), however the gain in the kinetic energy of the rocket will be considerably greater.

Yea, it'll be twice as big. But the important part here is efficiency. Here's a graph:

Inertia equals mass times velocity, so we have a straight line there.

Energy is the surface area below the inertia vs velocity plot. If we tossed out a potato or something from our ship its velocity change would be equal relative to us regardless of what velocity we travel (exhaust velocity or dV in the graph). As you can see the energy difference from tossing it is much bigger when the exhaust velocity is small compared to the spaceship's velocity.

Normally nobody would care because this is all dependent on reference frames. If you looked at it from the PoV of the spaceship you'd get the same energy every time for example. But when you're in orbit this allows you to increase orbital energy much more efficiently.

Edited December 3, 2013 by Ralathon

[kahlzun]

What about retrograde burns? Do those also get bonuses from high-speed situations?

[LLlAMnYP]

In terms of dV spent to increase/decrease the energy of your orbit by a certain amount - yes, it is most efficient to burn at high speeds. It's a reversible process.

[Ralathon]

What about retrograde burns? Do those also get bonuses from high-speed situations?

Yep, exact same story. If you are planning to get captured by, for example, the mun, it is best to get your periapsis as low as possible. You can then get captured into an elliptic orbit for very little dV. If you are orbiting a body with an atmosphere you can then rely on atmospheric drag to bring you down into a circular orbit. If you read about NASA missions that involved capture maneuvers you'll always notice the periapsis is very close above the surface for this exact reason.

[Idobox]

About the kinetic energy lost by the exhaust:

The exhaust speed is the ISP divided by g0 (acceleration at Earth's surface). The faster your exhaust, the more deltaV you get per kg of fuel. The idea of having the exhaust at rest doesn't really mean anything, because there isn't a reference frame that is preferred.

What you want is to move it as fast as possible from you, because you gain the same amount of momentum (speed times mass) in the reference frame of the rocket. Of course moving your exhaust twice as fast requires 4 times more energy, and that's why real rockets use highly energetic fuel.

If you have constant power and a fixed amount of propellant, like with an ion engine, you will get the best deltaV by expelling a small trickle of propellant at max speed for a very long time (you also spend the most energy that way).

If you have limited energy and fixed amount of propellant (chemical rockets), you will get the best efficiency by dumping all your fuel at once. In practice, you have also a limit on power (energy/time), so you burn as fast as you can without melting/destroying the engine or killing your crew with g force.

And by the way, if your exhaust velocity is 1000m/s, and your orbital speed is higher than that (it almost always is if you have a stable orbit) when you burn, the propellant will move in the same direction as you, albeit slower, from the point of view of observers on the planet. It's like driving a motorbike on a train, when you accelerate, you make the train slow down a little, but to the cow looking at you, both the train and the bike go in the same direction.

[K^2]

The idea of having the exhaust at rest doesn't really mean anything, because there isn't a reference frame that is preferred.

Actually, there sort of is when you are dealing with central potentials. Energy is conserved in a system where Hamiltonian is time-independent. If you choose a coordinate system in which source of gravity is moving, Hamiltonian is time-dependent, and energy is not conserved. So when we are talking about a ship in orbit of a planet, choice of coordinate system in which planet is at rest is a "better" coordinate system. Naturally, we can still describe the problem from any other coordinate system, but then we lose conservation of energy, and math becomes way more complicated.

[LLlAMnYP]

That's the problem though! The mechanical energy of the system is not conserved, regardless of coordinate system as chemical energy of the fuel is being converted into kinetic energy.

[K^2]

So? It just means we have energy source and losses to heat. We know exactly how much energy is released per unit of fuel burned and how much of that energy becomes kinetic energy of the exhaust. You don't have to work with a closed system to make use of the conservation laws. You just need to know how much is being added and how much is being lost.

If we choose an arbitrary coordinate system, then we don't know how much is being added by time-dependent interactions without solving equations of motion, which are going to be quite complex in a general coordinate system.

[LLlAMnYP]

No need to go about time-dependent interactions. Let's just stick to infinitesimally short burns and differential equations :-) That's what I did right now and came to quite a surprising result. I tried to find the optimal exhaust velocity of the propellant in order to get the largest energy increase of the orbit (i.e. the increase of the kinetic energy of the vessel in the reference frame of the parent body) compared to the energy extracted from the propellant (for simplicity's sake I disregarded waste heat and such as we're only interested in the kinetic stuff anyway).

It turns out, the smaller the exhaust velocity, the better (for a given amount of chemical energy used). And, of course, the larger the velocity of the rocket, the better too. Obviously, with smaller exhaust velocity you will need to expel a quadratically larger amount of propellant though. So this leads me to the question "what exactly are we trying to optimize here?".

Chemical energy of propellant to energy increase of orbit? Delta-v budget to energy increase? Or something entirely different?

[K^2]

Because you minimized energy for given amount of momentum. What you are neglecting is the fact that this also increases amount of propellant used.

[LLlAMnYP]

Not quite. I minimized useful chemical energy for a given increase in orbital energy (though that probably leads to the same thing). And I clearly pointed out, that that would lead to an increase in propellant used.

However, you ignored my last question. In your scenario you suggest to have exhaust velocity equal to orbital velocity. To get a result like that, what should one be seeking to optimize?