How does the Oberth effect work?

[travis575757]

Well I have looked across the internet and can't find any good answers.

1. W = F * D This is an equation, Yes I'm going faster so there is work being done but that doesn't make sense because if I am in a rocket it is not different. For example propellant is expelled at 500m/s while on the ground that is -500 m/s. Well what if I'm in space going 20,000m/s? Well I am obviously someone on earth will look and see rocket going 20,000m/s but in my mind on the rocket i am not moving the earth is just going around me. When my rocket burns since I'm going 0m/s (from my reference frame) the propellant is also being expelled at 500m/s so -500m/s

2.If it is simply a trade off kinetic energy and potential energy where does my energy go if i burn at the apoapsis of my inclined orbit and make my orbit a circle. Is it that I'm always maintaining that potential energy by being at a higher altitude away from the planet?

Please don't give me large math equations to explain stuff because they don't seem to explain anything.

EDIT: Okay so i just thought of something that makes W = F * D make even less sense to give a good reason in this situation. Lets say I'm orbiting and a giant ruler is next to me while I'm orbiting. The ruler is then 0 m/s relative to me If i burn for lets say 10 meters in my reference frame isn't that no different then burning 10 meters on earth from a stand still?

EDIT 2: so basically how Ralathon said

Say your rocket has 10m/s of delta V. If start out stationary you start with 0 m/s and end with 10 m/s. Thus giving you 100 units of kinetic energy.

Now imagine that your rocket starts out at 100 m/s. After your burn you're moving with 110 m/s. This means your burn gave you 110^2 - 100^2 = 2100 units of energy.

If i burn at the perapsis which is a higher speed i gain more kinetic energy then burning at the apoapsis at a low speed. Since i have more energy this way to conserve it my orbit must have greater change.

Edited March 27, 2014 by travis575757

[Skyler4856]

Yeah, I always understood it as the inability to access kinetic energy given from the orbit at apoapsis.

[K^2]

1. W = F * D This is an equation, Yes I'm going faster so there is work being done but that doesn't make sense because if I am in a rocket it is not different.

You are making a standard mistake, assuming that energy conservation works regardless of choice of coordinate system. That is simply not true. Energy is only conserved in an inertial frame of reference. If you consider everything from perspective of the rocket, you are in an accelerated frame of reference, and energy is not conserved. So your analysis is faulty. If you want to analyze it from perspective of energy, you should be considering motion relative to the star/planet, as that's, effectively the relevant inertial system. In that system, you are moving faster when you are closer to star/planet, and therefore, you get more work out of the same amount of fuel due to traveling a longer distance while burning it.

Edited March 27, 2014 by K^2

[Ralathon]

Oberth is rather hard to explain without indulging in some math. So please bear with me here, I'll try to keep it simple.

Basically, when you move your kinetic energy is equivalent to your velocity squared. So if you go 10 times as fast you contain 100 times as much kinetic energy. This is where the Oberth effect gets important, since the rocket always has the same amount of delta V regardless of velocity. As you say, the rocket doesn't care how fast it goes.

Say your rocket has 10m/s of delta V. If start out stationary you start with 0 m/s and end with 10 m/s. Thus giving you 100 units of kinetic energy.

Now imagine that your rocket starts out at 100 m/s. After your burn you're moving with 110 m/s. This means your burn gave you 110^2 - 100^2 = 2100 units of energy.

As you can see you gained way more energy than from the stationary burn. Normally this is just a matter of reference frames, from the perspective of a train moving at 100m/s along your rocket you still only gained 100 units of energy. But when you include a gravitational field things get interesting because an orbit is a constant exchange of potential energy and kinetic energy. As you fall closer to the planet you lose potential energy and gain kinetic energy and vica versa. Your orbital energy is always constant and equal to your kinetic energy relative to the planet and your potential energy. So if you want the maximum bang for your buck you need to fire your engines when you go fastest since this increases your kinetic energy the most.

A

would be the old party trick with a tennisball and a basketball that drop at the same time. When the basketball hits the ground it bounces up and hits the tennisball causing it to fly much higher than your initial drop height. When you go down (Fall to periapsis) your spacecraft consists of both the basketball (fuel) and the cargo (tennis ball). When it goes back up it leaves behind the basketball meaning that all the kinetic energy ends up in the tennis ball, causing it to move much higher.

[travis575757]

Say your rocket has 10m/s of delta V. If start out stationary you start with 0 m/s and end with 10 m/s. Thus giving you 100 units of kinetic energy.

Now imagine that your rocket starts out at 100 m/s. After your burn you're moving with 110 m/s. This means your burn gave you 110^2 - 100^2 = 2100 units of energy.

Okay I think that does a good job at explaining it. Funny because i said not to use math but this probably made the most sense, thanks.

[Sillychris]

If you consider everything from perspective of the rocket, you are in an inertial frame of reference, and energy is not conserved..

I think you meant to say "you are not in an inertial frame of reference" since the rocket is accelerating.

[K^2]

I think you meant to say "you are not in an inertial frame of reference" since the rocket is accelerating.

Yup. Thanks for catching that.

[Doozler]

For a "where does the energy go" perspective, imagine a rocket with exhaust that fires at 100 m/s. When the rocket is going slowly, the exhaust ends up as travelling at 100 m/s (i.e. it has lots of energy).

If the rocket is travelling at 100 m/s, the exhaust ends up with 0 kinetic energy, so there is more for the rocket.

[shynung]

...imagine a rocket with exhaust that fires at 100 m/s. When the rocket is going slowly, the exhaust ends up as travelling at 100 m/s (i.e. it has lots of energy).

If the rocket is travelling at 100 m/s, the exhaust ends up with 0 kinetic energy, so there is more for the rocket.

If the rocket is travelling at 110 m/s, the exhaust ending up travelling at 10 m/s in the same direction as the rocket, will the Oberth effect still apply?

[Ralathon]

If the rocket is travelling at 110 m/s, the exhaust ending up travelling at 10 m/s in the same direction as the rocket, will the Oberth effect still apply?

110^2 - 10^2 = 12000 > 100^2 = 10000

Yes, it will.

[magnemoe]

Just an example of how strong it is.

Going to Jool from LKO cost 2000 m/s. Going from minmus orbit cost 2500

[Lukaszenko]

Kinetic energy is always measured in relation to something. So, indeed if you measure it in relation to your rocket, it doesn't matter how fast you are moving. In fact, the rocket's kinetic energy in relation to the rocket is zero. But, when flying a rocket you are usually interested in changing your orbit or escaping a planet or a star, and in order to do that, you need to measure kinetic energy in relation to that planet or star. That's when the V^2 portion of the kinetic energy equation matters.

[Doozler]

If the rocket is travelling at 110 m/s, the exhaust ending up travelling at 10 m/s in the same direction as the rocket, will the Oberth effect still apply?

Yes, and the point here is that the exhaust (the fuel) started off at 100 m/s and ended up at 10 m/s.