Why doesn't conservation of momentum conflict with the Oberth Effect?

[beamthegreat]

After reading countless posts about of the Oberth Effect I still cannot understand how this concept works.

Suppose there are 2 rockets named A and B. It is situated 100km above the surface of earth and the main propulsion of this rocket works by ejecting mass.

Rocket A is moving at 4000 m/s relative to Earth

Rocket B is moving a 0 m/s relative to earth

Momentum must be always be conserved, so no matter how fast the rocket is moving the change in momentum must equal 0.

Now lets assume the following:

Mass of rocket = 10 kg

Mass ejected = 5 kg

Speed of mass ejected = 10 m/s

Using m1*v1=m2*v2,

5kg * 10m/s =10kg * v2

v2 = 5m/s

The speed of both rockets must change by 5m/s regardless of how fast they are travelling.

Obviously this should be wrong since the oberth effect states that rockets that contain more kenetic energy releases extra energy, but can someone point out why this example is flawed?

Thanks.

[Sillychris]

In a nutshell, you're forgetting about the initial momentum of the propellant. A rocket whipping forward at 1000 m/s is full of propellant whipping forward at 1000 m/s. It's here that you get the extra momentum to reverse.

[Ralathon]

Oh you're absolutely right. Both rockets will have the same dV. But that's not what the Oberth effect is about.

The Oberth effect says that it is a more efficient use of dV if you burn while going fast. It has to do with orbital energy, to quote myself on the matter:

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.

[Yasmy]

You are right on that conservation of momentum says you get the same dv regardless of your velocity.

Specific orbital energy determines the orbit: E = 1/2 v^2 - G M / r

Make a small velocity change dv: E = 1/2 (v+dv)^2 - G M / r

So the change in specific orbital energy is dE = 1/2 (v+dv)^2 - 1/2 v^2 = v dv + 1/2 dv^2

For a small dv, with dv << v, the 1/2 dv^2 term can be ignored because it is small compared to v dv. Thus dE is approximately v * dv.

This is the meaning of the Oberth effect: The change in orbital energy for a given dv is proportional to your velocity: dE = v * dv.

[jwenting]

and remember that both bodies affect each other. So your imparting a small momentum on the planet as well. So small the planet never really notices it because it gets dampened out by everything else pulling on it, but it is there.

[Lukaszenko]

Adding 5 m/s on top of 4000 takes MUCH more energy than adding 5 m/s from standstill.

[Red Iron Crown]

and remember that both bodies affect each other. So your imparting a small momentum on the planet as well. So small the planet never really notices it because it gets dampened out by everything else pulling on it, but it is there.

I think you're referring to gravity slingshots, not the Oberth effect.

[Red Iron Crown]

The Oberth effect is really counterintuitive. It's not really an effect at all, but instead is an observation based on two things:

#1. Kinetic energy is proportional to the square of speed.

#2. Reaction drives that carry their reaction mass with them have a fixed amount of velocity change rather than a fixed amount of energy.

The observation is: The faster you are going, the more a change in speed changes kinetic energy.

#1 should be familiar to anyone who's done high school physics. For most Earth-bound drives, it means that the faster you are going, the harder it is to accelerate further. Going from 10m/s to 11m/s costs 10.5 joules/kilogram; going from 100m/s to 101m/s costs 100.5 j/kg.

#2 is what changes everything, and confused even rocket scientists at first. If you can exchange a given amount of fuel for a given velocity change, you change the kinetic energy by a larger amount if you do the velocity change at greater speed. Using the above numbers, a 1m/s prograde burn adds 10.5 j/kg at 10m/s, but adds 100.5 j/kg at 100m/s.

We are used to thinking about delta-V in our mission planning in KSP, but the truth is that it's really specific orbital energy change that is important, that gets us to our destinations. The amount of delta-V a given vessel has is determines by its mass ratio and engine efficiency, but we can extract more orbital energy change from that delta-V through clever use of the Oberth effect.

Think of delta-V as the currency we use to buy a change in specific orbital energy; the Oberth effect lets us buy it at a discount.

Conservation of energy is observed because the expended propellant (exhaust) changes kinetic energy by exactly the opposite amount of the ship. The Oberth effect allows more energy to be exchanged between ship and propellant.

[TheGatesofLogic]

efficiently, it takes more energy to do it EFFICIENTLY.