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]

A very big thank you, Jay. That cleared it up a lot.

Lurky

Don't you just LOVE the nitty-gritty details of rocket science!!?

[JayUtah]

A very big thank you, Jay. That cleared it up a lot.

Lurky

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.

[ka9q]

The effect of the Effect is simple enough; the faster the rocket is going, more delta-V it gets from a kg propellant.o any constructive

This is actually incorrect; you get the same delta-v regardless of the rocket's velocity relative to the observer (ignoring relativistic effects). What Jay said is correct; you get more kinetic energy from each kg of propellant when you're going faster.

Some years ago I worked out the powers and energies for each stage of the Saturn V. It turns out that during much of the third stage burn more mechanical power (force times velocity) is being given to the Apollo spacecraft than the heat power released in the J-2's combustion chamber. This is not "free energy"; that extra energy is coming from the kinetic energy that had been stored in the J-2's propellants before they were burned. That energy was put there by the lower stages of the rocket. So rockets aren't quite as inefficient as they seem early in flight when they're furiously burning propellant and still moving fairly slowly. Only a small fraction of the released energy may be going into the payload but much more is being "invested" into the kinetic energy of the as-yet-unburned propellants, some of which will be recovered later.

The most energy-efficient rocket is one that increases its Isp (effective exhaust velocity) linearly with speed so that its exhaust is always stationary in the inertial frame of the launch site. Of course this would be a much more massive rocket because energy efficiency is usually not the primary consideration in rocket design. Size and weight are more important, which is why a highly energetic propellant like liquid hydrogen is so attractive even though it costs a lot of energy to make.

[bknight]

The effect of the Effect is simple enough; the faster the rocket is going, more delta-V it gets from a kg propellant.o any constructive

This is actually incorrect; you get the same delta-v regardless of the rocket's velocity relative to the observer (ignoring relativistic effects). What Jay said is correct; you get more kinetic energy from each kg of propellant when you're going faster.

Some years ago I worked out the powers and energies for each stage of the Saturn V. It turns out that during much of the third stage burn more mechanical power (force times velocity) is being given to the Apollo spacecraft than the heat power released in the J-2's combustion chamber. This is not "free energy"; that extra energy is coming from the kinetic energy that had been stored in the J-2's propellants before they were burned. That energy was put there by the lower stages of the rocket. So rockets aren't quite as inefficient as they seem early in flight when they're furiously burning propellant and still moving fairly slowly. Only a small fraction of the released energy may be going into the payload but much more is being "invested" into the kinetic energy of the as-yet-unburned propellants, some of which will be recovered later.

The most energy-efficient rocket is one that increases its Isp (effective exhaust velocity) linearly with speed so that its exhaust is always stationary in the inertial frame of the launch site. Of course this would be a much more massive rocket because energy efficiency is usually not the primary consideration in rocket design. Size and weight are more important, which is why a highly energetic propellant like liquid hydrogen is so attractive even though it costs a lot of energy to make.

With Blue Origin and mow SpaceX opting for methane is because it is easier to handle than liquid H₂?

[ka9q]

The reasons are all practical tradeoffs.

Methane/LOX has roughly the same performance as RP-1/LOX, so that's a wash.

Methane is less dense than RP-1, requiring a larger tank, and liquid methane is a cryogenic liquid, unlike RP-1 (but not nearly as cold as LH2). Those are strikes against methane.

Methane doesn't coke (release elemental carbon) as easily at high temperatures. That's an advantage, as it makes it easier to reuse a regeneratively cooled engine. More complex (and efficient) engine cycles (like preburning) may be possible. That's another advantage.

Methane can supposedly be made from raw materials on Mars.

So while methane has had no particular advantages (only disadvantages) until now, you can see why BlueOrigin and SpaceX are interested in it.

[JayUtah]

Plumbing fittings for liquid hydrogen are hard to engineer.  It's a very tiny molecule, and it likes to sneak past seals.

[jfb]

IINM, methane and oxygen have similar boiling points, so the tanks can share a common bulkhead, which saves some mass.  But I think the real attraction is that methane can be synthesized on Mars relatively easily.