Discussion of The Rocket Equation and Different Types of Rocket Propulsion

[Ben W]

This is how I think of the Oberth Effect; it seems like free energy - despite the explanations.

All of the above explanations of the Oberth Effect are pretty much equivalent; it's just a question of which appeals most to your intuition. It's not free energy; energy and momentum of the entire system (rocket + exhaust + planet) are always conserved.

Another way to think about it is in terms of tidal forces. Pretend the rocket splits in half at closest approach to the planet, with the forward section gaining 50m/s and the aft section losing 50ms. As the two rocket pieces escape the gravitational well, tidal forces will cause the rocket system to "stretch" and for the two pieces to get further apart faster than they would based on momentum alone; their relative speed increases from 100m/s to something larger. (In the limiting case where the forward section just barely achieves escape velocity, it will continue outward forever, while the aft section eventually falls back to the planet at high speed, so the relative velocity will become extreme.) With a rocket burn at periapsis, the propellant functions as the "aft end" of the rocket in this scenario. The other key is that a given rocket burn imparts a fixed amount of momentum, not a fixed amount of energy. A change in momentum at high speed will impart much more energy than the same change in momentum at low speed. (And the propellant loses correspondingly more energy at high speed, so everything balances out.)

Another way to think about it: for an outbound object, gravity causes a fixed loss of momentum per unit time. If you can escape a gravitational well in less time, you will lose less momentum getting out. (The energy loss is the same regardless of speed, but the momentum loss is not.) Counterintuitive, but this is the basis for the Oberth effect! No perpetual motion machines required

Yet another way to think about it: it's roughly equivalent to "pumping" a swingset at the bottom of the swing. This is a much more efficient way to gain amplitude than trying to pump at the top of the swing. The higher velocity at the bottom of the swing means that the changes in momentum (pumping the swing) adds a lot more energy when done there. It's not exactly analogous because there's no propellant involved [depending what you ate for lunch], but there are conceptual similarities.

Incidentally, the most efficient way to gain escape velocity from our solar system is as follows: First fly to Jupiter, then use a gravitational assist to lose most of your angular momentum with respect to the sun. Then, fall (almost) directly into the sun, performing an Oberth maneuver at perihelion while being whipped around the sun ridiculously fast. If you can avoid getting fried to a crisp, this little maneuver will slingshot you out of the solar system at ludicrous speed. This wonderful short story illustrates the concept.

 

[ecarfan]

it smacks of cold fusion claims

It certainly does.

Note how energy is "consumed", but it still gains more k.e. than is spent powering it. That's free energy right there.

A conclusion based on baseless assumptions. Is the slashdot commentator seriously saying that?

 

[ecarfan]

This wonderful short story illustrates the concept.

Thank you. What a great story!

 

[Ben W]

I'm throwing this in here because it smacks of cold fusion claims and I don't know where else I'd put it, but also because of a comment that was made on slashdot about this

NASA Veteran’s Propellantless Propulsion Drive That Physics Says Shouldn’t Work Just Produced Enough Thrust to Overcome Earth’s Gravity - The Debrief

A veteran NASA scientist says his company has tested a propellantless propulsion drive technology that produced one Earth gravity of thrust.

thedebrief.org

This is how I think of the Oberth Effect; it seems like free energy - despite the explanations.

The difference is that the Oberth Effect cannot be leveraged to create a perpetual motion machine, but the reactionless drive can. The Oberth Effect doesn't violate conservation of energy / momentum, but the reactionless drive does.

The reason I'm skeptical of the reactionless drive is this: every particle-physics interaction we've ever observed, precisely preserves both energy and momentum. And none of the particle interactions inside the "reactionless drive" are occurring in an environment that hasn't been exhaustively studied in particle physics experiments before. If you add a bunch of zeroes together, the result will still be zero. Yet the claim is that all the particle interactions inside the reactionless drive are somehow summing to something strongly non-zero, in terms of creating energy / momentum. If true, this implies there must be much simpler particle interactions that should also violate conservation of energy / momentum. Yet, it beggars belief that a century of particle physics experiments under an extraordinarily wide range of conditions has somehow missed this. The far more likely explanation is that the reactionless-drive experimenters are wrong, in their measurements and/or assumptions. (I do hope they're right, but I'm quite sure they're wrong.)

 

[scaesare]

I like the imagery, but I don't know that the explanation matches the physics. Yes, the outbound transit time is less, but does that actually mean anything?
Potential energy (Pe) is mass*height*gravity, there is no accounting for the time it took to get to that altitude.
Work is force*distance, in this case height. Crossing the distance faster requires greater power, but net energy of altitude change is the same. f=m*a, d=h, same as Pe.

The lesser speed decrease in the second situation is due to kinetic energy being proportional to the square of velocity.
150 vs 100 is 2.25 the inital energy. The ascent converts the same amount to potential energy in both cases, but in the second would still have 1.25 times case one's kinetic energy left over, meaning the velocity at the balloon altitude would still be greater than the original drop's final velocity.

I think the gravity well is a red herring because potential energy cancels out, and the root difference is the increase in energy imparted to the rocket by burns at higher speeds.
Ke=1/2*m*v², so the same delta-v requires a greater increase in kinetic energy the faster the rocket is going:
Kef=1/2*m*(Vi+Vd)² = 1/2*m*(Vi²+2Vi*Vd+Vd²)
Net Ke change: 1/2*m*(2*Vi*Vd+Vd²)

Given a rocket's delta-v budget doesn't care about intermediate velocities (note: rocket centric, not gravity assist trajectory planning), the topic de-jour (as I understand it) is the method by which the same engine burn can produce wildly different amounts of energy change in the rocket.

Kinetic energy's square term: If one is decelerating a moving object with a constant force, the object's velocity determines the distance over which the force is applied, twice the speed means twice the time to decelerate during which it is traveling twice as fast, thus 4x the distance.
position=-1/2*a*t² + Vi*t

Stopping time = Vi/a
Stopping distance = -1/2*Vi²/a + Vi²/a = 1/2*Vi²/a
f=ma, a=f/m
d = 1/2*Vi²*m/f
Work=f*d
Work=f*1/2*Vi²*m/f = 1/2*m*Vi² = Ke

Which does lend itself to the line of thought that at higher velocity the same burn duration with the same engine force occurs over a longer distance thus work done to the rocket is greater. This then requires the energy lost by the exhaust to be greater. So maybe both ways of looking at it are correct along with being numerically equivilent @scaesare ?

And they may both link back to position/ height having a square term for time, thus Pe does internally for a trajectory.
Calc I did to try and get my mind aligned:
In a vacuum, potential energy plus kinetic is a constant, regardless of the time in flight: Pe= m*h*a, h=-1/2*a*t² + t*Vi =m*(-1/2a²t²+atVi) Ke= 1/2*m*v², v=Vi-a*t = 1/2*m*(Vi²-2atVi+a²t²) Pe+Ke=m*(1/2*Vi²-atVi+1/2*a²t² -1/2*a²t² + atVi) =1/2*m*Vi² = inital Ke = final Pe at t=Vi/a, h=1/2*Vi²/a (ignoring gravity falling off with distance)

 

[scaesare]

I'm throwing this in here because it smacks of cold fusion claims and I don't know where else I'd put it, but also because of a comment that was made on slashdot about this

NASA Veteran’s Propellantless Propulsion Drive That Physics Says Shouldn’t Work Just Produced Enough Thrust to Overcome Earth’s Gravity - The Debrief

A veteran NASA scientist says his company has tested a propellantless propulsion drive technology that produced one Earth gravity of thrust.

thedebrief.org

Here's the slashdot comment:



This is how I think of the Oberth Effect; it seems like free energy - despite the explanations.

Wow... that article claims:

discovery of a fundamental new force

That's... uh.. signficant.

It seems to say electrostatics are involved, but that's not a fundamental new force, and:

The electrostatic force is also known as the Coulomb force or Coulomb interaction. It's the attractive or repulsive force between two electrically charged objects.

So in space, what's that gonna push against?

 

[ecarfan]

I would not describe it as the “discovery of a fundamental new force” when no explanation is provided as to what the force is.

If there is a “new force” I’ll believe it when a plausible theoretical explanation is made and it is then verified independently by multiple researchers with peer review.

 

[JB47394]

"New forces" have been "experimental error" for a while now. "According to known physics, it shouldn't do this, but it does, so that means there is new physics involved instead of experimental error."

With the cost of mass to orbit about to crater, all these crazy experiments that need to be demonstrated in zero gravity may get a chance to show their stuff.

 

[Grendal]

This wonderful short story illustrates the concept.

The writer, Vernor Vinge, passed away just a month ago. Thanks for sharing that little gem.