Discussion of The Rocket Equation and Different Types of Rocket Propulsion

[JB47394]

The work section of that article helps bridge the difference.

So the difference is that the rocket velocity at that point increases by the same amount as ever, but the rocket's kinetic energy jumps by much more. The rocket then retains that kinetic energy as it leaves periapsis, so it's going faster.

So this is like an assisted gravity assist? I'm taking advantage of the potential energy imparted to my rocket by having fallen into the gravity field, and then keeping some of it by changing my velocity via my engines while in the bottom of the gravity field?

I'm trying to wrap my head around where the extra energy is coming from, and I think it's the planet.

How does this work out for a rocket that accumulates lots of velocity on its own as it fires its engines? For example, an ion drive that pushes a vehicle to three times its original velocity. Is the kinetic energy of the rocket being increased by nine times as much? Does that mean that it's gaining nine times as much effective delta-V?

Does this mean that I could theoretically use a large spinning wheel with rocket motors on the perimeter to gain additional kinetic energy? I'd fire an engine when it lines up with wherever I want to thrust. Only the kinetic energy of the engines would increase, but I'd pack as much of my vehicles mass into those engines. Or is this like a perpetual motion machine?

 

[scaesare]

So the difference is that the rocket velocity at that point increases by the same amount as ever, but the rocket's kinetic energy jumps by much more. The rocket then retains that kinetic energy as it leaves periapsis, so it's going faster.

So this is like an assisted gravity assist? I'm taking advantage of the potential energy imparted to my rocket by having fallen into the gravity field, and then keeping some of it by changing my velocity via my engines while in the bottom of the gravity field?

I'm trying to wrap my head around where the extra energy is coming from, and I think it's the planet.

How does this work out for a rocket that accumulates lots of velocity on its own as it fires its engines? For example, an ion drive that pushes a vehicle to three times its original velocity. Is the kinetic energy of the rocket being increased by nine times as much? Does that mean that it's gaining nine times as much effective delta-V?

Does this mean that I could theoretically use a large spinning wheel with rocket motors on the perimeter to gain additional kinetic energy? I'd fire an engine when it lines up with wherever I want to thrust. Only the kinetic energy of the engines would increase, but I'd pack as much of my vehicles mass into those engines. Or is this like a perpetual motion machine?

So the formula for kinetic energy is 1/2mv^2.

So the increase in kinetic energy goes up with the square of velocity. So, if your mass =100 you can add 1Km/sec of and go from 10 to 11 Km/sec, then your kinetic energy increase is (1/2*100*11^2 )-(1/2*100*10^2) = 1050.

If you add the same 1 km/sec at 100KM/sec then (1/2*100*101^2)-(1/2*100*100^2) = 10,050

So in this case (if my math is right), adding the same 1Km/sec delta-V is ~10X the kinetic energy, if you add it to a starting velocity that's 10X as much.

So, if there's anything you can use (like a nearby gravity well) to accelerate you, it makes sense to do your own acceleration once you've gotten as fast as possible with the assist.

(This is all new to me and I'm trying to talk thru my understanding of it)

I think this means the answer to this:

Does that mean that it's gaining nine times as much effective delta-V?

is "no".

 

[JB47394]

(This is all new to me and I'm trying to talk thru my understanding of it)

Sure. The math says that things work a certain way. That's fine. I want to have an intuitive sense of why it would be true.

For example, consider two equal masses, but the second one has a rocket engine. The approach each other, and accelerate without hitting each other. While at periapsis, the second one fires its rocket engine. It ends up with extra velocity courtesy of bumping its kinetic energy so much during that pass. Where does that energy come from, and what happens to the other object? Does a fast rocket slow down a planet more than a slow rocket? What's going on?

I think this means the answer to this: is "no".

Why? What's the difference between moving fast because of an orbital maneuver and moving fast because of running a rocket engine? So long as I'm moving faster, I'm going to get the kinetic energy squaring bump.

 

[mongo]

the kinetic energy of the rocket being increased by nine times as much? Does that mean that it's gaining nine times as much effective delta-V?

The additional delta-V provided by the engine is the same regardless of the gravity assist.
Edit: scratch that, net delta-V at the same distance from the object is greater than the input delta-V, but my math is too rough to say by how much.

For example, consider two equal masses, but the second one has a rocket engine. The approach each other, and accelerate without hitting each other. While at periapsis, the second one fires its rocket engine. It ends up with extra velocity courtesy of bumping its kinetic energy so much during that pass. Where does that energy come from, and what happens to the other object? Does a fast rocket slow down a planet more than a slow rocket? What's going on?

They transfer momentum, the larger object slows a little and the smaller speeds up proportionately faster.
A faster flyby at the same distance reduces the transfer due to less time in proximity, but a faster rocket can also approach closer thus increasing peak gravitational force and increasing transfer.

Why? What's the difference between moving fast because of an orbital maneuver and moving fast because of running a rocket engine? So long as I'm moving faster, I'm going to get the kinetic energy squaring bump.

The difference is expending propellant to get to that faster speed versus the rocket and propellant being accelerated by the object.
Also, when accelerated by an object on approach, you will be decelerated on departure but by a lesser amount (if you are moving faster due to momentum transfer [or engine burn]).

 

[JB47394]

I appreciate the try, but this is only confusing things even more. The math says that the ship's kinetic energy is proportional to the square of the ship's velocity. So the ship falls into the gravity well to make a pass, increasing its velocity, reaching a maximum at periapsis. That means that the kinetic energy is at a maximum as well. The ship fires its engines, imparting a change of velocity.

The statement is that the ship will gain more total velocity upon leaving that system by firing its engines at periapsis than it would if it fired them before periapsis. Math has been presented as an explanation for that, which didn't resonate with me. So now I'm looking for an ELI5 as to why it gains so much velocity.

You mentioned that the approach acceleration and the departure acceleration are different. That certainly makes sense, but is that actually the way that the delta-V gets magnified? That the ship is accelerated into periapsis and then has a decision. If it does nothing, then it flies by and is decelerated by the same amount on the way out. No gain. But if it fires its engine at periapsis, then it will create an asymmetry, and depart faster than it arrived. That means that it escapes the gravity field more quickly, and it isn't penalized as much.

Is that the actual source of the increased velocity change, and the math is just expressing it quantifiably?

 

[mongo]

I appreciate the try, but this is only confusing things even more. The math says that the ship's kinetic energy is proportional to the square of the ship's velocity. So the ship falls into the gravity well to make a pass, increasing its velocity, reaching a maximum at periapsis. That means that the kinetic energy is at a maximum as well. The ship fires its engines, imparting a change of velocity.

The statement is that the ship will gain more total velocity upon leaving that system by firing its engines at periapsis than it would if it fired them before periapsis. Math has been presented as an explanation for that, which didn't resonate with me. So now I'm looking for an ELI5 as to why it gains so much velocity.

You mentioned that the approach acceleration and the departure acceleration are different. That certainly makes sense, but is that actually the way that the delta-V gets magnified? That the ship is accelerated into periapsis and then has a decision. If it does nothing, then it flies by and is decelerated by the same amount on the way out. No gain. But if it fires its engine at periapsis, then it will create an asymmetry, and depart faster than it arrived. That means that it escapes the gravity field more quickly, and it isn't penalized as much.

Is that the actual source of the increased velocity change, and the math is just expressing it quantifiably?

Except it's not the time that matters, I was wrong on that (and may be again). Work is force*distance and gravity and altitude are the same so work is the same regardless of speed or travel time. Power would be different, but that's irrelevant (I think). If we were dealing with lift off, there would be gravity losses at lower speeds, but this is all orbital.

Because kinetic energy goes by the square of velocity a ball thrown vertically twice as fast doesn't go twice as high, it goes 4 times as high with double the time of flight. To throw it twice as fast requires 4 times the work over the throw distance, that means 4 times the force if distance is fixed. In the case of a rocket, the same thrust over four times the distance is four times the work. If the ship is going 4x the speed, then the same burn time gives 4x the work...

Making it way too simple, potential energy at altitude A is the same inbound and outbound, but the burn boosts the kinetic energy which is carried onward so there is more Ke at that altitude and that Ke gain corresponds to a higher delta-V than the burn itself.
Mass cancels out but call it 1, and 0v at 50high
Pe=50 = Ke@0alt = 1/2*1*(10)² add 1 velocity
Ke' = 1/2*1*(10+1)² = 60.5 - Pe (50) = 10.5 = .5*1*v², v=sqrt(21) = 4.6

Slingshot manuvers don't need a burn and rely on capturing some of the larger body's momentum. Sort of like Charlie Chaplin making a quick turn by hooking his cane around a lamppost, but in this situation the lamppost is moving. (I think?)

 

[JB47394]

Because kinetic energy goes by the square of velocity

The answer that I'm looking for is one you've already given, which is that the kinetic energy of the propellant is higher at periapsis, and that makes it better as a propellant. That's the part that I don't understand. That the rocket moving faster means that the propellant has an improved ability to push the rocket seems absurd to me.

Oh well. I think this is one I'm just not going to get. Thanks for the college try.

 

[mongo]

That the rocket moving faster means that the propellant has an improved ability to push the rocket seems absurd to me.

Yeah that would be wacky, fortunately it doesn't. From the rocket's frame of reference, the propellant has the exact same impact.
The weirdness comes in because the rocket/ propellant system is moving relative to the orbiting body.

Similarly, that is how the delta-v budget stays the same regardless of the rocket's travel velocity. A burn at higher velocities must impart more energy into the rocket to produce the same change in velocity. Therfore, the exhaust must also lose more energy.

More words:
Say you are on a flatbed train car, and you are playing catch with another traveler. Regardless of the train's speed, tossing the ball takes the same effort. Propellant burn has same effect regardless of inital velocity).
But if you throw the ball off the train, it hits the scenery with an energy based on the sum of the train and throw velocities.

Now picture you on a zero friction surface on the train car throwing balls backwards. Each throw accelerates you the same amount regardless of inital train speed. However, the energy you gain relative to the scenery per throw is higher the faster the train is going. Simultaneously, the energy the ball loses increases with the train velocity.

For simplicity ball mass = person mass = m, equivilent velocities (Vb=Vy=Vd)
Ke0 (external frame): 1/2*2m*(Vt)²
Ke_throw (in frame) = 1/2*m(Vb)²

Ke_ball = 1/2*m(Vt-Vb)² = 1/2*m*(Vt²-2VbVt+Vb²)
Ke_you = 1/2*m(Vt+Vy)² = 1/2*m*(Vt²+2VyVt+Vy²)
Ke total = 1/2*m*(2Vt²+2Vb²)
= 1/2*2m*(Vt²+Vb²)
Delta Ke (work)= 1/2*2m*Vb²
Which is just Ke of both objects, work is invariant of train velocity

Out of frame:
Your initial
Ke= 1/2*m*(Vt)²
Post throw
Ke = 1/2*m(Vt+Vy)² = 1/2*m*(Vt²+2VyVt+Vy²)
Delta:
1/2*m*(2VyVt+Vy²)
Additional gain due to train:
1/2*m*2VyVt = m*Vy*Vt
So of the 1/2*2m*Vy² work, you recieved 1/2*m*(2VyVt + Vy²)
As a ratio: 1/2*(2VyVt+Vy²)/(Vy²) = 1/2*(2Vt/Vy+1) = Vt/Vy+1/2
Which shows that if the rocket is initially traveling at half the speed of the exhaust, then the rocket gets all the work. Which lines up since in that case the exhaust would end up with zero external frame velocity.

Equivalent additional velocity if original energy is removed (potential energy from gravity well):
m*Vy*Vt = 1/2*m*(Vf)²
Vf=sqrt(2*Vy*Vt)

 

[scaesare]

So... this is a fascinating discussion (thanks @Grendal for carving out a sperate thread for it). And I'm in the same boat in that I believe the math, but don't find it intuitive.

I feel like the issue boils down to Work = Force * Distance. That always made sense in that if your engine moved your vehicle twice as far you did 2X the work. And work is the delta in kinetic energy.

However, it seems the Oberth effect is saying that if your engine pushes on your vehicle with the same force for the same period, but it moves 4X the distance as before (because it was rolling down hill, was at the bottom of a gravity well, Faries were also pushing on it, etc...).... then you've done more work in that period of time, because the vehicle has gone farther in that time you've pushed on it.

I'm I grasping this correctly?

And, thanks to @JB47394 I now have "ELI5" to add to my bag O' acronyms.

 

[mongo]

So... this is a fascinating discussion (thanks @Grendal for carving out a sperate thread for it). And I'm in the same boat in that I believe the math, but don't find it intuitive.

I feel like the issue boils down to Work = Force * Distance. That always made sense in that if your engine moved your vehicle twice as far you did 2X the work. And work is the delta in kinetic energy.

However, it seems the Oberth effect is saying that if your engine pushes on your vehicle with the same force for the same period, but it moves 4X the distance as before (because it was rolling down hill, was at the bottom of a gravity well, Faries were also pushing on it, etc...).... then you've done more work in that period of time, because the vehicle has gone farther in that time you've pushed on it.

I'm I grasping this correctly?

And, thanks to @JB47394 I now have "ELI5" to add to my bag O' acronyms.

The more I think about it, the less I like that analogy. Work is a quantity, that we can get the same value doesn't mean the narrative is correct.

The above math barf shows the change is in the extenal energy of the exhaust and rocket and the amount of work apportioned to each.
Two stationary objects that push apart with final velocity v each
Versus
Two objects moving at velocity v that push apart resulting in one going 0 and one going 2v:
Same work is done, but in the first system
Inital energy: 0
Final total energy: 2*1/2*m*(v)² = m*v²
Added energy (work) = m*v²
Outbound object energy = 1/2*m*v²

Moving:
Inital energy: 2*1/2*m*v² = m*v²
Final energy: 0 + 1/2*m*(2v)² = 2*m*v²
Added energy (work) = m*v²
Outbound object energy = 2*m*v²
Outbound energy gain = 2*m*v² - 1/2*m*v² = 3/2*m*v²
Outbound gained all the work plus the exhaust's energy

 

[scaesare]

The more I think about it, the less I like that analogy.

Well crap. I tried.

 

[mongo]

Well crap. I tried.

Wait, I thought I made it up, that comment wasn't direct at you.

 

[scaesare]

Wait, I thought I made it up, that comment wasn't direct at you.

Hehe... all good... I my response was tongue-in-cheek...

Plus analogies are always imperfect.

But is the crux of this the idea that if you are going faster, you are thus able to push on the object over a longer distance and therefore do more work?

 

[JB47394]

Say you are on a flatbed train car, and you are playing catch with another traveler. Regardless of the train's speed, tossing the ball takes the same effort. Propellant burn has same effect regardless of inital velocity).
But if you throw the ball off the train, it hits the scenery with an energy based on the sum of the train and throw velocities.

Now picture you on a zero friction surface on the train car throwing balls backwards. Each throw accelerates you the same amount regardless of inital train speed. However, the energy you gain relative to the scenery per throw is higher the faster the train is going. Simultaneously, the energy the ball loses increases with the train velocity.

Oh, I can do the math. That's not the problem.

Before approaching the planet, my 1 kg rocket is moving 10 m/s. Kinetic energy of 50 Joules.
At periapsis, my 1 kg rocket is moving at 100 m/s. Kinetic energy of 5000 Joules. A gain of 4950 Joules.
I accelerate by 1 m/s. Kinetic energy of 5100.5 Joules. A gain of 100.5 Joules.
I depart the planet and lose the same energy that I gained . Kinetic energy of 150.5 Joules. A loss of 4950 Joules.

The net gain from prior to encountering the planet is 100.5 Joules, leaving me with 14 m/s of velocity as a result of my initial 10 m/s velocity and a 1 m/s burn at perigee.

Great. Unfortunately, I still don't know why a change in velocity at higher speed imparts extra energy to the rocket. The math isn't the reason, it's only a qualitative description of what happens. Perhaps I'm trying to do the equivalent of asking for an explanation of how gravity or magnetism works.

 

[mongo]

Great. Unfortunately, I still don't know why a change in velocity at higher speed imparts extra energy to the rocket. The math isn't the reason, it's only a qualitative description of what happens. Perhaps I'm trying to do the equivalent of asking for an explanation of how gravity or magnetism works.

There isn't more energy, just a different distribution. In the external frame, the exhaust lost more energy and more of the work was imparted to the rocket.

>>Before approaching the planet, my 1 kg rocket is moving 10 m/s.
>>Kinetic energy of 50 Joules.
Call it half propellant. 25 joules in rocket, 25 in propellant

>>At periapsis, my 1 kg rocket is moving at 100 m/s. >>Kinetic energy of 5000 Joules. A gain of 4950 Joules.
Rocket 2500, gain of 2475, same for propellant

>>I accelerate by 1 m/s.
By expelling exhaust in the other direction

>>Kinetic energy of 5100.5 Joules. A gain of 100.5 Joules.
Rocket 2550.25, Exhaust 2450.25, total delta of 100.5 from preburn

>>I depart the planet and lose the same energy that I gained . Kinetic energy of 150.5 Joules. A loss of 4950 Joules.
For rocket it's 75.25, exhaust -24.75 (doesn't make it that high)

>>The net gain from prior to encountering the planet is 100.5 Joules, leaving me with 14 m/s of velocity as a result of my initial 10 m/s velocity and a 1 m/s burn at perigee.
>>Rocket gains 50.25 and is at 17.35 m/s

 

[mongo]

Hehe... all good... I my response was tongue-in-cheek...

Plus analogies are always imperfect.

But is the crux of this the idea that if you are going faster, you are thus able to push on the object over a longer distance and therefore do more work?

I don't think so since the net work done stays the same. The rocket just gets more of the energy.
Also, it's a momentum transfer and not exhaust as a single thing exerting force, but many small pushes. Do collisions have a distance?

Or I could just go read the answer...

 

[JB47394]

In the external frame, the exhaust lost more energy

I'll have to ponder that.

 

[Ben W]

Here’s how I think about it.

Suppose you drop a rock from a hot air balloon. Acceleration acts on it while it’s falling, and it hits the ground at say 100mph. (Ignore air resistance.) Then throw the rock upward from the ground at 100mph. It will stop when it reaches the height of the hot air balloon.

Now, _throw_ a rock _downward_ from the hot air balloon at 100mph. Gravity will have a lot less time to act on it (to accelerate it) before it hits the ground, so it will hit the ground going somewhat slower than 200mph; say 150mph. Now reverse this; throw the rock upward from the ground at 150mph. When the rock passes the hot air balloon, it will be travelling 100mph upward. The 50mph difference in upward thrown speed at ground level (150mph vs 100mph) translates to a 100mph difference when the rock reaches balloon level (100mph vs 0mph). In other words, the ground-level "burn" is twice as effective as the balloon-level "burn". This is essentially how the Oberth effect works. A small delta-V burn deep in a gravity well can translate to a much larger delta-V outside the gravity well, since gravity will have less time to slow the object down on its way out.

 

[mongo]

Here’s how I think about it.

Suppose you drop a rock from a hot air balloon. Acceleration acts on it while it’s falling, and it hits the ground at say 100mph. (Ignore air resistance.) Then throw the rock upward from the ground at 100mph. It will stop when it reaches the height of the hot air balloon.

Now, _throw_ a rock _downward_ from the hot air balloon at 100mph. Gravity will have a lot less time to act on it (to accelerate it) before it hits the ground, so it will hit the ground going somewhat slower than 200mph; say 150mph. Now reverse this; throw the rock upward from the ground at 150mph. When the rock passes the hot air balloon, it will be travelling 100mph upward. The 50mph difference in upward thrown speed at ground level (150mph vs 100mph) translates to a 100mph difference when the rock reaches balloon level (100mph vs 0mph). In other words, the ground-level "burn" is twice as effective as the balloon-level "burn". This is essentially how the Oberth effect works. A small delta-V burn deep in a gravity well can translate to a much larger delta-V outside the gravity well, since gravity will have less time to slow the object down on its way out.

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)

 

[JB47394]

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:

Here's the thing. Assume you have a drive that produces 1N per Watt of power with no propellant. Basically, 1W goes in, 1N of force comes out by unexplained physics.

Let's assume for sake of argument that the drive and battery weigh 1kG, so the entire assembly when turned on will accelerate at 1m/s/s

After 1 seconds, the kinetic energy will be .5 m v^2, i.e. .5J, but you'll have expended 1J to power it. So far so good.

After 10 seconds, the kinetic energy will be .5 m v^2, i.e. 50J, but you'll have expended only 10J to power it. that's odd.

After 100 seconds, then k.e. will be 5000J, after expending only 100J to power it. That's bad.

Note how energy is "consumed", but it still gains more k.e. than is spent powering it. That's free energy right there. If you can close the loop and bleed off some of the free energy to power it, e.g. by attaching it to a wheel and getting it to spin a generator, then it will spew out energy from nothing.

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