Free Energy Puzzle

Apples and oranges I think. If A = 100 kg object and KE = 500,000j, and B = 100 kg object and KE = 5,000j. Then if A bumps in B while traveling through space( A+B), it is not the same as increasing the speed of A+.

Does this make sense? Regards JD.

Hi JD. "Then if A bumps in B while traveling through space( A+B), it is not the same as increasing the speed of A+."

No, surely not if A and B were traveling in the same direction, because the speed difference when they bump into each other will be 90 m/s, which does not represent the puzzle.

Jorrie

Exactly, A+B is a single entity, and A accelerated to 110m/s is not two separate parts, A = 5000,000 + C = 105,000 (A+C). therefore C-B = 100,100 makes no sense? as C is part of a single entity.

Still trying? Regards JD.

Hi JD, but you have not answered the puzzle (yet). The spacecraft achieved a genuine 105,000 joule increase in its kinetic energy relative to Earth. However, the booster that we (temporarily) attached to the craft only added 5,000 joule of kinetic energy to the craft (and only used fuel accordingly).

The puzzle is: how could the craft gain 105,000 joule of energy for an energy input of only 5,000 joule?

Jorrie

Jorrie, maybe this is similar to having two batteries; one of 100 volt and one of 10 volt. Connect each of these batteries to its own 1 kΩ load and you will get power 100²/1000 = 10 watt from the 100v battery and only 0.1 watt from the 10v battery, for a total of 10.1 watt.

Now connect the two batteries in series with one 1 kΩ load. You will get power 110²/1000 = 12.1 watt, a full 20% more than you get out of the two batteries separately. Since energy is power x time, doesn't this boil down to the same thing as your spacecraft puzzle?

Hi Guest. Your battery analogy is not quite equivalent, because the currents through the various setups differ. If you made the currents through all 3 setups equal, the power output would be equal.

Jorrie

Hello again Jorrie

Having had another think about this and seen others' comments - I can't agree with your "craft only added 5,000 joule of kinetic energy to the craft (and only used fuel accordingly)." This is energy needed to accelerate 100 kg from 0 to 10 m/s. The force needed to accelerate it from 100 to 110 m/s (over a given time) is the same as from 0 to 10 m/s, but in former case the force goes through a greater distance, hence transferring more energy.

Codey

Hi Codey, you wrote: "The force needed to accelerate it from 100 to 110 m/s (over a given time) is the same as from 0 to 10 m/s, but in former case the force goes through a greater distance, hence transferring more energy."

But, isn't a booster that provides a specific thrust profile for a specific time always delivering the same amount of energy? I mean, the chemical energy that it converts stays the same, not so?

Jorrie

Hi Jorrie, you said But, isn't a booster that provides a specific thrust profile for a specific time always delivering the same amount of energy? No I don't believe it is. It delivers the same change of momentum (= force x time). But for a given time, the distance is longer to accelerate from 100 to 110m/s than from 0 - 10m/s. So energy (= force x distance) is higher.

If the booster only had enough fuel to deliver 5000 joule I think it would run out way before it reached 110m/s.

Cheers....Codey

Hi Codey, you wrote: "If the booster only had enough fuel to deliver 5000 joule I think it would run out way before it reached 110m/s."

I beg to differ! My booster is designed to give a 10 m/s boost to a 100 kg spacecraft under any free space conditions, expending only 5000 joule of net energy to do so. I've tested it many times and it seems to work just fine.

Jorrie

Hi Jorrie

If we agree that A = 100 m/s and B = 10 m/s and they are relative to an observation point on the earth, let say in orbit? and for them to be observed they will need to be two separate entities so that they can be compared. Then put that into an equation that represents A and B. Then A = B + 90. then to keep the left and right side of the equation in balance, what you add to the left you must also add the the right, therefore A+10 = B+90+10. You can not then compare isolated parts of the equation to draw conclusions?

Regards JD.

Hi Jorrie

No. it's not free energy (Gibbs or otherwise). I think it's simple algebra. ½.m.(V + ΔV)² ≠½.m.ΔV²

½.m.(V + ΔV)² = ½.m.(V² + 2.V.ΔV + ΔV²) so the change in KE = ½.m.(2.V.ΔV + ΔV²) and the difference = ½.m.(2.V.ΔV) = m.V.ΔV = 100000 joule.

BTW, is the ref to inertial cruising in free space there to put folks off the track? As far as I can see situation is same for a body moving on Earth's surface (in ideal case neglecting friction etc).

Cheers.....Codey

Hi Codey. Yep, that were the sums that I also made, but it does not explain where the 'extra' energy is coming from. How can 5,000 j of added kinetic energy produce 105,000 j of additional kinetic energy?

Jorrie

KEdelta = 1/2 x 100 x (110^2 - 100^2) and not 1/2 x 100 x (110-100)^2

factorised = (100 + 110) x (110 - 100) = 210 x 10 = 2100 == (x50) == 105000

Hi Hendrik.

Your sum is right, but how do you explain the 5,000 j input for a 105,000 j gain?

Jorrie

No. Look at the formula. It is not a linear equation, therefore, not an accumulative process. You can't add the delta energy to the current energy before the increase in velocity. The velocity term is squared.

I should add that the velocity term, being squared is the issue. A simple proof:

9 squared is 81.

7+2 = 9

7 squared = 49

2 squared = 4

so

9 squared not equal to (7 squared) + (2 squared)

That is why the KE is not additive.

Hi Hero. You are also right, but nobody (except Codey to some extent) seems to want to explain the paradox. How can it be that I use the energy for a 5,000 j increase in kinetic energy and be rewarded by 105,000 j increase in kinetic energy?

Jorrie

Wait. Isn't it obvious through observation? It's velocity squared. You start out with the second burn at a high velocity, add more power, Scotty, and end up at a higher velocity. Or is it obvious?

There is no free energy, of course, but there is a misperception and the way it is worded reinforces that misperception because your implied thinking is based on the wrong frame of reference.

Why? Because your question switches the inertial frame of reference! You start out with an Earth-based reference, then you talk about the delta V and delta KE relative to the SV's 100 m/s initial velocity. Then you switch back to the Earth-base reference and ask why the discrepancy.

It's not really a paradox at all, but a preverbal 'slight of hand'.

Nicely done!

Ditto!

Anonymous Hero hit the nail on the head: because velocity is squared in the KE equation, the relationship between velocity and energy is exponential, not linear. Therefore, the alleged "discrepancy" about which Jorrie asks is a misconception. Comparing the two energies is invalid since they derive from exponential functions of velocities.

Hi svengali; you wrote "Comparing the two energies is invalid since they derive from exponential functions of velocities"

What "exponential functions"?

Read Anonymous Hero's reply #23 and you'll see that he came to a slightly different (and better) conclusion.

Jorrie

There is no "free energy" (but there is some bad math). ΔKE will be 1/2*m*(v1²-v2²) which is not the same as 1/2*m*(v1-v2)²

funny,

César

cesar.gutierrez_perez@mabe.com.mx

Hi Guest. "There is no "free energy" (but there is some bad math). ΔKE will be 1/2*m*(v1²-v2²) which is not the same as 1/2*m*(v1-v2)²"

Like you, most respondents got the sums right, but like them, you did not explain the paradox. I can assure you, that booster only delivered 5,000 j of KE to the spacecraft - because it was specified so that did not have the fuel to deliver more. Your own equation gives 105,000 j of additional KE. How does it fit?

Jorrie

Slick change in frame of reference . . .

I believe you have eloquently stated it.

Ooo! Ooo! I know, but I ain't tellin'!!!

I kept up with you and Gavilan!

Aside from the discrepancy of the KE, burning a booster on an orbital device in a forward direction of a satellite's travel will actually result in a lower velocity of the craft. The added kinetic energy is converted to a higher orbit, and a lower velocity, and your kinetic energy does not result in an increased velocity overall, except for the eccentricity of the orbital path, which is usually not a desired result. In order to speed the spacecraft up in a stable circular orbit, one would have to remove the existing kinetic energy by applying a reverse thrust to the satellite resulting in a lower, faster orbit due to the conversion of altitude to velocity. This doesn't change the velocity/joule/KE puzzle any, except that the net kinetic energy, combination of altitude ( height) and velocity, must be decreased to have an increase in velocity.

The only means of adding 10 m/s to an orbiting craft and not change the altitude would be to have a lateral application of force.

I know it's not the answer you are looking for, but it's an added bonus solution.

RichH

I bet you cheated!!!

I've always applied my retro to speed things up. Haven't you?

RichH

I use retro to wear cloths I've had hanging in the closet for at least ten years.

Hi NoSciFi, you said: "burning a booster on an orbital device in a forward direction of a satellite's travel will actually result in a lower velocity of the craft."

You're right, so that's why I placed the 'puzzle craft', far from Earth in free space, normally interpreted to mean no gravitational influence.

BTW, 100 m/s speed won't do an orbiting craft any good, unless it's an asteroid it's orbiting!

Jorrie

Newton was a drunkard. Forget him. Let's move on.

If you think this is hard, try reconciling kinetic energy against momentum factoring elastic and inelastic collisions.

I know! Kinetic wants the summer house, while momentum want full custody of the children! What's a lawyer to do?!

(A+B) square is greater than A square plus B square.

The error is in the calculation of the amount of energy required to increase the velocity. If you calculate in terms of KE=MA in order to find the energy required to change the velocity then the quantities will balance. Also, to attempt to use the formulas you have indicated would be the same as to attempt to say that 10^2 + 1^2 = 11^2. The power functions are not commutative.

You are wrong!!! But you don't know it.

So are you going to keep my error a secret?

Are you saying that 101=121?

Come on Vermin, you can't just say I'm wrong and run off like that...

Besides, saying that I'm wrong and I don't know it is redundant or at the very least illogical. Why would I post if I knew I were wrong?

Ok, was my error in naming the correct mathematical property. Maybe that should have been the distributive property. Non the less, you can't equate A^2 + B^2 to (A+B)^2 and that is what is just part of what is wrong with the puzzle. You cannot calculate the added/required energy by using the difference in velocity and plugging it into the KE=1/2MV formula.

However, you could plug the difference of the squares of the two velocities (12,100-10,000) in and you would get a correct answer.

Jorrie, where are you?

Hi rcapper, I'm sitting back, just enjoying this for now.

Like vermin said, you're wrong, but I must add, only in the sense that the mathematical issues alone do not 'explain' the puzzle properly. So, let's puzzle some more...

Jorrie

I think there are 2 different systems. From the point of view of the spacecraft and referring to it, the speed is 10m/s due to an energy injection of 5000joules.

An observer on earth sees quiet a different situation. He sees the initial push to 100m/s and the additional 5000 joule stored in the booster. The energy conservation works in each system independent.

Hi meirb, you said: "An observer on earth sees quiet a different situation. He sees the initial push to 100m/s and the additional 5000 joule stored in the booster."

Does this mean you say the Earth observer sees only 500,000 + 5,000 = 505,000 joule kinetic energy after the boost?

Jorrie

Exactly. Lets consider an other planet from which the spaceship speed is 1000m/s.

The booster starts and increases the speed to 1010m/s. The energy I have to inject from the space ship point of view is the same 5000Joule. However an observer on that planet calculates 1005000 joules difference.

Hi again. You now said: "The booster starts and increases the speed to 1010m/s. The energy I have to inject from the space ship point of view is the same 5000Joule. However an observer on that planet calculates 1005000 joules difference."

But this is not quite what you said in reply #40, as I read the second half of that post.

Jorrie

Hi Jorrie,

What I am saying, there is no energy discrepancy. The system is the spaceship, and ,having no other energy sources, that increase in speed of 10m/s is all you can buy for 5000 Joule.

jorrie

no eneargy in world you get free it is possible after applying some force you can ratate the sattelitefor looooong time i.e. till its life but it take eneargy from earth gravitational force.

all abject rotating and moving above earth field require are require less almost no eneargy that point is escape point. hence may your craft is at that point.

because in space craft designer says its limit its limit in 1000kml but it can move to 10000kml as after escape it require less eneargy anly matter is reentry technology for which you have to designe material.

jarunmagnet

OK, I'll try give you a piece of my mind....

1 - the problem is not linear, so it is not mathematically correct to consider the mere variation of velocity for the calculation of the kinetic energy variation (other friends stressed this concept out before)

2 - The velocity range is important for the definition of the energy variation. Actually, the momentum is only depending on the delta velocity between initial and final conditions. But the transition between these conditions will occur on a different path length, depending on velocity: the higher starting velocity, the longest piece of trajectory. And, being the work the product of force and distance, different paths produce different energy variations, being applied forces equal.

For example, by assuming that the applied force of the booster F is constant, and therefore the induced acceleration a, the time required to change velocity from V₀ to V_f is:

T = (V₀ - V_f ) / a

The time interval T is independent from velocity range, but only on its variation. The applied force of the booster is given by the momentum expression.

F = m *(V₀ - V_f ) / T

And this is again depending on velocity variation only. But the run distance is expressed as

S = (V_{0 +} V_f ) * T / 2 (uniformly accelerated motion)

and this actually depends on the starting velocity. This explain why the variation of energy, calculated by neglecting V₀, is not correct (obviously it is correct only when V₀ = 0). As a final confirmation, the variation of the spacecraft energy calculated with above expressions is:

E = F * S = m *(V₀² - V_f²) / 2

This is not about energy gain, this is about mathematical error.

deltaKE = KE' - KE.

deltaKE = 1/2 m *V^2 is an incorrect statement.

Let us take the humble resistor. 100ohms with 100amps passing through it.

Power P = I^2 * R

= 100 * 100 * 100

=1MW

increase the current to 110A

P' = 110 * 110 *100

= 1.21MW

yet by incorrect mathematical assumption delta P = 10 * 10 * 100

= 10KW so

deltaP = P' - P = 210KW

vy jste ale blbci

Actually there is no net gain in energy. The discrepancy 'appears' because of the erroneous formula used for calculating change in kinetic energy which should be =1/2x100x(110^2-100^2) instead of 1/2x100x10^2.

I'd love to know where you can get a booster that doesnt change mass and can still accelerate a spacecraft.

I think the problem comes from how you are defining your frame of references.

We can say ΔKE=(m/2)*(V_f-V_i)

In the earth's reference frame this equals (m/2)*(110²-100²) = 605000

However in the ships frame of reference, the initial velocity appears to an observer to be zero, giving ΔKE=(m/2)*(10²-0²) = 5000.

Both answers are correct for the frame of reference they were measured in.

Another example might be if I was in my car traveling at 100 MPH, and a car traveling in the opposite direction passes me at 100 MPH, it would appear to me that that car was traveling at 200 MPH, and my car is stationary when in fact it really isnt.

Good answer, guest, but your "I'd love to know where you can get a booster that doesnt change mass and can still accelerate a spacecraft." spoiled it all.

The puzzle said that it's the spacecraft's mass that remained the same, not the external booster's mass.

doesnt matter, its a coupled system

100 kg X 10 m/s = 1,000 kgm/s = 1,000,000 gm/s

1 gram @ 1,000,000 m/s = 1,000 kgm/s

1 gram of propellant expelled at 1,000,000 m/s will accelerate 100 kg 10 m/s.

10⁶ m/sec is about the range of an ion engine.

The loss of mass is 0.001%. I'd say for the purposes of education, the loss of mass expelled for acceleration is negligible.

Hi Jorrie,

Mass is energy. The trick is when saying the mass of the spacecraft does not change. In fact, the booster has to have first the same speed (100m/s), make the junction, ignite the motor up to 110m/s and detach. But the booster has a dead mass and the mass of all materials to be ejected to get the reaction force. You calculate the gain in kinetic energy of the spacecraft alone without mentioning that is attached to the massy booster.

Hi Hottech, you wrote: "You calculate the gain in kinetic energy of the spacecraft alone without mentioning that is attached to the massy booster."

Ignore the booster and just give that spacecraft a 10 m/s 'kick in the butt', i.e., just give it a push from behind. The problem is about the spacecraft alone, not the booster.

Jorrie

If you give that rocket a kick up the butt, and you have very little mass: in order to change its velocity by 10 m/s you will end up going in the opposite direction VERY fast.

I've done a little spread sheet in which I've simplified things quite a lot:-

For simplicity let's assume that the external booster has no mass other than that of the fuel, and, that all the fuel is used and expelled "backwards" at the same speed relative to the earth. (Yep I've also ignored the very slight change in the mass of the fuel).

These are screen shots of total energy needed to accelerate both the rocket and the fuel for different initial velocities relative to the earth.

Initial V 100 m/s

Initial V 50 m/s

Initial V 0m/s

You can see that the total amount of energy used is dependant on the mass of fuel required not the initial velocity relative to the earth.

Hi Randall, great stuff! I think your calcs are right.

However, the intention of the puzzle was that you don't need to worry about the fuel and the booster at all. The spaceship simply got a boost in velocity by whatever means and has a new kinetic energy.

Jorrie

Ah, but if you don't worry about the fuel and booster, etc, you do not have a conservative system, mmm?

Could it be the extra KE goes into accelerating the booster? You said it does not change mass, you did not say it is mass-less.

DDM

KE1-KE2=.5M(V2-V1)**2

(V2-V1)**2 does not equal V2**2-V1**2 but V2**2 - 2V2V1+V1**2 basic algebra

(which took awhile to remember) change in KE is 5000joules.

If you want to just subtract without using algebra, use momentum calculations.

DDM

Hi,

KE= 1/2mv^2 (velocity << c)

is the measurement of the amount of work necessary to accelerate a known mass at rest to a particular velocity. That energy was added by whatever external force was expended to get it to that velocity and is carried with it unless acted on by another external force in any direction.

It takes 5000 joules of energy to accelerate the 100kg mass from rest to 10 m/2 but takes 605,000 joules of energy to accelerate the same mass from rest to 110 m/2.

No free energy....?

The answer to this puzzle is obviously that kinetic energies in two different frames of reference are compared and they will always differ. JD first hinted at it (reply #1) and then Bill (reply #19) hammered it home very subtly. There were many others that realized that the reference frames are the issue.

Kinetic energy in Newtonian dynamics ('moving mass' in relativistic dynamics) is not invariant under coordinate transformations, i.e. it changes if you change frames of reference and it does so non-linearly. The 'rest energy' is invariant under coordinate transformations.

The amount of energy (or fuel) needed to give an object a small velocity boost (fortunately) depends only on its rest energy, i.e. on it's mass and speed in its own frame of reference. Its speed is obviously zero in its own frame of reference; hence it has zero Newtonian kinetic energy in its own frame.

So, the only 'correct', i.e. frame independent value for the energy needed to change the speed of the 100 kg spacecraft by 10 m/s is 5,000 joule.

Jorrie

Hi jorrie,

I understand the different frames of reference. Call me thick but help me understand this:

If I'm traveling along with the mass at the same speed and I reach over and give it a kick with my foot and increase its speed by 10 m/s then that is the speed at which the mass would be moving in reference to me and I would measure its KE at 5000 joules. But if I were standing on earth (not taking into account the earths moving thru space) and watched it flying thru space at 110 m/s and I knew that its mass was 100kg wouldn't I then calculate its KE as 605,000 joules?

Hi snygolfgs. "If I'm traveling along with the mass at the same speed and I reach over and give it a kick with my foot and increase its speed by 10 m/s then that is the speed at which the mass would be moving in reference to me and I would measure its KE at 5000 joules."

Yep. In your original reference frame, you will obviously recoil and be moving in the opposite direction to direction you kicked the mass into. However, if the relative speed between you and the 100 kg mass settles as 10 m/s, its KE is 5000 joule in your reference frame.

"But if I were standing on earth (not taking into account the earths moving thru space) and watched it flying thru space at 110 m/s and I knew that its mass was 100kg wouldn't I then calculate its KE as 605,000 joules?"

Correct. KE is reference frame dependent, but this puzzle illustrates that ΔKE only makes sense if you measure it from the perspective of a 'co-moving observer', meaning like you were before you gave the 100 kg mass a kick. The Earth observer's 105,000 joule ΔKE calculation has nothing to do with how hard you had to kick!

Jorrie

Hi jorrie,

Thank you for the reply and the lesson.

I have it know.

I appreciate it.

Jeff

"Vanna, tell our contestants what they've won!"

Hello again Jorrie. Codey here, not logged in.

I'm having trouble getting my head round this. I can understand that according to an observer who stays at 10 m/s the KE = 5000 J, but I don't see how it can reach 110 m/s with that energy input.

I envisage 2 scenarios (on Earth's surface, as it helps me visualise, but it's not essential). Both assume ideal frictionless motion.

1) Body moving at 100 m/s. Accelerate to 110 m/s. Can we agree this needs energy (additional) 105000 J?

2) Body stationary on a train doing 100 m/s (no power needed for train, in ideal situation). Accelerate body to 10 m/s relative to train. An observer on the train says KE = 5000 J. But the force needed to accelerate the body causes a reaction on the train, and energy 100000 J has to be supplied to keep it at 100 m/s.

Either scenario needs 105000 J input. It seems to me 1) is pretty much equivalent to your original puzzle. I still don't see how it can go from 100 to 110 m/s and use only 5000 J (your post #26).

Cheers…..Codey

Hi again Codey.

Your 1) is correct in the reference frame of an observer static on Earth.

Your 2) is wrong on this: "But the force needed to accelerate the body causes a reaction on the train, and energy 100000 J has to be supplied to keep it at 100 m/s."

Why a reaction on the train? Place the unattached 100 kg mass on the roof and use a rocket to add the 10 m/s velocity to the mass. The amount of energy used is just 5000 joules in a frictionless environment, irrespective of the speed of the train.

Another way to wrap one's head around it is this: say you car needs 1000 joule of energy to accelerate from 50 mph to 60 mph, with friction and other losses. The guy that drives at 50 mph in the opposite direction 'sees' you accelerate from 100 to 110 mph relative to him. Does that make any difference to the amount of energy your car uses for the acceleration?

Likewise, what a stationary observer 'sees' has no bearing on the amount of fuel your car uses for the acceleration. In a frictionless environment it is only the 10 mph that counts.

Jorrie

Hello Jorrie

I'm afraid I still can't agree.

My 2 scenarios were to try to calculate energy by 2 routes as a cross-check.

I don't think it makes any difference whether the body is accelerated relative to the train by something which puts a reaction on the train (my scenario 2) or by a rocket which doesn't. The rocket still has to do work. With a rocket the situation is the same as scenario 1 (going at steady 100 m/s then accelerating) and you agreed with 105000 J for that.

The original puzzle is based on velocities relative to Earth. Saying it only needs 5000 J to go from 100 to 110 m/s is a bit like driving at 60 mph, calling that my inertial frame, and thinking I can accelerate to 90 mph but only use same fuel as going 0 - 30 mph. Perhaps I'm missing something but I don't think it works. Or if I'm then stopped by a speed cop, telling him I'm doing 30 mph in a 60 mph inertial frame .

Also you wrote "Another way to wrap one's head around it is this: say you car needs 1000 joule of energy to accelerate from 50 mph to 60 mph, with friction and other losses. The guy that drives at 50 mph in the opposite direction 'sees' you accelerate from 100 to 110 mph relative to him. Does that make any difference to the amount of energy your car uses for the acceleration?" No, it doesn't, but he's in an inertial frame different from Earth's. Just as in the puzzle somebody who stays at 100 m/s while the craft accelerates sees things differently (10 m/s, 5000 J) from one in the Earth's inertial frame.

We may have to agree to differ on this!

Cheers........Codey

Simply put, from a static position relative to the earth, the spaceship has has a total expenditure of 11 seonds @ 10,000 j, each second producing 10 m/s acceleration. The total output of the spaceship, via its ion engine, is 110,000 j. If you accept the 500 kj at 100 m/s, then the 605 kj at 110 m/s is the only acceptable solution. So do you also dispute the 500 kj at 199 m/s?

RichH

Hi Codey.

"Saying it only needs 5000 J to go from 100 to 110 m/s is a bit like driving at 60 mph, calling that my inertial frame, and thinking I can accelerate to 90 mph but only use same fuel as going 0 - 30 mph."

If it was an ideal frictionless, gravity-free environment and your car's efficiency was the same at 60-90 mph than at 0-30 mph, you would use the same amount of energy (fuel) for the two scenarios. Maybe it's the fact that in a real car's 'draggy' environment it doesn't hold that obscures the issue here.

Remember that the issue is that kinetic energy is frame dependent, so every different inertial frame will get a different result for the ΔKE. The proper amount of energy needed for acceleration of a constant mass in a frictionless, gravity-free environment is fixed for a given ΔV, in that ΔKE = ½mΔV², provided that ΔV << c. "Proper amount" means it is measured in the frame where it is applied, i.e. the car or the spacecraft, on the spot.

It is sort-of obvious that the real amount of fuel needed cannot depend on which inertial frame happens to make the ΔKE measurement. This tells us that initial speed relative to whoever (Earth observers or others) cannot influence the amount of fuel needed in free space. It is different in gravitational fields, where the speed and direction (velocity vector) may make a big difference, but that's a long story.

Jorrie

PS: Before we agree to differ, maybe we should try to get other engineers and physicists in on this 'dispute', because you are not the only one that differ...

Jorrie

Hello Jorrie

You've surprised me now! You wrote "If it was an ideal frictionless, gravity-free environment and your car's efficiency was the same at 60-90 mph than at 0-30 mph, you would use the same amount of energy (fuel) for the two scenarios."

But KE goes as V², and 90² - 60² = 4500, while 30² - 0² = 900, even in ideal environment. Several posters have made similar point. As a check, for a given time of acceleration (which cancels) it's easy enough to work out the force hence the work done.

Back to original puzzle, I'm not disagreeing that according to an observer who stays at 100 m/s the KE = 5000 J. But I don't think that's the energy needed.

To follow tkot's 3^rd para - if the spacecraft were considered at rest, the Earth doing -100 m/s (and having a large calculated KE) I contend it would still take 105000 J to accelerate the spacecraft from 0 - 10 m/s, as the before and after velocities relative to the mutual centre of mass of spacecraft and Earth are unchanged.

Codey

"But KE goes as V², and 90² - 60² = 4500, while 30² - 0² = 900, even in ideal environment." No Codey, No!!

In the ideal environment, your car 'sees' (90 - 60)² = 900 m²/s² and (30 - 0)² = 900 m²/s². Why? If your car is cruising inertially, it has zero speed in it's own frame of reference, no matter how fast it moves relative to the ideal, friction-less Earth. The moment it accelerates, it 'knows' it is accelerating and all it can be certain of is that its speed has changed by 30 mph. How?

It simply measures the acceleration by means of an accelerometer and time the duration of the acceleration by means of a clock and walla, it finds ΔV = a x t = 30 mph, by virtue of its on-board computer. I'm sure that computer will not argue with me about it, neither would the fuel gauge...

Anyway, I'm enjoying this little encounter...

Jorrie

Hello again Jorrie

Re-reading your #88 I see your "Err....5 kj I guess" was to correct the 5j in #86, not an answer to his momentum question! Apologies for the mistake.

But replying to this post, I'm sure there's a flaw in your argument somewhere! I can see that in a frame of reference doing 60 mph the car after accelerating thinks it's doing 30 and calculates KE accordingly. But I don't think you can change your frame at will, part way through a (thought) experiment. We on Earth see ΔKE 90² - 60². I can't agree with your position that the amount of fuel used corresponds to 30² - 0². If frames of reference weren't brought into it, but we just considered motion and KE for a body moving (frictionlessly) on Earth, I'm pretty sure we'd calculate the energy changes and fuel consumption the way I have. We have to add the fuel in our frame, on Earth

You wrote "It simply measures the acceleration by means of an accelerometer and time the duration of the acceleration by means of a clock and walla, it finds ΔV = a x t = 30 mph, by virtue of its on-board computer. I'm sure that computer will not argue with me about it, neither would the fuel gauge..." I'd only disagree with the last bit about the fuel. A given force for a given time produces a ΔV, irrespective of initial velocity. But if you start at 60 mph the force has to go through greater distance (measured in either frame) than if you start from rest, so does more work (as I mentioned in earlier post). That's where I think the flaw is.

Cheers.......Codey

Hi Codey, you wrote: "We have to add the fuel in our frame, on Earth "

Yep, that's generally true. But once that fuel and engine is already moving at 60 mph, it's no longer in our frame, on Earth... It's in it's own frame, moving frictionless and with no lateral forces acting on it - more or less in an inertial reference frame.

Now fire up that engine and let it consume some of that fuel and give us a 30 mph speed in that inertial frame in any direction. That engine will only consume fuel in accordance with the 0-30 mph speed increment that it 'sees'. How else in a frictionless environment?

Jorrie

OK Jorrie, let me try this on you!

We're on Frame A. Frame B is moving by at 100 m/s with the 100 kg spacecraft S stationary on it. S accelerates to 10 m/s on Frame B and uses 5 kJ fuel. No problem with that, we can see from Frame A the level drop in the fuel tank. We on Frame A say it's doing 110 m/s and has KE 605 kJ. We have a rope wound on a drum and we throw the end, catch S and slow it to 100 m/s, 500 kJ (in our Frame A) recovering 105 kJ from the drum. We then lob S back on to Frame B (no velocity or KE change). We then repeat the above as often as we like.

There's clearly something wrong as this violates conservation of energy, but it seems to represent your position (please correct me if I'm wrong about that). On the other hand, if we reach out from Frame A and apply the force to accelerate S, we have to input the "right" energy and everything is OK. (same situation as my body on a train in #76).

The original puzzle referred to Earth's reference frame and 100 m/s relative to Earth. The spacecraft started in Earth's frame and accelerated to 100 m/s. I don't think it's legit to bring another frame into it, leastways not if energy is exchanged between them. What I think I'm saying is the spacecraft isn't in a frame of reference independent of Earth's. I still don't see the difference between the spacecraft's situation and a body moving on Earth (or any flat plane).

Have a good weekend.....Codey

Codemaster,

Try this. I'm going to reframe the question slightly. The problem for me, and possibly others, is keeping track of the fuel and what it does. There's all kind of hand waving and vague allusions to the earth, etc, but I think I know a way to keep it straight. I'm going to use u = speed since all motion will be in the +x direction.

So, we have an observer in O. Another frame O' is moving to the right (+x direction) at 100 m/s. There is a 100 kg mass, m', at the origin of O'; it is currently stationary in O'. There is an observer in O', not part of the mass m'.

In O', m' has KE'_O'= (0.5)(100 kg)(0 m/s)² = 0 J

In O, m' has KE'_O = (0.5)(100 kg)(100 m/s)² = 500000 J

[the subscripts refer to the measuring frame]

Now, we add a third frame O'' for the sole purpose of keeping all fuel there so that none of us have to do any bookkeeping. In O'' there is a wonderful gun that fires 100 kg balls out at a certain speed depending on the amount of fuel it uses. There is no observer in O'' to measure that speed so we always have to measure the speed (and corresponding energy) from one of the other frames. This should be OK since we are used to having x be different in the different frames and KE is, after all, proportional to x.

So, the gun in O'' has a 1 liter fuel load and fires. The 100 kg ball m'' comes out in the +x direction, perfectly level with the origin in O'. The observers in O and O' promptly jump to their instruments and measure the speed of m". The observer at O' measures 10 m/s and the observer in O measures 110 m/s. This is completely consistent.

The observers then calculate the KE''. The observer in O' finds KE''_O' = (0.5)(100 kg)(10 m/s)² = 5000 J. The observer in O finds KE''_O = (0.5)(100 kg)(110 m/s)² = 605000 J

As m'' nears the origin of O', the observer at O' calculates the new total energy in O' as 0 J + 5000 J = 5000 J. The observer at O calculates the new total energy in O' as 500000 J + 605000 J = 1105000 J.

Now m'' strikes m' in an elestic collision. Since their masses are equal they exchange momenta and speeds. [Note: so, OK, I made the masses the same to make this easy].

Now, the observer in O' sees u'_O' = 10 m/s and u''_O' = 0 m/s. He further calculates the energies as KE'_O' = (0.5)(100 kg)(10 m/s)² = 5000 J and KE''_O' = (0.5)(100 kg)(0 m/s)² = 0 J. He calculates the total energy in O' as 5000 J + 0 J = 5000 J. Energy has been conserved and so, obviously, has been momentum.

The observer at O sees u'_O = 110 m/s and u''_O = 100 m/s. he calculates the energies as KE'_O = (0.5)(100 kg)(110 m/s)² = 605000 J and KE''_O = (0.5)(100 kg)(100 m/s)² = 500000 J. He calculates the total energy as 605000 J + 50000 J = 1105000 J. Energy has been conserved and so has momentum.

Note that I did not calculate the momenta. I didn't want to make this any more cumbersome than it is. I hope it is obvious that, with equal masses and speed exchange, momentum is conserved.

Now we know everything about O', both from the point of view of the observer in O and the one in O'. All energy has been conserved. All momenta has been conserved. We don't know what happened in O'' and we don't care. There was probably some acceleration and some recoil and other stuff that only makes my head hurt.

Hopefully, this helps get away from the hurdle of how fuel is measured in different frames.

Tom

Hi Codey, you wrote: "We have a rope wound on a drum and we throw the end, catch S and slow it to 100 m/s, 500 kJ (in our Frame A) recovering 105 kJ from the drum."

Nope, we can only recover 5 kJ (if at 100% efficiency) by this means, so energy is conserved. The energy required (or extracted) for any change in KE can only be measured/calculated/used in a frame that is initially (or finally) at rest relative to the test subject.

I think Bill has it perfectly right in his last reply, so I'm not going into it deeper for now.

Jorrie

PS: Watch the 'Potential energy puzzle' for real 'free energy'.

Correction: I wrote: "I think Bill has it perfectly right in his last reply, so I'm not going into it deeper for now."

I meant Tom has it right (reply 116 above.) Sorry Tom!

His analysis makes my head spin a bit (especially on a weekend), but I'll have a good look at at.

Jorrie

Jorrie,

I apologize in advance if this is a double post; I am completely befuddled by the "BACK" button.

I agree with your last post. Kinetic Energy does not transfer under these circumstances. And, your scenario completely satisfies classic relativity.

But, you asked where the extra 100,000J came from and I think you didn't answer that.

Tom

Hi Tom. You wrote: "But, you asked where the extra 100,000J came from and I think you didn't answer that."

I tried to answer that in post #61 above. It is comparing kinetic energies of an object in two different inertial frames and they will always be different. The difference has no real meaning however, i.e., it does not relate to 'real energy'. In the case of the spacecraft, it is only the energy change as measured on-board (or locally) that has any 'real energy' meaning.

Jorrie

Hi Jorrie

I think I just grasped what we were getting at with that puzzle. If I'm not wrong, what all boils down to is that when one sees a ΔE = 1/2m(v+Δv)² -1/2mv² another sees a ΔE = 1/2mΔv² . These are obviously not equal, so the paradox is how the difference is explained.

I have a feeling that even one not expert in physics (and I am certainly not one!) can see that kinetic energy is very reference specific. One can say something similar about potential energy, it all depends on where you take your reference. So one must understand that there is no absolute energy when we meddle with potential or kinetic energy. What counts is how this energy is exchanged through various interactions, or if you like, how it remains stable within a closed system. Energy can only have an absolute significance, when we are talking about heat energy for example. (In fact, heat energy is again a mean of the kinetic energies of the molecules, but mere statistics leave the center of mass unchanged.)

Think another example, of why kinetic energy cannot be taken as an absolute value: One gets up and starts walking at 5km/h. Suddenly he sees the whole universe coming against him at a speed of 5km/h !!! I would find it absurd, if he wondered where the hell did the whole cosmos got this kinetic energy from... Instead, he should realise that at the time he was laying low he belonged to another reference system, than the one he assumed when he started walking; consequently he has to chose where he likes to do his maths. Whatever reference system he chooses, he will get consistent results about energy and momentum, after he interacts with stuff around him.

Hi tkot, excellent summary and example!

Jorrie

Yes I understand it now. Regards JD.

You guys kinda lost me a Lil ... follow the basics though But I just had to add something ............. So what i want to know is how you managed to burn the booster with no change in the mass unless the booster detached at the end of the burn and your no longer figuring it into the equation. And if so did it use a explosive device to do so. And how do you know there are no other outside influences like gravity or solar winds or who knows what else.

Hi wgh71. You asked: "... how you managed to burn the booster with no change in the mass unless the booster detached at the end of the burn..."

I think it has been said in some reply that the booster is attached, it burns and is detached again. The operative issue was that the 100 kg spacecraft got a 10 m/s boost in speed without a change in its own mass - we do not care what happened to the booster or actually how it got the boost.

"And how do you know there are no other outside influences like gravity or solar winds or who knows what else."

Free space, far from any source of gravity usually means that inertial movement is possible, free from any influences. Hey, and remember, this was just a thought experiment to illustrate an important and often misunderstood point.

Jorrie

Hey, Jorie,

How much momentum does the earth gain to the observer on the spaceship when the velocity changes by 10 m/s? All for the same 5j.

Rich

Hi NoSciFi: "How much momentum does the earth gain to the observer on the spaceship when the velocity changes by 10 m/s? All for the same 5j."

Err... 5 kj, I guess.

Answer is: Earth gains a lot of momentum and KE in the frame of the observer on the spaceship. Terra-terra-terrajoules, I guess. And what's more, if an Earth-like mass smacks into the spaceship, it will 'experience' that added energy and momentum.

It is similar to the point 'tkot' made in post #81 above.

Jorrie

Hello Jorrie

Haven't had time yet to look at all the latest postings, but I don't agree with your 5 kJ. Momentum is not measured in joule. Units are kg.m/s (or equivalently, N.s).

I think you have to look at momentum changes from a frame moving with the mutual centre of mass. This is of course much nearer to Earth than to the spacecraft. Then ΔV of spacecraft = 10 m/s (very close to) and Δmomentum = 100kg.10m/s = 1000 kg.m/s. ΔV of Earth is very small but mass is large so Δmomentum is the same (but opposite sign).

From the spacecraft viewpoint, Δmomentum of Earth is large, while for spacecraft it's zero. In that case momentum is not conserved so there must be something wrong!

Cheers........Codey

Hi Codey, you wrote: "...but I don't agree with your 5 kJ. Momentum is not measured in joule."

The 5 kj referred to the original energy input on the spacecraft, not the momentum.

Total momentum of a closed system remains constant when the parts of the system act on one another. In the case of a spacecraft being boosted (or kicked, or whatever) in free space, the closed system comprises out of the spacecraft the booster and the exhaust gasses, etc, but does not include Earth.

One can surely calculate Earth's momentum in any frame of reference and it will be different for most of them, so it cannot be a conserved quantity. A closed system like the Sun with its planets and their satellites will conserve momentum, more particularly angular momentum.

Jorrie

OK Jorrie, I thought you were answering the question in #88 - "How much momentum........." It seemed an odd mistake for a man of your erudition!

I'll try to respond to other points, trouble is having to work gets in the way of CR4.

Codey

Is the explination not in good old fashion relativity?

Hi Scotty. "Is the explination not in good old fashion relativity?"

If you mean good old-fashioned Galilean/Newtonian relativity, the answer is yes. Due to the low speeds and no gravity, it is not Einsteinian relativity though.

Jorrie

That's what threw me. I didn't realize there was such a profound effect for such a low speed & mass.

Thanks Jorrie for the puzzle!

I think that 2 pints of import are included. If that spaceship were on a collision course with the planet earth, and were say a Kuiper object, accelerated by an out-gassing, the extra 105,000 joule of momentum would be very demonstrable.

Second, low power engines (ion engines) can have a tremendous impact.

By the way, the acceleration of the ions is a momentum transfer, and my earlier post would have required only 1 mg of propellant, so nobdy wanted to kick my butt? I guess I can understand the Hubble and that Mars mission screw-ups better now that my mind is a little less...whatever. And, it was a momentum puzzle. My bad.

RichH

Perhaps the mud would be more transparent if we did not have to concern ourselves with the position of the observer.

Where the impulse is parallel to the velocity vector, the energy input required to change the kinetic potential of a mass will be Input E= KE_f - KE_i.

There is no free energy, and yes there are methods of propulsion other than Newtonian Mass Reaction Propulsion. But since the scientific and engineering inertia is focused on chemical propellants whether it be LOX/LOH, hydrazine or Xenon Ions, that is where we will focus until we can get those same scientists and engineers to sit down and experiment with a simple hand crank generator and a pair of long bar magnets.

Where m is the fuel mass - I.E. - ion or any other mass and V is the exhaust velocity. The energy imparted in Newtonian Mass Reaction Propulsion where the exhaust velocity of the action mass is constant is E=.5(Δm) V²_ex

Because of the limits specific impulse our space craft will not be going anywhere very fast using Newtonian propulsion.

Once the status quo of engineering inertia begins to change and we begin to enter the coming age of field reaction propulsion and electro-dynamic braking the -ΔV in dynamic braking will probably be derived as

-ΔV=√(2Pt/m) where P is dissipated electrical power, t is time in process, and m is the system mass.

In propulsion, in a boosted gravity assist, the increase in Hyperbolic Excess Speed relative to the assisting planet will also be ΔV=√(2Pt/m).

In stepping up the orbit, acting against the prime focus magnetic field, the increase in gravitational potential will approximate the effective applied power * time.

http://www.bestsyndication.com/?q=072407_solar-power-sailing-in-outer-space-extend-long-distance-travel.htm

Hi Gavilan. "Where the impulse is parallel to the velocity vector, the energy input required to change the kinetic potential of a mass will be Input E= KE_f - KE_i."

Note that E= KE_f - KE_i has no meaning unless you specify the frame of reference; unfortunately, the observer's position must be specified to alleviate the 'mud'. Only a co-moving observer can directly measure the input energy required for a particular ΔV.

For the rest I agree with you; just turn that propulsion so that it works with the direction of movement (not radial) and gain (compared to your calculations) an enormous amount of 'free energy'!

Jorrie

PS: The next puzzle in this 'series' will be more applicable to your issue - this one was set in free space, just to try and remove the misconceptions about KE.