Reference Frame, Difference in Kinetic Energy, Fuel Consumed

[MattGeo]

Say 2 cars are traveling side by side at 10 m/s in some flat, wide open space. Relative to each other they are stationary. Relative to someone on the ground they are both moving at 10 m/s. Now say you're in 1 of the cars and you see the other car accelerate, changing his velocity by 10 m/s in the forward direction. In your reference frame he is now traveling at 10 m/s but according to the observer on the ground he just accelerated from 10 m/s to 20 m/s. The same amount of gasoline is consumed in both reference frames.

I know that one could say that according to the observer on the ground that the car provides a force over a longer distance and hence more work must have been done in that reference frame to accelerate, and hence, more energy. But if you're in 1 of the cars watching the other car accelerate from 0 to 10 m/s according to you, wouldn't that be the same if the stationary observer had watched the same car accelerate from 0 to 10 m/s? the change in energy would be the same, but the fuel consumed can't be the same for each of those cases for the stationary observer. He'd see less fuel consumed from in the 0 to 10 m/s interval than in the 10 to 20 m/s interval.

If I am in the car moving at 10 m/s next to the other car doing the same, and I watch him accelerate "from rest" to 10 m/s, wouldn't that be more fuel used than if the stationary observer had watched the car accelerate from rest to 10 m/s? I can't really make sense of where I am going wrong and keep confusing myself.

 

[jbriggs444]

Say 2 cars are traveling side by side at 10 m/s in some flat, wide open space. Relative to each other they are stationary. Relative to someone on the ground they are both moving at 10 m/s. Now say you're in 1 of the cars and you see the other car accelerate, changing his velocity by 10 m/s in the forward direction. In your reference frame he is now traveling at 10 m/s but according to the observer on the ground he just accelerated from 10 m/s to 20 m/s. The same amount of gasoline is consumed in both reference frames.

Be careful with the energy accounting. You need to account for all of the energy flows. In the rest frame of the observer's car the Earth is moving. The force of the wheels on the Earth does work on the Earth.

The amount of work done by a force depends on the frame of reference you adopt.
The total amount of work done by a third law force pair does not.

 

[MattGeo]

Be careful with the energy accounting. You need to account for all of the energy flows. In the rest frame of the observer's car the Earth is moving. The force of the wheels on the Earth does work on the Earth.

The amount of work done by a force depends on the frame of reference you adopt.
The total amount of work done by a third law force pair does not

The last thing you said makes sense to me but I am not sure how to apply it in this situation.

If we consider the 2 cars moving at 10 m/s relative to the ground, but at rest to each other, and 1 of them accelerates from rest in that frame to 10 m/s, that is a greater amount of fuel used than if the observer who is stationary relative to the ground were to watch one of the same identical cars accelerate from 0 to 10 m/s relative to the ground, right? Because in the frame of the 2 moving cars who are at rest relative to each other, one sees the other change his velocity from 0 to 10 m/s, but the stationary observer on the ground sees it change from 10 to 20 m/s. If I am in 1 of the 2 cars watching the other accelerate I measure the same change in kinetic energy that the stationary observer would measure if he watched a the car accelerate from 0 to 10 in his stationary frame. 0 to 10 in stationary frame is less fuel than 10 to 20 in the stationary frame, which makes intuitive sense. So how can 0 to 10 in the frame of 1 of the 2 co-moving cars actually consume more gasoline? It has to though. I hope I am making sense because it is somewhat hard to articulate clearly.

 

[jbriggs444]

The last thing you said makes sense to me but I am not sure how to apply it in this situation.

If we consider the 2 cars moving at 10 m/s relative to the ground, but at rest to each other, and 1 of them accelerates from rest in that frame to 10 m/s, that is a greater amount of fuel used than if the observer who is stationary relative to the ground were to watch one of the same identical cars accelerate from 0 to 10 m/s relative to the ground, right?

In this comparison you are comparing two different physical scenarios, not two different frames of reference.

You are comparing the gas used to accelerate from 0 to 10 m/s relative to the ground upon which the car's wheels are pushing with the gas used to accelerate from 10 to 20 m/s relative to the ground upon which the car's wheels are pushing.

Your conclusion is correct. It takes more energy to accelerate from 10 to 20 m/s relative to the ground than from 0 to 10 m/s relative to the ground.

Because in the frame of the 2 moving cars who are at rest relative to each other, one sees the other change his velocity from 0 to 10 m/s, but the stationary observer on the ground sees it change from 10 to 20 m/s.

Yes, indeed.

So the moving observer sees an increase in kinetic energy of $\frac{1}{2}m(10)^2 - \frac{1}{2}m(0)^2$ = 50m.

Meanwhile the stationary observer sees an increase in kinetic energy of $\frac{1}{2}m(20)^2-\frac{1}{2}m(10)^2$ = 150m.

If I am in 1 of the 2 cars watching the other accelerate I measure the same change in kinetic energy that the stationary observer would measure if he watched a the car accelerate from 0 to 10 in his stationary frame.

Yes, a moving observer would measure a kinetic energy increase of $50m$ for his scenario.
Meanwhile a stationary observer would measure a kinetic energy increase of $50m$ for his scenario.

This is comparing two different scenarios, of course.

0 to 10 in stationary frame is less fuel than 10 to 20 in the stationary frame, which makes intuitive sense. So how can 0 to 10 in the frame of 1 of the 2 co-moving cars actually consume more gasoline?

Now we get to it. This is the important part. One scenario and two seemingly incompatible results for an energy accounting.

We have burned enough fuel to produce an energy increase of 150m (as assessed in the ground frame). But we have only accounted for an energy increase of 50m (as assessed in the moving frame). This is definitely an affront to our intuitions. We must ferret out the missing 100m worth of energy.

Let us use the work energy theorem. The kinetic energy increase in an isolated system must match the work done by internal forces within that system. If we want to have an isolated system, we have to consider the Earth plus the car. The only relevant force that does work then is the force between car and road.

[Let us sweep the messy details of the drive train under the carpet. They add nothing to the problem and would force us to trace a chain of internal contact forces that all do zero net work until we finally get to the expanding combustion gasses in the cylinders which actually do non-zero net work].

We have the car pushing backward on the Earth. We have the Earth pushing forward on the car. For convenience, let us assume that the car is 1 kg in mass and that the force between Earth and car is 1 Newton.

From the moving frame...

It takes 10 seconds for the car to accelerate from 0 m/s to 10 m/s.
During this time the car moves at an average velocity of 5 m/s. So it displaces by +50 meters. 1 Newton times 50 meters is 50 Joules.

During this time the Earth moves at a constant velocity of -10 m/s. So it displaces by -100 meters. -1 Newton times -100 meters is 100 Joules.

We see an energy increase of 150 Joules total for this accounting.

From the rest frame...

It takes 10 seconds for the car to accelerate from 10 m/s to 20 m/s
During this time the car moves at an average velocity of 15 m/s. So it displaces by 150 meters. 1 Newton times 150 meters is 150 Joules.

During this time the Earth is stationary. So its displacement is zero. Zero work is done on the Earth.

We see an energy increase of 150 Joules total for this accounting.

All is right with the world. Gas used and total kinetic energy gained are both invariants. They do not depend on the reference frame you choose for the analysis.

 

[MattGeo]

In this comparison you are comparing two different physical scenarios, not two different frames of reference.

You are comparing the gas used to accelerate from 0 to 10 m/s relative to the ground upon which the car's wheels are pushing with the gas used to accelerate from 10 to 20 m/s relative to the ground upon which the car's wheels are pushing.

Your conclusion is correct. It takes more energy to accelerate from 10 to 20 m/s relative to the ground than from 0 to 10 m/s relative to the ground.

Yes, indeed.

So the moving observer sees an increase in kinetic energy of $\frac{1}{2}m(10)^2 - \frac{1}{2}m(0)^2$ = 50m.

Meanwhile the stationary observer sees an increase in kinetic energy of $\frac{1}{2}m(20)^2-\frac{1}{2}m(10)^2$ = 150m.

Yes, a moving observer would measure a kinetic energy increase of $50m$ for his scenario.
Meanwhile a stationary observer would measure a kinetic energy increase of $50m$ for his scenario.

This is comparing two different scenarios, of course.

Now we get to it. This is the important part. An energy accounting in the moving frame. We have burned enough fuel to produce an energy increase of 150m (as assessed in the ground frame). But we have only accounted for an energy increase of 50m (as assessed in the moving frame). This is definitely an affront to our intuitions. We must ferret out the extra 100m worth of energy.

Let us use the work energy theorem. The kinetic energy increase in an isolated system must match the work done by internal forces within that system. If we want to have an isolated system, we have to consider the Earth plus the car. The only relevant force that does work is the force between car and road.

[Let us sweep the messy details of the drive train under the carpet. They add nothing to the problem and would force us to trace a chain of internal contact forces that all do zero net work until we finally get to the expanding combustion gasses in the cylinders which actually do non-zero net work].

We have the car pushing backward on the Earth. We have the Earth pushing forward on the car. For convenience, let us assume that the car is 1 kg in mass and that the force between Earth and car is 1 Newton.

From the moving frame...

It takes 10 seconds for the car to accelerate from 0 m/s to 10 m/s.
During this time the car moves at an average velocity of 5 m/s. So it displaces by +50 meters. 1 Newton times 50 meters is 50 Joules.

During this time the Earth moves at a constant velocity of -10 m/s. So it displaces by -100 meters. -1 Newton times -100 meters is 100 Joules.

We see an energy increase of 150 Joules total for this accounting.

From the rest frame...

It takes 10 seconds for the car to accelerate from 10 m/s to 20 m/s
During this time the car moves at an average velocity of 15 m/s. So it displaces by 150 meters. 1 Newton times 150 meters is 150 Joules.
During this time the Earth is stationary. So its displacement is zero. Zero work is done on the Earth.

We see an energy increase of 150 Joules total for this accounting.

All is right with the world. Gas used and total kinetic energy gained are both invariants. They do not depend on the reference frame you choose for the analysis.

Wow thank you, this was a great and helpful response.

 

[Dale]

the change in energy would be the same, but the fuel consumed can't be the same for each of those cases for the stationary observer.

You can just work through these problems, but as @jbriggs444 mentioned, you need to pay attention to the Earth as well as the cars.

So assume that the car is 1000 kg, then at 0 m/s it has 0 J of KE, at 10 m/s it has 50 kJ of energy, and at 20 m/s it has 200 kJ. Assume the Earth is 1 Yg (10^24 g) mass. Then the Earth also gains KE due to the force acting on it. It gains a velocity of 10 am/s and 20 am/s in the opposite direction of the car. These speeds correspond to utterly negligible KE’s of 50 fJ and 200 fJ respectively.

Importantly, the fuel consumption is about 50 kJ from 0 m/s to 10 m/s and 150 kJ from 10 m/s to 20 m/s.

Now, suppose that we start in the frame moving at 10 m/s. In that frame the car goes from -10 m/s to 0 m/s to 10 m/s. Let’s focus on the second part. In going from 0 m/s to 10 m/s the car gains 50 kJ of KE, but has fuel consumption of 150 kJ. So there is 100 kJ of missing energy. In this frame the Earth goes from -10 m/s to -(10 m/s + 10 am/s). This is an increase of 100 kJ.

So in both cases the fuel consumption is the same. In one case all of the energy goes into the car and in the other case a good portion of the energy goes into the earth.

 

[jbriggs444]

10 am/s

Go, @Dale -- attoboy. (I'd neglected that bit of energy).

 

[Dale]

Go, @Dale -- attoboy. (I'd neglected that bit of energy).

Very reasonable to neglect.

 

[MattGeo]

Now, suppose that we start in the frame moving at 10 m/s. In that frame the car goes from -10 m/s to 0 m/s to 10 m/s.

I was curious about this part here. Why did you say that in the moving frame the car goes from -10m/s to 0 and then to 10m/s?

 

[Dale]

I was curious about this part here. Why did you say that in the moving frame the car goes from -10m/s to 0 and then to 10m/s?

In the frame moving at 10 m/s with respect to the ground, all velocities are 10 m/s less than in the ground frame. So subtracting 10 means that 0 to 10 to 20 becomes -10 to 0 to 10.

 

[A.T.]

I was curious about this part here. Why did you say that in the moving frame the car goes from -10m/s to 0 and then to 10m/s?

https://en.wikipedia.org/wiki/Galilean_transformation

 

[MattGeo]

In this comparison you are comparing two different physical scenarios, not two different frames of reference.

You are comparing the gas used to accelerate from 0 to 10 m/s relative to the ground upon which the car's wheels are pushing with the gas used to accelerate from 10 to 20 m/s relative to the ground upon which the car's wheels are pushing.

Your conclusion is correct. It takes more energy to accelerate from 10 to 20 m/s relative to the ground than from 0 to 10 m/s relative to the ground.

Yes, indeed.

So the moving observer sees an increase in kinetic energy of $\frac{1}{2}m(10)^2 - \frac{1}{2}m(0)^2$ = 50m.

Meanwhile the stationary observer sees an increase in kinetic energy of $\frac{1}{2}m(20)^2-\frac{1}{2}m(10)^2$ = 150m.

Yes, a moving observer would measure a kinetic energy increase of $50m$ for his scenario.
Meanwhile a stationary observer would measure a kinetic energy increase of $50m$ for his scenario.

This is comparing two different scenarios, of course.

Now we get to it. This is the important part. One scenario and two seemingly incompatible results for an energy accounting.

We have burned enough fuel to produce an energy increase of 150m (as assessed in the ground frame). But we have only accounted for an energy increase of 50m (as assessed in the moving frame). This is definitely an affront to our intuitions. We must ferret out the missing 100m worth of energy.

Let us use the work energy theorem. The kinetic energy increase in an isolated system must match the work done by internal forces within that system. If we want to have an isolated system, we have to consider the Earth plus the car. The only relevant force that does work then is the force between car and road.

[Let us sweep the messy details of the drive train under the carpet. They add nothing to the problem and would force us to trace a chain of internal contact forces that all do zero net work until we finally get to the expanding combustion gasses in the cylinders which actually do non-zero net work].

We have the car pushing backward on the Earth. We have the Earth pushing forward on the car. For convenience, let us assume that the car is 1 kg in mass and that the force between Earth and car is 1 Newton.

From the moving frame...

It takes 10 seconds for the car to accelerate from 0 m/s to 10 m/s.
During this time the car moves at an average velocity of 5 m/s. So it displaces by +50 meters. 1 Newton times 50 meters is 50 Joules.

During this time the Earth moves at a constant velocity of -10 m/s. So it displaces by -100 meters. -1 Newton times -100 meters is 100 Joules.

We see an energy increase of 150 Joules total for this accounting.

From the rest frame...

It takes 10 seconds for the car to accelerate from 10 m/s to 20 m/s
During this time the car moves at an average velocity of 15 m/s. So it displaces by 150 meters. 1 Newton times 150 meters is 150 Joules.

During this time the Earth is stationary. So its displacement is zero. Zero work is done on the Earth.

We see an energy increase of 150 Joules total for this accounting.

All is right with the world. Gas used and total kinetic energy gained are both invariants. They do not depend on the reference frame you choose for the analysis.

So I was wondering something new after this problem that you elucidated for me finally clicked. How is this situation related to (or unrelated to) the Oberth Effect? Like suppose a car were traveling in a situation where there was zero air resistance and we could safely ignore internal mechanical inefficiencies in the gears and tires, an ideal situation if you will. Would the car experience energetic efficiencies by expending fuel at high velocities like a rocket would? It seems not right.

A rocket is a variable mass system with a reactive engine which is throwing exhaust out behind it, to drive it forward. But in a sense, the car is pushing the Earth behind it. Is there a connection between the two scenarios? It seems like it is increasingly difficult to gain speed in a car because more energy is required per unit of additional velocity gained (and more fuel consumption), but is that fundamental or does it only stem from the car needing more power to overcome internal energy losses?

 

[PeroK]

A rocket is a variable mass system with a reactive engine which is throwing exhaust out behind it, to drive it forward. But in a sense, the car is pushing the Earth behind it. Is there a connection between the two scenarios? It seems like it is increasingly difficult to gain speed in a car because more energy is required by unit of additional velocity gained, but is that fundamental or does it only stem from the car needing more power to overcome internal energy losses?

The answer ultimately depends on the propulsion mechanism. If a car accelerates by pushing on the road, then the force acts over a distance that increases with the speed of the car. The faster the car is going (relative to the road), the greater the power to maintain the same force.

If, however, a vehicle were to accelerate by ejecting mass, then the acceleration is independent of the speed. The greater increase of KE of the car (in the original rest frame) is balanced by the greater loss of KE of the expellant. In principle, such a mechanism should be able to sustain at least constant acceleration until the propellant runs out.

The Oberth effect, ultimately, works by using the gravitation of the planet - which is an added complication.

 

[PeroK]

PS you could get a neat inverse-Oberth effect as follows:

You have a vehicle hovering above a surface and two propulsion mechanisms:

1) You may reach down and push against the surface (expending a certain amount of energy in the process).

2) You may throw an object out of the back of the vehicle. Again expending a certain amount of energy which amounts to a fixed relative velocity between the vehicle and the expellant.

It is more efficient to push against the surface first and then use the expellant than it is to use the expellant first. This is because the first mechanism results in a fixed increase in KE and the second in a fixed acceleration. Assuming that the objective is to gain greatest speed relative to the surface. And assuming that the expellant has a small mass and high exhaust speed.

 

[Dale]

But in a sense, the car is pushing the Earth behind it. Is there a connection between the two scenarios?

The connection between the two scenarios is that if you want to correctly analyze the scenarios in multiple frames then you need to consider both energy and momentum as well as the reaction mass, whether it is rocket exhaust or the earth.

 

[MattGeo]

PS you could get a neat inverse-Oberth effect as follows:

You have a vehicle hovering above a surface and two propulsion mechanisms:

1) You may reach down and push against the surface (expending a certain amount of energy in the process).

2) You may throw an object out of the back of the vehicle. Again expending a certain amount of energy which amounts to a fixed relative velocity between the vehicle and the expellant.

It is more efficient to push against the surface first and then use the expellant than it is to use the expellant first. This is because the first mechanism results in a fixed increase in KE and the second in a fixed acceleration. Assuming that the objective is to gain greatest speed relative to the surface. And assuming that the expellant has a small mass and high exhaust speed.

I don't really understand why the first mechanism results in a fixed increase in KE and the second mechanism results in a fixed acceleration. Could you elaborate on that? How are they "fixed" and how are they different?

 

[PeroK]

I don't really understand why the first mechanism results in a fixed increase in KE and the second mechanism results in a fixed acceleration. Could you elaborate on that? How are they "fixed" and how are they different?

Are you able to do any calculations? Using conservation of momentum etc.

 

[MattGeo]

Are you able to do any calculations? Using conservation of momentum etc.

If I accelerated by using the excellent first and then I apply the same force to the surface after already moving, how could the energy be the same as having just pushed off the surface first with the same force. I guess it's conceptually a stumbling block in the first place that prevents me from setting up conservation equations.

 

[PeroK]

I guess it's conceptually a stumbling block in the first place that prevents me from setting up conservation equations.

It's a calculational stumbling block. No one could have answered these questions before Newton formulated the laws of motion. In general, it's only after you calculate the answer that you see why it must be so.

 

[jbriggs444]

I apply the same force

Over a different displacement. So it is a different amount of energy (aka work).

Or over a different time. So it is a different amount of momentum (aka impulse).

A "force" in the sense that the term is used in a first year physics classroom is a sustained push or pull between two objects. It is not an instantaneous interaction that can be completely characterized by a single number.