# Energy Definition

#### shlosmem

Let's say one gallon of fuel can make a spaceship rocket engine with a mass M to run for 1 hour and to create a constant acceleration A.
If we will use the full gallon starting from zero velocity, we will reach a velocity of A*1=V, the kinetic energy of the spaceship then will be Ek=0.5MV^2 .
We can use this gallon to run two spaceships for half an hour, is that case they will reach a speed of A*0.5=0.5V , the total energy of the system will then be Ek'=0.5*2M*(V/2)^2.
Ek/Ek'=2, in plain words, the first system of one spaceship has the twice kinetic energy than the second system of two spaceships.
What is the actual potential energy of one gallon of fuel?

#### ChipB

PHF Helper
You are correct that the KE of the two ships combined in the second scenario is less than the KE of the single ship in the first scenario. The reason for this "discrepancy" is that you are leaving out the KE of the rocket exhaust in the two scenarios. To see the complete picture of how the potential energy of the rocket fuel is turned into kinetic energy you must include the mass of the exhaust gasses and the velocity that the gasses are expelled.

• topsquark and Global Tutor

#### shlosmem

I'm not sure how the rocket exhaust is relevant, in both cases we have one hour of engine work. Anyway this is just an example, the same question can be asked on a battery of electric car with 100% efficiency and 0 exhaust .

#### shlosmem

We can think of it from a different angle. let's say we have 2 cars, one is accelerating on a road from 0 to velocity V, the second accelerates on a platform until it reached a velocity V but the platform itself is also moving in velocity V, so the second car is actually reaching a velocity of 2V. if we measure the difference of the kinetic energy of each car before and after the acceleration, we will get that the difference in the kinetic energy of the second car is 3 times then the change in the first car, (0.5M(2V)^2 - 0.5MV^2 = 1.5MV^2). According to the book, the sconed car engine has 3 times the power of the first one, this is because we measuring energy according to the distance the object moves, if for example V=50, the average speed of the first car was 25, when the average speed of the sconed car is 75, so the second car does 3 times the distance that the first car does, also they both consume the same energy to do so. It of course not relevant to the type of engine or to exhaust etc. it's just the way we are measuring energy.
It seems that we can't just ask about the potential energy of a battery or a gallon of fuel, because it depends on the way we are using it. it basically can be any number. But we may ask instead about the Impuls size we can produce from it which will give us a straight answer.

#### ChipB

PHF Helper
I'm not sure how the rocket exhaust is relevant, in both cases we have one hour of engine work. Anyway this is just an example, the same question can be asked on a battery of electric car with 100% efficiency and 0 exhaust .
The exhaust gas is high;y relevant, as it has mass and has been accelerated by the energy of the rocket fuel, and therefore has had work done on it. You must consider the entire mechanical system, or you will get problems such as the missing energy of your example.

As for an electric car, it would work like this:

The energy contained in a battery is measured in watt-hours. If a particular electric car of mass M has a battery that has sufficient potential energy to accelerate the car to velocity V, then the KE of the car will be 1/2 MV^2. Now let's cut that battery in half, resulting in two batteries each with half the energy content of the original (as measured in watt-hours), and place each into a car of mass M. What velocity will each car be accelerated to? The answer to that is not V/2; it's actually V/sqrt(2), and the KE of each car will become 1/2 M (V/sqrt(2)^2 = (1/4)MV^2, or half of he KE of the original car. Hence the total KE of two of these cars is, again, 1/2MV^2. The fundamental issue here is that a battery that has half the energy content does not accelerate the car to half the velocity. Consider that the energy required to accelerate the car from 0 to (1/2)V is less than the energy required to accelerate the car from (1/2)V to V. In this exampe a battery of half the energy content will accelerate to V/sqrt(2), which is greater than V/2.

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• donglebox

#### shlosmem

Consider that the energy required to accelerate the car from 0 to (1/2)V is less than the energy required to accelerate the car from (1/2)V to V.
Is this assumption is not conflicted with the principle of relativity?

#### ChipB

PHF Helper
Is this assumption is not conflicted with the principle of relativity?
No, there is no conflict. I assume you're talking about classical relativity (Newtonian), as opposed to Einstein's theories of special or general relativity which may come into play for objects's moving at velocities approaching the speed of light. An object's KE is dependent on its velocity relative to the observer. KE is NOT invariant. Two different observers who are in motion relative to each other will calculate different values for KE for a given object, and as the speed of the object changes they will calculate different values for the object's change in KE. For example, consider an object of mass 1 Kg traveling through space at a velocity of 100 m/s as measured by observer A. That observer calculates the object's KE as (1/2) x 1 Kg x (100m/s)^2 = 5000 J. Meanwhile a second observer is traveling along side the object, and so measure's the objects relative velocity as 0 m/s, and determine's the objects KE is 0 J. Which observer is correct? Answer: they both are.

In the case of the spaceship keep in mind that as it uses fuel to increase its velocity the potential energy of its fuel is converted into KE of both the spaceship and its exhaust. To a stationary observer at first most of the fuel PE is converted to KE of the exhaust and just a little goes to KE of the ship. But as the ship gains speed more of the PE of the fuel is converted to KE of the ship and less to the exhaust (because the velocity of the exhaust relative to the observer decreases as the ship gains speed). Thus while the rate for fuel usage may remain constant per unit time, the rate of change of KE of the ship increases, and the rate of change of KE of the exhaust decreases a corresponding amount. The sum of rate of change of KE for the ship plus the exhaust remains constant, for all observers, and is equal to the rate of decrease of PE of the fuel on board.

#### shlosmem

Now let's cut that battery in half, resulting in two batteries each with half the energy content of the original (as measured in watt-hours), and place each into a car of mass M. What velocity will each car be accelerated to? The answer to that is not V/2....
According to the classical principle of relativity, the same amount of fuel or battery power is needed to accelerate from v1 to v2 regardless of the value of v1 and v2. This is true for all observers as long they agree on the value of |v1-v2|. Otherwise, 2 observers, one is standing and one is already moving are not going to agree on the amount of fuel that remains in the tank. If this is true, why half of a battery will not accelerate a car to V/2?

#### ChipB

PHF Helper
shlosman said:
According to the classical principle of relativity, the same amount of fuel or battery power is needed to accelerate from v1 to v2 regardless of the value of v1 and v2.
This is incorrect. Please go back and reread my earlier posts. And let me try a different explanation, which should help clarify your confusion over the electric car example.

The power produced by an electric motor is equal to the torque it produces times the rotational velocity of the motor. Torque is what accelerated the car. Ignoring the transmission gearing for a moment, as the car goes faster the speed of the motor must increase. If the motor is drawing a constant amount of energy per second from the batteries (in other words it's drawing constant power), then the power put out by the motor is constant. Which means as the rotational velocity of the motor's drive shaft increases the torque it produces decreases, leading to slower acceleration of the car. The net result is that it takes longer to accelerate from 0 to v1 than it does to accelerate from V1 to 2 x V1. And the power drawn from the battery to go from 0 to V1 is less than the power required to go from v1 to 2 x V1. One major difference from the rocket ship example is that for an electric car there is an absolute velocity than can be agreed upon by all observers - namely the speed of the car relative to the road. So all observers can agree that the electrical energy required to accelerate the car relative to the road increases with the cars speed.

#### Woody

What ChipB is pointing out is that you have to include all features of the whole system,
(or at least all features that are not negligible).
if you simplify your model too far, you start to miss out important energy drains, which then give rise to apparent discrepancies.
For example, with the rockets, you have to include the energy of the ejected mass of the rocket plumes.
(the energy loss of the transmission gearing in the cars can, probably, be considered minor, compared with the other energy paths of that system).