KE of system / different reference frames question

Humber

There is a change of 100kJ for the Earth in the second case, but the car gains only 50kJ.
Does the remaining 50kJ go as heat?

DaleSpam

If the KERS works ideally then the remaining 50 kJ goes into the battery. Of course, nothing is 100% efficient, so some fraction will be lost to heat. It would only go all to heat if you were using traditional non-regenerative braking.

Humber

Differences in velocity are frame invariant? No, velocity is relative, so is frame dependent.

No, in the first case, the Earth is considered to be stationary, and the car changes velocity by 10m/s, and the Earth moves in the opposite direction - 1e-21m/s, so the difference, the relative velocity, is 10m/s + 1e-21m/s

In the second case the the car remains at 0m/s, and the Earth changes by -9.999999999999999999999m/s. which is not the same as the first case.

No, in the first case, the Earth gains 5e-18J.
That will hold if the acceleration is from 0m/s to 10m/s or 10m/s to 0m/s

Car loses p
Earth gains -p

Total energy = p^2/2m_car + p^2/2m_earth
When the car accelerates the sign of the force changes, so the sign of the second term changes, but remains negligible.

Another worry is the mass of the Earth.
The math is circular, where a velocity of 10m/s is posited. The total p is calculated, and then 10000J subtracted, and then the KE is re-calculated using the mass, which was used to derive the momentum on the basis that the initial velocity of the Earth was 0m/s.
The result is always 10000J. Try it use m= 1e26kg.

ETA: That is in contrast with the case where actual values are used, and the same method.

Vearth ~ 29.8km/s = 29800m/s
mearth ~ 6e24kg
pi = 178800000000000000000000000000J
KEi = 2.66412e+33
subtract 10000kg.m/s
KEf 2.664119999999999999999999702e+33
dKE =298000000J

Humber

That's true, but I understood that it was idealised. The loss from the Earth of 50kJ, must be dissipated at the source, which is the Earth. Which does not happen in the first case.

Ken G

This is incorrect, and will lead you astray. Changes in energy are not p²/2m, they are changes in p²/2m, so you must not treat p, and changes in p, as if they were the same thing. They aren't! Let the change in p be denoted dp, as in calculus. Then a change in p²/2m looks like (p+dp)²/2m - p²/2m, and if dp << p, this is very close to p*dp/m or v*dp. That's why when KERS kicks in and there is some dp across some relative v between the car and the Earth, the energy change of the Earth, in the car-stationary frame, is the same as the energy change of the car, in the Earth-stationary frame: it is because either way the v and the dp are the same numbers. Work it out, and spend no time on the irrelevant dp² terms.

Humber

Momentum is frame independent, otherwise, there could be violations of conservation.
There would be no means of correcting after the fact, so violations simply don't occur.
That alone guarantees frame independence. If follows that changes are also frame independent.

That is not correct, because "change" in momentum dp/dt = Force. KE is not frame independent, and that is also accommodated.
KE = p^2/2m = m^2v^2/2m = 1/2mv^2.

Total energy = p^2/2mcar + p^2/2mearth is correct. If the energy change of the car were the same as the change in the ground, p^2/2mcar - p^2/2mearth = 0, then obviously the KERS could not store any energy.

DaleSpam

The quantity you are describing, [itex]v_{i,c}+v_{f,e}[/itex], is not the relative velocity. See: http://en.wikipedia.org/wiki/Relative_velocity

The fact that some random sum of velocities is not frame invariant is hardly surprising, nor is it troubling.

I recommend that you work through the math completely and post your work like I did. Unfortunately, I will be travelling the next couple of days so I won't be able to do it myself.

The fact that the result is independent of the mass of the earth is interesting, but not troubling. Many times you get a scenario where something cancels out. I will try to do it algebraically to see where and why it drops out, but not for a few days.

Oh, that's a mess. The units of p should be kg m/s, not J. You cannot subtract a quantity in units of kg m/s from a quantity in units of J. I am not sure what you were trying to get at here.

DaleSpam

What makes you say that?

Humber

Velocity is frame dependent, and sums linearly. Random velocities sum as the RMS of each, but none are random in this case.

I did that. There are several problems, including independence of Earth's mass.
That is the result of simply assigning a velocity to the Earth that is the same as the car's.

(Δp_c=1e4kg.m/s)
p_i,e = M.v
KE_i,e = p_i,e²/2M
KE_f,e = (p_i,e)²-Δp_c/2M
KE_i,e - KE_f,e = 100kJ

V_earth ~ 29.8km/s = 29800m/s (orbital velocity of the Earth)
m_earth ~ 6e24kg (mass of the Earth)
p_i,e = 178800000000000000000000000000 kg.m/s (initial momentum mv)
KE_i,e = 2.66412e+33J ( Initial KE of the Earth)

subtract Δ_pc= 10000kg.m/s ( The applied impulse from the car)
KE_f,e 2.664119999999999999999999702e+33J ( Final KE of Earth)
dKE =298000000J (change in KE of Earth)

The same method as you employed, but where the initial Earth velocity is not zero.

Humber

The change in the Earth's KE is 100kJ. The car gains 50kJ. The remainder must go somewhere, and heat at the source seems to be the most likely. The actual reason is that there is no 100kJ

Total energy = p^2/2m_car + p^2/2_mearth
Applies to energy to and from the cart as the case may be.

Humber

When a car brakes, momentum is transferred from the car to the ground.
From the conservation of momentum;

Car gains momentum -p
Earth gains momentum p

The car loses KE = p²/2m_car
The Earth gains KE = p²/2m_earth

The total energy is therefore;
p²/2m_car+ p²/2m_earth

Because m_earth is very large ,~6e24kg, the second term is negligible.

All of that energy will have come from the fuel of the F1 car. It could be batteries, and the result is the same.

In a conventionally braked vehicle, the KE of the car's mass is dissipated as heat in the brakes. KERS does not affect momentum transfer to the ground, but stores the KE that is otherwise lost to heat.

That energy may be transferred to a flywheel;
The angular momentum L, of a particle, and the moment of inertia, I
L = r.p = r.mv
L =I.ω
KE = 1/2.I.ω²

There is no significant transfer of energy to or from the F1 and the ground, as a result of KERS.

DaleSpam

They may not be random, but they are not relative velocity either.

I don't know why you think that is a problem.

This is not the same method I employed. I never subtracted a momentum from an energy, it is an invalid operation. The units don't work out.

DaleSpam

The remainder goes into the battery. Where else would the energy that goes into the battery come from?

Tea Jay

I'd like to think that if the car's acceleration using the earth for traction makes the world spin faster, etc, when the car eventually stops, it will surely slow the earth back down by that same (infinitesimal) amount.

;)

Humber

vi,c =initial velocity of car
vf,e = final velocity of earth
The difference is therefore the relative velocity of car and Earth.
They are not the same in each case.

Then you don't have a case for your calculations.

The reduction of the Earth's momentum, is the basis of your KE claim.
It applies to the second case too. Which is why you will always get 100kJ, regardless of mass. It is an error.
The result is indenpendet of the Earth's mass because it is the equivalent of

Δp_c*v_f,c
= -10000kg.m/s*10m/s = 100000 kg.m²/s²
= -100kJ

Humber

You have calculated 100kJ energy from the ground of which 50kJ is stored in the battery.
That leaves 50kJ unaccounted for.

Which does not allow 50kJ for the car and 50kJ for the battery, because the system is a Kinetic Energy Recovery System. And you would still have a difference of 50kJ, when compared with the first case.

Humber

Infinitesimal it is. With the given numbers, 5e-18J and a velocity change of 1e-21m/s. That is well wide of the 100kJ claimed, and of the 100W of the o.p.
In reality, the Earth is not an infinitly rigid sphere, and will dissipate energy, just as it does for Earthquakes and nuclear weapons tests.