Energy Misdefined

Science Errors

Gary Novak

Energy Home

1. quick proof
2. rockets
3. history
4. definitions
5. collisions
6. falling objects
7. engines
8. levers

Second Proof
math explained
complete math
Consise Math
Joule's constant
potential energy   



Second Proof

Power gets Absurd at the High Velocity of a Rocket

The erroneous definition of energy has a velocity problem. First, squaring velocity creates a problem in the kinetic energy formula (½mv²). Then power becomes absurd at high velocities.

power curve

Power of rocket payload as rate of change in ½mv²

The horizontal line is the power of the rocket engine.

Rockets are usually constant powered, meaning they use fuel at a constant rate. Power is rate of energy addition.

Mathematically, the power of the rocket mass and exhaust mass are added to create the constant power. But there is a problem with that analysis. As velocity increases, it reaches a point where the power added to the payload is greater power than the engine produces.

When the definition of energy is corrected, power is proportional to force only, which is constant, as fuel use is constant.


mo = mass of rocket at start = 21kg

m' = rate of mass loss = -0.01kg/s

mt = mass of rocket at time T = mo + m't

ve = separation velocity of exhaust = 103m/s (pos)

F = force = -m've = -(-0.01)(103) = 10 newtons

v't = acceleration at time T = F/mt

v = velocity = v'dt = 10/(mo - 0.01t)dt =

     103/(100mo - t)dt = -103ln(100mo - t) + c

     (-c is quantity at T = 0)

     c = 103ln(100m0) = 7650

Shortcut Formula for Velocity:

Δv = veln(m0/mt)

= (1000)ln(21/mt)

With the existing (erroneous) definition of energy, power as rate of energy addition reduces to force times velocity. With simple math, the logic is this: mv/t = mv/tv = mav = Fv. The one half is dropped, because it was for averaging.

More precisely: the derivative of mv with respect to t = m(v)' + (m)'(v).
(v)' = 2vdv/dt = 2vv'.
(m)' = dm/dt
The total is m(2vv') + m'(v) = mv'v + m'v

But when evaluating the payload only, there is no change in mass. So the power is force times velocity:

KE'p = m(2vv') + 0(v) = mv'v = mav = Fv

The force acting upon the payload (Fp) is the force of the engine (10 newtons) times the mass ratio (mp/mt).

m = 21kg total, payload = 1kg

vr Fp KE'p
200 100 0.526 52.6
400 211 0.588 124
600 336 0.667 224
800 480 0.769 369
1000 647 0.909 588
1200 847 1.11 941
1400 1099 1.43 1,570
1600 1435 2.00 2,870
1800 1946 3.33 6,487

The Logic Problem

The reason why the rocket payload acquires energy as power at a higher rate than the power of the rocket engine is because velocity is an element of power, when energy is defined as ½mv². This means that the rocket engine creates velocity, and then velocity becomes a source of power.
Velocity is not really a source of power. Yet the incorrect definition of energy makes it a source of power. The graph above shows why velocity should not be a source of power. When velocity increases, the total power for an object which does not lose mass, such as the rocket payload, gets larger than that produced by the real source of power—the engine.