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   



Math Explained

x' = rate of change for x = dx/dt (' is rate of change)

mo = mass of rocket at start = 20kg plus payload (21 or 24 kg)

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

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The only complexities are derivation of formulas for velocity of rocket, time of burn for rocket and fraction of time for payload. With these derived, a high school kid could plug in the numbers. If there were the slightest error in these equations, the result would be nonsense, not the exact ratios of 2/1 and 1/1 for end results.

The derived equations are these (explained below):

Velocity of rocket = -103ln(100mo - t) + c

Time of burn for rocket = mo/0.01 - [(c-v)/10³]ex

Fraction of time for payload = -100mpln(100mo - t) + c


The rocket is arbitrarily given its basic characteristics. Then velocity is calculated from them. The analysis of a rocket's motion is based upon the fact that mass propelled from the exhaust produces a force in proportion to its acceleration; and an equal and opposite force acts upon the rocket.

The force is based upon mass times acceleration (F=ma), which is the same as rate of mass loss (m'=0.01kg/s) times separation velocity of exhaust (ve=103m/s) (both were arbitrarily chosen), because time in the denominator can be moved from one element to the other, which is how the rocket analysts do it.


  v't = acceleration at time T = F/mt

The acceleration of the rocket mass is force over mass (a=F/m). Since acceleration is velocity over time, it is denoted as v'. (The apostrophe is being used to denote division by time.)

For any particular instant in time, acceleration is force over mass at that particular time.

Mass Loss

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

Since mass is being lost through the exhaust, the rocket mass varies with time. The mass at any particular time (mt) equals the starting mass (mo) plus the rate of mass loss (m') times the elapsed time (t).


Velocity is derived from acceleration. Since acceleration is velocity over time, velocity is acceleration times time. In calculus, acceleration is the derivative of velocity with respect to time, which is the same thing as dividing something by time. The question is the opposite, knowing acceleration, what is the velocity. It is the anti-derivative of acceleration, which is the integration dt.

Acceleration which changes with time (v't) is F/mt or 10/(mo - 0.01t).

Velocity is the antiderivative of it, which is v'dt.

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

For this mass (mo), we will use 20 kg to improve clarity.

So we need to know this:   10/(20 - x)dx

The antiderivative is:

     10/(20 - x)dx = -10-dx/(20 - x)

     = -10ln(20 - x) + c

The generic antiderivative for that type of equation is this:

     du/u = ln(u) + c

     u = (20 - x)    du = -dx

"ln" is natural log base e.

Before the actual equation is converted, it must be factored as follows:

    10/(mo - 0.01t) = 103/(100mo - t)

Its antiderivative (same as velocity) is

    -103ln(100mo - t) + c

The quantity for c is calculated by using zero for time and changing the sign.

The net result is this:

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

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

  (-c is quantity at T = 0)

This formula is factored for time to get the desired quantity. Time is the only unknown, because the velocity is determined in the falling object analysis.


   v = -103ln(100mo - t) + c

   t = mo/0.01 - [(c-v)/103]ex

ex is inverse of natural log for the preceding quantity.

This equation for time is tailored in form and sequence for a calculator.

The method of factoring velocity to get time is this:

First, move a few things around to get this: (c-v)/103 = ln(100mo -t)

Apply this principle: If a = lnb, then exa = b

It says: If a is the natural log of b, then inverse of natural log of a equals b

Now have a equal (c-v)/103 and b equal (100mo -t)

Then inverse natural log of (c-v)/103 equals (100mo -t)

   ex[(c-v)/103] = 100mo -t

Changing the signs and relocating is this:

   t = 100mo - ex[(c-v)/103]

For sequencial entries on a calculator, it looks like this:

   t = mo/0.01 - [(c-v)/103]ex

Fraction of Time for Payload

The ratio of test mass to total mass is integrated with time. It is then averaged, and the average is multiplied times total time. The mathematical procedure for averaging nonlinear change is to integrate with the desired variable and then divide by the interval for that variable.

The mass ratio integrated with time is this:

    m/(mo - 0.01t) dt =

   -100mpln(100mo - t) + c

This quantity is divided by burn time to determine average mass fraction. It is then multiplied times burn time to determine the amount of time attributed to the test mass. Since dividing and then multiplying by the same number is unnecessary, the integration quantity is in itself the exact quantity desired. It represents the amount of time attributed to the test mass.

Example for Exact Time

For example, in the first test case, m = 4kg. The integration is this:

   -400ln(2400 - 10.6072) + c

   -c is quantity at t = 0

   it is = 3,113.2896

The integration quantity is 1.771779, which is exact time for test mass.

Shortcut Formula for Burn-Time

This can be applied to payload only:

   From Ft = mv

   t = mtvt/m've


1. Energy: Historical Development of the Concept. R. Bruce Lindsay. 1975. p345. Dowden, Hutchison & Ross. 396pp. ISBN-10: 0470538813. ISBN-13: 978-0470538814