Gary Novak Background Second Proof math explained Consise Math Joule's constant potential energy Contradictions

Consise Math
The proof is that only mv is conserved under all conditions, while ½mv² is not. The test is to use a rocket to replace the energy of falling objects and calculate the amount of fuel used for the added mass only. The rocket replaces or duplicates gravity to create the same velocities as dropping does. Then various conditions are compared. (Rockets are used to calculate force, which is independent of the definition of energy.) Here's a summary of the test:
To determine the amount of energy in the falling objects, a mathematical model of a rocket can be used to add the energy. The total energetics can be calculated without doing an experiment. The rocket is constant powered, as rockets usually are. This means fuel is used at a constant rate. For this reason, burn time can replace actual fuel use. In other words, a two second burn uses twice as much fuel as a one second burn. Therefore, the method is to determine the time of burn required to give each of the objects their final velocities. The burn time is then adjusted for the payload only by subtracting the proportion of the mass due to the rocket. The rocket mass changes as fuel is burned; so an average mass is determined through calculus techniques. The Rocket Equations: The rocket is arbitrarily given three properties, and the rest is calculated. Its mass is 20kg; its rate of mass loss is 0.01kg/s; and the separation velocity of the exhaust is 1000m/s. x' = rate of change for x = dx/dt (' is rate of change) m_{o} = mass of rocket at start = 20kg plus payload (21 or 24 kg) m' = rate of mass loss = 0.01kg/sec m_{t} = mass of rocket at time T = m_{o} + m't v_{e} = separation velocity of exhaust = 10³m/sec (pos) F = force = m'v_{e} = (0.01)(10³) = 10 newtons v'_{t} = acceleration at time T = F/m_{t} v = velocity = ∫v'dt = ∫10/(m_{o}  0.01t)dt = ∫10³/(100m_{o}  t)dt = 10³ln(100m_{o}  t) + c (c is quantity at T = 0) c = 10³ln(100m_{o}) burn time = t = m_{o}/0.01  [(cv)/10³]e^{x} e^{x} is inverse of natural log for the preceding quantity. mass ratio averaged for loss of mass = ∫m_{p}/(m_{o}  0.01t) dt = 100m_{p}ln(100m_{o}  t) + c c is quantity at t = 0 c = 100m_{p}ln(100m_{o}) Example: four kilograms dropped one meter. g = acceleration of gravity = 9.81 m/sec (rounded throughout) s = distance down = 1m t = time dropping = (2s/g)^{½} = 0.451523641 sec v = velocity after dropping = gt = 4.429446918 m/sec Constant C in burn time: c = 10³ln(100m_{o}) = 10³ln[100(24)] = 7,783.224016 Total Time of Burn: t = m_{o}/0.01  [(cv)/10³]e^{x} = 24kg/0.01  [(7,783.2240164.429446918)/103]e^{x} = 10.6071633 sec Fraction for Payload: m_{p} = mass of payload = 4 kg t_{p} = time for payload including mass ratio integrated with time = ∫m_{p}/(m_{o}  0.01t) dt = 100m_{p}ln(100m_{o}  t) + c c ( quantity at t = 0) = 100(4kg)ln[100(24kg)] = 3,113.289607 t_{p} = 100(4kg)ln[100(24kg)  10.6071633] + 3,113.289607 = 1.7717788 sec Explanation: This last step might look a little too magical even to an expert. Actually it is. By logical analysis, this last number would be divided by burn time to determine average mass fraction (Calculus averaging divides by the interval.). It would then be multiplied times burn time to determine the amount of time attributed to the payload. 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 payload. Numbers for each object: A  4 kg dropped 1 m B  1 kg dropped 4 m C  4 kg dropped 1 sec D  1 kg dropped 4 sec Time Dropping: A: t = [2(1m)/9.81]^{½} = 0.45152364098573 sec Velocity Dropping: A: v = 9.81(0.45152364098573) = 4.42944691807002 m/s Constant C in burn time: velocity of rocket = A: c = 10^{3}ln[100(24)] = 7783.22401633603 Total Time of Burn: t = m_{o}/0.01  [(cv)/10^{3}]e^{x} A: 24kg/0.01  [(7783.224016336034.42944691807002)/10^{3}]e^{x} = 10.6071633272070 sec Fraction for payload: A: c = 100(4kg)ln[100(24kg)] = 3113.28960653441 Time of Burn for Payload Only: A: t = 100(4kg)ln[100(24kg)  10.6071633272070] + 3113.28960653441 = 1.77177876722800 sec Ratios:
