Energy Misdefined in Physics Kinetic energy is being defined as mass times velocity squared. Velocity squared does not represent nature, because nothing moves at velocity squared. Mathematical proof shows that kinetic energy is really mass times velocity nonsquared, which is called momentum. Momentum is kinetic energy. Here's the problem: If a small rocket engine adds a one pound force for one second to a spacecraft moving at 25,000 miles per hour, a lot of energy is added when using the erroneous definition of energy; but if the same rocket engine adds a one pound force for one second to a roller-skate, a lot less energy is added. Same transformation; different energy. With the corrected definition of energy, the same amount of energy is added in both cases. The Test This model, involving falling objects, was the original basis for the definition of kinetic energy. Rockets are used to calculate force, which is independent of the definition of energy. Part 1: Kinetic energy is not ½mv². A 4kg object dropped 1m (meter) has the same amount of ½mv² as a 1kg object dropped 4m, because force times distance equals ½mv² for an accelerating mass. But a rocket accelerating the masses to those velocities requires twice as much energy as fuel for the large mass as for the small one. Therefore, both masses do not have the same energy; the rocket does not transform energy in proportion to ½mv²; ½mv² is not kinetic energy; and a gallon of fuel does not produce a consistent amount of ½mv². Part 2: Kinetic energy is mv. A 4kg object dropped for 1s (second) has the same amount of mv (momentum) as a 1kg object dropped for 4s, because force times time equals mv for an accelerating mass. A rocket accelerating the masses to those velocities uses the same amount of energy as fuel for both masses. Therefore, both masses have the same amount of energy; the rocket transforms energy in proportion to mv; mv is kinetic energy; and a gallon of fuel produces a consistent amount of mv. Rocket Equations: 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. 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. A shortcut formula (shown at bottom) uses the force acting upon the payload only. The longer procedure shows the validity of the process. 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) mass ratio averaged for loss of mass = Velocities: Formulas: v = gt Numbers for each object: 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 = -c = velocity at t=0 c = 10^{3}ln(100m_{0}) A: c = 10^{3}ln[100(24)] = 7783.22401633603 Total Time of Burn: t = m_{o}/0.01 - [(c-v)/10^{3}]e^{x} (e^{x} is inverse natural log for preceding quantity.) A: 24kg/0.01 - [(7783.22401633603-4.42944691807002)/10^{3}]e^{x} = B: 21kg/0.01 - [(7649.69262371151-8.85889383614004)/10^{3}]e^{x} = C: 24kg/0.01 - [(7783.22401633603-9.81)/10^{3}]e^{x} = D: 21kg/0.01 - [(7649.69262371151-39.24)/10^{3}]e^{x} = Fraction for payload: equals test mass divided by average total mass (loses 0.01kg/s). The mass ratio integrated with time is this: ∫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_{0}) Constant C in Mass Fraction: 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 B: t = -100(1kg)ln[100(21kg) - 18.5215158540212] + 764.969262371151 = 0.885889383614004 sec C: t = -100(4kg)ln[100(24kg - 23.4288933861318] + 3113.28960653441 = 3.92400000000000 sec D: t = -100(1kg)ln[100(21kg - 80.8081749880098] + 764.969262371151 = 3.92400000000000 sec Shortcut Formula for Burn-Time This can be applied to payload only:
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