Gary Novak 1. quick proof

Math Explained
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/s m_{t} = mass of rocket at time T = m_{o} + m't v_{e} = separation velocity of exhaust = 10^{3}m/s (pos) F = force = m'v_{e} = (0.01)(10^{3}) = 10 newtons v'_{t} = acceleration at time T = F/m_{t} ◆ ◆ ◆ ◆ ◆ ◆ ◆ ◆ ◆ ◆ ◆ Force 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 ( Acceleration v'_{t} = acceleration at time T = F/m_{t} 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 m_{t} = mass of rocket at time T = m_{o} + m't Since mass is being lost through the exhaust, the rocket mass varies with time. The mass at any particular time Velocity 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 antiderivative of acceleration, which is the integration dt. Acceleration which changes with time (v'_{t}) is Velocity is the antiderivative of it, which is ∫v'dt. v = velocity = ∫v'dt = ∫10/(m_{o}  0.01t)dt For this mass (m_{o}), 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/(m_{o}  0.01t) = 10^{3}/(100m_{o}  t) Its antiderivative (same as velocity) is 10^{3}ln(100m_{o}  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/(m_{o}  0.01t)dt = ∫10^{3}/(100m_{o}  t)dt = 10^{3}ln(100m_{o}  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. Time v = 10^{3}ln(100m_{o}  t) + c t = m_{o}/0.01  [(cv)/10^{3}]e^{x} e^{x} 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: (cv)/10^{3} = ln(100m_{o} t) Apply this principle: If a = lnb, then e^{x}a = b It says: If a is the natural log of b, then inverse of natural log of a equals b Now have a equal (cv)/10^{3} and b equal (100m_{o} t) Then inverse natural log of (cv)/10^{3} equals (100m_{o} t) e^{x}[(cv)/10^{3}] = 100m_{o} t Changing the signs and relocating is this: t = 100m_{o}  e^{x}[(cv)/10^{3}] For sequencial entries on a calculator, it looks like this: t = m_{o}/0.01  [(cv)/10^{3}]e^{x} 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/(m_{o}  0.01t) dt = 100mln(100m_{o}  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 = 3,113.2896The integration quantity is 1.771779, which is exact time for test mass. Shortcut Formula for BurnTimeThis can be applied to payload only: From Ft = mvt = m_{t}v_{t}/m'v_{e}  1. Energy: Historical Development of the Concept. R. Bruce Lindsay. 1975. p345. Dowden, Hutchison & Ross. 396pp. ISBN10: 0470538813. ISBN13: 9780470538814
