This rocket (on rocket page) is used for the mathematical proof based upon falling objects, as summarized on the "quick summary" page and explained in more detail on the "falling object" page.

Time

t = m_o/0.01 - [(c-v)/10³]e^x

e^x is inverse of natural log for the preceding quantity.

The amount of fuel that the rocket burns while accelerating the masses is determined as time of burn, since the fuel burns at a constant rate. The time formula is derived from velocity. Velocity is derived from acceleration based on simple calculus.

Force

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

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 going out 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 (v_e=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.

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 (m_t) equals the starting mass (m_o) plus the rate of mass loss (m') times the elapsed time (t).

Velocity

  v = velocity = òv'dt = ò10/(m_o - 0.01t)dt

  = ò10³/(100m_o - t)dt = -10³ln(100m_o - t) + c

  (-c is quantity at T = 0)

Velocity is the antiderivative of acceleration, which is a calculus determination. Acceleration which changes with time is F/m_t or 10/(m_o - 0.01t). The velocity is the antiderivative of it.

It is therefore necessary to know the antiderivative for this equation:

     a = 10/(m_o - 0.01t)

Generically it is    a = 10/(20 - x)

The antiderivative is:

     ò10/(20 - x)dx = -10ò-dx/(20 - x)

     = -10ln(20 - x) + c

The generic antiderivative for that type of equation is this:

     òdu/u = ln(u) + c

     u = (20 - x)    du = -dx

If this symbol fails to show properly, it is the integral sign: ò

"ln" is natural log base e.

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

    10/(m_o - 0.01t) = 10³/(100m_o - t)

Its antiderivative (same as velocity) is

    -10³ln(100m_o - t) + c

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

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.

Fraction of time

  fraction = test mass/(total mass - 0.01t/2).

The total time of burn represents the total energy for both the rocket mass and attached mass. A fraction of the total time was attributed to the attached mass, and average mass loss corrected for the change in mass. This method left a slight error in the calculation. An exact procedure which has no error is the following:

Exact Procedure

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:

ò 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 = 3,113.2896

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

Summary

In summary, the purpose is to determine how much energy the falling objects have when they acquire their final velocities. The amount of energy that a rocket uses to produce the same velocity is an objective measurement of that energy.

The energy which the rocket uses is determined by the amount of time that it burns fuel. So time indicates the energy which is used.

When the rocket has to burn for the same amount of time for two objects, they have the same kinetic energy at their final velocities.

This conclusion is based upon the fact that as the rocket burns fuel it must be adding a fixed amount of kinetic energy to a mass per unit of time. This result is exact and invariable, because the propulsion system consists of mass accelerating in the exhaust to create a force which accelerates the mass in the rocket with the same amount of force, because forces must be equal and opposite. The force must create kinetic energy. The example assumes an absence of friction or gravity influencing the result of the rocket.

If this technical statement is mentally challenging, I'll say it in a less exact and easier to understand way. Energy of rocket exhaust equals energy added to total mass of rocket. This is because forces must be equal and opposite. Force is the only product of the rocket, so equal and opposite forces create equal and opposite kinetic energies.

Notice this key point: force is the only product of the rocket which is relevant, because the force produced by the rocket replaces the force of gravity. Force is equal to rate of mass loss times separation velocity of exhaust, as indicated in rocket characteristics.

The chemical energy of the rocket fuel is not relevant (after separation velocity is determined—given in advance), because the kinetic energy (not the chemical energy) of the exhaust determines the force. Since chemical energy is not relevant, efficiency of the rocket motor is not relevant.