Energy Misdefined
  

      

Gary Novak

Energy Home

 

          

▼▼▼  

Math Explained

 

 

The End Result Is This:

Part 1: Kinetic energy is not ½mv²

A 4kg object dropped 1 meter has the same amount of ½mv² as a 1kg object dropped 4 meters, 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.

Rocket burn time:
large mass: 1.771778 seconds
small mass: 0.8858893 seconds

Part 2: Kinetic energy is mv

A 4kg object dropped for 1 second has the same amount of mv (momentum) as a 1kg object dropped for 4 seconds, 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.

Rocket burn time:
large mass: 3.924000 seconds
small mass: 3.924000 seconds

  

 
The Basic Rocket Equations Provided By The Jet Propulsion Laboratory:

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)   1

m' = rate of mass loss = -0.01kg/sec   2

ve = separation velocity of exhaust = 103m/sec (pos)   3

mt = mass of rocket at time T = mo + m't = (mo -0.01t)   4

F = force = -m've = -(-0.01kg/sec)(103m/sec) = 10 newtons   5

v't = acceleration at time T = F/mt   6   (= 5 /4)

◆ ◆ ◆ ◆ ◆ ◆ ◆ ◆ ◆ ◆ ◆

The rocket is arbitrarily given three properties, and the rest is calculated. Its mass is 20kg; its rate of mass loss is 0.01kg/sec; and the separation velocity of the exhaust is 1000m/sec. Added is a payload of 1kg or 4kg.

The only complexities are derivation of formulas for velocity of rocket, time of burn for rocket and fraction of time for payload.

Force is based upon mass times acceleration (F=ma), which is the same as rate of mass loss (m' = 0.01kg/sec) times separation velocity of exhaust (ve = 103m/sec) (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 as given by the Jet Propulsion laboratory.

In other words, acceleration equals velocity over time, so F=ma=mv/t. Rocket experts start with mass divided by time and multiply times velocity of exhaust to get force. I state a uniform rate of mass loss (m') as 0.01kg/sec and separation velocity of exhaust as 1000m/sec. So force equals m've — which is (0.01kg/sec)(1000m/sec) = 10 newtons of force.

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

Four Tests

Force Times Distance Test:
A) 4 kg object dropped 1 meter
B) 1 kg object dropped 4 meters

Force Times Time Test:
C) 4 kg object dropped 1 second
D) 1 kg object dropped 4 seconds

Time Dropping:
A) t = 0.45152 seconds
    t = [2(1m)/9.81]½
B) t = 0.90304 seconds
    t = [2(4m)/9.81]½
C) t = 1 second
D) t = 4 seconds

Calculations:

Velocity dropping = acceleration x time, v=at, v=gt

Acceleration of gravity = g = 9.81 m/s (rounded throughout)

Distance (s) = ½at² = ½gt²

Time (t) = (2s/g)½

Velocity Dropping:

Velocity = acceleration x time, v=at, v=gt

A) v = 4.4294 m/sec
    v = 9.81 m/sec x (0.45152 sec)

B) v = 8.8588 m/sec
    v = 9.81 m/sec x (0.90304 sec)

C) v = 9.8100 m/sec
    v = 9.81 m/sec x (1 sec)

D) v = 39.240 m/sec
    v = 9.81 m/sec x (4 sec)
 

Velocity of Rocket

Needed for factoring burn time.

Velocity is the anti-derivative of acceleration (a).

Acceleration (v') equals force over changing mass (a=F/m').

  v't = acceleration at time T = F/mt   6 = 5 / 4

Acceleration which changes with time (v't) is F/mt

     F = 10N   5

mt = (mo -0.01t)   4

a (6) = 10N / (mo - 0.01t)   (5 / 4)

Velocity is the antiderivative of a, 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

   v = -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.

Time of Rocket Burn

   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 in v 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 sequential entries on a calculator, it looks like this:

   t = mo / 0.01 - [(c-v)/103]ex
 

Constant C in burn time:

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

   -c = velocity at t=0

   c = 103ln(100mo)

   mo = 20kg
      1kg payload or 4kg payload.

A) c = 7783.2
    c = 103ln[100(24 kg)]

B) c = 7649.6 is.
    c = 103ln[100(21 kg)]

C) c = 7783.2
    c = 103ln[100(24 kg)]

D) c = 7649.6
    c = 103ln[100(21 kg)]
 

Total Time of Burn:

   t = mo / 0.01 - [(c-v)/103]ex

   (ex is inverse natural log for preceding quantity.)

A) t = 10.607 sec
    t = 24kg/0.01 - [(7783.2-4.4294)/103]ex

B) t = 18.521 sec
    t = 21kg/0.01 - [(7649.6-8.8588)/103]ex

C) t = 23.428 sec
    t = 24kg/0.01 - [(7783.2-9.81)/103]ex

D) t = 80.808 sec
    t = 21kg/0.01 - [(7649.6-39.24)/103]ex
 

Fraction for payload:

Payload proportionality equals payload mass (mp) divided by total mass, which is decreasing at the rate of mass loss (0.01 kg/sec).

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:

   ∫ mp / (mo - 0.01t) dt =

   ∫ 100mp / (100mo -t) 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.

mp = mass of payload = 1 kg or 4 kg

tp = time for payload including mass ratio integrated with time

It equals test mass divided by average total mass (loses 0.01kg/sec).

The mass ratio integrated with time is this:

   ∫ mp / (mo - 0.01t) dt =

   -100mpln(100mo - t) + c

   -c is quantity at t = 0

   c = 100mpln(100mo)

Constant C in Mass Fraction:

A) c = 3113.2
    c = 100(4kg)ln[100(24kg)]

B) c = 764.9
    c = 100(1kg)ln[100(21kg)]

C) c = 3113.2
    c = 100(4kg)ln[100(24kg)]

D) c = 764.9
    c = 100(1kg)ln[100(21kg)]

Time of Burn for Payload Only:

A) t = 1.7717 sec
    t = -100(4kg)ln[100(24kg) - 10.607] + 3113.2

B) t = 0.88588 sec
    t = -100(1kg)ln[100(21kg) - 18.521] + 764.96

C) t = 3.9240 sec
    t = -100(4kg)ln[100(24kg - 23.428] + 3113.2

D) t = 3.9240 sec
    t = -100(1kg)ln[100(21kg - 80.808] + 764.96

Ratios:
A/B – 2.0000
C/D – 1.0000

Top   

    top