Sunday, May 20, 2007

Ref: Momentum Basics

LIST OF TABLES

Momentum Basics

Momentum is mass times velocity. It typically uses symbol: "p".

p = m * v

p = (2.0 kg)(4.0 m/s)

p = 8.0 kg-m/s


Thus, we calculate this object's momentum to be 8.0 kg-m/s. Momentum's dimensions are kg-m/s, pronounced: 'kilogram meter per second'.

Like velocity, momentum is a vector; it has a magnitude as well as a direction. In fact, they both have the same direction. That is, if an object has a velocity due north, then its momentum will also be due north. However, momentum is both mass times velocity.

Momentum is directly proportional to velocity. Change the velocity of an object by a factor of 1/4, then momentum would also change by 1/4.

Momentum is also directly proportional to mass. With constant velocity, momentum varies with the mass. Example: triple the mass of an object, then the momentum also triples.

Momentum is a conserved quantity. Within a closed system of interacting objects, total momentum of entire system maintains a constant. When object A collides with object B in an isolated system, the total momentum of the two objects before the collision is the same as the total momentum after the collision. Any momentum lost by object A goes to object B.

Thus, total momentum of a system of objects is "conserved"; total amount of mass times velocity is static. To further understand momentum conservation, consider Newton's Law of Momentum Conservation. When two objects collide, the forces acting between the two objects are equal in magnitude and opposite in direction. In equation form, m*v = -M * V.

The forces between the two objects are equal in magnitude and opposite in direction, which leads to Newton's Law of Momentum Conservation. In a collision, the momentum change of object A is equal and opposite to the momentum change of object B. That is, the momentum lost by object A is equal to the momentum gained by object B. In a collision between two objects, one object slows down and loses momentum while the other object speeds up and gains momentum. If object A loses 75 units of momentum, then object B gains 75 units of momentum. Yet, the total momentum of the two objects (object A plus object B) is the same before the collision as it is after the collision; the total momentum of the system (the collection of two objects) is conserved.

For any collision occurring in an isolated system, momentum is conserved - the total amount of momentum of the collection of objects in the system is the same before the collision as after the collision.

Momentum can be readily observed by watching billiard balls collide.



Similarly, high velocity gas goes in one direction, and rocket goes in opposite direction. Action causes reaction.

Recall momentum exchange: M * V = m * v

Rocket and jets propel themselves due to momentum transfer where high speed exhaust particles move a much larger vehicle in the opposite direction to a much slower but still useful speed. This momentum exchange between fast moving fuel particles and a slow moving spaceship can be expressed:

Mship * Vship = mfuel * vfuel

First, we choose to increase the speed of a large space vessel by 10 m/s in one second. Thus, set Vship be a constant 10 m/s. This relates to the approximate value of g = 10 m/s2 acceleration due to gravity near Earth’s surface. Thus, assume our notional spaceship can expel enough high speed particles to increase velocity an additional 10 m/s for every second of spaceflight. The resultant force will simulate Earth-like gravity for ship and contents.

2nd, we further choose to expend one gram of particles in one second. I.e., set mfuel to be a constant 1.0 gram for convenience.

3rd, our thot experiment uses an onboard particle accelerator to give these fuel particles extremely high speeds. Since current devices easily accelerate particles to 99% of light speed, our thought experiment will compare sustainable fuel exhaust velocity, vfuel, to light speed, c. Furthermore, we now choose to name vExh to more clearly indicate exhaust velocity of particles exiting vessel. Finally, we choose to independently vary vExh by incrementally increasing from row to row in the following tables. Thus, vExh is our Independent Variable (IV).

4th, solve for the remaining term, Mship, mass of ship, to determine how much ship can be propelled 10 m/s by a certain amount of fuel mass exiting vehicle at high speed in opposite direction of travel. Since Mship depends on the value of vExh, Mship is our Dependent Variable (DV).

Mship = mfuel * vExh / Vship

(Note: Following tables will consider only Newtonian concepts and will disregard relativistic mass increase for now. Relativistic mass growth will be considered later.)



TABLE-1. Consistent units.

Mship = mfuel * vExh / Vship

Mship

Vship

mfuel

vExh

Ship's Mass

ship's vel

fuel mass

exhaust vel

Grams

m/s

grams

m/s

1,000,000

10.0

1.0

10,000,000

2,000,000

10.0

1.0

20,000,000

3,000,000

10.0

1.0

30,000,000

DV

Const.

Const.

IV

Grams and m/s are consistently used on both sides of equation.

While above calculations are straight forward, two terms have a lot of digits which might prove inconvenient. We can streamline this process through a conversion contant.


TABLE-2. Express vExh as dc * c; decimal c times light speed

For convenience, use a mix different velocity units. In particular, we'd like to continue expressing ship's velocity increase ,Vship, as 10 meters per second; but we'd like to change vExh from m/s to decimal light speed, dc c, where dc is decimal component of particle exhaust speed expressed in terms of light speed.

Recall light speed is approximately 300 million meters per second, c = 300,000,000 m/s.
Thus, Table-1 proposes example value for vExh of .1 c, 30,000,000 m/s = 10% c. (Note: Disregard relativity effects for now; we'll discuss that later).

vExh as decimal c

Mship = mfuel * K2 * vExh/ Vship

K1 = 300,000,000

Mship

V ship

m fuel

vExh

Ship's Mass

ship's vel

fuel mass

exhaust vel

Grams

m/s

grams

dc c

3,000,000

10.0

1.0

.1 c

6,000,000

10.0

1.0

.2 c

9,000,000

10.0

1.0

.3 c

12,000,000

10.0

1.0

.4 c

DV

Const.

Const.

IV

Consistent values requires conversion constant. Note that VExh has equivalent values in last row of Table 1. and first row of Table 2. For consistent Mship, determine conversion constant K2 such VExh can be expressed as decimal light speed.

 
 


TABLE-3. Convert Mship to metric Tonnes (mT)

By definition: 1 mT = 1,000,000 gms; thus, example value of MShip=3,000,000 gms = 3 mT.

K3=Mship * Vship
mfuel * vExh
= 3mT * 10m/s
1 gm * .1c
= 3
c
Thus, k3 = 3/c, given following set of dimensions:

  • mfuel as grams (gm)
  • vfuel as dc c
  • Mship as metric Tonnes (mT)
  • Vship as meters per second (m/s)

    Mship = K3 * mfuel * vExh / Vship

    Mship in mT

    K3 = 3/c

    Mship

    Vship

    mfuel

    vExh

    Ship's Mass

    ship's vel

    fuel mass

    exhaust vel

    metric tons

    m/s

    grams

    dc c

    3.0

    10.0

    1.0

    .10 c

    4.5

    10.0

    1.0

    .15 c

    6.0

    10.0

    1.0

    .20 c

    7.5

    10.0

    1.0

    .25 c

    9.0

    10.0

    1.0

    .30 c

    10.5

    10.0

    1.0

    .35 c

    12.5

    10.0

    1.0

    .40 c

    13.5

    10.0

    1.0

    .45 c

    DV

    Const.

    Const.

    IV



    TABLE-4. Convert to Rates

    Changing to mixed units affects the terms: vExh and Mship; thus, unaffected terms (Vship and mfuel) keep the same dimensions. However, those terms (Vship and mfuel) can be usefully converted to rates.

    Start with the original ship-fuel momentum equation:

    Mship * Vship= mfuel * vExh

    and divide both sides by one second:

    Mship * Vship
    1 second
    = mfuel * vExh
    1 second

    Next, we'll arbitrarily rearrange as follows:
    Mship * Vship
    1 sec
    = mfuel
    1 sec
    * vExh

    Since definition of acceleration is velocity increase per time, Vship per second is an acceleration. We've chosen Vship to be 10 m/s, same value as g, acceleration due to Earth's surface gravity. Thus, we can rearrange equation as follows:

    Mship * g = (mfuel/1 sec) * vExh

    Further, we can introduce new term, fuel flow per sec (ffsec), and substitute for the expression: mfuel /sec. Thus, far we've arbitrarily chosen the value of 1 gram per second for convenience, but we will eventually use this term for any value.

    Mship * g = ffsec * vExh




    Mship * 
    Vship
    1 sec
    =300 * vExh
    mfuel
    1 sec
    Using previous logic:

    • restate Vship/sec as g, gravity caused acceleration.
    • restate mfuel/sec as ffsec, fuel flow per second;
    • restate vExh as dcc, decimal light speed.
    • rearrange:

    Mship = k * ffsec * dc c
    g

    Mship = K * ffsec * vExh / g

    K = 30 g/c

    M ship

    g

    ffsec

    vExh

    Ship's Mass

    Gravity

    fuel flow

    exhaust vel

    metric tons

    m/s2

    gm/sec

    dc c

    6.0

    10.0

    1.0

    .20 c

    7.5

    10.0

    1.0

    .25 c

    9.0

    10.0

    1.0

    .30 c

    10.5

    10.0

    1.0

    .35 c

    12.0

    10.0

    1.0

    .40 c

    13.5

    10.0

    1.0

    .45 c

    15.0

    10.0

    1.0

    .50 c

    16.5

    10.0

    1.0

    .55 c

    DV

    Const.

    Const.

    IV

    k =Mship * g

    ffsec * dcc
    = 3 mT * g

    1 gm/sec * .1c
    = 30 g

    c
    Thus, we've taken the Momentum Conservation equation and applied it for a duration of one second. For space travel to become routine, technology will have to accomplish two things:
    1. Micro-scale. Spread the amount of fuel ejected in a smooth continuous manner throughout the entire second.
    2. Macro-scale. Reliably repeat above item for each of the 86,400 seconds during each day of space flight.
    Both above items are easily done now in various modes of transportation, but can they be done for particles accelerated to large portions of c, speed of light???



    Summary

    Consistent Units
    Give terms following values:
    • mfuel = 1.0 gram
    • Vship = 10 meters / second
    • vExh= 30,000,000 m/s
    Then
    Mship = (1.0 gm * 30,000,000 m/s) / (10 m/s)
    Mship= 3,000,000 gm
    Basic Momentum
    M = mfuel * vExh/ Vship

    Mship

    mfuel

    Vship

    vfuel

    gm

    gm

    m/sec

    m/s

    3,000,000

    1.0

    10

    30,000,000



    Mixed Units

    Conversion Constant: 300
    Mship = (300)(mfuel * vExh / Vship)

    Mship

    mfuel

    Vship

    vExh

    mT

    gm

    m/sec

    dc c

    3

    1.0

    10

    .1 c

    6

    1.0

    10

    .2 c

    1,000,000 gms = 1,000 kgms = 1.0 metric Tonnes = 1 mT
    Thus, Mship = 3,000,000 gm = 3.0 mT

    300,000,000 meters /sec = light speed = c
    Thus, VExh= 30,000,000 m/s = .1c

    It’s more convenient to express Mship in mT and vExhin decimal c.

    To do this, introduce a conversion constant, k3, as follows:

    Mship = k3 * mfuel * vExh / Vship
    where k3 = 300

    Rates
    Convert to ffsec
    Fuel Flow/sec
     

    g

    ffsec

    vExh

    Mship

    con

    con

    IV

    DV

    m/s2

    gm/sec

    dc c

    mT

    10

    1

    0.1 c

    3

    10

    1

    0.2 c

    6

    10

    1

    0.3 c

    9

    Particle velocity, vExh, of .1 c contributes sufficient momentum so that one gram of exhaust particles can propel a 3 ton vessel to an additional 10 m/sec.


    Since momentum is directly proportional to velocity, increasing vExh another .1 c will propel another 3mT of ship's mass.

    Recall: mfuel * vExh = Mship * Vship.

    Divide both sides by one sec:mfuel * vExh
    1 sec
    = MShip * vShip
    1 sec
    Rearrange:mfuel
    1 sec
    * vExh = MShip * vShip
    1 sec

    Recall that we set Vship equal to 10 m/s; thus, Vship/ 1sec = 10m/s/s = g, acceleration due to gravity near Earth’s surface.

    Introduce new term, fuel flow per second: ffsec = mfuel/sec.

    Thus, we substitute to get: ffsec* vExh = Mship * g

    To continue expressing vExh as decimal c and Mship in metric Tonnes, we'll introduce conversion constant, k = 300/c. Thus, k * vExh = 300 * dc.

    Mship= 300 * ffsec * dc
    g





    Further Study, Related Concepts: Impulse and Specific Impulse



    ImpulseThe Physics Classroom In physics, the quantity Force*time is known as the impulse. And since the quantity m*v is the momentum, the quantity m*"Delta "v must be the change in ...www.physicsclassroom.com/Class/momentum/U4L1b.html - 48k - Cached - Similar pages Physics - Impulse - Martin Baker Physics - Impulse. Impulse is the integral of force over time, it is measured in Newton-seconds. For instance a force of one Newton applied over one second ...www.euclideanspace.com/physics/dynamics/collision/impulse/index.htm - 26k - Cached - Similar pages SparkNotes: SAT Physics: Impulse SAT II Physics may also present you with a force vs. time graph, and ask you to calculate the impulse. There is a single, simple rule to bear in mind for ...www.sparknotes.com/testprep/books/sat2/physics/chapter9section2.rhtml - 32k - Cached - Similar pages Impulse -- from Eric Weisstein's World of Physics An impulse is an instantaneous change in momentum. which can be found by integrating a force F over a characteristic time , giving ...scienceworld.wolfram.com/physics/Impulse.html - 12k - Cached - Similar pages Impulse In everyday language, an impulse is something you have - "I just had an impulse, so I bought a new coat." In physics, an impulse is something else (you ...www.batesville.k12.in.us/physics/PHYNET/Mechanics/Momentum/impulse.htm - 6k - Cached - Similar pages Momentum Lesson 1: The Impulse Momentum Change Theorem ... Momentum is a physics term; it refers to the quantity of motion that an object has. ...www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/momentum/u4l1a.html - 15k - Cached - Similar pages Physics Zone: Impulse (When Push Comes to Shove) The product of average net force and change in time is called Impulse. The formula for it actually comes from a little manipulation of Newton's Second Law. ...regentsprep.org/Regents/physics/phys01/impulse/default.htm - 3k - Cached - Similar pages Physics Help: Impulse Impulse is change in momentum and results from force acting over a period of time. ... Impulse in Physics Problems. Impulse and Momentum ...www.physics247.com/physics-help/impulse.shtml - 11k - Cached - Similar pages Impulse Physics Hive Assault: Impulse Physics. Dynamic Animation Calculations in Hive Assault Real-time Calculation of Bone Impulse Reactions from Weapons ...www.owlnet.rice.edu/~comp460/PRSite/tech/impulse.htm - 9k - Cached - Similar pages FHSST Physics Momentum:Impulse - Wikibooks, collection of open ... FHSST Physics Momentum:Impulse ... [edit] Worked Example 36 Impulse and Change in momentum. Question: A 150 N resultant force acts on a 300 kg object. ...en.wikibooks.org/wiki/FHSST_Physics_Momentum:Impulse - 28k - Cached - Similar pages



    Specific Impulse
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    Peripheral Energy Requirements
    While this seems like a great way to quickly increase inflight performance, this comes at the cost of enormous energy. It's a fact of nature; it's not necessarily free. Consider a gasoline powered automobile. The fuel converts to energy, but not all the energy rotates the wheels; a significant portion of the energy from the exploding gasoline pushes the pistons and rotates the drive shafe. Eventually, some of the gas's energy results in rotary motion pushing the wheels which touch the road and propel the auto forward. Furthermore, some energy must be diverted for peripheral uses: radio, air conditioning, etc.. However, the essential diversion of energy goes to the auto's generator (or alternator) which converts motion into electricity to the spark plugs which explode subsequent gasoline vapors to produce more motion and so on. We can call this an energy cycle.In like manner, our spaceship's propulsion system must have a similar energy cycle. It must be designed so the system diverts some energy from the accelerator's output to create more plasma and charge the magnets to propel more ions to keep the cycle continuing.Thus far, we've taken the Momentum Conservation equation and applied it for a duration of one second. For space travel to become routine, technology will have to accomplish two things: Micro-scale. Spread the amount of fuel ejected in a smooth continuous manner throughout the entire second. Macro-scale. Reliably repeat above item for each of the 86,400 seconds during each day of space flight. Both above items are easily done now in various modes of transportation, (for example, the typical plane, train, and automobile have done this for many years). However, can micro/macro scale energy conversion be done for exhaust particles accelerated to large portions of c, speed of light??? Note: for current state of nuclear propulsion, see excellent book,Nuclear Space Power and Propulsion Systems (Progress in Astronautics and Aeronautics) by Claudio Bruno(Editor) --------------------------------------------------------------------------------On the other hand, thought experiment further assumes onboard fuel is consumed at fuel's original size. Thus, inflight gross weight of the vehicle will be ever decreasing due to continuous fuel consumption which will be computed by time, t, times original fuel flow, ffsec.

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