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Variable acceleration terminal velocity

  1. May 28, 2007 #1
    4 conditions
    - initial velocity = 1000 m/s
    - height from earth = 100 km
    - total mass = 3000 kg
    - rocket capabilities (thrust) = 100 kg/s at MACH 5

    i was told that to use viscous and laminar drag as two options and to keep mass constant (to simplify things a little).

    the space ship is under variable acceleration but constant gravitational pull.

    it doesnt matter if up or down is negative or positive so long as the derivation and integration work.

    *** the question is will the space ship have a safe landing plan? or will it be able to approach the ground with out crashing therefore when displacement = 100 000 m, velocity = 0****

    can some one please help me? any help is greatly appreciated...
     
  2. jcsd
  3. May 28, 2007 #2

    Danger

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    Welcome to PF, Kieran.
    Initial velocity 1,000 m/s on what vector?
     
  4. May 28, 2007 #3
    assume that it is done in the same dimension (same plane)

    i was given

    mg-kv-u=m*a

    where m=mass
    g=gravity (10)
    k=drag coefficient
    u=thrust
    a=acceleration

    hope that helps a bit... if that helps
    i found that when t = 0 v=1000 and therefore i can find k (and i assume that this stays constant)

    i need this expression to be integrated (mg-kv-u=m*a) using either a=dv/dt or a=v*dv/ds

    once this is integrated, we can find what t is equal to at anytime.
    and then get v on one side and then find terminal velocity buy letting t approach infinity which makes e^(?)= 1

    using that i can find the terminal velocity but its all the algebra that is confusing me...

    try not to think about the world spinning around and in three dimensions

    thanks for the reply
     
  5. May 28, 2007 #4
    mg is on the y axis downwards but it has a positive value and up is negative... kv and u are both up on the y axis
     
  6. May 28, 2007 #5

    Danger

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    Good stuff. Unfortunately, I can be of absolutely no help to you. My math ability is limited to occassionally being able to balance my chequebook. I just had to make sure that all parameters were clearly defined so that the real scientists here wouldn't have to ask you themselves. The lack of direction for that initial velocity just struck me as being a stumbling block. :redface:
    Anyhow, someone who can help will probably be along shortly. I'm going to bed.
     
  7. May 28, 2007 #6
    hey thanks for that though... please pass it on to someone that will be able to help me with the algebra

    thanks again bye
     
  8. May 28, 2007 #7

    rcgldr

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    edit - changed speed of sound to 344m/s for 21 degress C, included thrust formula.

    Ok, 100kg of expended fuel is ejected from the rocket engine at Mach 5 every second.

    Thrust = (100 kg x 1720 m / sec^2) = 172,000 Newtons, agreed.

    There's no coefficient of drag or frontal area given, so there's not enough information to figure out the drag data. In real life, because the equation for density of the air is complicated, this problem can't be converted into an equation. Instead, numerical integration is done (in very small increments of time) to predict ballistic paths.

    On a side note:
    Energy gained by the fuel in this case is 1/2 m v^2 so energy would be 1/2 (100 kg) (1720m/s)^2 = 147,920 kilo joules. Since this is done every second, this tranlates into 147,920 kilo watts (about 86.75 times as much power as the previous case).
     
    Last edited: May 29, 2007
  9. May 28, 2007 #8
    you're thinking about it way too literally...

    the thrust is just saying that there is a force up of 172 000 N
    im just a year 12 student in australia... dont expect me to now all this stuff lol...

    all i know is i have to integrate mg-u-kv=m*a
    -m/k ln (mg-u-kv) = t+c which i think is wrong my calculation for terminal velocity is 5 times the initial velocity
    and from that get v on one side and find the terminal velocity when t approaches infinity...

    value of v which i have as
    v=(mg+u-ue^-t/100)/30

    when t-->inifinity
    e^-t/100=1
    therefore
    v=6733.33333 m/s
     
  10. May 28, 2007 #9

    rcgldr

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    Update

    Thrust = (100 kg x 1720 m / sec^2) = 172,000 Newtons, agreed.
     
    Last edited: May 29, 2007
  11. May 28, 2007 #10
    thrust=dm/dt x V0=344x5 x 1000 = 172000N. Im talking about a basic scenario here, the model is very basic and alot of assumptions need to be made. K is a constant, m is constant and g is constant. Density at 100km about sea level is negligable and MACH5 is constant no matter what altitude.
     
  12. May 29, 2007 #11

    rcgldr

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    > "(thrust)" = 100kg / sec at Mach 5

    dm/dt = 100kg / sec. Speed is 344x5 m/s if temperature is 21C.

    Thrust = (100 kg x 1720 m / sec^2) = 172,000 Newtons, agreed.

    (I was confused when you wrote 344x5x1000 verus 344x5x100)...

    > mach 5 is a constant

    It becomes slower as altitude increases due to decrease in temperature, at altitude, temperature can be -55 C, and speed of sound 295 m/s. Temperature does start increasing again at very high altitudes though. You could have just stated the the exhaust velocity of the gasses is 1720 meters / second and dropped the reference to mach.

    > the space ship is under variable acceleration but constant gravitational pull.

    What are the factors for the variation in accleration? You mentioned that a reduction in mass was going to be ignored (although since fuel to total weight for a typical rocket is around 91%, I'm not sure why this is being ignored). Are you taking into account that the pull of gravity will decrease with altitude? Are you taking into account that aerodynamic drag will decrease with altitude (assuming you're planing on landing the rocket on the earth).

    > will it be able to approach the ground with out crashing therefore when initial conditions are altitude = 100km, velocity = 0.

    Well if it works with no atmosphere, it will work with atmosphere. For the no atmoshpere case, it's probably easier to calculate if the rocket could launch, spend all it's fuel, and reach 100km (note the rocket continues upwards after fuel runs out).
     
    Last edited: May 29, 2007
  13. May 29, 2007 #12
    Despite the open ended question you are really limited in the way you can go by a solution. Take into account the air resistance, acceleration due to gravity and the thrust. Then break that into variable and constant mass. I can't do the maths for you though :)
     
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