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Question about exhaust velocity and terminal velocity

  1. Jul 30, 2008 #1
    I had made a model assuming a relativistic rocket accelerating with constant power,variable acceleration,variable total mass the result however is somehow counterintuitive. The less exhaust velocity the rocket have, the higher terminal velocity it will have.

    Is that means I did something wrong ?
  2. jcsd
  3. Jul 30, 2008 #2


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    Staff: Mentor

    Terminal velocity is the final equilibrium velocity of an object against a drag force, typically in freefall. A rocket in space doesn't have a terminal velocity - it will continue to accelerate as long as it has fuel. The closest thing to it you can get is an abitrarily large rocket, regardless of the specifics of the rocket, will have a maximum velocity that approaches C.

    Are you trying to calculate an acceleration profile in Relativity? My guess is that your error isn't with the relativity part, it's with the rocket exhaust velocity vs acceleration part. Could you give us the equation you are using...?
  4. Jul 30, 2008 #3
    I am assuming you are asking if a rocket with a given total mass and a given fuel payload will reach a higher terminal velocity by the time it runs out of fuel, by using a lower constant exhaust velocity?

    You have to specify what you mean by exhaust velocity. Do you mean proper exhaust velocity as measured by the observer on the rocket or exhaust velocity as measured by an inertial observer? If you mean the latter, you should be aware that the exhaust velocity will be getting less over time due to time dilation as a function of the rockets instantaneous relative velocity. In Newtonian calculations in a gravitational field, burning all the fuel as rapidly as possible is the most effecient way to launch a rocket and achieve the maximum terminal velocity for a given fuel load. It can be seen that the opposite extreme, is burning all the fuel so slowly that the rocket is only hovering and its final velocity when the fuel is exhausted is zero. From the Baez FAQ here http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken] it can be deduced that the final terminal velocity v/c of a photon drive (exhaust velocity=c) rocket is:

    v/c = (F+2m)/(F+2m+2m2/F) where F is fuel load of the rocket and m is the payload of the rocket (total mass of the rocket less the fuel load).

    Generally speaking the higher the exhaust velocity the greater the efficiency of converting fuel to momentum and the higher the terminal velocity of the rocket. You would have to make it clear about exactly what circumstances you are considering.
    Last edited by a moderator: May 3, 2017
  5. Jul 31, 2008 #4
    Yes I used Baez Relativistic Rocket FAQ to make my model.

    I have to modify the equations though, since what I want is a model with :
    - constant wattage
    - constant exhaust velocity relative to the ship (is this the one called proper exhaust velocity)
    - variable proper acceleration
    - variable coordinate acceleration

    The equations used is attached as 2 jpg images :

    Mass Energy Conservation equation :
    PBT*EPM*t = (MShip+MFuel-FBT*t)*c^2/(1-v^2/c^2)^(1/2)-(MShip+MFuel-FBT*t)*c^2+EXE
    inside massenergyconservation.JPG

    Linear Momentum Conservation equation :
    (MShip+MFuel-FBT*t)*v/(1-v^2/c^2)^(1/2) = EXE/EXV
    inside linearmomentumconservation.JPG

    The variables used is as follow :

    PBT : Mass of Fuel burned per unit time
    EPM : Fuel Potential Energy per unit mass
    MShip : Rest Mass of the Ship
    MFuel : Rest Fuel Initial Mass
    EXE : EXhaust Energy
    EXV : EXhaust Velocity

    The complete derivations I used is attached as mws file.

    Thank you.

    Attached Files:

    Last edited by a moderator: May 3, 2017
  6. Jul 31, 2008 #5
    Oh I just remember that. Guess I will have to learn more English Physics Vocabularies.

    Btw what I mean is the highest velocity the rocket would reach after using some part of its fuel to accelerate itself, provided that the rocket should have enough fuel to decelerate and arrive at its destination.

    Yes it is given in the mws file I attached on my other reply above.
  7. Jul 31, 2008 #6
    I think it is inappropriate for me to give the chaotic mws file like the first one I attached. This is the new mws file which have been divided to section. I hope this one is more readable.

    Thank you

    Attached Files:

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