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On Rocket Motion

  1. Aug 16, 2010 #1
    In the consideration of rocket motion we always use the principle of the Conservation of Linear Momentum.This principle holds true only for isolated systems.But the initial stage of rocket motion is strongly affected by gravity-----which is definitely an external force [if we consider the rocket and the ejected fuel as the "system",as it is normally done in most typical calculations].
    Do we expect to get accurate results after all this?
     
  2. jcsd
  3. Aug 16, 2010 #2

    russ_watters

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    Sure - all you have to do is subtract 1g from your acceleration.
     
  4. Aug 18, 2010 #3
    It is true that from the theoretical stand point it would be incorrect to apply the principle of conservation of momentum.In such a case[ if we apply the above stated principle] we should either make an estimate of the amount of error we incur or we induce corrections to take care of the error . A third option would be to try out the calculations without using the conservation of momentum principle right from the beginning.

    Let me try out the third option
    Let F(th) be the thrust on the rocket due to fuel ejection.
    Therefore,
    F(th) – mg=(d/dt)(mv)
    F(th) – mg=m(dv/dt) + v(dm/dt)

    We assume dm/dt=constant=-k,k>0
    That is, m=m0-kt, where m0 is the initial mass of the rocket.
    We now have,
    F(th)-(m0-kt)g=(m0-kt) dv/dt – kv
    dv/dt-k/(m0-kt) v=[F(th)-m0g+kgt]/(m0-kt)

    We pose the question ---"How would the velocity change if the thrust were to remain constant? The thrust would indeed remain constant if the rate of ejection(dm/dt) and the relative speed of ejection remain constant.

    The linear equation yields the solution:
    V=[2[F(th)-m0g]t+kgt^2]/2(m0-kt)

    dv/dt=[2m0[F(Th)-m0g]+ktg[2m0-kt]]/2[m0-kt]^2

    It is to be noted m0-kt>0
    and(consequently) 2m0-kt>0
    For the acceleration to be positive at the initial instant we should have,

    F(Th)>m0g
    This is quite natural.

    The thrust produced by the ejection[at the initial moment] should be greater than the initial mass of the rocket to produce an acceleration.

    With the increase in time the acceleration should increase(provided the thrust ie,F(Th) remains constant).
    Reason:
    1)[2[F(th)-m0g]/[m0-kt]^2 increases with time.
    2)kgt[2m0-kt]/[m0-kt]^2 is an increasing function of time and initially it is positive.
    This is natural because the mass of the rocket is decreasing but the thrust is remaining constant.And of course we have taken "g" as a constant.
     
    Last edited: Aug 18, 2010
  5. Aug 18, 2010 #4

    Filip Larsen

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    As you may know, the gravitational loss you describe in your post is a major reason for launch vehicles to turn away from the vertical as soon after launch as it is aerodynamically possible.
     
  6. Aug 19, 2010 #5
    We have to be very careful with the following equation in relation to variable mass systems:

    F=d(mv)/dt

    The term on the left side should include external forces(like gravity) and also the "reaction forces" due to the ejected masses. In my calculations[Thread #3 ]I have taken care of the "reaction force " by the term F(th).I have also assumed F(Th) to be constant. For my method to succeed F(Th) has to be kept constant by technological means.
    In a statement in thread [Thread #3]I have said
    "The thrust would indeed remain constant if the rate of ejection(dm/dt) and the relative speed of ejection remain constant."
    But this concept[Thrust=v(rel)dm/dt] is usually derived by applying the Conservation of Momentum Principle.
    To make my ideas logically acceptable I have to modify the statement and rephrase it by saying " We must keep the Thrust(F(Th) somehow by technological means[to consider the calculations]"

    Now to the next point:
    In calculations using the conservation of momentum,if we minus "g" from the expression for acceleration,[as suggested by Russ_Watters] we get in the final expression:
    a=[v(rel)/m dm/dt]-g

    If Thrust=v(rel)dm/dt=constant
    We have,
    a=dv/dt=F(Th)/m-g ,F(th) treated as constant
    a=dv/dt=F(Th)/[m0-kt]-g

    In the last equation I have taken v(rel) and dm/dt constant independently.

    In any case these results are different from what have been obtained in thread #3. Is it important in any way to have an account of such differences?
     
  7. Aug 20, 2010 #6

    russ_watters

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    No, actually, it isn't. The trajectory is due to the fact that the rockets have a lot more speeding up to do than they have altitude to gain.
     
  8. Aug 20, 2010 #7

    D H

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    Yes, actually, it is. The gravity losses over the course of a launch are [itex]\int_{t_0}^{t_f} g\,\sin \gamma \, dt[/itex]. A gravity turn trajectory minimizes that gravity loss. While a vehicle that is launched vertically cannot follow a true gravity turn trajectory, it can come close, particularly if it turns from this vertical orientation very soon after launch.
     
  9. Aug 20, 2010 #8

    Filip Larsen

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    The reason a launcher picks up more speed thrusting away from the vertical is because the gravitational loss then is a lot less or even zero if thrusting is made in the horizontal, as can be seen from its dependency on [itex]\sin\gamma[/itex] in DH's post. For big launchers the fractional gain in acceleration can be considerable. For instance, the space shuttle would right after launch be able to accelerate around 2.8 times more in the horizontal than in the vertical, everything else being equal.

    As I understand it, the gravity turn, apart from being a neat and simple way to tilt a rocket in a controlled fashion, is the "only possible" ascent profile when aerodynamical forces limits the angle of attack to be very close to zero. Without this limit on angle of attack, like for the ascent of the Apollo lunar ascent module, I believe the optimal profile would be somewhat different, that is, optimal thrusting would in this case not necessarily give rise to a gravity turn trajectory.
     
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