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Homework Help: Can a rocket go faster than it's terminal velocity?

  1. Nov 25, 2007 #1
    1. The problem statement, all variables and given/known data

    u ---> gas speed (relative to the rocket)

    v ---> rocket speed (relative to the stars)

    v - u ---> speed of gas relative to the stars

    [tex] v - v_0 = u ln \frac{M_0}{M} [/tex]

    Where M0 is the initial mass of the rocket, and M is the final mass.

    Part a) If a rocket, initially at rest, is to attain a terminal velocity (relative to the stars and when the fuel is all burnt up) of a magnitude that happens to equal the magnitude of the exhaust velocity (which is fixed relative to the rocket), what fraction of the initial mass must be fuel? Hint: Examine the situation when we have reached v = u.

    Part b) Can a rocket go faster than this terminal velocity? What is its limit?

    Part c ) Describe the motion of the overall original system CM, starting from the initial moment (when the rocket first began to blast off) to the attained final velocity in part a.

    2. Relevant equations

    Given above.

    3. The attempt at a solution

    Part a) The inital speed is zero, and u equals the final speed so:

    [tex] v - 0 = v ln \frac{M_0}{M} [/tex]

    [tex] 1 = ln \frac{M_0}{M} [/tex]

    Take the exponential of boths sides:

    [tex] e = \frac{M_0}{M} [/tex]

    I'm not sure if the question is asking for the ratio of the inital mass to the final mass, or if it's asking what the inital mass must be in terms of the final mass...

    Part b) The hint given is: What we mean is given that you had no
    constraint on the fraction of the rocket that can be used for fuel, is there
    an upper limit on the final speed you can the rocket up to?

    I have no idea on how to find this. I haven't tried to tackle part c yet because I don't have a handle on parts a and b.

    Any help is appreciated.
  2. jcsd
  3. Nov 25, 2007 #2

    D H

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    Neither of those.

    Your lack of understanding lies in not delving into the meaning of these two masses. You are given that all of the fuel has been exhausted at the end. The final mass [itex]M=M_v[/itex] is the mass of the empty vehicle and the payload. At the start, the rocket was loaded with fuel. The initial mass is just the final mass plus the initial fuel load, [itex]M_0=M_v+M_{f_\mathrm{init}}[/itex].

    For part a, you are asked "what fraction of the initial mass must be fuel". If you represent the initial mass as I suggested you should be able to answer the question at hand.

    What does the hint mean in terms of [itex]M_{f_\mathrm{init}}[/itex]? What does that mean in terms of the final velocity?
  4. Nov 25, 2007 #3
    Thank you for the reply.

    If I understand what you're saying, I'd have:

    [tex] M_0 = M_{final} + M_{fuel}, M_{fuel} = M_0 - M_{final} [/tex]

    [tex] \frac{M_{final} + M_{fuel}}{M_{final}} = e [/tex]

    [tex] M_{fuel} = eM_{final} - M_{final} = M_{final}(e - 1) [/tex]

    Do I follow correctly?

    I assume the hint is implying what would happen if the mass of the fuel approached infinity, but wouldn't that just give infinity for the final velocity?
  5. Nov 25, 2007 #4

    D H

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    You still are not answering the question at hand: "what fraction of the initial mass must be fuel?" In other words, what is [itex]M_{fuel}/M_0[/itex]?

    That's correct. So what does that say regarding a limit on the rocket's speed?
  6. Nov 25, 2007 #5
    Ah I see, ok so I'd have:

    [tex] \frac{M_{fuel}}{M_0} = \frac{M_0 - M_{fuel}}{M_0} = \frac{M_0 - M_{final}(e - 1)}{M_0} [/tex]

    I believe it means if the mass of the fuel goes to infinity, the rocket does not have a limit on speed, and it has a terminal velocity that approaches infinity.

    Also for part c, is it reasonable to assume that since the momentum of the rocket system (rocket, fuel, and exhaust) is conserved, that the velocity of the center of mass is zero?

    Thanks for your help.
  7. Nov 25, 2007 #6

    D H

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    You have not answered the question yet. Expand [itex]M_0[/itex]. You should get a nice, simple answer.
  8. Nov 26, 2007 #7
    Hmmm I don't know where I got the equation above, I must have been a bit tired last night. (Edit: I do know, I tried to solve for [tex] M_{final} / M_0 [/tex] instead of [tex] M_{fuel} / M_0 [/tex]).This is my reasoning:

    I first solve for [tex] M_{fuel} [/tex]:

    [tex] \frac{M_{final} + M_{fuel}}{M_{final}} = e [/tex]

    [tex] M_{fuel} = M_{final}(e - 1) [/tex]

    I'm looking for [tex] M_{fuel} / M_0 [/tex], so I have:

    [tex] \frac{M_{fuel}}{M_0} = \frac{M_{final}(e - 1)}{M_0} [/tex]

    Do I follow correctly?
  9. Nov 26, 2007 #8

    D H

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    You need to completely eliminate the masses in that ratio. The final answer should be nice and simple.
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