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Speed of the rocket

  1. Oct 28, 2016 #1
    • Poster is reminded to show their Attempt at a Solution in schoolwork threads
    1. The problem statement, all variables and given/known data
    Hi, I would like to apologie for my bad english, but I will try to write my problem as good as I can.
    I have a homework...


    From one stage rocket with initinial speed [m[/0] Ciolky´s number C are escaping gasses with the speed u.
    Let´s assume that weight of the rocket is changing by the formula m=m0*e^(-kt).
    Calculate the speed of the rocket in general distance r from the middle of the Earth. Gravity acceleration is changing by the formula a=gR^2/r^2

    2. Relevant equations
    To be honest, i really have no idea where to start, and I would really apreciate your help.
    3. The attempt at a solution
     
    Last edited by a moderator: Oct 28, 2016
  2. jcsd
  3. Oct 28, 2016 #2

    haruspex

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    I assume this is a reference to Ciolkowski, but I am unable to find a definition of a number named after him. Is it perhaps the log of the mass ratio, initial to final?
     
  4. Oct 28, 2016 #3
    Yes, it is. The result should be

    Bez_n_zvu.png
    But I have no idea how to get that. I tried it through the Ciolkowski formula, but ..I just do not know..
     
  5. Oct 28, 2016 #4

    haruspex

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    Ok. But there seems to be something missing here:
    Should it say "initial speed u0, initial mass m0"?
    When at distance r, what are the forces acting on it? What is the net force? Write some differential equations for distance and mass.
     
  6. Oct 28, 2016 #5
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    Yes, my bad, initial mass, sorry.

    The forces acting on it? Gravity force, I guess. ..I am not pretty sure what do you mean by the "net force". And some differential equations..

    I got through d/dt=-km0e^(-kt) to m(dv/dt)=-mg+kmu, but, i am lagged now.
     
  7. Oct 28, 2016 #6

    haruspex

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    There is a gravitational force and a thrust from the engine. The vector sum of these is the net force.
    how do you get kmu?
     
  8. Oct 28, 2016 #7
    Maybe I am totally already useless today, but I got similiar example in my study textbook (its from my university, and in Czech language, but i try to post it)
    Bez_n_zvu.png

    so I tried to solve it in this way, but I am stuck. Hope you will understant that, its just series of the formulas, and the "v" should be the result. But when I try to make my example by that way, i get nowhere. In my example is just lambda=k, and vr=u.
     
  9. Oct 28, 2016 #8

    haruspex

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    Ok, I just wanted to see how you got m(dv/dt)=-mg+kmu.
    Now, that g is the gravitational acceleration at radius r, so should really be written gr to distinguish it from g at Earth's surface. Rewrite it in terms of the surface g, Earth's radius R, an the rocket's radius r.
     
  10. Oct 28, 2016 #9
    I really appreciate that you help me, but I dont understand you now. I can rewrite ag as ag=H(Mz)/R^2 ...where H is gravity constant and Mz is weight of the earth.

    I understand why result looks like it looks. Except the "2".I have no idea where I got it from. But I do not know how to get that result, or from what.
     
    Last edited: Oct 28, 2016
  11. Oct 28, 2016 #10

    haruspex

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    No.
    You wrote that the gravitational force on the rocket is mg. What did you mean by g there? If you meant the gravitational acceleration at Earth's surface then that is wrong. The force will diminish as r increases. What will it be in general?
     
  12. Oct 28, 2016 #11
    Here is 2am, and I am mad at myself, sorry. I know it will diminish as r increase. I can write is an (r-R), assuming R is the middle of Earth. Do you mean this?
     
  13. Oct 28, 2016 #12

    haruspex

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    No. For a uniform sphere mass M radius R, at a distance r>R from its centre the gravitational field is as though it were a point mass. What is the field at distance r, according to Newton?
     
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