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Rocket motion under gravity

  1. Jan 21, 2017 #1
    1. The problem statement, all variables and given/known data
    A rocket with initial mass of m0. The engine that can burn gas at a rate defined by m(t)=m0-αt, and expel gas at speed (relative to the rocket) of u(t)=u0-βt. Here, m0, α, u0, and β are all constants. Assume the lift-off from ground is immediate

    a) The rocket speed v(t)=?

    b) The rocket height (from ground) y(t)=?

    2. Relevant equations
    F = dp/dt

    3. The attempt at a solution

    So F = dp/dt → -mg = u dm/dt + m dv/dt since dm/dt = -α → -g = (-α/m)u + dv/dt

    → ∫(-g + α(u0 - βt)/(m0 - αt) dt) = v

    so Im not really too sure about how to solve the above integral
     
  2. jcsd
  3. Jan 21, 2017 #2

    haruspex

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    You need to be careful using dp/dt =mdv/dt+vdm/dt. That equation is as though the mass can magically change without the total momentum changing. E.g. consider a car moving at speed v, not burning fuel, just leaking it. With no external force, dp/dt =0. Since dm/dt<0, you would conclude that the car must be accelerating. In short, it is only correct in a reference frame where the gained or lost mass now has zero velocity.
     
    Last edited: Jan 21, 2017
  4. Jan 21, 2017 #3
    I don't see what the problem is using dp/dt in the way i have here is?
     
  5. Jan 21, 2017 #4

    haruspex

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    The lost mass, the burnt fuel, does not have zero velocity in the reference frame.

    Edit: Maybe your final equation is ok, though. I haven't checked that. I will now.

    Edit 2: My mistake. I misread this line: u dm/dt + m dv/dt as v dm/dt + m dv/dt.
    Sorry about the confusion.

    To solve the integral, expand the (u0-βt)/(m0-αt) using partial fractions.
     
    Last edited: Jan 21, 2017
  6. Jan 21, 2017 #5
    So then how can i take into account the changing mass? I thought your argument was just about when there is zero external force
     
  7. Jan 21, 2017 #6

    haruspex

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    Please see my later edit.
     
  8. Jan 21, 2017 #7
    Ah
    ah ok, i see. anyways, i was still using the equation without knowing why it was right. So it is valid because "u" is measured relative to the rocket?
     
  9. Jan 21, 2017 #8

    haruspex

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    On this forum, we often see dp/dt = d(mv)/dt = mdv/dt+vdm/dt, using the product rule. That's why I misread what you posted.
    I'm not sure how you figured out to use mdv/dt+udm/dt, where u is the relative velocity of the exhaust. Maybe you did it using the frame of reference of the instantaneous velocity of the rocket.
     
  10. Jan 21, 2017 #9
    I just noticed that you can do the following also
    Fext dt = u dm + m dv, since dm/dt = -α
    -mg dt = -uαdt + m dv
    -g dt + (u/m) α dt = dv
    (-g + (u/m)α)dt = dv. Then using the same relation from before

    (-g + (u/m)α) (-dm/α) = dv
    (g/α)(m-m0) + u ln(m0/m) = v, then using m - m0 = -αt
    -gt + u ln (m0/m) = v
    Isn't this contradictory to my previous result?
     
  11. Jan 21, 2017 #10

    haruspex

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    Doesn't that step treat u as constant?
     
  12. Jan 22, 2017 #11
    It is a constant with respect to m isn't it?
     
  13. Jan 22, 2017 #12

    haruspex

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    No. Both m and u vary over time.
     
  14. Jan 22, 2017 #13
    Not sure if you worked out the whole problem, but if you did, could you take a look at what I got and let me know if it's ok?
     

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  15. Jan 22, 2017 #14

    haruspex

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    Unfortunately your image is sideways, making it hard to read.
    I get the same expression for v, except that you can simplify it by recognising that the two ln() functions are the same, just opposite sign.
    I end up with
    ##y=\frac 12 (\beta-g)t^2-(\beta-\frac{u_0\alpha}{m_0})((1-\frac{\alpha}{m_0}t)\ln(1-\frac{\alpha}{m_0}t)-\frac{\alpha}{m_0}t))##
    I'll leave you to decide whether that matches your result. If it doesn't, try differentiating both to get back to v.
     
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