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Variable mass problem -signs

  1. Jan 12, 2014 #1
    I'm having some issues with the rocket equation. I'm deriving the velocity as a function of time for a descending rocket (so the rocket is accelerating upwards in order to slow the descent). The result I should obtain is

    v=v0+gt+uln(mf/mi)

    where mf is the final mass, mi is the initial mass, u is the relative speed between rocket and the fuel it releases, v0 is the the speed at t=0, and v is the speed after a time t.

    Now I can obtain this if:
    I take positive downwards and consider the scene from an inertial frame on, say, the ground. At time t consider a rocket with mass M and speed v downwards. At time t+Δt consider the rocket with mass M-Δm and speed v+Δv downwards. A small piece of fuel has mass Δm and speed u+v downwards. Then let ΔP be the change in momentum, so

    ΔP=MΔv+uΔm.

    Divide through by Δt and let t→0 so

    dP/dt=Mdv/dt+udm/dt.

    dm/dt=-dM/dt so

    dP/dt=Mdv/dt-udM/dt.

    But dP/dt=Mg so this becomes

    Mg=Mdv/dt-udM/dt
    dv/dt=g+(u/M)dM/dt

    Solving by integrating between t=0 when the speed is v0 and the mass mi to a time t when the velocity is v and the mass mf, and we get the desired result.

    However, if I instead say that when the rocket emits the fuel at t+Δt, its velocity becomes v-Δv, then the whole thing messes up... I can't see why this shouldn't be a valid method of solving it, as in reality the rockets velocity really does decrease if I have positive downwards, rather than increase as in my above method.
     
  2. jcsd
  3. Jan 12, 2014 #2

    mfb

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    You have to be consistent with your signs - the change of the velocity, evaluated in a small time step, gives v+Δv afterwards: a positive change (positive Δv) increases the velocity. This Δv turns out to be negative - so what?

    v-Δv would mean "a positive change in velocity decreases the velocity".
     
  4. Jan 12, 2014 #3
    Oh right, so is it a case that my solution would be correct if I used v-Δv provided I interpret it in the context of solving it. So the solution would tell me the speed is increasing, so v-Δv>v, meaning Δv<0. Δv is a small change in the velocity, and because this is negative, the rocket must be slowing. Obviously it doesn't give the required result but is this the correct way of thinking about it?
     
  5. Jan 12, 2014 #4

    mfb

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    You can work with v-Δv if you interpret Δv as "pointing in the opposite direction" and reverse its sign everywhere. That is an easy way to get really confused, so I would avoid that.
     
  6. Jan 12, 2014 #5
    Ok, thanks.

    Similarly I've also been getting confused with the sign of Δm in this derivation - I've seen two derivations, each using either Δm (where Δm is a positive quantity) or -Δm (where Δm is a negative quantity) as the mass of the fuel packet released.

    The first is similar to mine above, so
    At time t, rocket has speed v and mass M. At time t+Δt the rocket has mass M-Δm, speed v+Δv and the fuel packet has mass Δm, and speed v-u with u the relative speed of fuel packet and rocket. This is all according to an observer in an inertial frame on the ground, say.
    Then if we're in outer space, ignoring any gravity or other external forces, the mometum of the system is conserved.
    We have MΔv=uΔm. Divide through by Δt, let Δt→0 and we have Mdv/dt=udm/dt. Then use the fact that dm/dt=-dM/dt (this is the bit that I can't quite seem to fit in with the second derivation), divide through by M and integrate from the initial to final time to get the usual result,
    v2-v1=uln(mi/mf).

    The second derivation is similar, but uses -Δm so
    At time t, rocket has speed v and mass m (this derivation uses m and Δm rather than M and Δm which I think might be significant maybe?). At time t+Δt the rocket has mass m+Δm, speed v+Δv and the fuel packet has mass -Δm, and speed v-u with u the relative speed of fuel packet and rocket. This is all according to an observer in an inertial frame on the ground, say.
    Then if we're in outer space, ignoring any gravity or other external forces, the mometum of the system is conserved.
    Now this gives -mΔv=uΔm. Now in this derivation, we simply do the integration from the initial time to the final time to get the above result.

    However, I'm a little confused about it because in the first derivation they had to use the fact dm/dt=-dM/dt, whilst here we don't relate the mass ejection rate of the fuel to the mass loss rate of the rocket - I guess this must mean they are equal but if they aren't equal above how can they be here (obviously the sign of Δm is key to this but I can't quite see it). Secondly the use of m and Δm and M and Δm - I guess that they could have used M instead of m, then said, dM/dt=dm/dt, but obviously starting with m avoids this. Could you point me the right way? Sorry if this is a little long...
     
  7. Jan 12, 2014 #6

    mfb

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    You have this equation in the first post.

    As far as I can see M and m mean the same thing, they are just different symbols.
    In the same way, you can swap some signs, the basic idea stays the same.
     
  8. Jan 12, 2014 #7
    Hmm I can't see the justifcation for 'swapping signs' though. We use dm/dt=-dM/dt in the first derivation - why don't we need to use this relation in the second derivation?
     
  9. Jan 12, 2014 #8

    mfb

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    I thought it was used in both.
    That's the same thing as the derivatives.
     
  10. Jan 12, 2014 #9
    I struggle to see how that is the same as the derivatives though...
     
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