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Questioning logic used for highschool rocket propulsion problem

  1. Dec 3, 2003 #1
    Standard rocket propulsion problem. We are supposed to derive an expression for the final velocity of a rocket being launched into space based upon the exhuast velocity (Ve), final mass (Mf), intial mass (Mo), g, and time. The only forces acting on the rocket throughout the motion is the force of thrust and gravity, with the force of gravity being constant.

    Now this is what my teacher has taught me, and this is the explanation that I have recieved on many websites. The derivation for the constant force of thrust is:

    Fthrust = Ve(dm/dt)

    This I understand. But then apparently when doing a summation of forces on the rocket it should be as follows:

    [tex]F_{thrust}[/tex] - [tex]F_{g}[/tex] = ma

    What I do not understand is why the summation of the forces can be set equal to ma. Obviously the velocity is increasing, so the rocket is experiencing an acceleration, but isn't the mass of the rocket itself changing? In this case the summation of forces should be:
    [tex]F_{thrust}[/tex] - [tex]F_{g}[/tex] = d(mv)/dt
    My physics teacher understands the point I am trying to make, and says he does not know for sure why the changing mass is disregarded. I have the feeling that my challenge is wrong, but I am not sure why. Can anyone help me with this problem?
  2. jcsd
  3. Dec 3, 2003 #2


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    Mass is a function of time. You're looking at a differential equation:

    [tex]v_e \frac{dm(t)}{dt} - m(t) g = m(t) a(t)[/tex]

    - Warren
  4. Dec 3, 2003 #3


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    Depending on how detailed you're getting, this is far from a standard problem.

    OK. That's right for the level of detail you're looking at.

    That's correct for any single point in time. You're right about the mass changing. You can't just plug and chug for a single acceleration value. This is why many launch vehicles throttle their engines during launch to keep acceleration levels reasonable on the craft.

    Warren's equation is correct here. I don't know if you're familiar with differential equations or not.

    You'll need to use the ideal rocket equation for this one (which is derived from Warren's formula).

    [tex] \Delta V = - V_e \ln{\frac{M_{final}}{M_{initial}}} [/tex]

    On top of that, you'll need to figure out how much delta V is given by gravity.

    Now, note that just shooting straight up isn't what really happens, and that makes the problem seriously non-trivial. A rocket in orbit is going sideways relative to the surface of the Earth, so the gravity and the thrust are not acting in opposite directions throughout the ascent.

    I don't know how (or if) you want to handle this.
    Last edited: Dec 3, 2003
  5. Dec 3, 2003 #4
    The problem itself is a ideal rocket problem that would use the equation that enigma stated. I guess my problem was trying to look beyond the equation towards the derivation that come's from Warren's equation which ( I assume ) would have answered my question if I fully understood it.

    Thank you for the quick reply... Now I've just got to let this stew in my head for a little bit.

    *breaks out dad's college calc book and starts trying to learn differential equations*
  6. Dec 4, 2003 #5


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    Here's how the derivation goes. I hope this'll make sense to you.

    Ignoring the gravity term:

    [tex]v_e \frac{dm}{dt} = m a[/tex]

    Now, since for your acceleration to be up, the exit velocity of the fuel is down, so you need a (-) sign in there. Dividing m to the other side to isolate a:

    [tex] - V_e \frac{1}{m} \frac{dm}{dt} = a[/tex]

    Now you want to integrate this from two points in time.

    An integral is the area under the curve. To give you an example you can relate to, if you have a constant acceleration and a set time, you multiply the two together giving you a 'square' of size a*t, and units m/s^2 * s = m/s.

    The integral just does this including non-constant values. A definite integral looks at the ending conditions and subtracts the initial conditions from the final conditions - effectively looking at the difference.

    Also, the integral of 1/x is ln(x), the integral of acceleration is velocity (acceleration is the rate of change of velocity), and constants can be pulled out of the integral, so:

    [tex]\int_{ti}^{tf} - V_e \frac{1}{m} \frac{dm}{dt} dt= \int_{t1}^{tf} a dt[/tex]

    [tex]- V_e \int_{ti}^{tf} \frac{1}{m} dm= \int_{t1}^{tf} a dt[/tex]

    [tex] -V_e * (\ln{m_f} - \ln{m_i}) = V_f - V_i [/tex]

    [tex] -V_e * \ln{\frac{m_f}{m_i}} = \Delta V [/tex]

    Hopefully you can understand what all that means, not having taken calculus. If you have any questions, just ask.
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