Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Aerospace Acceleration, velocity and position of a rocket

  1. Feb 19, 2012 #1
    Hi everyone,

    Quick question I may just not be thinking right here but I was trying to find the acceleration, velocity, and position of a rocket as a function of time. I started with acceleration:

    [itex]a =\frac{T}{m-\dot{m}t} - g[/itex] where T is the Thrust, m is the initial mass, mdot is the mass flow rate, and g is gravity. This equation seems to work out with dimensional analysis and logically it seems to make sense, but maybe I'm wrong there. So from there I integrated wrt to time to get the velocity:

    [itex]v = -\frac{T}{\dot{m}} ln(m-\dot{m}t) -gt + v_0 [/itex] Here is where the problem comes in, while I'm pretty sure I did my integration right, the units don't work out properly and velocity doesn't start out at 0 either, unless you set the [itex] v_0 [/itex] term to some value. Finally I tried to get position by integrating v:

    [itex]s = \frac{T}{\dot{m}^2} ((m-\dot{m}t)ln(m-\dot{m}t) - (m-\dot{m}t)) - 0.5gt^2 + v_0t + s_0 [/itex] Again the units don't work out properly.

    I'm just considering where the rocket goes vertical for now, no horizontal components.

    What am I not seeing here? This should be fairly straight forward. Thanks in advance for any help.
  2. jcsd
  3. Feb 20, 2012 #2


    User Avatar
    Science Advisor
    Homework Helper

    Hi Belginator! :smile:
    No, the units are fine.

    ln has no units (like sin) …

    your ln(m-m't) + vo is really ln((m-m't)/(mo)) for some constant mo :wink:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook