1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A rocket accelerating upwards

  1. Nov 30, 2012 #1
    1. The problem statement, all variables and given/known data

    A rocket is accelerating vertically, and it is ejecting gas downwards (which allows it to move upwards) at a rate of B. The gas is ejected at a velocity w relative to the rocket.
    What is the velocity and acceleration of the rocket as a function of time?

    2. Relevant equations



    3. The attempt at a solution

    My teacher arrived at a slightly different solution than mine, so I was hoping you could tell me what I am doing wrong. Maybe mixing up the directions somehow?

    This is my solution:
    We should look at a small section of time, from t to t+dt.
    So ∫ Fexternal dt = P[t+dt]-P(t), and since the time is really short, Fexternal is basically constant, so we can take him out of the integral:

    Fexternal * ∫dt = P[t+dt]-P[t]. The limits of integration are t to t+dt so:
    Fexternal*(t+dt-t) = P[t+dt]-P[t].
    Fexternal*dt = P[t+dt]-P[t].

    Vg will signify the velocity of gas, and V will signify the velocity of the rocket. So Vg[t]=V[t]-w.

    The momentum of the gas at time t is Bt*Vg[t], and at time t+dt is Bt*Vg[t]+B*dt*Vg[t+dt].
    The momentum of the rocket at time t is m[t]v[t] and at time t+dt is m[t+dt]v[t+dt].

    So:
    Fexternal*dt = (m[t+dt]v[t+dt]+Bt*Vg[t]+B*dt*Vg[t+dt])-(Bt*Vg[t]+m[t]v[t]).
    Fexternal*dt = m[t+dt]v[t+dt]+Bt*Vg[t]+B*dt*Vg[t+dt]-Bt*Vg[t]-m[t]v[t].
    Fexternal*dt = m[t+dt]v[t+dt]+B*dt*Vg[t+dt]-m[t]v[t].

    m[t+dt]=m[t]-Bdt, and Vg[t+dt]=V[t+dt]-w, so:

    Fexternal*dt = (m[t]-Bdt)v[t+dt]+B*dt*(v[t+dt]-w)-m[t]v[t].
    Dividing by dt:
    Fexternal = (m[t]/dt - B)v[t+dt]+B*(v[t+dt]-w)-m[t]v[t]/dt.
    Fexternal = v[t+dt]*m[t]/dt -B*v[t+dt]+B*V[t+dt]-Bw-m[t]v[t]/dt.
    Fexternal = (m[t]/dt) * (v[t+dt] - v[t]) -Bw
    Fexternal = (m[t]/dt) * dv[t] -Bw
    Fexternal+Bw = (m[t]/dt) * dv[t]
    (Fexternal+Bw)/m[t] = dv[t]/dt
    (Fexternal+Bw)/m[t] = dv[t]/dt

    Since the only external force acting on the rocket is mg downwards, then:
    (-m[t]g+Bw)/m[t] = dv[t]/dt
    -g+Bw/m[t] = dv[t]/dt

    a[t]=-g+Bw/(m[0]-Bt)

    The positive direction is upwards, so w is a negative. which means the acceleration will always be downwards...

    What am I doing wrong?
     
  2. jcsd
  3. Nov 30, 2012 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    You already took into account the direction of w when you wrote the minus sign in front of w for the velocity of the gas. So, w is a postive number.
     
    Last edited: Nov 30, 2012
  4. Dec 1, 2012 #3
    Of course... thank you!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook