A rocket accelerating upwards

1. Nov 30, 2012

sapsapz

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. Nov 30, 2012

TSny

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
3. Dec 1, 2012

sapsapz

Of course... thank you!