How Does Rocket Ejection Rate Affect Its Acceleration?

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SUMMARY

The discussion centers on the relationship between rocket ejection rate and its acceleration, specifically analyzing the equations governing the rocket's motion. The participant's calculations involve external forces, momentum, and the effects of gas ejection velocity (w) on the rocket's acceleration. The correct formulation indicates that the acceleration of the rocket is given by a[t] = -g + Bw/(m[0] - Bt), where g is the gravitational acceleration, B is the ejection rate, and m[0] is the initial mass. A critical point clarified in the discussion is the treatment of the gas ejection velocity (w) as a positive value, impacting the overall direction of acceleration.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with momentum conservation principles
  • Knowledge of calculus, specifically integration and differentiation
  • Basic concepts of rocket propulsion and thrust dynamics
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sapsapz
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Homework Statement



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?

Homework Equations


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

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?

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 positive number.
 
Last edited:
Of course... thank you!
 

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