Rocket Acceleration in a Uniform Gravitational Field: Analyzing Momentum Change

Click For Summary
SUMMARY

The discussion centers on the dynamics of a rocket accelerating in a uniform gravitational field, specifically analyzing momentum changes. The equation derived is mdv/dt = kv(ext) - mg, where m(t) is the mass of the rocket, k is the constant fuel consumption rate, and vex is the exhaust speed. The participants confirm that if k is very large, the results align with previous class findings that disregard gravity. The focus is on deriving the speed v as a function of the remaining mass m.

PREREQUISITES
  • Understanding of classical mechanics, particularly Newton's laws of motion.
  • Familiarity with the concepts of momentum and mass flow rate.
  • Knowledge of calculus, specifically integration techniques.
  • Basic principles of rocket propulsion and exhaust velocity.
NEXT STEPS
  • Study the derivation of the Tsiolkovsky rocket equation for a deeper understanding of rocket motion.
  • Explore the implications of variable mass systems in classical mechanics.
  • Learn about the effects of gravitational forces on rocket trajectories.
  • Investigate the relationship between exhaust velocity and specific impulse in rocket design.
USEFUL FOR

Students and professionals in aerospace engineering, physics enthusiasts, and anyone interested in the principles of rocket dynamics and propulsion systems.

Gogsey
Messages
153
Reaction score
0
A rocket is accelerating upwards from rest in a uniform gravitational field g. Notation: m(t) is the mass of the rocket plus remaining fuel, m0 is the initial total mass, vex is the exhaust speed (relative to the rocket), and k is the rate, in kg/s, at which fuel is consumed. By considering momentum changes in a short time dt, show that mdv/dt = kv(ext) - mg(upward direction is positive). Assuming k is constant, find the speed v as a function of the
remaining mass m. Show that, if k is very large, it agrees with the result derived in class (without gravity).

Ok, so i have already proved the first part.

P(t)=mv and P(t+dt) = (m+dm)(v+dv)-dM(v-v(ext))

=mv+vdm+dmv(ext)

The a couple of steps later we get: -mgdt=mdv+(ext)(dm/dt)

so we get -mg=mdv/dt -kv(ext), where -k=dm/dt

and therefore mdv/dt=kv(ext) - mg.

Ok, so I'm a little stuck on the second part. Not really sure where to start off. Any hints?
 
Physics news on Phys.org
Assuming that k=dm/dt is constant isolate dv then integrate. You have to assume the exhaust velocity, v(ext) is constant.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
4K
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
2K
Replies
6
Views
10K
  • · Replies 42 ·
2
Replies
42
Views
6K