Deriving Thrust from Newton II for Rocket Motors

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SUMMARY

The discussion focuses on deriving the thrust produced by a rocket motor using Newton's Second Law. The thrust is defined as the force generated by the rocket exhaust, which ejects material at a constant velocity (u) and a mass ejection rate (a). The correct formulation involves expressing thrust as F = a(vf - vi)/ln(M/(m - at)), where M is the initial mass and m is the mass at time t. Key principles include the application of Newton's laws and the conservation of momentum.

PREREQUISITES
  • Understanding of Newton's Second Law of Motion
  • Familiarity with rocket propulsion principles
  • Knowledge of momentum and mass flow rates
  • Basic calculus for derivatives and logarithmic functions
NEXT STEPS
  • Study the derivation of thrust equations in rocket propulsion
  • Learn about the conservation of momentum in closed systems
  • Explore the implications of mass ejection rates on rocket performance
  • Investigate the application of differential equations in physics
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Students in physics or engineering, aerospace engineers, and anyone interested in understanding the mechanics of rocket propulsion and thrust generation.

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Homework Statement



Derive from Newton II the thrust (the force produced by the rocket exhaust) achieved by a rocket motor which ejects material with constant velocity u and mass ejection rate a.


Homework Equations





The Attempt at a Solution



Not sure...I know how to derive vf = vi + u ln(mi/mf) but that's not what its asking for..

So my attempt

F = mdv/dt

then using what we have for m i come to

F = a(vf-vi)/ln(M/m-at)

is this right?
 
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bon said:

Homework Statement



Derive from Newton II the thrust (the force produced by the rocket exhaust) achieved by a rocket motor which ejects material with constant velocity u and mass ejection rate a.

Homework Equations


The Attempt at a Solution



Not sure...I know how to derive vf = vi + u ln(mi/mf) but that's not what its asking for..

So my attempt

F = mdv/dt

then using what we have for m i come to

F = a(vf-vi)/ln(M/m-at)

is this right?
Newton's second and third laws of motion apply. Rather than F = ma, write the second law as:

\vec{F} = \frac{d\vec{p}}{dt}

The third law for an object on which no external forces act can be written:

\sum \vec{F_i} = \sum {\frac{d\vec{p_i}}{dt} = 0

which in this case reduces to:

\frac{d\vec{p_{rocket}}}{dt} = - \frac{d\vec{p_{gas}}}{dt}

For the gases fired out the end of the rocket, the velocity relative to the rocket is constant. What gives it a non-zero time rate of change of momentum?

AM
 

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