Derivation of the Rocket Propulsion Equation
Physics Solver
Objective
To derive the rocket propulsion equation using the principle of conservation of momentum,
with a focus on the role of relative velocity.
Assumptions
• The rocket is moving vertically in a vacuum (1D motion).
• No external horizontal forces (neglect air resistance).
• The only external force in the vertical direction is gravity.
• Exhaust gas is ejected at constant velocity relative to the rocket, vex .
Step-by-Step Derivation
Let:
• m: Mass of the rocket at time t
• v: Velocity of the rocket at time t (relative to Earth)
• ∆m: Mass of exhaust ejected in small time ∆t
• vex : Exhaust velocity relative to the rocket
Initial Momentum (before ∆t)
pinitial = mv
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Final Momentum (after ejecting ∆m)
• Rocket now has mass m − ∆m
• Rocket’s velocity increases to v + ∆v
• The exhaust mass ∆m is moving at velocity:
vgas = v − vex
(relative to Earth)
Total final momentum:
pfinal = (m − ∆m)(v + ∆v) + ∆m(v − vex )
Apply Conservation of Momentum
Assuming no external force (ignoring gravity for now):
pfinal = pinitial
Substitute and expand:
(m − ∆m)(v + ∆v) + ∆m(v − vex ) = mv
Expanding:
mv + m∆v − ∆mv − ∆m∆v + ∆mv − ∆mvex = mv
Simplify:
m∆v − ∆mvex = 0 ⇒ m∆v = ∆mvex
Divide by ∆t and Take the Limit
m
dv
∆m
dm
∆v
= vex
⇒ m = vex
∆t
∆t
dt
dt
Solve for Acceleration
dv
vex dm
=
dt
m dt
This gives the acceleration due to thrust. Since the rocket is moving vertically, we now
include the external force of gravity:
a=
a=
vex dm
−g
m dt
Conclusion
This equation shows that a rocket accelerates due to the momentum it gains from ejecting
mass backward (relative to itself). The term vex must be interpreted as the relative velocity
of exhaust gas with respect to the rocket.
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