AcademyMomentum Systems
Academy
Rocket Motion
Level 1 - Physics topic page in Momentum Systems.
Principle
Rocket speed changes because expelled mass carries momentum opposite the rocket's gain.
Notation
\(m\)
instantaneous rocket mass
\(\mathrm{kg}\)
\(v\)
rocket speed
\(\mathrm{m\,s^{-1}}\)
\(v_e\)
exhaust speed relative to rocket
\(\mathrm{m\,s^{-1}}\)
\(\alpha\)
positive mass flow rate
\(\mathrm{kg\,s^{-1}}\)
\(T\)
thrust
\(\mathrm{N}\)
\(F_{\mathrm{ext}}\)
external force along motion
\(\mathrm{N}\)
Method
With no external impulse, the rocket-exhaust system keeps total momentum while the rocket mass decreases.
Mass sign
\[dm<0,\qquad \alpha=-\frac{dm}{dt}>0\]
Momentum balance
\[m\,dv=-v_e\,dm\]
The expelled mass has velocity opposite the rocket's speed gain in the rocket frame.
Integrate mass
\[\Delta v=v_e\int_{m_f}^{m_0}\frac{dm}{m}=v_e\ln\left(\frac{m_0}{m_f}\right)\]
Include forces
\[m\frac{dv}{dt}=\alpha v_e+F_{\mathrm{ext}}\]
The free-body diagram separates thrust from external forces such as weight.
In vertical flight, gravity enters through \(F_\{\\mathrm\{ext\}}\); it is not part of the ideal rocket equation.
Rules
These are the compact ideal and forced rocket relations.
Thrust
\[T=\alpha v_e\]
Rocket equation
\[\Delta v=v_e\ln\left(\frac{m_0}{m_f}\right)\]
With external force
\[m\frac{dv}{dt}=T+F_{\mathrm{ext}}\]
Mass ratio
\[\frac{m_0}{m_f}=e^{\Delta v/v_e}\]
Examples
Question
A deep-space rocket has
\[v_e=3000\,\mathrm{m\,s^{-1}}\]
and mass ratio \[m_0/m_f=2.5\]
Find ideal \[\Delta v\]
Answer
\[\Delta v=3000\ln2.5=2.75\times10^3\,\mathrm{m\,s^{-1}}\]
Checks
- Rocket mass decreases during burn.
- Exhaust speed is measured relative to the rocket.
- Constant exhaust speed gives a logarithmic speed gain.
- Gravity or drag requires the external-force term.