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Deriving the Rocket Equation: Where Thrust Comes From

If thrust is not an applied force at all, where does it come from — and why does it point forward?


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Your summary note

    1. 1

      System choice forcing Δmr=−Δmf\Delta m_r = -\Delta m_fΔmr​=−Δmf​

      Take the system as the rocket's mass at time ttt, equate the system mass at both instants, and write the exhaust velocity u⃗+v⃗r(t+Δt)\vec{u} + \vec{v}_r(t+\Delta t)u+vr​(t+Δt) in the fixed frame.

    2. 2

      F⃗ext=mr(t)dv⃗rdt−dmrdtu⃗\vec{F}_{ext} = m_r(t)\frac{d\vec{v}_r}{dt} - \frac{dm_r}{dt}\vec{u}Fext​=mr​(t)dtdvr​​−dtdmr​​u

      Assemble p⃗sys\vec{p}_{sys}p​sys​ at ttt and t+Δtt+\Delta tt+Δt, take the limit to reach this equation, and write the component form Fext,x=mrdvr,xdt+dmrdtuF_{ext,x} = m_r\frac{dv_{r,x}}{dt} + \frac{dm_r}{dt}uFext,x​=mr​dtdvr,x​​+dtdmr​​u for u⃗=−ui^\vec{u} = -u\hat{i}u=−ui^.

    3. 3

      Thrust Fthrust,x=−dmrdtu=dmfdtuF_{thrust,x} = -\frac{dm_r}{dt}u = \frac{dm_f}{dt}uFthrust,x​=−dtdmr​​u=dtdmf​​u

      Move the exhaust term to the force side, record the sign of dmrdt\frac{dm_r}{dt}dtdmr​​ against the positive burn rate dmfdt\frac{dm_f}{dt}dtdmf​​, and note thrust as forward recoil rather than an applied force.

    Attempt 1 of 2