Past exam of the mathematics course of the University of Cambridge / 2009 / ia / Paper 4 / 3A (Dynamics and Relativity)

Take upward as positive and let $dm<0$ be the change in rocket mass during $dt$. The exhaust emitted has mass $-dm$ and, to first order, upward velocity $v-U$. The linear momentum of the rocket and just-emitted gas changes byThe external gravitational impulse is $-mgdt$ to first order. Applying the momentum balance to this material system, rather than applying a fixed-mass formula to the rocket alone, gives the rocket equationThe term $-U\dot{m}$ is the upward thrust.

For the specified mass loss and exhaust law, $\dot{m}=-\alpha$ andThus the condition for strictly positive upward acceleration at launch isAt equality the initial thrust just balances the weight; under this particular exhaust law the acceleration becomes positive immediately afterwards. If the initial thrust is smaller than the weight and the rocket initially rests on a launch surface, a nonzero supporting force persists until a later lift-off, so the free-flight equation does not describe its earlier supported motion.

For immediate free flight, integrate the rocket equation with $v(0)=0$:This applies while $0\le t<m_0/\alpha$ and the nonrelativistic approximation remains valid. The formal divergence as the model's remaining mass approaches zero is not a prediction within that approximation.

For dimensional analysis, write the mass, length and time dimensions as $\mathsf{M},\mathsf{L},\mathsf{T}$. Then $[m_0]=[m]=\mathsf{M}$, $[\alpha ]=\mathsf{M}{\mathsf{T}}^{-1}$, $[t]=\mathsf{T}$, $[U_0]=[U]=[v]=\mathsf{L}{\mathsf{T}}^{-1}$, and $[g]=\mathsf{L}{\mathsf{T}}^{-2}$. Both $m_0$ and $\alpha t$ have mass dimension, so $\alpha t/(m_0-\alpha t)$ is dimensionless; both terms in the boxed velocity have dimension $\mathsf{L}{\mathsf{T}}^{-1}$. The launch inequality compares two forces, each of dimension $\mathsf{ML}{\mathsf{T}}^{-2}$.