Past exam of the physics 1 paper of the University of Oxford / 2017 resit / 8

Choose downward velocity $v$ positive. The remaining rocket mass is $m(t)=m_0-\alpha t$, and the upward rocket thrust has magnitude $\alpha u$, where $u>0$ is the relative rocket exhaust speed. The rocket equation including gravity isThis follows by applying momentum conservation to the rocket and its just-expelled rocket propellant, with gravity supplying the external impulse. Applying $d(mv)/dt=mg$ to the rocket alone would omit the exhaust momentum flux.

Integrating the rocket equation with $v(0)=v_0$ givesThis formula holds during the constant-flow burn, while $m_0-\alpha t>0$ and fuel remains. The negative logarithmic term represents the slowing due to upward rocket thrust.

Height decreases at rate $\dot{h}=-v$. Set $m=m_0-\alpha t$; an antiderivative needed for the rocket equation isConsequentlyDifferentiating this expression recovers $\dot{h}=-v$ and substituting $t=0$ gives $h(0)=h_0$, which check both the sign and the integration constant.

The specified rocket fuel fraction gives $m(T)=2m_0/3$, so $\alpha T=m_0/3$. Imposing $v(T)=0$ yieldsThen $h(T)=0$ fixesNumerically,The associated constant mass loss rate is $\alpha =m_0/(3T)\simeq 88.78\mathrm{kg}{\mathrm{s}}^{-1}$. The initial upward thrust acceleration already exceeds $g$ and increases as mass falls, so the downward speed decreases monotonically to zero. This calculation assumes constant lunar gravitational acceleration, constant relative rocket exhaust speed, and negligible drag.