Past exam of the physics 1 paper of the University of Oxford / 2021 resit / 7

Powered ascent. Use upward signed velocity, constant gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, constant relative rocket exhaust speed $u=2500\mathrm{m}{\mathrm{s}}^{-1}$, vertical flight from rest, no drag force, and no impulse at stage separation. The constant-$g$ approximation is an explicit modelling assumption; a varying Newtonian gravitational field would require integrating a different ascent equation. Put $m_0=540000\mathrm{kg}$ and $q=2500\mathrm{kg}{\mathrm{s}}^{-1}$. During ascent $m(t)=m_0-qt$, and momentum balance gives the rocket equation with gravity:The launch rocket thrust $uq=6.25\times 10^6\mathrm{N}$ exceeds $m_{0}g$, so immediate lift-off is consistent. At ascent burnout the entire attached vehicle still has mass $m_b=140000\mathrm{kg}$, not $40000\mathrm{kg}$; the latter is the separated first stage. The burn duration and endpoint speed are

Coasting height. Integrating the ascent velocity from zero height givesFor example this follows by substituting $m=m_0-qt$ in ${\int }_0^t\ln [m_0/(m_0-qs)]ds$. At separation,The separated stage then undergoes free fall in a constant gravitational field with initial upward speed $v_b$. Conservation of mechanical energy during this coast givesIts smaller mass after separation does not change that coasting height.

Reserve burn. On descending to $z_i=100000\mathrm{m}$, conservation of mechanical energy gives downward speedReset time to $s=0$ at reserve ignition. Now $m_i=40000\mathrm{kg}$, $m_f=14000\mathrm{kg}$ and $q_r=26000/92\mathrm{kg}{\mathrm{s}}^{-1}$. Downward ejection supplies upward rocket thrust. The upward signed velocity isThus the formal free-flight burnout value iscorresponding to a formal downward speed of $2.26\mathrm{m}{\mathrm{s}}^{-1}$. Keeping $g=9.8\mathrm{m}{\mathrm{s}}^{-2}$ throughout instead gives $3.82\mathrm{m}{\mathrm{s}}^{-1}$ downward; premature rounding is especially damaging because three large velocity terms almost cancel.

The free-flight endpoint is below ground. Integrating the same reserve-burn velocity givesAt the stated endpoint this gives $z_r(92)=-167.27\mathrm{m}$ for $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. The surface is crossed at approximately $s=89.15\mathrm{s}$, before reserve burnout. The stage has already hit the surface, so a physical post-impact burnout speed cannot be inferred without a ground-contact model. The boxed velocity is the intended ideal continuation, not an achievable free-flight landing result under these exact assumptions. In particular, the given numbers do not describe an exact soft landing. The constant-rate retro-rocket landing must satisfy both its velocity and height conditions.