= Solution
<Newton's first law> identifies <inertial frames> in which a force-free body maintains constant <velocity>. The second is $\dot{\mathbf P}=\mathbf F_{\rm net}$ for a closed system; for fixed <mass> it becomes $m\mathbf a=\mathbf F$. The third gives equal and opposite pair <forces>. Summing the second law over a closed collection cancels internal <force> pairs, so total <momentum> is conserved when the resultant external <force> vanishes. A rocket alone is an open variable-mass subsystem, so one must include the <momentum> carried by exhaust before applying this reasoning.
Choose downward <velocity> as positive and write $M(t)=M_0-\alpha t$. During $dt$, exhaust <mass> $dm=\alpha dt$ leaves downward with <speed> $v+v_E$ in the surface frame. To first order, <momentum> balance for rocket plus this exhaust gives
$$
Mv+Mg\,dt=(M-dm)(v+dv)+dm(v+v_E),\qquad \dot v=g-\frac{\alpha v_E}{M_0-\alpha t}.
$$
Integration, with $v(0)=v_0$, yields
$$
\boxed{v(t)=v_0+gt+v_E\log\frac{M_0-\alpha t}{M_0}.}
$$
Setting $g=0$ gives the gravity-free result. This assumes a constant downward relative exhaust <speed>, constant mass-loss rate and $M_0-\alpha t>0$ throughout the burn; the negative logarithm supplies the upward thrust effect.
Height satisfies $\dot h=-v$. Put $u(t)=1-\alpha t/M_0$. Integrating the logarithm and choosing $h(0)=h_0$ gives
$$
\boxed{h(t)=h_0-v_0t-\frac12gt^2+\frac{M_0v_E}{\alpha}[u\log u-u+1].}
$$
Equivalently, the last term is $v_E[t+(M_0-\alpha t)\log u/\alpha]$. Differentiating either form recovers $-v$ and checks the sign.
For this <constant-rate retro-rocket landing>, at touchdown $u_L=3/4$ and $v_L=0$, hence
$$
\boxed{t_L=\frac{v_E\log(4/3)-v_0}{g}=12.481\,\mathrm s.}
$$
The fuel condition gives $\alpha=M_0/(4t_L)=200.30$ kg/s. Requiring $h(t_L)=0$ then gives
$$
\boxed{h_0=v_0t_L+\frac12gt_L^2-v_Et_L[1+3\log(3/4)]=275.4\,\mathrm m.}
$$
In this burn the downward <speed> decreases monotonically to zero: the initial thrust acceleration already exceeds <gravity> and increases as fuel is consumed. The result is the idealized vertical landing, without atmospheric drag, changing <gravity> or a separate terminal burn <phase>.
Back to article page
