# Rocket Equations

A rocket of initial mass $M_0$ burns fuel at a constant rate of $\alpha$. The fuel ejected travels at a speed of $w$ in the rocket’s frame. Assume the rocket travels vertically away from Earth’s centre and does not experience air resistance. Assume also that the rocket can actually launch, that is

$\alpha w>M_0 g$

Question 1: Using Newton’s second law, show that the rocket’s equation of motion is

$\displaystyle (M_0-\alpha t) \frac{dv}{dt}=\alpha w-(M_0-\alpha t)g$

for $0\leqslant t \leqslant \frac{M_0}{\alpha}$ where $g$ is acceleration due to gravity.

Question 2: Taking $g$ to be constant throughout its flight, show that the vertical height $h$ of the rocket above Earth’s surface is given by

$\displaystyle h=wt-\Big(\frac{w}{\alpha}\Big) \Big(M_0-\alpha t\Big)\ln\bigg(\frac{M_0}{M_0-\alpha t}\bigg)-\frac{1}{2}\,gt^2$

Suppose the mass of fuel on-board the rocket is $nM_0$ where $0.

Question 3: By taking the (partial) derivative of $x$ with respect to $\alpha$, show that the optimum rate at which to burn fuel $\alpha_o$ is given by

$\displaystyle \alpha_o=\Big(\frac{W_0}{w}\Big)\bigg(\frac{ n^2}{(1-n)\ln(1-n)+n}\bigg)$

where $W_0$ is rocket’s initial weight.

Question 4: Show that the greatest height $H$ reached by the rocket just as the last of its fuel is burnt is given by

$H=\displaystyle \lambda(n)\bigg(\frac{w^2}{2g}\bigg)$

where $\lambda(n)$ is the function

$\lambda(n)=\displaystyle \bigg(\frac{n+(1-n)\ln(1-n)}{n}\bigg)^2$

Question 5: Using the expressions for $v$ and $H$, deduce the maximum height above the Earth’s surface the rocket reaches in this case.

According to this model, does $\alpha=\alpha_0$ give the maximum possible height the rocket can reach  for fixed values of $M_0, n, w$ and $g$?