Rocket Equations

Puzzle 9 RocketA 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<n<1.

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?


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