Question 11.1: A sounding rocket of initial mass mo and mass mf after all p...
A sounding rocket of initial mass m_{o} and mass m_{f} after all propellant is consumed is launched vertically (γ = 90°). The propellant mass flow rate \dot{m}_{e} is constant.
(a) Neglecting drag and the variation of gravity with altitude, calculate the maximum height h attained by the rocket.
(b) For what flow rate is the greatest altitude reached?
The vehicle mass as a function of time, up to burnout, is
m=m_{o}-\dot{m}_{e} t (a)
At burnout, m=m_{f}, so the burnout time t_{b o} is
t_{b o}=\frac{m_{o}-m_{f}}{\dot{m}_{e}} (b)
The drag loss is assumed zero, and the gravity loss is
\Delta v_{G}=\int\limits_{0}^{t_{b o}} g_{o} \sin \left(90^{\circ}\right) d t=g_{o} tRecalling that I_{s p} g_{o}=c and using (a), it follows from Equation 11.26 that, up to burnout, the velocity as a function of time is
\Delta v=I_{s p} g_{o} \ln \frac{m_{o}}{m_{f}}-\Delta v_{D}-\Delta v_{G} (11.26)
v=c \ln \frac{m_{o}}{m_{o}-\dot{m}_{e} t}-g_{o} t (c)
Since dh/dt = v, the altitude as a function of time is
h=\int\limits_{0}^{t} v d t=\int\limits_{0}^{t}\left\lgroup c \ln \frac{m_{o}}{m_{o}-\dot{m}_{e} t}-g_{o} t \right\rgroup d t=\frac{c}{\dot{m}_{e}}\left[\left(m_{o}-\dot{m}_{e} t\right) \ln \frac{m_{o}-\dot{m}_{e} t}{m_{o}}+\dot{m}_{e} t\right]-\frac{1}{2} g_{o} t^{2} (d)
The height at burnout h_{b o} is found by substituting (b) into this expression,
h_{b o}=\frac{c}{\dot{m}_{e}}\left\lgroup m_{f} \ln \frac{m_{f}}{m_{o}}+m_{o}-m_{f} \right\rgroup -\frac{1}{2}\left\lgroup \frac{m_{o}-m_{f}}{\dot{m}_{e}} \right\rgroup ^{2} g_{o} (e)
Likewise, the burnout velocity is obtained by substituting (b) into (c),
v_{b o}=c \ln \frac{m_{o}}{m_{f}}-\frac{g_{o}}{\dot{m}_{e}}\left(m_{o}-m_{f}\right) (f)
After burnout, the rocket coasts upward with the constant downward acceleration of gravity,
v=v_{b o}-g_{o}\left(t-t_{b o}\right)
h=h_{b o}+v_{b o}\left(t-t_{b o}\right)-\frac{1}{2} g_{o}\left(t-t_{b o}\right)^{2}
Substituting (b), (e) and (f) into these expressions yields, for t>t_{b o},
v=c \ln \frac{m_{o}}{m_{f}}-g_{o} t
h=\frac{c}{\dot{m}_{e}}\left\lgroup m_{o} \ln \frac{m_{f}}{m_{o}}+m_{o}-m_{f} \right\rgroup +c t \ln \frac{m_{o}}{m_{f}}-\frac{1}{2} g_{0} t^{2} (g)
The maximum height h_{\max } is reached when v = 0,
c \ln \frac{m_{o}}{m_{f}}-g_{o} t_{\max }=0 \Rightarrow t_{\max }=\frac{c}{g_{o}} \ln \frac{m_{o}}{m_{f}} (h)
Substituting t_{\max } into (g) leads to our result,
h_{\max }=\frac{1}{2} \frac{c^{2}}{g_{o}} \ln ^{2} n-\frac{c m_{o}}{\dot{m}_{e}} \frac{n \ln n-(n-1)}{n} (i)
where n is the mass ratio (n > 1). Since nln n is greater than n – 1, it follows that the second term in this expression is positive. Hence, h_{\max } can be increased by increasing the mass flow rate \dot{m}_{e}. In fact,
the greatest height is achieved when \dot{m}_{e} \longrightarrow \infty
In that extreme, all of the propellant is expended at once, like a mortar shell.