Question 11.1: A sounding rocket of initial mass m0 and mass mf after all p...
A sounding rocket of initial mass m_{0} and mass m_{f} after all propellant is consumed is launched vertically (γ = 90°). The propellant mass flow rate \dot{m}_{e} is constant. Neglecting drag and the variation of gravity with altitude, calculate the maximum height h attained by the rocket. For what flow rate is the greatest altitude reached?
The vehicle mass as a function of time, up to burnout, is
m = m_{0}−\dot{m}_{e}t (a)
At burnout, m =m_{f}, so the burnout time t_{bo} is
t_{bo}=\frac{m_{0}-m_{f}}{\dot{m}_{e}} (b)
The drag loss is assumed to be zero, and the gravity loss is
\Delta v_{G}=\int_{0}^{t_{bo}}g_{0} \sin(90°)dt = g_{0}t_{bo}
Recalling that I_{sp}g0 = 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_{sp}g_{0}\ln \frac{m_{0}}{m_{f}}-\Delta v_{D}-\Delta v_{G} (11.26)
v=c\ln \frac{m_{0}}{m_{0}- \dot{m}_{e}t}-g_{0}t (c)
Since dh/dt = v, the altitude as a function of time is
h=\int_{0}^{t}v dt=\int_{0}^{t}(c\ln \frac{m_{0}}{m_{0}- \dot{m}_{e}t}-g_{0}t)dt
=\frac{c}{\dot{m}_{e}}[(m_{0}-\dot{m}_{e}t)\ln \frac{m_{0}-bt}{m_{0}}+\dot{m}_{e}t ] -\frac{1}{2}g_{0}t^{2} (d)
The height at burnout h_{bo} is found by substituting (b) into this expression,
h_{bo}=\frac{c}{\dot{m}_{e}} \left(m_{f}\ln \frac{m_{f}}{m_{0}}+m_{0}-m_{f}\right) -\frac{1}{2}\left(\frac{m_{0}-m_{f}}{\dot{m}_{e}} \right)^{2}g (e)
Likewise, the burnout velocity is obtained by substituting (b) into (c),
v_{bo}=c\ln \frac{m_{0}}{m_{f}}-\frac{g_{0}}{\dot{m}_{e}}(m_{0}-m_{f}) (f)
After burnout, the rocket coasts upward with the constant downward acceleration of gravity,
v = v_{bo} − g_{0}(t − t_{bo})
h = h_{bo} + v_{bo}(t − t_{bo})-\frac{1}{2}g_{0}(t-t_{bo})^{2}
Substituting (b), (e) and (f) into these expressions yields, for t > t_{bo},
v = c \ln \frac{m_{0}}{m_{f}}-g_{0}t
h=\frac{c}{\dot{m}_{e}}\left(m_{0}\ln \frac{m_{f}}{m_{0}}+m_{0}-m_{f}\right)+ct \ln\frac{m_{0}}{m_{f}}-\frac{1}{2}g_{0}t^{2} (g)
The maximum height h_{max} is reached when v = 0,
c\ln \frac{m_{0}}{m_{f}}-g_{0}t_{max}=0\Rightarrow t_{max}=\frac{c}{g_{0}}\ln \frac{m_{0}}{m_{f}}
Substituting t_{max} into (g) leads to our result,
\underline{h_{max}=\frac{cm_{0}}{\dot{m}_{e}}(1+\ln n-n)+\frac{1}{2}\frac{c^{2}}{g_{0}}\ln ^{2}n}
where n is the mass ratio (n > 1). Since n > (1 + \ln n), it follows that (1 + \ln n − n) is negative. 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, i.e., all of the propellant is expended at once, like a mortar shell.