A sounding rocket of initial mass mo and mass mf after all propellant is consumed is launched vertically ( γ = 90°). The propellant mass flow rate me is constant. (a) Neglecting drag and the variation of gravity with altitude, calculate the maximum height h attained by the rocket. (b) For what

Question

A sounding rocket of initial mass [latex]m_{o}[/latex] and mass [latex]m_{f}[/latex] after all propellant is consumed is launched vertically (γ = 90°). The propellant mass flow rate [latex]\dot{m}_{e}[/latex] 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?

Accepted Answer

The vehicle mass as a function of time, up to burnout, is

[latex]m=m_{o}-\dot{m}_{e} t[/latex] (a)

At burnout, [latex]m=m_{f}[/latex], so the burnout time [latex]t_{b o}[/latex] is

[latex]t_{b o}=\frac{m_{o}-m_{f}}{\dot{m}_{e}}[/latex] (b)

The drag loss is assumed zero, and the gravity loss is

[latex]\Delta v_{G}=\int\limits_{0}^{t_{b o}} g_{o} \sin \left(90^{\circ}\right) d t=g_{o} t[/latex]

Recalling that [latex]I_{s p} g_{o}=c[/latex] and using (a), it follows from Equation 11.26 that, up to burnout, the velocity as a function of time is

[latex]\Delta v=I_{s p} g_{o} \ln \frac{m_{o}}{m_{f}}-\Delta v_{D}-\Delta v_{G}[/latex] (11.26)

[latex]v=c \ln \frac{m_{o}}{m_{o}-\dot{m}_{e} t}-g_{o} t[/latex] (c)

Since dh/dt = v, the altitude as a function of time is

[latex]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}[/latex] (d)

The height at burnout [latex]h_{b o}[/latex] is found by substituting (b) into this expression,

[latex]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}[/latex] (e)

Likewise, the burnout velocity is obtained by substituting (b) into (c),

[latex]v_{b o}=c \ln \frac{m_{o}}{m_{f}}-\frac{g_{o}}{\dot{m}_{e}}\left(m_{o}-m_{f}\right)[/latex] (f)

After burnout, the rocket coasts upward with the constant downward acceleration of gravity,

[latex]v=v_{b o}-g_{o}\left(t-t_{b o}\right)[/latex]
[latex]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}[/latex]

Substituting (b), (e) and (f) into these expressions yields, for [latex]t>t_{b o}[/latex],

[latex]v=c \ln \frac{m_{o}}{m_{f}}-g_{o} t[/latex]
[latex]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}[/latex] (g)

The maximum height [latex]h_{\max }[/latex] is reached when v = 0,

[latex]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}}[/latex] (h)

Substituting [latex]t_{\max }[/latex] into (g) leads to our result,

[latex]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}[/latex] (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, [latex]h_{\max }[/latex] can be increased by increasing the mass flow rate [latex]\dot{m}_{e}[/latex]. In fact,

the greatest height is achieved when [latex]\dot{m}_{e} \longrightarrow \infty[/latex]

In that extreme, all of the propellant is expended at once, like a mortar shell.

Question 11.1: A sounding rocket of initial mass mo and mass mf after all p...

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