Question 11.5: Repeat Example 11.4 for the restricted three-stage launch ve...
Repeat Example 11.4 for the restricted three-stage launch vehicle.
Equation 11.56 gives the burnout velocity for three stages,
v_{b o 3-\text { stage }}=3 I_{s p} g_{o} \ln n_{3-\text { stage }}=I_{s p} g_{o} \ln \left\lgroup \frac{1}{\pi_{P L}^{1 / 3}(1-\varepsilon)+\varepsilon} \right\rgroup ^{3} (11.56)
v_{b o}=350 \cdot 0.00981 \cdot \ln \left\lgroup \frac{1}{0.05^{1 / 3}(1-0.15)+0.15} \right\rgroup ^{3}=7.928 km / sSubstituting m_{P L}=10,000 kg , \pi_{ PL }=0.05 \text { and } \varepsilon=0.15 into Equations 11.57 and 11.58 yields
m_{E 1}=\frac{\left(1-\pi_{P L}^{1 / 3}\right) \varepsilon}{\pi_{P L}} m_{P L} \quad m_{E 2}=\frac{\left(1-\pi_{P L}^{1 / 3}\right) \varepsilon}{\pi_{P L}^{2 / 3}} m_{P L} \quad m_{E 3}=\frac{\left(1-\pi_{P L}^{1 / 3}\right) \varepsilon}{\pi_{P L}^{1 / 3}} m_{P L} (11.57)
\begin{aligned}m_{p 1} &=\frac{\left(1-\pi_{P L}^{1 / 3}\right)(1-\varepsilon)}{\pi_{P L}} m_{P L} \\m_{p 2} &=\frac{\left(1-\pi_{P L}^{1 / 3}\right)(1-\varepsilon)}{\pi_{P L}^{2 / 3}} m_{P L} \\m_{p 3} &=\frac{\left(1-\pi_{P L}^{1 / 3}\right)(1-\varepsilon)}{\pi_{P L}^{1 / 3}} m_{P L}\end{aligned} (11.58)
m_{E 1}=18,948 kg \quad m_{E 2}=6980 kg \quad m_{E 3}=2572 kg
m_{p 1}=107,370 kg \quad m_{p 2}=39,556 kg \quad m_{p 3}=14,573 kg
Again, the total empty mass and total propellant mass are the same as for the single and two-stage vehicles. Notice that the velocity increase over the two-stage rocket is just 7%, which is much less than the advantage the two-stage had over the single stage vehicle.