Question 11.3: Repeat Example 11.2 for the restricted three-stage launch ve...
Repeat Example 11.2 for the restricted three-stage launch vehicle.
Equation 11.56 gives the burnout velocity for three stages,
v_{bo_{3-stage}}=3I_{sp}g_{0}\ln n_{3-stage}= I_{sp}g_{0}\ln \left(\frac{1}{\pi _{PL}^{\frac{1}{3}}(1-\varepsilon)+\varepsilon } \right)^{3} (11.56)
v_{bo}= 350 · 0.00981 · \ln\left(\frac{1}{0.05^{\frac{1}{3} }(1-0.15)+0.15}\right)^{3}= \underline{7.928 km/s}
Substituting m_{PL} = 10 000 kg,π_{PL} = 0.05 and \varepsilon= 0.15 into Equations 11.57 and 11.58
yields
m_{p_{1}}=\frac{(1-\pi ^{\frac{1}{3} }_{PL})(1-\varepsilon )}{\pi_{PL}}{m_{PL}}
m_{p_{1}}=\frac{(1-\pi ^{\frac{1}{3} }_{PL})(1-\varepsilon )}{\pi ^{\frac{2}{3} }_{PL}}{m_{PL}} (11.58)
m_{p_{3}}=\frac{(1-\pi ^{\frac{1}{3} }_{PL})(1-\varepsilon )}{\pi ^{\frac{1}{3} }_{PL}}{m_{PL}}
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 percent, which is much less than the advantage the two stage had over the single stage vehicle.