Question 11.2: The following data is given mPL = 10 000 kg πPL = 0.05 ε = 0...

(a) From Equation 11.43 we find
v_{bo}=I_{sp}g_{0}\ln \frac{1}{\pi _{PL}(1-\varepsilon)+\varepsilon}                                   (11.43)
v_{bo} = 350 · 0.00981\ln \frac{1}{0.05(1 + 0.15) + 0.15}=\underline{5.657  km/s}
Equation 11.40 yields the gross mass
\pi _{PL}=\frac{m_{PL}}{m_{0}}                                         (11.40)
m_{0}=\frac{10000}{0.05}=200000  kg
from which we obtain the empty mass using Equation 11.34,
\varepsilon=\frac{m_{E}}{m_{E}+m_{P}}=\frac{m_{E}}{m_{0}-m_{PL}}                    (11.34)
m_{E} = \varepsilon(m_{0} − m_{PL}) = 0.15(200 000 − 10 000) =\underline{28 500  kg}
The mass of propellant is
m_{p} = m_{0} − m_{E} − m_{PL} = 200 000 − 28 500 − 10 000 =\underline{161 500  kg}
(b) For a restricted two-stage vehicle, the burnout speed is given by Equation 11.50,
v_{bo_{2-stage}}= I_{sp}g_{0} \ln\left[\frac{1}{\pi _{PL}^{\frac{1}{2}}(1-\varepsilon )+\varepsilon }\right]^{2}                                          (11.50)
v_{bo_{2-stage}}= 350 · 0.00981 \ln\left[\frac{1}{0.05^{\frac{1}{2} }(1-0.15)+0.15} \right]^{2}=\underline{7.407  km/s}
The empty mass of each stage is found using Equations 11.51,
\begin{matrix}m_{E_{1}}=\frac{\left(1-\pi _{PL}^{\frac{1}{2}}\right)^{\varepsilon}}{\pi _{PL}}m_{PL} & m_{E_{2}}=\frac{\left(1-\pi _{PL}^{\frac{1}{2}}\right)^{\varepsilon}}{\pi_{PL}^{\frac{1}{2}}}m_{PL} \end{matrix}                           (11.51)
m_{E_{1}}=\frac{(1-0.05^{\frac{1}{2}})· 0.15}{0.15}· 10 000 =\underline{23 292 kg}
m_{E_{2}}=\frac{(1-0.05^{\frac{1}{2}})· 0.15}{0.15^{\frac{1}{2}}}· 10 000 =\underline{5208  kg}
For the propellant masses, we turn to Equations 11.53
\begin{matrix}m_{P_{1}}=\frac{\left(1-\pi _{PL}^{\frac{1}{2}}\right)(1-\varepsilon)}{\pi _{PL}}m_{PL} & m_{P_{2}}=\frac{\left(1-\pi _{PL}^{\frac{1}{2}}\right)(1-\varepsilon)}{\pi_{PL}^{\frac{1}{2}}}m_{PL} \end{matrix}                    (11.53)
m_{p_{1}} =\frac{(1-0.05^{\frac{1}{2}})·(1- 0.15)}{0.15}· 10 000 =\underline{131990  kg}
m_{p_{1}} =\frac{(1-0.05^{\frac{1}{2}})·(1- 0.15)}{0.15^{\frac{1}{2}}}· 10 000 =\underline{29513  kg}
The total empty mass, m_{E} = m_{E_{1}} + m_{E_{2}}, and the total propellant mass, m_{p} = m_{p_{1}}+m_{p_{2}}, are the same as for the single stage rocket. The mass of the second stage, including the payload, is 22.4 percent of the total vehicle mass.