Question 11.2: The data for a single stage rocket are as follows: Launch ma...

(a) From Example 11.1(b) the burnout time t_{b o} is

t_{b o}=\frac{m_{o}-m_{f}}{\dot{m}_{e}}                 (a)

The burnout mass m_{f} is obtained from Equation 11.24,

m_{f}=\frac{m_{o}}{n}=\frac{68,000}{15}=4533.3 kg             (b)

The propellant mass flow rate \dot{m}_{e} is given by Equation 11.20,

I_{s p}=\frac{T}{\dot{m}_{e} g_{o}}                    (11.20)

\dot{m}_{e}=\frac{T}{I_{s p} g_{o}}=\frac{933,913}{390 \cdot 9.81}=244.10 kg / s            (c)

Substituting (b), (c) and m_{o}=68,000 kg into (a) yields the burnout time,

(b) The burnout altitude is given by Example 11.1(e)

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}                (d)

The exhaust velocity c is found in Equation 11.21,

c=I_{s p} g_{o}=390 \cdot 9.81=3825.9 m / s              (e)

Substituting (b), (c) and (e), along with m_{o}=68,000 kg \text { and } g_{o}=9.81 m / s ^{2} into (d), we get

h_{b o}=\frac{3825.9}{244.1}\left\lgroup 4533.3 \ln \frac{4533.3}{68,000}+68,000-4533.3 \right\rgroup -\frac{1}{2}\left\lgroup \frac{68,000-4533.3}{244.1} \right\rgroup ^{2} \cdot 9.81
h_{b o}=470.74 km

(c) From Example 11.1(f) we find

v_{b o}=c \ln \frac{m_{o}}{m_{f}}-\frac{g_{o}}{\dot{m}_{e}}\left(m_{o}-m_{f}\right)=3825.9 \ln \frac{68,000}{4533.3}-\frac{9.81}{244.1}(68,000-4533.3)
v_{b o}=7.810 km / s

(d) To find h_{\max }, where the speed of the rocket falls to zero, we use Example 11.1(i),

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}=\frac{1}{2} \frac{3825.9^{2}}{9.81} \ln ^{2} 15-\frac{3825.9 \cdot 68,000}{244.1} \frac{15 \ln 15-(15-1)}{15}
h_{\max }=3579.7 km

Notice that the rocket coasts to a height more than seven times the burnout altitude.