Question 11.3: The rocket in Example 11.2 has a diameter of 5 m. It is to b...
The rocket in Example 11.2 has a diameter of 5 m. It is to be launched on a gravity turn trajectory. Pitchover begins at an altitude of 130 m with an initial flight path angle \gamma o of 89.85°. What are the altitude h and speed v of the rocket at burnout \left(t_{b o}=260 s \right)? What are the velocity losses due to drag and gravity (cf. Equations 11.27)?
\Delta v_{D}=\int\limits_{t_{o}}^{t_{f}} \frac{D}{m} d t \quad \Delta v_{G}=\int\limits_{t_{o}}^{t_{f}} g \sin \gamma d t (11.27)
The \text { MATLAB }^{\circledR} program Example_11_03.m in Appendix D.40 finds the speed v, the flight path angle γ , the altitude h and the downrange distance x as a function of time. It does so by using the ordinary differential equation solver rkf_45.m (Appendix D.4) to numerically integrate Equations 11.6 through 11.8, namely
\frac{d v}{d t}=\frac{T}{m}-\frac{D}{m}-g \sin \gamma (a)
\frac{d \gamma}{d t}=-\frac{1}{v}\left\lgroup g-\frac{v^{2}}{R_{E}+h} \right\rgroup \cos \gamma (b)
\frac{d h}{d t}=v \sin \gamma (c)
\frac{d x}{d t}=\frac{R_{E}}{R_{E}+h} v \cos \gamma (d)
The variable mass m is given in terms of the initial mass m_{o}=68,000 kg and the constant mass flow rate \dot{m}_{e} by Equation 11.11,
m=m_{o}-\dot{m}_{e} t (e)
The thrust T = 933.913 kN is assumed constant, and \dot{m}_{e} is obtained from T and the specific impulse I_{s p}=390 by means of Equation 11.20,
\dot{m}_{e}=\frac{T}{I_{s p} g_{o}} (f)
The drag force D in (a) is given by Equation 11.1,
D=q A C_{D} (11.1)
D=\frac{1}{2} \rho v^{2} A C_{D} (g)
The drag coefficient is assumed to have the constant value C_{D}=0.5. The frontal area A = πd²/4 is found from the rocket diameter d = 5 m. The atmospheric density profile is assumed exponential,
\rho=\rho_{o} e^{-h / h_{o}} (h)
where \rho_{o}=1.225 kg / m ^{3} is the sea level atmospheric density and h_{o}=7.5 km is the scale height of the atmosphere. (The scale height is the altitude at which the density of the atmosphere is about 37% of its sea level value.)
Finally, the acceleration of gravity varies with altitude h according to Equation 1.36,
g=g_{0} \frac{R_{E}^{2}}{\left(R_{E}+z\right)^{2}}=\frac{g_{0}}{\left(1+ z / R_{E}\right)^{2}} (1.36)
g=\frac{g_{o}}{\left(1+h / R_{E}\right)^{2}} \quad\left(R_{E}=6378 km , g_{o}=9.91 m / s ^{2}\right) (i)
The drag loss and gravity loss are found by numerically integrating Equations 11.27.
Between lift off and pitchover, the flight path angle γ is held at 90°. Pitchover begins at the altitude h_{p}=130 m with the flight path angle set at \gamma_{o}=89.85^{\circ}.
For the input data described above, the output of Example_11_03.m is as follows. The solution is very sensitive to the values of h_{p} \text { and } \gamma_{o}.
Initial flight path angle = 89.85 deg
Pitchover altitude = 130 m
Burn time = 260 s
Final speed = 8.62116 km/s
Final flight path angle = 8.80161 deg
Altitude = 133.211 km
Downrange distance = 462.318 km
Drag loss = 0.298199 km/s
Gravity loss = 1.44096 km/s
Thus, at burnout
Altitude = 133.2 km
Speed = 8.621 km/s
The speed losses are
Due to drag: 0.2982 km/s
Due to gravity: 1.441 km/s
Figure 11.3 shows the gravity-turn trajectory.
