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 γo of 89.85° . What are the altitude h and speed v of the rocket at burnout (tbo = 260 s)? What are the velocity losses

Question

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)

[latex]\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[/latex] (11.27)

Accepted Answer

The [latex] ext { MATLAB }^{\circledR}[/latex] 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

[latex]\frac{d v}{d t}=\frac{T}{m}-\frac{D}{m}-g \sin \gamma[/latex] (a)

[latex]\frac{d \gamma}{d t}=-\frac{1}{v}\left\lgroup g-\frac{v^{2}}{R_{E}+h} ight group \cos \gamma[/latex] (b)

[latex]\frac{d h}{d t}=v \sin \gamma[/latex] (c)

[latex]\frac{d x}{d t}=\frac{R_{E}}{R_{E}+h} v \cos \gamma[/latex] (d)

The variable mass m is given in terms of the initial mass [latex]m_{o}=68,000 kg[/latex] and the constant mass flow rate [latex]\dot{m}_{e}[/latex] by Equation 11.11,

[latex]m=m_{o}-\dot{m}_{e} t[/latex] (e)

The thrust T = 933.913 kN is assumed constant, and [latex]\dot{m}_{e}[/latex] is obtained from T and the specific impulse [latex]I_{s p}=390[/latex] by means of Equation 11.20,

[latex]\dot{m}_{e}=\frac{T}{I_{s p} g_{o}}[/latex] (f)

The drag force D in (a) is given by Equation 11.1,

[latex]D=q A C_{D}[/latex] (11.1)

[latex]D=\frac{1}{2} ho v^{2} A C_{D}[/latex] (g)

The drag coefficient is assumed to have the constant value [latex]C_{D}=0.5[/latex]. The frontal area A = πd²/4 is found from the rocket diameter d = 5 m. The atmospheric density profile is assumed exponential,

[latex] ho= ho_{o} e^{-h / h_{o}}[/latex] (h)

where [latex] ho_{o}=1.225 kg / m ^{3}[/latex] is the sea level atmospheric density and [latex]h_{o}=7.5 km[/latex] 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,

[latex]g=g_{0} \frac{R_{E}^{2}}{\left(R_{E}+z ight)^{2}}=\frac{g_{0}}{\left(1+ z / R_{E} ight)^{2}}[/latex] (1.36)

[latex]g=\frac{g_{o}}{\left(1+h / R_{E} ight)^{2}} \quad\left(R_{E}=6378 km , g_{o}=9.91 m / s ^{2} ight)[/latex] (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 [latex]h_{p}=130 m[/latex] with the flight path angle set at [latex]\gamma_{o}=89.85^{\circ}[/latex].
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 [latex]h_{p} ext { and } \gamma_{o}[/latex].

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.

Question 11.3: The rocket in Example 11.2 has a diameter of 5 m. It is to b...

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