Question 11.3: A winged entry vehicle with lift-to-drag ratio L/D = 1.5 ent...

a) First, we compute the ballistic coefficient for this winged vehicle:

C_{B}=\frac{m}{S C_{D}}=357.143 \mathrm{~kg} / \mathrm{m}^{2}

Next, we use Eq. (11.57) to compute density at the pull-up altitude

\rho_{\text {pullup }}=\frac{2 \beta C_{B}}{(L / D) \cos \phi}\left(1-\cos \gamma_{\mathrm{EI}}\right)=8.9929\left(10^{-5}\right) \mathrm{kg} / \mathrm{m}^{3}

\rho_{\mathrm{pullup}}=\rho_{0}e^{-\beta h_{\mathrm{pullup}}}={\frac{2\beta C_{B}}{\left(L/D\right)\cos\phi}}\left(1-\cos\gamma_{\mathrm{EI}}\right) (11.57)

The reader should remember that inverse scale height \beta must be in units of per meter in the previous expression because it is multiplied by ballistic coefficient C_{B} (which has units of \mathrm{kg} / \mathrm{m}^{2} ). Next, we use Eq. (11.58) (with surface density \rho_{0}=1.225 \mathrm{~kg} / \mathrm{m}^{3} and \beta= 0.1378 \mathrm{~km}^{-1} ) to determine the pull-up altitude in \mathrm{km} :

h_{\text {pullup }}=\frac{-1}{\beta} \ln \left(\frac{\rho_{\text {pullup }}}{\rho_{0}}\right)=69.08 \mathrm{~km}

We use Eq. (11.63) to determine the pull-up velocity:

\nu_{\text {pullup }}=\nu_{\mathrm{EI}} \exp \left[\frac{\gamma_{\mathrm{EI}}}{(L / D) \cos \phi}\right]=7.532 \mathrm{~km} / \mathrm{s}

b) Peak deceleration along the flight path is determined using Eq. (11.68):

a_{t_{\max }}=\frac{\beta\left(1-\cos \gamma_{\mathrm{EI}}\right)}{(L / D) \cos \phi} \nu_{\mathrm{EI}}^{2} \exp \left[\frac{2 \gamma_{\mathrm{EI}}}{(L / D) \cos \phi}\right]=7.143 \mathrm{~m} / \mathrm{s}^{2}

(it is useful here to use \beta=0.1378\left(10^{-3}\right) \mathrm{m}^{-1} and \nu_{\mathrm{EI}}=7,800 \mathrm{~m} / \mathrm{s} ). The peak deceleration in units of Earth g_{0} (using g_{0}=9.80665 \mathrm{~m} / \mathrm{s}^{2} ) is

a_{t_{\max }}=0.728 g_{0}

We use Eq. (11.69) to determine the peak acceleration normal to the flight path:

a_{n_{\max }}=\beta\left(1-\cos \gamma_{\mathrm{EI}}\right) \nu_{\mathrm{EI}}^{2} \exp \left[\frac{2 \gamma_{\mathrm{EI}}}{(L / D) \cos \phi}\right]=10.715 \mathrm{~m} / \mathrm{s}^{2}

Or, in Earth g_{0}: a_{n_{\max }}=1.093 g_{0}

c) Exit flight-path angle is simply \gamma_{\text {exit }}=-\gamma_{\mathrm{EI}}=3^{\circ}

We use Eq. (11.64) to determine exit velocity:

\nu_{\mathrm{exit}}=\nu_{\mathrm{EI}} \exp \left[\frac{2 \gamma_{\mathrm{EI}}}{(L / D) \cos \phi}\right]=7.274 \mathrm{~km} / \mathrm{s}

d) The vehicle follows a Keplerian orbit after atmospheric skip-out. We compute the energy of the orbit using \nu_{\text {exit }} and r_{\mathrm{EI}}=122 \mathrm{~km}+R_{E}=6,500 \mathrm{~km} :

\xi=\frac{\nu_{\mathrm{exit}}^{2}}{2}-\frac{\mu}{r_{\mathrm{EI}}}=-34.8673 \mathrm{~km}^{2} / \mathrm{s}^{2}=-\frac{\mu}{2 a}

Therefore, semimajor axis is a=5,715.95 \mathrm{~km}. Angular momentum is

h=r_{\mathrm{EI}} \nu_{\mathrm{exit}} \cos \gamma_{\mathrm{exit}}=47,216.4 \mathrm{~km}^{2} / \mathrm{s}=\sqrt{p \mu}

So, parameter is p=5,593.05 \mathrm{~km}. Eccentricity of the orbit is

e=\sqrt{1-\frac{p}{a}}=0.1466

Finally, the apogee radius is

r_{a}=\frac{p}{1-e}=6,554.13 \mathrm{~km}

Hence, the peak altitude after the skip is r_{a}-R_{E}=176.13 \mathrm{~km}, or 54 \mathrm{~km} above EI altitude.

e) We determine the analytical skip entry using the following steps: first, use Eq. (11.54) to compute flight-path angle for a given altitude (repeated below):

\cos \gamma=\cos \gamma_{\mathrm{EI}}+\frac{\rho_{0} e^{-\beta h}}{2 \beta C_{B}}(L / D) \cos \phi

Flight-path angle for the “down” phase is computed for altitudes descending from h_{\mathrm{EI}}(122 \mathrm{~km}) to pull-up altitude h_{\text {pullup }}=69.08 \mathrm{~km}. The flight-path angle profile for the “up” phase \left(h>h_{\text {pullup }}\right) is a mirror image of the “down” phase except with a sign change because \dot{h}>0. Next, the velocity profile is computed using Eq. (11.62) (repeated below)

\nu=\nu_{\mathrm{EI}} \exp \left[\frac{\gamma_{\mathrm{EI}}-\gamma}{(L / D) \cos \phi}\right]

where flight-path angle (determined by the previous step) ranges from \gamma_{\mathrm{EI}}<\gamma<\gamma_{\text {exit }}. After competing these steps, we may plot velocity vs. altitude. Although velocity is the dependent variable and altitude is the independent variable here, it is (perhaps) easier to visualize their relationship by plotting altitude on the y-axis and velocity on the x-axis (after all, altitude is the vertical direction).

Equation (11.66) determines the normal acceleration (repeated below):

a_{n}=\frac{\rho_{0} e^{-\beta h} \nu^{2}}{2 C_{B}}(L / D) \cos \phi

Here we use the “down-up” altitude profile and corresponding analytical velocity profile from the previous calculations. Finally, we compare the analytical skip entry to a numerically integrated skip trajectory obtained by using MATLAB’s ode45.m. As with Examples 11.1 and 11.2, we numerically integrate the governing nonlinear equations [Eqs. (11.1), (11.2), and (11.7)]

\begin{aligned} & \dot{\nu}=-\frac{D}{m}-g \sin \gamma & (11.1)\\ & \nu \dot{\gamma}=\frac{L \cos \phi}{m}-\left(g-\frac{\nu^2}{r}\right) \cos \gamma & (11.2) \end{aligned}

{\dot{h}}=\nu~{\mathrm{sin}}\gamma        (11.7)

using the given entry conditions and vehicle parameters.

Figure 11.12 shows the analytical and numerical skip trajectories (altitude vs. velocity). The analytical solution accurately portrays the first skip maneuver: both solutions show essentially the same pull-up altitude (about 69 \mathrm{~km} ), pull-up velocity (about 7.55 \mathrm{~km} / \mathrm{s} ), and exit velocity (between 7.2 and 7.3 \mathrm{~km} / \mathrm{s} ). The winged vehicle re-enters the atmosphere at a sub-circular velocity (equal to the first exit velocity) and performs a second skip maneuver. Because the second entry speed is sub-circular, the vehicle does not achieve a full skip-out exit, and only reaches a peak altitude of about 95 \mathrm{~km}. However, the analytical solution (by definition) assumes a full skip-out to EI as shown in Figure 11.12. Subsequent “reduced skips” with diminishing peak altitudes follow (see the third numerically integrated skip in Figure 11.12), but the analytical firstorder method does not accurately predict these skips because the centrifugal-gravity balance no longer holds. In addition, note that the velocity profile of the numerically integrated skip entry in Figure 11.12 shifts to the left (slows down) as altitude increases during skip out, and shifts to the right (speeds up) as altitude decreases during re-entry. This velocity shift vs. altitude is, of course, due to gravity. The analytical skip solution does not include the effect of gravity, and hence the skip-out and reentry velocity vs. altitude profiles shown in Figure 11.12 appear as vertical lines.

Figure 11.13 shows normal acceleration computed by the analytical and numerical methods. The first-order method accurately predicts normal acceleration for the first two skips but thereafter becomes inaccurate for the reasons previously mentioned. The reader should note that the “true” numerically integrated skip entry shows small negative normal acceleration when the vehicle is at its apogee. This small negative normal acceleration is due to gravity turning the velocity vector downward. Recall that the first-order method neglects gravity altogether.