Question 11.6: Using the vehicle data and entry conditions from Example 11....
Using the vehicle data and entry conditions from Example 11.3, determine the critical flight-path angle, altitude, and velocity corresponding to the peak aerodynamic heating rate for a skip entry.
The winged entry vehicle in Example 11.3 has L / D=1.5, zero bank angle, and enters the atmosphere with flight-path angle \gamma_{\mathrm{EI}}=-3^{\circ}. Equation (11.92) allows us to compute the critical flight-path angle for peak heat rate during the skip entry:
\gamma_{\text {crit }} \cong \frac{-3 \gamma_{\mathrm{EI}}^{2}}{2(L / D) \cos \phi}=-0.002742 \mathrm{rad}=-0.157^{\circ}
The critical flight-path angle is very shallow and hence very close to (but prior to) the pull-up altitude. The ballistic coefficient for the winged vehicle is
C_{B}=\frac{m}{S C_{D}}=357.143 \mathrm{~kg} / \mathrm{m}^{2}
Using Eq. (11.93), the critical altitude for peak heating during the skip entry is
h_{\text {crit }}=\frac{-1}{\beta} \ln \left[\frac{2 \beta C_{B}\left(\cos \gamma_{\text {crit }}-\cos \gamma_{\mathrm{EI}}\right)}{\rho_{0}(L / D) \cos \phi}\right]=69.10 \mathrm{~km}
Finally, we use Eq. (11.94) to find the critical velocity for peak heating:
\nu_{\text {crit }}=\nu_{\mathrm{EI}} \exp \left[\frac{\gamma_{\mathrm{EI}}-\gamma_{\text {crit }}}{(L / D) \cos \phi}\right]=7.546 \mathrm{~km} / \mathrm{s}
The pull-up altitude and velocity in Example 11.3 are 69.08 \mathrm{~km} and 7.532 \mathrm{~km} / \mathrm{s}, respectively. The peak heating rate occurs just before the pull-up altitude where \gamma=0.