Question 11.2: The Space Shuttle is in a 300-km altitude circular orbit whe...

a) Figure 11.8 shows that the 90 \mathrm{~m} / \mathrm{s} de-orbit burn (anti-tangent to LEO) creates a transfer orbit with an apogee that is tangent to LEO. The apogee velocity on the transfer orbit is

\nu_{a}=\sqrt{\frac{\mu}{r_{\mathrm{LEO}}}}-\Delta \nu=7.726-0.090=7.636 \mathrm{~km} / \mathrm{s}

where r_{\mathrm{LEO}}=6,678 \mathrm{~km}. The energy, semimajor axis, angular momentum, parameter, and eccentricity of the transfer orbit are

\text { Energy: } \quad \xi=\frac{\nu_{a}^{2}}{2}-\frac{\mu}{r_{\text {LEO }}}=\frac{-\mu}{2 a}=-30.536 \mathrm{~km}^{2} / \mathrm{s}^{2}

\rightarrow \text { semimajor axis is } a=6,526.8 \mathrm{~km}

Angular momentum and parameter: h=r_{\mathrm{LEO}} \nu_{a}=\sqrt{p \mu}=50,992 \mathrm{~km}^{2} / \mathrm{s}

\rightarrow parameter is p=6,523.3 \mathrm{~km}

\text { Eccentricity: } e=\sqrt{1-\frac{p}{a}}=0.0232

The perigee radius of the transfer orbit is

r_{p}=\frac{p}{1+e}=6,375.6 \mathrm{~km} \approx r_{0}

We see that perigee radius of the Keplerian transfer is nearly equal to the Earth’s mean radius as shown in Figure 11.8 (of course, the perigee radius is fictitious because the Shuttle does not follow a Keplerian orbit beyond EI). Velocity at EI is obtained from the energy equation using r_{\mathrm{EI}}=R_{E}+122 \mathrm{~km}=6,500 \mathrm{~km}

\xi=\frac{\nu_{\mathrm{EI}}^{2}}{2}-\frac{\mu}{r_{\mathrm{EI}}}=-30.536 \mathrm{~km}^{2} / \mathrm{s}^{2} \rightarrow \nu_{\mathrm{EI}}=7.847 \mathrm{~km} / \mathrm{s}

We can compute flight-path angle at EI from angular momentum (h), r_{\mathrm{EI}}, and \nu_{\mathrm{EI}}

\cos \gamma_{\mathrm{EI}}=\frac{h}{r_{\mathrm{EI}} \nu_{\mathrm{EI}}}=0.99974 \rightarrow \gamma_{\mathrm{EI}}=-1.306^{\circ}

Of course, flight-path angle at EI is negative because the Shuttle is approaching Earth. We can use the transfer orbit elements to compute the true anomaly of the Shuttle at EI: \theta_{\mathrm{EI}}=278.9^{\circ} as shown in Figure 11.8 .

b) We use Eq. (11.38)

\nu=\sqrt{\frac{g_0 r_0}{1+(L / D) \frac{r_0 \cos \phi}{2 C_B} \rho_0 e^{-\beta h}}}         (11.38)

to compute the velocity of the Shuttle on an equilibrium glide. We need the Shuttle’s ballistic coefficient; using Eq. (11.11) we obtain

C_{B}=\frac{m}{S C_{D}}=410 \mathrm{~kg} / \mathrm{m}^{2}

Using Earth’s atmosphere parameters \left(\beta\right. and \left.\rho_{0}\right) from Table 11.1, L / D=1.1, \phi=50^{\circ}, r_{0}=6,375,416 \mathrm{~m}, and g_{0}=9.80665 \mathrm{~m} / \mathrm{s}^{2}, we may now use Eq. (11.38) to compute the Shuttle’s velocity on the equilibrium glide at the three altitudes:

At 80 \mathrm{~km} altitude: \nu=\sqrt{\frac{g_{0} r_{0}}{1+(L / D) \frac{r_{0} \cos \phi}{2 C_{B}} \rho_{0} e^{-\beta h}}}=7,506 \mathrm{~m} / \mathrm{s}

At 60 \mathrm{~km} altitude: \nu=4,787 \mathrm{~m} / \mathrm{s}

At 40 \mathrm{~km} altitude: \nu=1,489 \mathrm{~m} / \mathrm{s}

Next, we use Eq. (11.42)

{\dot{\nu}}={\frac{{\frac{\nu^{2}}{r_{0}}}-g_{0}}{(L/D)\cos\phi}}     (11.42)

to compute the deceleration (a=-\dot{\nu}) in Earth g_{0}

\frac{a}{g_{0}}=\frac{g_{0}-\frac{\nu^{2}}{r_{0}}}{g_{0}(L / D) \cos \phi}

Deceleration is

c) We can compute velocity and deceleration (in g_{0} ) using Eqs. (11.38) and (11.42) evaluated at altitudes ranging from h_{\mathrm{EI}}=122 to 30 \mathrm{~km}. In addition, we determine a more precise entry trajectory by numerically integrating Eqs. (11.1), (11.2), (11.7), and (11.8)

\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)

{\dot{s}}={\frac{r_{0}}{r}}\nu\cos\gamma\,      (11.8)

with constant C_{D}=0.8, constant C_{L}=(L / D) C_{D}=0.88, constant bank angle \phi=50^{\circ}, inverse-square gravity and an exponential atmosphere modeled by Eq. (11.6).

\rho=\rho_{0}e^{-\beta h}      (11.6)

We use MATLAB’s ode4 5 . m for the numerical integration process. Figure 11.9 shows velocity vs. altitude for the analytical and numerical entry solutions. Note that the velocity of the numerically integrated gliding entry “oscillates” about the analytical equilibrium-glide solution. These oscillations occur because altitude increases and decreases during the “true” gliding entry with fixed bank \phi=50^{\circ} (note that altitude oscillates left-to-right in Figure 11.9 because h is the x-axis). Therefore, the numerically integrated glide entry is not a “true” equilibrium glide: the vertical lift and centrifugal forces do not always balance gravity and the flight-path angle experiences both positive and negative rates during the gliding entry. Figure 11.10 shows the Shuttle’s deceleration vs. velocity. As with Figure 11.9, deceleration from the numerically integrated glide exhibits oscillations about the analytical solution. Again, these oscillations occur because the numerically integrated glide experiences regions where the Shuttle rises above (\dot{\gamma}>0) and dips below (\dot{\gamma}<0) a true equilibrium glide. Below a velocity of about 3 \mathrm{~km} / \mathrm{s}, the analytical equilibrium-glide solution overestimates the deceleration. Nevertheless, the analytical solutions illustrate the general trends of the gliding entry. It is interesting to note that the analytical solution predicts a final velocity of 758 \mathrm{~m} / \mathrm{s} (at an altitude of 30 \mathrm{~km} ), which shows a good match with the velocity at the end of the Space Shuttle’s entry phase (about Mach 2.5).

d) We use Eq. (11.47) to predict the total ground-track range:

s=\frac{r_{0}}{2}(L / D) \cos \phi \ln \left(\frac{\nu_{s}^{2}-\nu_{2}^{2}}{\nu_{s}^{2}-\nu_{1}^{2}}\right)

The initial velocity \nu_{1} is the velocity at \mathrm{EI}, or \nu_{\mathrm{EI}}=7,847 \mathrm{~m} / \mathrm{s}, and the final velocity at altitude h=30 \mathrm{~km} is \nu_{2}=758 \mathrm{~m} / \mathrm{s} [as determined by Eq. (11.38); see Figure 11.9]. The circular orbital velocity at the Earth’s surface is

\nu_{s}=\sqrt{\mu / r_{0}}=\sqrt{g_{0} r_{0}}=7,907 \mathrm{~m} / \mathrm{s}

where g_{0}=9.80665 \mathrm{~m} / \mathrm{s}^{2} and r_{0}=6,375,416 \mathrm{~m}. Using these values, the total groundtrack range is

s=9,425 \mathrm{~km}

The total ground-track range from the numerically integrated glide is 9,130 \mathrm{~km}, and therefore the range error is about 3 \%.