Question 11.1: The Soviet Union’s Vostok spacecraft was used for the first ...
The Soviet Union’s Vostok spacecraft was used for the first manned space flights in the early 1960s. It was spherical in shape (with drag coefficient C_{D}= 2.0 ), and had a mass of 2,460 \mathrm{kg} and diameter of 2.3 \mathrm{~m}. The Vostok capsule departs a low-Earth orbit and reaches entry interface (EI) altitude h_{\mathrm{EI}}=122 \mathrm{~km} with velocity \nu_{\mathrm{EI}}=7.74 \mathrm{~km} / \mathrm{s} and flight-path angle \gamma_{\mathrm{EI}}=-3.2^{\circ}. It follows a ballistic entry.
a) Compute the velocity, flight-path angle, and deceleration (in Earth g_{0} ) at altitudes of 80,60 , and 40 \mathrm{~km}.
b) Compute the peak deceleration and associated critical altitude and velocity at peak deceleration.
c) Create plots of velocity and deceleration vs. altitude for the analytical ballistic entry and a numerically integrated ballistic entry profile.
a) The analytical ballistic entry equations require the constants C_{B} and B. Ballistic coefficient for the Vostok capsule is
C_{B}=\frac{m}{S C_{D}}=296.05 \mathrm{~kg} / \mathrm{m}^{2}
where the Vostok reference area is S=\pi(2.3 / 2)^{2}=4.1548 \mathrm{~m}^{2}. The dimensionless coefficient is
B=\frac{\rho_{0}}{2 \beta C_{B} \sin \gamma_{\mathrm{EI}}}=-268.9650
where the values for Earth’s atmospheric constants \rho_{0} and \beta are taken from Table 11.1 (note that \beta must be expressed in units of \mathrm{m}^{-1} if surface density \rho_{0} is in units of \mathrm{kg} / \mathrm{m}^{3} ). Equation (11.18) provides velocity during the ballistic entry as a function of altitude:
\nu=\nu_{\mathrm{EI}} \exp \left[B e^{-\beta h}\right]
The velocities at the required altitudes are
\begin{aligned} & h=80 \mathrm{~km}: \nu=7.7062 \mathrm{~km} / \mathrm{s} \\ & h=60 \mathrm{~km}: \nu=7.2239 \mathrm{~km} / \mathrm{s} \\ & h=40 \mathrm{~km}: \nu=2.6126 \mathrm{~km} / \mathrm{s} \end{aligned}
Note that the Vostok’s velocity changes very little from EI to h=80 \mathrm{~km} (due to low drag) but dramatically changes between 60 and 40 \mathrm{~km} altitude.
The analytical ballistic solution for flight-path angle assumes that it remains constant during the entire entry. Therefore, \gamma=\gamma_{E I}=-3.2^{\circ} at h=80,60, and 40 \mathrm{~km}.
Equation (11.20) determines the vehicle’s acceleration
\dot{\nu}=-\beta B \nu_{\mathrm{EI}}^{2} \sin \gamma_{\mathrm{EI}} e^{-\beta h} \exp \left[2 B e^{-\beta h}\right]
We compute deceleration in units of meters per second squared by expressing \beta in per meter, \nu_{\mathrm{EI}} in meters per second, and h in meters. The values of vehicle deceleration \left(a=-\dot{\nu}\right., in \left.g_{0}\right) at the three altitudes are
\begin{aligned}& h=80 \mathrm{~km}: \quad a=2.003 \mathrm{~m} / \mathrm{s}^{2}=0.204 g_{0} \\& h=60 \mathrm{~km}: \quad a=27.704 \mathrm{~m} / \mathrm{s}^{2}=2.825 g_{0} \\& h=40 \mathrm{~km}: a=57.021 \mathrm{~m} / \mathrm{s}^{2}=5.814 g_{0}\end{aligned}
where we have used g_{0}=9.80665 \mathrm{~m} / \mathrm{s}^{2}.
b) We use Eq. (11.27) to determine the peak deceleration of the Vostok capsule
a_{\max }=\frac{-\beta \nu_{\mathrm{EI}}^{2} \sin \gamma_{\mathrm{EI}}}{2 e}=84.763 \mathrm{~m} / \mathrm{s}^{2}=8.643 g_{0}
Critical altitude for peak deceleration is computed using Eq. (11.26)
h_{\mathrm{crit}}=\frac{\ln (-2 B)}{\beta}=45.629 \mathrm{~km}
We use Eq. (11.30) to predict the critical velocity for peak deceleration:
\nu_{\text {crit }}=0.6065 \nu_{\mathrm{EI}}=4.694 \mathrm{~km} / \mathrm{s}
c) Finally, we can compute velocity and deceleration (in g_{0} ) using Eqs. (11.18) and (11.20) determined at altitudes ranging from h_{\mathrm{EI}} to about zero. We compare the analytical ballistic entry results with a more precise entry trajectory obtained by numerically integrating 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)
with constant C_{D}, C_{L}=0 (zero lift), inverse-square gravity (11.3), and an exponential atmosphere modeled by Eq. (11.6).
\rho=\rho_{0}e^{-\beta h} (11.6)
We employ MATLAB’s M-file ode45.m to numerically integrate the three nonlinear ODEs. Figure 11.5 shows velocity vs. altitude for the analytical and numerical entry solutions. The Vostok entry trajectory begins in the upper right corner of Figure 11.5 (at h_{\mathrm{EI}}=122 \mathrm{~km} ), and progresses to the left as both altitude and velocity become smaller (note the arrows on the plots). The analytical velocity solution shows a good match with the “true” numerically integrated trajectory despite the major assumption that flight-path angle remains constant. Figure 11.6 presents the Vostok’s deceleration for the analytical and numerical ballistic trajectories. The analytical method accurately predicts deceleration until altitude reaches 50 \mathrm{~km}, which is slightly before the critical altitude of 45.6 \mathrm{~km}. The peak deceleration of the analytical method is about 2.7 g_{0} lower than the “true” peak deceleration from the numerically integrated trajectory.
Figure 11.7 shows how the Vostok spacecraft’s flight-path angle changes with altitude for the numerically integrated ballistic entry. Flight-path angle remains nearly constant for the high-altitude portion of the entry (i.e., the centrifugal force approximately cancels gravity) until an altitude of about 50 \mathrm{~km}. However, the Vostok capsule continues to decelerate (see Figure 11.5), and below 50 \mathrm{~km} gravity dominates the \nu^{2} / r term and consequently the flight-path angle turns downward toward the vertical direction. Below 6 \mathrm{~km} altitude the spacecraft is essentially dropping vertically toward the ground with \gamma=-90^{\circ}. Of course, the first-order analytical method assumes constant flight-path angle (also shown in Figure 11.7) which is not an accurate approximation for h<50 \mathrm{~km}.
