Question 7 of 7: Rocket Staging, Ballistic Re-entry, Re-entry Heating and Escape Velocity

Question 7: Rocket Staging, Ballistic Re-entry, Re-entry Heating and Escape Velocity (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Velocity achieved by the two-stage vehicle

Given. Two identical stages, each of initial mass 1500 kg, each with an initial-to-final mass ratio of 6.7, and each with an exhaust velocity of 3800 m/s. Gravity and drag are to be ignored.

Find. The final velocity of the vehicle after both stages have burnt out.

Approach. Apply the Tsiolkovsky rocket equation to each stage in turn, taking care that the mass ratio used in each burn is the ratio of the vehicle masses at the start and end of that burn — not the stage's own mass ratio, except for the final stage.

1. Break each stage into propellant and structure. A stage of initial mass 1500 kg with an initial-to-final mass ratio of 6.7 burns out at $1500/6.7 = 223.9\ \text{kg}$, so $$m_{propellant} = 1500 - 223.9 = 1276.1\ \text{kg}, \qquad m_{structure} = 223.9\ \text{kg}.$$
2. First burn. At lift-off the vehicle is both stages together, $m_0 = 2(1500) = 3000\ \text{kg}$. When the first stage's propellant is exhausted the vehicle mass is $3000 - 1276.1 = 1723.9\ \text{kg}$. The rocket equation gives $$\Delta V_1 = V_e \ln\!\left(\frac{m_0}{m_1}\right) = 3800\,\ln\!\left(\frac{3000}{1723.9}\right) = 3800(0.5540) = 2105\ \text{m}/\text{s}.$$
3. Staging. The spent first-stage structure of 223.9 kg is jettisoned, leaving the intact second stage of 1500 kg. No velocity change accompanies this event.
4. Second burn. The second stage now flies alone, so its own mass ratio applies in full: $$\Delta V_2 = V_e \ln(6.7) = 3800(1.9021) = 7228\ \text{m}/\text{s}.$$
5. Total velocity. With no gravity or drag losses the increments simply add: $$V = \Delta V_1 + \Delta V_2 = 2105 + 7228 = \boxed{9333\ \text{m}/\text{s}}$$

The value of staging is visible in the comparison. A single-stage vehicle of the same 3000 kg launch mass and the same 223.9 kg of surviving structure would reach $3800\ln(3000/223.9) = 9862\ \text{m}/\text{s}$ — higher, but only because that hypothetical vehicle magically discards structure continuously. Compare instead with what the two-stage vehicle would achieve if it could not stage at all, i.e. if it had to carry both structures to the end: $3800\ln(3000/447.8) = 7228\ \text{m}/\text{s}$, some 2100 m/s less. That 2100 m/s is the payoff of throwing the first stage away, and it is why every launch vehicle that has reached orbit has been staged.

(b) Maximum deceleration during ballistic re-entry

Given. The entry data are collected below.

Find. The maximum deceleration experienced during entry (and, to close the answer, the altitude at which it occurs).

Approach. This is the classical Allen–Eggers ballistic-entry problem: a non-lifting body entering an exponential atmosphere on a straight path, with gravity neglected compared with the very large drag deceleration. The entry speed is the circular orbital speed at 550 km.

1. Entry velocity. For a circular orbit of radius $r = R_E + h = 6378 + 550 = 6928\ \text{km}$, with $\mu = GM_E = 3.986\times10^{14}\ \text{m}^3/\text{s}^2$, $$V_E = \sqrt{\frac{\mu}{r}} = \sqrt{\frac{3.986\times10^{14}}{6.928\times10^{6}}} = 7585\ \text{m}/\text{s}.$$
2. Allen–Eggers velocity profile. Along a straight path inclined at $\gamma$ to the horizontal, with $dh = -\sin\gamma\,ds$, integrating $m\,V\,dV/ds = -\tfrac{1}{2}\rho V^2 C_D A$ with $\rho = \rho_0 e^{-\beta h}$ gives $$V = V_E \exp\!\left(-K e^{-\beta h}\right), \qquad K = \frac{C_D A \rho_0}{2 m \beta \sin\gamma}.$$
3. Evaluate the entry parameter. With $\beta = 0.00012\ \text{m}^{-1}$ and $\sin(10.5^\circ) = 0.18224$, $$K = \frac{1.05(5)(1.225)}{2(2200)(0.00012)(0.18224)} = \frac{6.431}{0.09622} = 66.84.$$
4. Locate the peak. The deceleration is $a = \tfrac{1}{2}\rho V^2 C_D A/m$; writing $x = K e^{-\beta h}$ it becomes $a = \beta V_E^2 \sin\gamma\; x\,e^{-2x}$, which is maximised at $x = 1/2$. Hence the peak occurs where $K e^{-\beta h} = 1/2$: $$h_{peak} = \frac{\ln(2K)}{\beta} = \frac{\ln(133.7)}{0.00012} = \boxed{40\,800\ \text{m} \approx 40.8\ \text{km}}$$ and the speed there is $V = V_E e^{-1/2} = 0.6065\,V_E = 4601\ \text{m}/\text{s}$.
5. Maximum deceleration. Substituting $x = 1/2$, $$a_{\max} = \frac{\beta V_E^2 \sin\gamma}{2e} = \frac{0.00012(7585)^2(0.18224)}{2(2.71828)} = \boxed{231\ \text{m}/\text{s}^2 = 23.6\,g}$$

Two features of this result are worth stating explicitly. First, $a_{\max}$ depends only on the entry speed, the path angle and the atmospheric scale — the mass, the drag coefficient and the frontal area have all cancelled. Those quantities affect only where the peak occurs, through $K$ in step 4. Second, 23.6 g is far beyond human tolerance (about 12 g for a few seconds, in the eyeballs-in direction), which is exactly why crewed vehicles enter at much shallower angles — roughly 1.5° for Apollo, giving under 7 g — or generate lift to stretch the entry, and why 10.5° is the sort of angle used for an unmanned ballistic capsule.

Finally, the validity of the no-gravity assumption should be checked at the low-altitude end. The sea-level terminal velocity of this body would be $\sqrt{2mg/(\rho_0 C_D A)} = 81.9\ \text{m}/\text{s}$, and the Allen–Eggers profile gives a surface speed of $V_E e^{-K} \approx 0$, so by the time the vehicle is low in the atmosphere the drag deceleration has become small compared with $g$ and the analysis has ceased to apply. That does not affect the answer, because the peak deceleration occurs at 41 km where the drag force is 23.6 times the weight; but the descent below about 20 km is a terminal-velocity problem, not a ballistic-entry one, which is why parachutes are deployed there.

(c) Re-entry heating and heat shields

Re-entry heating is the conversion of a re-entering vehicle's enormous kinetic energy into thermal energy in the surrounding air, and thence partly into the vehicle. A body entering at 7.6 km/s carries about 29 MJ/kg of kinetic energy — several times the energy released by burning a kilogram of TNT — and essentially all of it must be disposed of before touchdown. The mechanism is not, as is often supposed, friction. The dominant process is compression: the bow shock ahead of a blunt body raises the stagnation temperature of the air to the order of $T_0 = T_\infty(1 + 0.2M^2)$, which at $M = 25$ is many thousands of kelvin — hot enough to dissociate and ionise the gas. Heat then reaches the surface by two paths: convective heating from the hot boundary layer, which scales roughly as $V^3\sqrt{\rho/R_n}$ and so is reduced by a large nose radius $R_n$; and radiative heating from the incandescent gas in the shock layer, which scales as a much higher power of velocity and dominates at lunar-return and interplanetary entry speeds.

The key design insight, again due to Allen and Eggers, is that a blunt body is far better than a slender one. A blunt shape produces a strong detached bow shock that dumps most of the entry energy into the air stream far from the surface, and it also decelerates the vehicle high in the atmosphere where the air is thin. Slenderness minimises drag, which is the opposite of what an entry vehicle wants.

A heat shield is the thermal protection system that keeps the underlying structure below its design temperature during this period. Three families are used. Ablative shields — phenolic-impregnated carbon or resin-loaded honeycomb, as on Apollo, Soyuz, Orion and every planetary entry probe — work by pyrolysing and vaporising the outer layer: the phase change absorbs enormous energy, and the outflow of pyrolysis gas blocks part of the convective heat flux by thickening the boundary layer. They are highly effective but are consumed and must be replaced. Radiative or "hot structure" systems, such as the reinforced carbon–carbon leading edges and silica tiles of the Space Shuttle orbiter, instead let the surface reach a high equilibrium temperature (1500 K or more) and re-radiate the incoming heat to space, with a low-conductivity insulator preventing it from reaching the airframe; these are reusable but fragile and maintenance-intensive. Transpiration or actively cooled systems, in which coolant is bled through a porous surface, have been studied for hypersonic cruise vehicles but are rarely used operationally.

(d) Escape velocity

Escape velocity is the minimum speed a body must be given at a specified point in a gravitational field so that, with no further propulsion, it will never return — it recedes indefinitely, arriving at infinite distance with exactly zero residual speed. It follows immediately from conservation of energy. The specific mechanical energy of a body at radius $r$ moving at speed $V$ is $\varepsilon = V^2/2 - \mu/r$, and escape corresponds to $\varepsilon = 0$ (a parabolic trajectory), giving

At the Earth's surface, with $\mu = 3.986\times10^{14}\ \text{m}^3/\text{s}^2$ and $R_E = 6378\ \text{km}$, this gives $V_{escape} = 11.2\ \text{km/s}$, compared with a hypothetical surface circular speed of 7.91 km/s. Three properties should be included in a complete answer. First, escape velocity is a speed, not a velocity in the vector sense: the direction does not matter (only that the path misses the planet), because gravitational potential energy is a scalar. Second, it falls with altitude as $1/\sqrt{r}$, so a vehicle already in a 550 km orbit at 7.59 km/s needs only $\sqrt{2}(7.59) - 7.59 = 3.14\ \text{km/s}$ more to escape — which is why departures are made from parking orbits. Third, it is not the speed a rocket must reach at lift-off; a continuously thrusting vehicle could in principle leave at any speed, and escape velocity is properly the ballistic requirement for an unpowered body. In practice, launch vehicles do not aim for it directly; they establish a parking orbit and then apply a trans-lunar or trans-planetary injection burn, both to allow checkout before commitment and to avoid the drag and structural loads of accelerating to 11.2 km/s in the lower atmosphere.