22-Mec-B7 Aero and Space Flight · December 2013

Question 7 of 7: Ballistic re-entry of an iron sphere — peak deceleration and impact speed

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, December 2013 — 07-Mec-B7 Aero and Space Flight. Three hours, OPEN BOOK, any non-communicating calculator permitted. Seven questions are printed and all carry equal value; the paper states that any six constitute a complete examination and that only the first six appearing in the answer book will be marked, so each question is worth 20 of the 120 marks a candidate can attempt. The printed marking key splits every question into its sub-parts, and those weights are reproduced in the headings below. All seven questions are solved here, because this set is a study resource rather than a timed sitting. The paper also instructs the candidate to state any assumption made where a required quantity has been omitted; that instruction is used explicitly in Questions 4 and 5, where the thrust lapse with altitude and the ground-run averaging method are not given.

Reference texts. Solutions follow the conventions of the texts recommended for this examination code:

J. D. Anderson Jr., Introduction to Flight, 9th ed. — the standard atmosphere and the four altitudes (Ch. 3), incompressible aerodynamics and the Pitot-static tube (§3.4, §4.11), airplane performance (Ch. 6), atmospheric entry (§8.11). This is the primary reference throughout.
J. D. Anderson Jr., Fundamentals of Aerodynamics, 6th ed. — finite-wing theory and induced drag (Ch. 5), transonic flow and drag divergence (Ch. 11).
B. N. Pamadi, Performance, Stability, Dynamics and Control of Airplanes, 3rd ed. — take-off and landing field lengths (Ch. 2), turning flight and the load factor (Ch. 2), static stability (Ch. 3).
W. F. Phillips, Mechanics of Flight, 2nd ed. — the parabolic drag polar, minimum-drag and maximum-climb speeds (Ch. 3).
P. H. Oosthuizen & W. E. Carscallen, Introduction to Compressible Fluid Flow, 2nd ed. — critical Mach number and compressibility effects (Ch. 1, Ch. 8).
H. J. Allen & A. J. Eggers, A Study of the Motion and Aerodynamic Heating of Ballistic Missiles Entering the Earth's Atmosphere at High Supersonic Speeds, NACA Report 1381 (1958) — the closed-form ballistic-entry solution used in Question 7.

SI units are used throughout, matching the paper. The International Standard Atmosphere constants are taken as $p_{0}=101.325\ \text{kPa}$, $T_{0}=288.15\ \text{K}$, $\rho_{0}=1.225\ \text{kg}\,\text{m}^{-3}$, lapse rate $L=0.0065\ \text{K}\,\text{m}^{-1}$ through the troposphere, $R=287.05\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$ and $g=9.80665\ \text{m}\,\text{s}^{-2}$ inside the atmosphere model. Aircraft weights use the rounded $g=9.81\ \text{m}\,\text{s}^{-2}$ that Canadian examination practice expects. All pressures are absolute.

Question 7: Ballistic re-entry of an iron sphere — peak deceleration and impact speed (20 marks — 5 marks each)

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.

Given.

Quantity	Value
sphere diameter and material density	$d=1.2\ \text{m}$, $\rho_{s}=7000\ \text{kg}\,\text{m}^{-3}$
entry velocity and path angle	$V_{E}=11\ \text{km}\,\text{s}^{-1}$, $\gamma=14^{\circ}$ below the local horizontal
drag coefficient on frontal area	$C_{D}=1$
atmosphere model	$\rho=\rho_{0}e^{-\beta h}$ with $\beta=0.00012\ \text{m}^{-1}$, $\rho_{0}=1.225\ \text{kg}\,\text{m}^{-3}$

Find. The altitude of peak deceleration, the peak deceleration itself, the speed at which it occurs, and the speed with which the sphere arrives at sea level.

Allen–Eggers geometry: the flight-path angle is held constant, so altitude and path length are locked together and the equation of motion reduces to one integrable ordinary differential equation.

Approach. Because the sphere generates no lift and the entry is steep enough for gravity to be negligible beside the drag, the path is a straight line at fixed $\gamma$; that lets altitude replace time as the independent variable and reduces the equation of motion to a single separable differential equation whose integral is the Allen–Eggers velocity law.

Compute the mass, frontal area and ballistic coefficient. $$\begin{aligned} m &= \rho_{s}\frac{\pi d^{3}}{6} \\ &= 7000\times\frac{\pi(1.2)^{3}}{6} \\ &= 6333\ \text{kg}, \\ A &= \frac{\pi d^{2}}{4}=1.131\ \text{m}^{2} \end{aligned}$$ so the ballistic coefficient is $m/(C_{D}A)=6333/1.131=5600\ \text{kg}\,\text{m}^{-2}$, a very high value: this is a dense, small, blunt object that will penetrate deeply before the air can slow it.
Write the equation of motion along the path and change variable to altitude. Neglecting gravity beside the drag, and with $s$ measured along the straight path, $$\begin{aligned} m\frac{\mathrm{d}V}{\mathrm{d}t} &= -\tfrac{1}{2}\rho V^{2}C_{D}A, \\ \frac{\mathrm{d}h}{\mathrm{d}t} &= -V\sin\gamma \end{aligned}$$ Dividing the first by the second removes time completely and leaves $$\frac{\mathrm{d}V}{V}=\frac{C_{D}A\rho_{0}}{2m\sin\gamma}\,e^{-\beta h}\,\mathrm{d}h$$
Integrate from the entry interface to a general altitude. Integrating the right-hand side from $h=\infty$, where the density vanishes and $V=V_{E}$, down to $h$ gives the Allen–Eggers result $$\begin{aligned} V &= V_{E}\exp\!\left(-K\,e^{-\beta h}\right), \\ K &= \frac{C_{D}A\rho_{0}}{2m\beta\sin\gamma} \end{aligned}$$ Numerically, with $\sin14^{\circ}=0.24192$, $$\begin{aligned} K &= \frac{1\times1.131\times1.225}{2\times6333\times0.00012\times0.24192} \\ &= \frac{1.3854}{0.36773} \\ &= 3.7676 \end{aligned}$$
Express the deceleration in terms of a single variable. Writing $u=K e^{-\beta h}$, so that $V=V_{E}e^{-u}$ and $\rho=\rho_{0}e^{-\beta h}=\rho_{0}u/K$, the deceleration becomes $$\left|\frac{\mathrm{d}V}{\mathrm{d}t}\right| =\frac{\rho C_{D}AV^{2}}{2m} =\beta V_{E}^{2}\sin\gamma\;u\,e^{-2u}$$ Every property of the sphere has cancelled from the prefactor except through $u$ itself — a remarkable and much-used result.
Part (a) — locate the maximum. The function $ue^{-2u}$ is stationary where $\mathrm{d}(ue^{-2u})/\mathrm{d}u=e^{-2u}(1-2u)=0$, that is at $u=\tfrac{1}{2}$. Hence $Ke^{-\beta h}=\tfrac{1}{2}$ and $$\boxed{\;h_{\max}=\frac{\ln(2K)}{\beta}=\frac{\ln(7.5352)}{0.00012} =16{,}831\ \text{m}\approx16.8\ \text{km}\;}$$
Part (b) — evaluate the peak deceleration. Substituting $u=\tfrac{1}{2}$, $$\boxed{\;\left|\frac{\mathrm{d}V}{\mathrm{d}t}\right|_{\max} =\frac{\beta V_{E}^{2}\sin\gamma}{2e} =\frac{0.00012\times(11\,000)^{2}\times0.24192}{2\times2.71828} =646\ \text{m}\,\text{s}^{-2}=65.9\,g\;}$$ The mass, diameter, drag coefficient and material density have all disappeared: the peak deceleration of any non-lifting entry body depends only on the entry speed, the path angle and the atmospheric scale height. The ballistic coefficient decides where the peak occurs, not how large it is.
Part (c) — the speed at the peak. Since $u=\tfrac{1}{2}$ there, $$\boxed{\;V=V_{E}e^{-1/2}=11\,000\times0.60653=6672\ \text{m}\,\text{s}^{-1} =6.67\ \text{km}\,\text{s}^{-1}\;}$$ that is $60.7\%$ of the entry speed — again a universal fraction, independent of the body.
Part (d) — the speed at sea level. Setting $h=0$ makes $u=K$, so $$\boxed{\;V_{\text{surface}}=V_{E}e^{-K}=11\,000\times e^{-3.7676} =11\,000\times0.023114=254\ \text{m}\,\text{s}^{-1}\;}$$ The sphere loses more than $97.5\%$ of its entry speed, yet arrives at roughly the speed of a rifle bullet, carrying about $205\ \text{MJ}$ of kinetic energy.

The deceleration is the product of a density that grows downward and a velocity squared that collapses downward; the product peaks where the speed has fallen to 1/√e of the entry value.

Check: gravity neglected near the ground. The Allen–Eggers solution assumes drag dominates gravity, which is excellent above about $10\ \text{km}$ but fails in the last few kilometres, where the sphere has slowed. Its unaccelerated terminal velocity is $$\begin{aligned} V_{t} &= \sqrt{\frac{2mg}{\rho_{0}C_{D}A}} \\ &= \sqrt{\frac{2(6333)(9.81)}{1.225\times1.131}} \\ &= 300\ \text{m}\,\text{s}^{-1} \end{aligned}$$ which is above the $254\ \text{m}\,\text{s}^{-1}$ just computed. In reality gravity would re-accelerate the sphere over the last kilometre or so and it would arrive close to $300\ \text{m}\,\text{s}^{-1}$. The answer requested is the Allen–Eggers value, $254\ \text{m}\,\text{s}^{-1}$; the terminal-velocity comparison should be quoted as a check on its validity.

Quantity	Result
mass / frontal area / ballistic coefficient	$6333\ \text{kg}$ / $1.131\ \text{m}^{2}$ / $5600\ \text{kg}\,\text{m}^{-2}$
(a) altitude of maximum deceleration	$16{,}831\ \text{m}$ ($16.8\ \text{km}$)
(b) maximum deceleration	$646\ \text{m}\,\text{s}^{-2}$ ($65.9\,g$)
(c) velocity at maximum deceleration	$6672\ \text{m}\,\text{s}^{-1}$ ($6.67\ \text{km}\,\text{s}^{-1}$)
(d) velocity at the earth's surface	$254\ \text{m}\,\text{s}^{-1}$

Topic. The Allen–Eggers closed-form solution for ballistic atmospheric entry.

Key relations. $V=V_{E}\exp(-Ke^{-\beta h})$ with $K=C_{D}A\rho_{0}/(2m\beta\sin\gamma)$; peak deceleration at $u=Ke^{-\beta h}=1/2$, hence $h_{\max}=\ln(2K)/\beta$; $|\mathrm{d}V/\mathrm{d}t|_{\max}=\beta V_{E}^{2}\sin\gamma/(2e)$; $V=V_{E}e^{-1/2}$ there; ballistic coefficient $m/(C_{D}A)$.

Why this works. Two approximations make the problem integrable and both are physically defensible for a steep, high-speed, non-lifting entry. First, the drag at these speeds is enormous compared with the weight — sixty-six times it at the peak — so gravity may be dropped from the equation of motion. Second, with no lift and negligible gravity the flight-path angle cannot change, so altitude and path length are proportional and time can be eliminated. What remains separates exactly. The physical content of the answer is the competition between two opposing exponentials: density grows as the body descends while speed collapses, and the product of density and speed squared must therefore peak somewhere in between. That the peak always occurs at exactly $1/\sqrt{e}$ of the entry speed, and that its magnitude is independent of the vehicle, is what makes this a design tool rather than a curiosity: it tells a designer at once that a shallow entry angle is the only way to reduce peak load, which is why crewed capsules enter at five to seven degrees and ballistic warheads at twenty or more.

Common pitfall. Using $\cos\gamma$ instead of $\sin\gamma$ because the angle is quoted to the local horizontal — the rate of altitude loss is $V\sin\gamma$, and the error changes $K$ by a factor of four. A second is dropping the factor $\rho_{0}$ from $K$ or forgetting that $\beta$ is per metre, not per kilometre. A third is stopping at the Allen–Eggers impact speed without checking it against the terminal velocity; whenever the two are comparable the no-gravity assumption has expired.

Source. Anderson, Introduction to Flight, 9th ed., §8.11 (atmospheric entry, the Allen–Eggers analysis, deceleration and heating); Allen & Eggers, NACA Report 1381 (1958) for the original derivation.

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