Level 4 — ApplicationRocket Flight Mechanics

Rocket Flight Mechanics

⏱ 60 minutes60 marksprintable — key stays hidden on paper

Level 4 (Application: novel problems, no hints) Time limit: 60 minutes Total marks: 60

Use $g_0 = 9.81\ \text{m/s}^2$ , Earth radius $R_E = 6371\ \text{km}$ , $\mu = 3.986\times10^{14}\ \text{m}^3/\text{s}^2$ where needed. State all assumptions.

Question 1 — Stability of a finned sounding rocket [12 marks]

A single-stage rocket has a body diameter $d = 0.15\ \text{m}$ . Barrowman analysis gives the following component normal-force derivatives (referenced to the nose tip) and centres of pressure:

Component	$C_{N\alpha}$	$\bar{x}$ (from nose tip, m)
Nose cone	2.0	0.35
Body/shoulder	0.0	—
Fins (4)	12.5	2.10

(a) Compute the total $C_{N\alpha}$ and the overall centre of pressure $X_{CP}$ from the nose. [4]

(b) At burnout the centre of gravity is at $X_{CG} = 1.55\ \text{m}$ from the nose. Compute the static margin in calibers. Is the rocket statically stable? [4]

(c) The engineer wishes the static margin to be exactly $2.0$ calibers at burnout without moving the fins. State two distinct physical modifications that achieve this and, for one of them, quantify the required change. [4]

Question 2 — Gravity-turn pitch rate [12 marks]

During a gravity turn the thrust is aligned with the velocity vector (angle of attack $\alpha = 0$ ), so pitch is driven only by gravity. The flight-path angle $\gamma$ (measured from local horizontal) evolves as

$\dot{\gamma} = -\frac{g\cos\gamma}{v}$

for a flat-Earth, no-lift model.

(a) A vehicle passes through $\gamma = 80^\circ$ at $v = 400\ \text{m/s}$ . Compute the instantaneous pitch rate $\dot{\gamma}$ in deg/s. [4]

(b) Explain physically why a higher velocity at pitch-over reduces gravity-turn losses, and why launching too shallow early causes excessive drag. [4]

(c) For a constant-speed approximation $v = 400\ \text{m/s}$ , integrate the equation to find the time taken to rotate from $\gamma = 80^\circ$ to $\gamma = 45^\circ$ . [4]

Question 3 — Max-Q determination [12 marks]

A rocket climbs vertically. Its speed and the local air density are approximated over the lower atmosphere by

$v(t) = 25\,t \ \ (\text{m/s}), \qquad \rho(h) = \rho_0 e^{-h/H}$

with $\rho_0 = 1.225\ \text{kg/m}^3$ , scale height $H = 8500\ \text{m}$ , and altitude $h(t) = \tfrac{1}{2}(25)t^2$ .

(a) Write dynamic pressure $q(t)$ as an explicit function of $t$ . [3]

(b) Find the time of maximum dynamic pressure (Max-Q) by setting $dq/dt = 0$ . [5]

Question 4 — Reentry ballistic coefficient & heating [14 marks]

A capsule has mass $m = 3800\ \text{kg}$ , drag coefficient $C_D = 1.4$ , and frontal area $A = 12\ \text{m}^2$ .

(a) Compute the ballistic coefficient $\beta = m/(C_D A)$ . State its physical significance for deceleration altitude. [4]

(b) Two capsules with different $\beta$ enter identically. Explain which decelerates higher in the atmosphere and why a lower $\beta$ is favourable for peak deceleration but not necessarily for total heat load. [4]

(c) The stagnation-point convective heat flux follows a Chapman-type relation $\dot{q}_s = C\sqrt{\rho}\,v^{3}$ . If, during descent, the velocity halves while density increases by a factor of 40, by what factor does $\dot{q}_s$ change? Comment on where along the trajectory peak heating occurs relative to peak deceleration. [6]

Question 5 — Suicide burn (terminal landing) [10 marks]

A lander descends vertically under gravity $g = 3.71\ \text{m/s}^2$ (Mars). At the ignition altitude $h$ it has downward speed $v_0 = 250\ \text{m/s}$ . Its engine provides a constant net (thrust − gravity) deceleration $a = 15\ \text{m/s}^2$ once lit, and the burn must bring it to rest exactly at the ground.

(a) Derive the required ignition altitude $h$ in terms of $v_0$ and $a$ , and evaluate it. [4]

(b) Explain why igniting late (below the ideal $h$ ) is catastrophic while igniting early merely wastes propellant, and relate this to why the suicide burn is control-sensitive. [3]

(c) If actuator delay causes ignition $0.4\ \text{s}$ late (vehicle keeps falling at constant $v_0$ during the delay before decel begins), find the residual impact speed. [3]

Answer keyMark scheme & solutions

Question 1

(a) Total normal-force derivative (body contributes 0): $C_{N\alpha,\text{tot}} = 2.0 + 12.5 = 14.5 \quad [1]$ CP is the $C_{N\alpha}$ -weighted mean of component CPs: $X_{CP} = \frac{\sum C_{N\alpha,i}\bar{x}_i}{\sum C_{N\alpha,i}} = \frac{2.0(0.35) + 12.5(2.10)}{14.5} \quad [2]$ $= \frac{0.70 + 26.25}{14.5} = \frac{26.95}{14.5} = 1.859\ \text{m} \quad [1]$

(b) Static margin: $SM = \frac{X_{CP} - X_{CG}}{d} = \frac{1.859 - 1.55}{0.15} = \frac{0.309}{0.15} = 2.06\ \text{cal} \quad [3]$ Since $SM > 0$ (and $> 1$ caliber), the rocket is statically stable — CP is aft of CG so a disturbance produces a restoring moment (weathercocking). [1]

(c) Need $SM = 2.0$ , i.e. $X_{CP} - X_{CG} = 2.0\times0.15 = 0.30\ \text{m}$ , so target $X_{CG} = 1.859 - 0.30 = 1.559\ \text{m}$ (must move CG aft by $1.559 - 1.55 = 0.009\ \text{m} = 9\ \text{mm}$ ). [2] Two valid modifications (any two): (i) add ballast mass near the tail to move CG aft; (ii) lengthen the body/add nose mass reduction; (iii) increase fin size to move CP aft (raising SM). Quantify (i): adding mass $\Delta m$ at distance $\ell$ aft of current CG shifts CG by $\Delta X_{CG}=\Delta m\,\ell/(m+\Delta m)$ ; choose $\Delta m,\ell$ to give 9 mm shift. [2]

Question 2

(a) $\dot\gamma = -\frac{g\cos\gamma}{v} = -\frac{9.81\cos80^\circ}{400} \quad [2]$ $\cos80^\circ = 0.17365$ : $\dot\gamma = -\frac{9.81(0.17365)}{400} = -4.259\times10^{-3}\ \text{rad/s} = -0.244^\circ/\text{s} \quad [2]$

(b) Since $\dot\gamma \propto 1/v$ , a higher velocity at pitch-over means gravity turns the vehicle slowly, so less thrust is spent fighting gravity while nearly vertical — but the key: gravity loss $\int g\sin\gamma\,dt$ is reduced by turning downrange sooner at adequate speed. Launching too shallow early puts the vehicle into denser atmosphere at high speed → high dynamic pressure → large drag losses and structural loads. [4]

(c) With $v$ constant: $\frac{d\gamma}{\cos\gamma} = -\frac{g}{v}dt$ $\int_{80^\circ}^{45^\circ}\sec\gamma\,d\gamma = -\frac{g}{v}\,t$ $\int\sec\gamma\,d\gamma = \ln|\sec\gamma+\tan\gamma|$ . Evaluate:

at $80^\circ$ : $\sec+\tan = 5.759+5.671=11.430$ , $\ln = 2.4364$
at $45^\circ$ : $\sec+\tan = 1.4142+1.0=2.4142$ , $\ln = 0.8814$

$\Delta = 0.8814 - 2.4364 = -1.5550 = -\frac{g}{v}t \quad [2]$ $t = \frac{1.5550\times400}{9.81} = 63.4\ \text{s} \quad [2]$

Question 3

(a) $h(t)=12.5t^2$ , so $q(t) = \tfrac12\rho v^2 = \tfrac12\rho_0 e^{-12.5t^2/8500}(25t)^2 = 382.8\,t^2 e^{-t^2/680}\ \text{Pa} \quad [3]$ (using $\tfrac12(1.225)(625)=382.8$ , and $12.5/8500 = 1/680$ ).

(b) Differentiate $q = K t^2 e^{-t^2/680}$ : $\frac{dq}{dt}=K\Big(2t\,e^{-t^2/680} + t^2 e^{-t^2/680}(-2t/680)\Big)=0 \quad [2]$ $2t - \frac{2t^3}{680}=0 \Rightarrow 1 = \frac{t^2}{680} \Rightarrow t^2 = 680 \quad [2]$ $t = 26.08\ \text{s} \quad [1]$

(c)

Altitude: $h = 12.5(680) = 8500\ \text{m}$ (= scale height, as expected). [1]
Speed: $v = 25(26.08) = 652.0\ \text{m/s}$ . [1]
$q_{max} = 382.8(680)e^{-1} = 260,300\,(0.3679) = 9.58\times10^4\ \text{Pa} \approx 95.7\ \text{kPa}$ . [2]

Question 4

(a) $\beta = \frac{m}{C_D A} = \frac{3800}{1.4\times12} = \frac{3800}{16.8} = 226.2\ \text{kg/m}^2 \quad [2]$ Physical significance: $\beta$ measures the ratio of inertia to drag; higher $\beta$ → the body penetrates deeper before decelerating (deceleration occurs lower/denser). A "heavy, small-area, slippery" body has high $\beta$ . [2]

(b) Lower- $\beta$ capsule decelerates higher in the atmosphere (less mass per unit drag area), so it experiences peak deceleration at higher altitude / lower density → lower peak $g$ . However total integrated heat load $\int\dot q\,dt$ can be larger for low $\beta$ because it spends longer decelerating in the atmosphere (longer exposure time), even though peak heat flux is lower. [4]

(c) $\dot q_s = C\sqrt\rho\,v^3$ . Ratio: $\frac{\dot q_2}{\dot q_1} = \sqrt{\frac{\rho_2}{\rho_1}}\left(\frac{v_2}{v_1}\right)^3 = \sqrt{40}\times(0.5)^3 \quad [3]$ $= 6.3246\times0.125 = 0.7906 \quad [2]$ So heat flux decreases to about $0.79\times$ (a ~21% drop). Because $\dot q\propto v^3$ but only $\rho^{1/2}$ , velocity dominates: peak heating occurs earlier (higher, faster) than peak deceleration, which is set by $\rho v^2$ and occurs deeper. [1]

Question 5

(a) Constant deceleration from $v_0$ to rest over distance $h$ : $0 = v_0^2 - 2a h \Rightarrow h = \frac{v_0^2}{2a} = \frac{250^2}{2(15)} = \frac{62500}{30} = 2083.3\ \text{m} \quad [4]$

(b) Igniting late means the vehicle is below the minimum stopping distance; even at full deceleration it cannot arrest before ground contact → crash (no recovery, since thrust is already maxed). Igniting early leaves excess altitude when velocity reaches zero — vehicle hovers/wastes propellant but lands safely. This asymmetry (catastrophic vs merely wasteful) makes the suicide burn extremely timing/control-sensitive. [3]

(c) During $0.4\ \text{s}$ delay it falls at $v_0 = 250$ , losing $250\times0.4 = 100\ \text{m}$ of margin, so only $h' = 2083.3 - 100 = 1983.3\ \text{m}$ remains for braking. Residual speed at ground: $v_f^2 = v_0^2 - 2a h' = 62500 - 2(15)(1983.3) = 62500 - 59500 = 3000 \quad [2]$ $v_f = 54.8\ \text{m/s} \quad [1]$ (Impact at ~55 m/s — destructive.)

[
  {"claim":"Q1 CP location = 1.859 m and static margin = 2.06 cal","code":"CN=2.0+12.5; XCP=(2.0*0.35+12.5*2.10)/CN; SM=(XCP-1.55)/0.15; result = (abs(XCP-1.85862)<1e-3) and (abs(SM-2.057)<1e-2)"},
  {"claim":"Q2c gravity-turn time approx 63.4 s","code":"import sympy as sp; g=sp.Function; sec=lambda x: 1/sp.cos(x); tan=sp.tan; a=sp.log(1/sp.cos(sp.rad(80))+sp.tan(sp.rad(80))); b=sp.log(1/sp.cos(sp.rad(45))+sp.tan(sp.rad(45))); t=float((a-b)*400/9.81); result = abs(t-63.4)<0.5"},
  {"claim":"Q3 Max-Q at t=26.08s, h=8500m, qmax approx 95.7 kPa","code":"import sympy as sp; t=sp.sqrt(680); h=12.5*680; v=25*float(t); q=382.8*680*sp.exp(-1); result = (abs(float(t)-26.077)<1e-2) and (h==8500) and (abs(float(q)-95740)<300)"},
  {"claim":"Q4c heat flux ratio = 0.7906","code":"import sympy as sp; r=sp.sqrt(40)*sp.Rational(1,8); result = abs(float(r)-0.79057)<1e-3"},
  {"claim":"Q5 ignition altitude 2083.3 m and delayed impact speed 54.8 m/s","code":"h=250**2/(2*15); hp=h-100; vf=(62500-2*15*hp)**0.5; result = (abs(h-2083.33)<0.1) and (abs(vf-54.77)<0.1)"}
]