Level 5 — MasteryRocket Flight Mechanics

Rocket Flight Mechanics

⏱ 90 minutes60 marksprintable — key stays hidden on paper

Time limit: 90 minutes Total marks: 60 Instructions: Answer all three questions. Show all derivations, state assumptions, and provide runnable pseudocode where requested. Use SI units throughout unless stated. Take Earth radius $R_E = 6371$ km, $g_0 = 9.81\ \mathrm{m/s^2}$ , $\mu = 3.986\times10^{14}\ \mathrm{m^3/s^2}$ .

Question 1 — Stability of a finned sounding rocket (20 marks)

A single-stage finned rocket has the following geometry (all lengths from the nose tip, body reference diameter $d = 0.20\ \mathrm{m}$ ):

Ogive nose cone, length $L_n = 0.50\ \mathrm{m}$
Cylindrical body, no boat-tail
Four trapezoidal fins at the base: root chord $C_r = 0.30\ \mathrm{m}$ , tip chord $C_t = 0.12\ \mathrm{m}$ , semi-span $s = 0.15\ \mathrm{m}$ , fin leading-edge sweep length (root LE to tip LE, axial) $x_r = 0.18\ \mathrm{m}$ . Fin root leading edge at $X_{fin} = 1.90\ \mathrm{m}$ from nose. Mid-chord line sweep length $m = 0.21\ \mathrm{m}$ .
Total rocket length $L = 2.20\ \mathrm{m}$ .

(a) Using the Barrowman equations, compute the normal-force coefficient derivative $C_{N\alpha}$ and centre of pressure location $X_{CP}$ for the nose and the fin set. State the standard Barrowman assumptions and note which are violated at high angle of attack. (9)

For a nose cone: $(C_{N\alpha})_n = 2$ , $\bar{X}_n = 0.466 L_n$ (ogive). For $N$ fins with body interference factor $K_{fb} = 1 + \frac{r}{s+r}$ (body radius $r$ ): $(C_{N\alpha})_f = K_{fb}\,\frac{4N (s/d)^2}{1 + \sqrt{1 + \left(\frac{2\ell_m}{C_r+C_t}\right)^2}},\quad \ell_m = m$ $\bar{X}_f = X_{fin} + \frac{x_r}{3}\frac{C_r + 2C_t}{C_r + C_t} + \frac{1}{6}\left(C_r + C_t - \frac{C_r C_t}{C_r + C_t}\right)$

(b) The loaded CG is at $X_{CG} = 1.35\ \mathrm{m}$ ; after burnout it shifts to $1.20\ \mathrm{m}$ (propellant near the base depleted). Compute the static margin (in calibers) at both instants and comment on whether the rocket is over- or under-stable at each. (6)

(c) Explain physically the coupling between the CG-shift computed in (b) and dynamic stability. Derive an expression for the pitch-damping contribution of the fins (per unit pitch rate $q$ ) and argue why a rocket that is statically over-stable can still exhibit poor coning behaviour. (5)

Question 2 — Gravity-turn ascent and Max-Q (20 marks)

A rocket ascends under a gravity turn (zero aerodynamic angle of attack throughout the atmospheric phase). The 3DOF point-mass equations in a flat-Earth planar frame, with flight-path angle $\gamma$ measured from horizontal, are:

$m\dot v = T - D - m g \sin\gamma, \qquad m v\dot\gamma = -m g\cos\gamma + L$

(a) For a pure gravity turn, state the aerodynamic condition that fixes $L$ , and hence derive the pitch-rate ODE $\dot\gamma = -\frac{g\cos\gamma}{v}$ . Explain why this makes the pitch program open-loop-free once initiated by a small pitch-over kick. (5)

(b) Using the exponential atmosphere $\rho = \rho_0 e^{-h/H}$ ( $\rho_0 = 1.225\ \mathrm{kg/m^3}$ , $H = 7200\ \mathrm{m}$ ), dynamic pressure is $q = \tfrac12\rho v^2$ . Show that at the Max-Q instant along a near-vertical trajectory, $\frac{1}{v}\frac{dv}{dt} = \frac{\dot h}{2 H}\quad\text{(with } \dot h \approx v\sin\gamma\text{)},$ and interpret what this balance says physically. (6)

(c) Write clean pseudocode (or Python using an explicit RK4 you outline) that integrates the 3DOF gravity-turn equations from lift-off to burnout and returns: burnout velocity, altitude, and the Max-Q value with its altitude. Specify the state vector, all inputs (constant $T$ , mass-flow $\dot m$ , $C_D$ , reference area $A$ ), the drag model, and the stopping condition. Identify the two dominant velocity-loss terms (gravity loss, drag loss) as integrals and state how you would extract them from the integration. (9)

Question 3 — Ballistic reentry, heating and the corridor (20 marks)

A capsule reenters with ballistic coefficient $\beta = m/(C_D A)$ . Using the Allen–Eggers approximation for a steep ballistic entry (constant flight-path angle $\gamma$ , exponential atmosphere $\rho=\rho_0 e^{-h/H}$ , gravity and lift neglected during the deceleration pulse):

$v(h) = v_e\,\exp\!\left[-\frac{\rho_0 H}{2\beta\,\sin\gamma}\,e^{-h/H}\right]$

(a) Derive this velocity–altitude relation from $m\dot v = -\tfrac12\rho v^2 C_D A$ starting from $\frac{dv}{dh}=\frac{\dot v}{\dot h}$ , $\dot h = -v\sin\gamma$ . State each assumption. (6)

(b) Show that the maximum deceleration magnitude is $a_{max} = \frac{v_e^2\,\sin\gamma}{2 e H},$ independent of $\beta$ , and find the altitude at which it occurs. Interpret why $\beta$ drops out. (7)

(c) Stagnation-point heat flux follows a Chapman-type relation $\dot q_s = C\,\sqrt{\rho/R_n}\,v^3$ ( $R_n$ = nose radius). Using $v(h)$ from part (a), determine the altitude of peak heating relative to the altitude of peak deceleration (qualitatively and with a derived condition), and explain the design trade-off between a high- $\beta$ (slender, small $R_n$ ) and low- $\beta$ (blunt, large $R_n$ ) reentry body in terms of the reentry corridor, peak-g, peak-heating, and communications blackout (plasma sheath). (7)

Answer keyMark scheme & solutions

Question 1

(a) Barrowman (9 marks)

Body radius $r = d/2 = 0.10$ m. Interference factor: $K_{fb} = 1 + \frac{r}{s+r} = 1 + \frac{0.10}{0.15+0.10} = 1 + 0.40 = 1.40.$ (1)

Fin term denominator: $\dfrac{2\ell_m}{C_r+C_t} = \dfrac{2(0.21)}{0.30+0.12} = \dfrac{0.42}{0.42}=1.0$ , so $1+\sqrt{1+1^2} = 1+\sqrt2 = 2.4142.$ (1)

$(s/d)^2 = (0.15/0.20)^2 = 0.5625$ , $N=4$ : $(C_{N\alpha})_f = 1.40\cdot\frac{4(4)(0.5625)}{2.4142} = 1.40\cdot\frac{9.0}{2.4142} = 1.40(3.7280) = 5.219.$ (2)

Nose: $(C_{N\alpha})_n = 2$ , $\bar X_n = 0.466(0.50) = 0.233$ m. (1)

Fin CP: $\bar X_f = 1.90 + \frac{0.18}{3}\cdot\frac{0.30+2(0.12)}{0.42} + \frac16\Big(0.42 - \frac{0.30\cdot0.12}{0.42}\Big).$ Second term: $0.06\cdot\frac{0.54}{0.42}=0.06(1.2857)=0.07714$ . Third term: $\frac16(0.42-0.08571)=\frac16(0.33429)=0.05571$ . $\bar X_f = 1.90+0.07714+0.05571 = 2.0329\ \text{m}.$ (2)

Combined: $C_{N\alpha}=2+5.219=7.219$ ; $X_{CP}=\frac{2(0.233)+5.219(2.0329)}{7.219}=\frac{0.466+10.610}{7.219}=\frac{11.076}{7.219}=1.5343\ \text{m}.$ (1)

Assumptions: small $\alpha$ (linear aero), incompressible/low subsonic, slender body, no body-lift on cylinder, no fin–fin/fin–body wake interaction beyond $K_{fb}$ . Violated at high $\alpha$ : nonlinear normal force (viscous crossflow), body carryover, fin stall — CP moves forward, invalidating linear result. (1)

(b) Static margin (6 marks)

$SM = (X_{CP}-X_{CG})/d$ . Loaded: $(1.5343-1.35)/0.20 = 0.1843/0.20 = 0.921$ cal. (2) Burnout: $(1.5343-1.20)/0.20 = 0.3343/0.20 = 1.671$ cal. (2) Loaded margin $<1$ caliber → marginally under-stable at launch (risk near rail exit / low speed). Burnout over-stable (1.67 cal) — strong weather-cocking. As propellant near base depletes, CG moves forward, increasing SM. (2)

(c) Dynamic stability (5 marks)

CG shift changes both the moment arm of the aero restoring force and the moment of inertia, hence the natural pitch frequency $\omega_n=\sqrt{C_{N\alpha}qS\,(X_{CP}-X_{CG})/I_{yy}}$ . (1) Fin pitch damping arises because a pitch rate $q$ induces a local $\alpha_{loc}=q(\bar X_f-X_{CG})/v$ at the fins, giving a damping moment $M_q = -C_{N\alpha,f}\,\bar q S\,(\bar X_f - X_{CG})^2\,\frac{q}{v},$ so damping $\propto (\bar X_f - X_{CG})^2/v$ . (2) A statically over-stable rocket ( $X_{CP}$ far aft) has high $\omega_n$ ; if damping is insufficient (high $v$ reduces $q/v$ effect, low $I$ recovery) oscillations can grow/persist and couple into coning (yaw–pitch cross-coupling from roll), so high static margin alone does not guarantee dynamic stability — over-stability can worsen weather-cocking response to gusts. (2)

Question 2

(a) (5) Gravity turn ⇒ $\alpha=0$ ⇒ aerodynamic lift $L=0$ (all aero force is axial drag). (2) Then $mv\dot\gamma=-mg\cos\gamma \Rightarrow \dot\gamma=-\dfrac{g\cos\gamma}{v}$ . (2) Because gravity itself supplies the turning moment through the velocity vector, no active pitch commands are needed after the initial pitch-over kick sets $\gamma$ slightly off vertical; the trajectory self-shapes, minimizing aerodynamic loads ( $\alpha\approx0$ ⇒ low normal force / bending). (1)

(b) (6) $q=\tfrac12\rho v^2$ , $\ln q=\ln\tfrac12+\ln\rho+2\ln v$ . Differentiate w.r.t. $t$ : $\frac{\dot q}{q}=\frac{\dot\rho}{\rho}+2\frac{\dot v}{v}.$ (2) With $\rho=\rho_0e^{-h/H}$ , $\dot\rho/\rho=-\dot h/H$ . At Max-Q, $\dot q=0$ : $0=-\frac{\dot h}{H}+2\frac{\dot v}{v}\Rightarrow \frac1v\frac{dv}{dt}=\frac{\dot h}{2H},\ \dot h\approx v\sin\gamma.$ (2) Interpretation: dynamic pressure peaks when the fractional rate of velocity gain ( $2\dot v/v$ ) exactly balances the fractional density loss from climbing ( $\dot h/H$ ). Below Max-Q velocity gain dominates (q rising); above it density falloff dominates (q falling). (2)

(c) (9) State $\mathbf{x}=[v,\gamma,h,x,m]$ . (1)

# constants: T, mdot, CD, A, g=9.81, rho0=1.225, H=7200, m_dry
def rho(h): return rho0*exp(-h/H)
def deriv(x):
    v,gam,h,xr,m = x
    D = 0.5*rho(h)*v*v*CD*A
    vdot   = (T - D)/m - g*sin(gam)      # thrust axial, drag opposes v
    gamdot = -g*cos(gam)/v               # gravity turn, L=0
    hdot   = v*sin(gam)
    xdot   = v*cos(gam)
    return [vdot, gamdot, hdot, xdot, -mdot]
 
# RK4 step: k1=deriv(x); k2=deriv(x+dt/2*k1); k3=deriv(x+dt/2*k2);
#           k4=deriv(x+dt*k3); x += dt/6*(k1+2k2+2k3+k4)
# init: v=v0(small), gam=pi/2 - kick, h=0, m=m0
# STOP when m <= m_dry (burnout)
# track q=0.5*rho(h)*v*v each step -> record max (maxQ, h_atQ)
# outputs: v_bo, h_bo, maxQ, h_maxQ

(RK4 outline 3; drag model + stopping 2; Max-Q tracking 1)

Velocity losses (integrate over burn time $t_b$ ): $\Delta v_{grav}=\int_0^{t_b} g\sin\gamma\,dt,\qquad \Delta v_{drag}=\int_0^{t_b}\frac{D}{m}\,dt.$ Extract by accumulating g*sin(gam)*dt and (D/m)*dt inside the integrator; ideal $\Delta v = v_{eq}\ln(m_0/m_{dry})$ and $v_{bo}=\Delta v_{ideal}-\Delta v_{grav}-\Delta v_{drag}$ . (2)

Question 3

(a) (6) $m\dot v=-\tfrac12\rho v^2C_DA$ . Chain rule $\dfrac{dv}{dh}=\dfrac{\dot v}{\dot h}$ with $\dot h=-v\sin\gamma$ : $\frac{dv}{dh}=\frac{-\tfrac12\rho v^2 C_DA/m}{-v\sin\gamma}=\frac{\rho v C_D A}{2m\sin\gamma}=\frac{\rho v}{2\beta\sin\gamma}.$ $\frac{dv}{v}=\frac{\rho_0 e^{-h/H}}{2\beta\sin\gamma}dh.$ Integrate from $h$ (below entry) up to $h_e\to\infty$ where $v=v_e$ : $\ln\frac{v}{v_e}=\frac{\rho_0}{2\beta\sin\gamma}\int_{\infty}^{h}(-)e^{-h'/H}dh'=-\frac{\rho_0 H}{2\beta\sin\gamma}e^{-h/H},$ $v=v_e\exp\!\Big[-\frac{\rho_0 H}{2\beta\sin\gamma}e^{-h/H}\Big].$ (4) Assumptions: constant $\gamma$ (steep entry), gravity and lift negligible vs drag pulse, exponential atmosphere, constant $C_D$ , $\beta$ . (2)

**(b) (