Question 7 of 7: Rocket Staging, Atmospheric Entry and Orbital Mechanics

Question 7: Rocket Staging, Atmospheric Entry and Orbital Mechanics (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.

Part (a) — maximum velocity of the two-stage rocket.

Given. A two-stage vehicle with no separately identified payload; the quoted "initial mass" of each stage is the mass of that stage alone, so the vehicle leaves the pad at the sum of the two.

Find. The maximum velocity the vehicle can reach, starting from rest.

Approach. Apply the Tsiolkovsky rocket equation once per stage and add the two velocity increments, taking care that the first burn must lift the whole vehicle — the second stage rides on top of it as payload — while the second burn carries only the second stage.

1. Split each stage into structure and propellant. The paper defines the mass ratio of each stage as initial mass divided by burnt-out structural mass. Stage 1: $m_{s1} = 2500/7.5 = 333.3\ \text{kg}$ and $m_{p1} = 2500 - 333.3 = 2166.7\ \text{kg}$. Stage 2: $m_{s2} = 1900/8 = 237.5\ \text{kg}$ and $m_{p2} = 1662.5\ \text{kg}$.
2. First burn: use the VEHICLE mass ratio, not the stage mass ratio. The vehicle starts at $M_0 = 4400\ \text{kg}$ and ends the first burn at $M_1 = 4400 - 2166.7 = 2233.3\ \text{kg}$, because the second stage is still attached and is being accelerated too. Hence $\Delta V_1 = V_e \ln (M_0/M_1) = 3200 \ln(4400/2233.3) = 3200(0.6781) = 2170\ \text{m}/\text{s}$.
3. Staging. The spent first stage, $333.3\ \text{kg}$ of structure, is jettisoned; the vehicle continues at $1900\ \text{kg}$. This is the entire point of staging — that mass is no longer carried by the remaining propellant.
4. Second burn: now the stage flies alone, so its own mass ratio applies. $\Delta V_2 = V_e \ln(1900/237.5) = 3200 \ln 8 = 3200(2.0794) = 6654\ \text{m}/\text{s}$.
5. Total burnout velocity. Starting from rest and ignoring gravity and drag losses, the increments add: $$\boxed{\ V_{max} = 2170 + 6654 = 8824\ \text{m}/\text{s} = 8.82\ \text{km}/\text{s}\ }$$

The value of staging can be quantified directly from the same data. If the identical propellant load were burned in a single vehicle that carried both structures all the way to burnout, the mass ratio would be $4400/(333.3 + 237.5) = 7.71$ and the burnout velocity would be $3200\ln 7.71 = 6535\ \text{m}/\text{s}$. Staging is therefore worth $8824 - 6535 = 2289\ \text{m}/\text{s}$ — a 35 per cent gain bought purely by throwing away $333\ \text{kg}$ of empty tankage at the right moment. That gain, repeated over two or three stages, is what makes orbit reachable at all: the required $\approx 9.4\ \text{km}/\text{s}$ including gravity and drag losses is simply not available from a single stage with realistic structural fractions.

Part (b) — maximum deceleration during ballistic re-entry.

Given. A non-lifting (ballistic) vehicle entering at $V_E = 10.5\ \text{km}/\text{s}$ on a path inclined at $\gamma = 10^\circ$ to the local horizontal, with $C_D = 1$ on a frontal area of $9\ \text{m}^2$, into an exponential atmosphere $\rho = \rho_0 e^{-\beta h}$ with $\beta = 0.000118\ \text{m}^{-1}$ and $\rho_0 = 1.225\ \text{kg}/\text{m}^3$. The vehicle mass is not stated.

Find. The maximum deceleration experienced during the entry.

Approach. Use the Allen–Eggers ballistic-entry solution: for a steep, non-lifting entry, gravity and the curvature of the path are negligible compared with the drag force over the region where the deceleration is large, so the path angle stays constant and the equation of motion integrates in closed form. Differentiating the resulting deceleration with respect to altitude locates the peak.

1. Equation of motion along the path. With gravity neglected against drag, $m\,dV/dt = -\tfrac{1}{2}\rho V^2 C_D A$. Changing the independent variable from time to altitude using $dh/dt = -V\sin\gamma$ (the path angle is measured to the local horizontal, so the rate of loss of altitude is $V\sin\gamma$, not $V\cos\gamma$) gives $$\frac{dV}{V} = \frac{C_D A \rho_0}{2m\sin\gamma}\,e^{-\beta h}\,dh.$$
2. Integrate from the top of the atmosphere. With $V = V_E$ as $h \to \infty$, $$V = V_E \exp\!\left(-K e^{-\beta h}\right), \qquad K = \frac{C_D A \rho_0}{2 m \beta \sin\gamma}.$$
3. Write the deceleration in terms of the single variable $x = K e^{-\beta h}$. The deceleration is $a = \rho C_D A V^2/(2m) = \beta\sin\gamma\,x\,V^2$, and substituting $V = V_E e^{-x}$ gives $a = \beta V_E^2 \sin\gamma\; x\,e^{-2x}$.
4. Locate the maximum. Setting $d(x e^{-2x})/dx = e^{-2x}(1-2x) = 0$ gives $x = 1/2$, so the peak occurs where $K e^{-\beta h} = \tfrac{1}{2}$, and at that point $x e^{-2x} = 1/(2e)$. Hence $$a_{max} = \frac{\beta V_E^2 \sin\gamma}{2e}.$$
5. Substitute the given data. $a_{max} = \dfrac{0.000118 \times (10\,500)^2 \times \sin 10^\circ}{2 \times 2.71828} = \dfrac{2258.9}{5.4366}$, so $$\boxed{\ a_{max} = 415.5\ \text{m}/\text{s}^2 = 42.4\,g\ }$$

1. Altitude of the peak. From $K e^{-\beta h} = 1/2$, $h_{peak} = \ln(2K)/\beta$. Taking an illustrative mass of $5000\ \text{kg}$ for a capsule of this frontal area, $K = \dfrac{1 \times 9 \times 1.225}{2 \times 5000 \times 0.000118 \times \sin 10^\circ} = 53.81$, so $h_{peak} = \ln(107.6)/0.000118 = 39.6\ \text{km}$.
2. Velocity at the peak. Since $x = 1/2$ there, $V = V_E e^{-1/2} = 10\,500 \times 0.6065 = 6369\ \text{m}/\text{s}$ — the vehicle has shed only 39 per cent of its entry speed when it meets the worst of the deceleration.

Two closing checks are worth making. First, on the validity of neglecting gravity: the peak deceleration is $42.4\,g$, so the omitted gravitational term is about 2 per cent of the drag force there and the approximation is sound in the region that matters — but it expires near the ground, where the drag has fallen back to the vehicle weight. For the illustrative $5000\ \text{kg}$ vehicle the terminal velocity is $\sqrt{2mg/(\rho_0 C_D A)} = 94\ \text{m}/\text{s}$, and the Allen–Eggers formula returns an essentially zero surface speed; its speed falls to that terminal value at about $24\ \text{km}$, so below roughly that altitude the vehicle is in a gravity-balanced descent that this analysis no longer describes; a parachute or a lifting entry takes over there. Second, on the engineering consequence: $42.4\,g$ is far beyond human tolerance, which is why crewed vehicles enter far shallower — at $\gamma = 1.5^\circ$ the same formula gives $63\ \text{m}/\text{s}^2$, about $6.4\,g$ — and generate lift to stretch the entry further still. The result also shows why the entry corridor is so narrow: peak deceleration scales directly with $\sin\gamma$ and with the square of the entry speed.

Part (c) — eccentricity and apogee velocity of the elliptical orbit.

Given. Perigee altitude $650\ \text{km}$, apogee altitude $1500\ \text{km}$, Earth radius $6400\ \text{km}$, so the orbital radii measured from the centre of the Earth are $r_p = 7050\ \text{km}$ and $r_a = 7900\ \text{km}$. The Earth’s gravitational parameter is taken as $\mu = GM = 3.986\times 10^{14}\ \text{m}^3/\text{s}^2$.

Find. The eccentricity of the orbit and the satellite’s velocity at apogee.

Approach. Perigee and apogee are the two ends of the major axis, so they give the semi-major axis and the eccentricity immediately from geometry; the vis-viva equation then gives the speed at any radius.

1. Convert altitudes to orbital radii. Orbital mechanics uses distance from the centre of the Earth, so $r_p = 6400 + 650 = 7050\ \text{km}$ and $r_a = 6400 + 1500 = 7900\ \text{km}$.
2. Semi-major axis. Perigee and apogee lie at opposite ends of the major axis, so $2a = r_p + r_a = 14\,950\ \text{km}$ and $a = 7475\ \text{km}$.
3. Eccentricity. The focus sits a distance $ae$ from the centre of the ellipse, so $r_p = a(1-e)$ and $r_a = a(1+e)$. Subtracting and dividing, $e = \dfrac{r_a - r_p}{r_a + r_p} = \dfrac{850}{14\,950}$, giving $$\boxed{\ e = 0.0569\ }$$
4. Apply the vis-viva equation. Conservation of energy for a two-body orbit gives $v^2 = \mu\left(\dfrac{2}{r} - \dfrac{1}{a}\right)$. At apogee, with $r_a = 7.900\times 10^{6}\ \text{m}$ and $a = 7.475\times 10^{6}\ \text{m}$, $v_a^2 = 3.986\times 10^{14}(2.5316\times 10^{-7} - 1.3378\times 10^{-7}) = 4.759\times 10^{7}$, so $$\boxed{\ v_a = 6898\ \text{m}/\text{s} = 6.90\ \text{km}/\text{s}\ }$$

Two checks confirm the answer. The same equation at perigee gives $v_p = 7730\ \text{m}/\text{s}$, and conservation of angular momentum requires $v_p r_p = v_a r_a$ at the apsides, where the velocity is perpendicular to the radius: $7730 \times 7050 = 6898 \times 7900 = 5.450\times 10^{7}\ \text{km}\cdot\text{m}/\text{s}$, which it does. And the orbital period follows from Kepler’s third law as $T = 2\pi\sqrt{a^3/\mu} = 6432\ \text{s} = 107.2\ \text{min}$, a wholly conventional low-Earth-orbit period. Note that if the Earth’s gravitational parameter is instead reconstructed from the given radius as $\mu = g R_E^2 = 9.81(6400\ \text{km})^2 = 4.018\times 10^{14}$, the apogee speed comes out as $6926\ \text{m}/\text{s}$ — a difference of 0.4 per cent, which reflects the fact that $6400\ \text{km}$ is a rounded Earth radius rather than a discrepancy in the method. The eccentricity of 0.057 is very small, so the orbit is nearly circular and the speed varies by only about 11 per cent around it.