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

Question 7: Rocket Performance, 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.

Given.

Find. The burnout velocity of the single-stage rocket; the peak deceleration during a ballistic entry; and the eccentricity and apogee velocity of the elliptical orbit.

Approach. Part (a) is the ideal rocket equation applied between ignition and burnout. Part (b) uses the Allen–Eggers ballistic-entry solution, which integrates the drag deceleration through an exponential atmosphere along a straight-line path and yields a peak deceleration that turns out to depend only on the entry speed, the entry angle and the atmospheric scale height. Part (c) uses the geometry of the ellipse to obtain the eccentricity and the vis-viva equation to obtain the apogee speed.

1. Part (a) — apply the ideal rocket equation. With gravity and drag ignored, conservation of momentum for a body expelling mass at constant relative speed $c$ integrates to Tsiolkovsky's result: $$\Delta V = c\,\ln\!\left(\frac{m_i}{m_f}\right)$$ Since the rocket starts from rest, the burnout speed is the whole velocity increment: $$\Delta V = 3300 \times \ln(8.3) = 3300 \times 2.1163 = \boxed{6984\ \text{m}\,\text{s}^{-1}}$$ that is 6.98 km/s.
2. Interpret the result against orbital requirements. Low Earth orbit requires roughly 7.8 km/s of orbital speed plus 1.5–2 km/s of gravity and drag losses, so about 9.3–9.5 km/s of ideal $\Delta V$. This single stage delivers 6.98 km/s and therefore cannot reach orbit, which is precisely the argument for staging: discarding structure as it becomes dead weight raises the effective mass ratio far beyond what a single tank and engine can achieve. Note also that the answer is an upper bound — including gravity loss $\int g\sin\theta\,\mathrm{d}t$ and drag would reduce it further.
3. Part (b) — set up the Allen–Eggers entry analysis. For a non-lifting vehicle on a steep, straight entry path the gravitational component along the path is small compared with drag once the vehicle is deep enough to decelerate, so the equation of motion along the path is $$m\frac{\mathrm{d}V}{\mathrm{d}t} = -\tfrac12 \rho V^{2} C_D A,\qquad \rho = \rho_0 e^{-\beta h}$$ With the flight-path angle $\gamma$ measured from the local horizontal, the altitude falls at $\mathrm{d}h/\mathrm{d}t = -V\sin\gamma$, which converts the time derivative to an altitude derivative and allows the equation to be integrated in closed form: $$V = V_E \exp\!\left[-K e^{-\beta h}\right],\qquad K = \frac{C_D A \rho_0}{2 m \beta \sin\gamma}$$
4. Differentiate to locate the peak deceleration. The deceleration is $$\left|\frac{\mathrm{d}V}{\mathrm{d}t}\right| = \beta V^{2} \sin\gamma \; K e^{-\beta h}$$ and setting its derivative with respect to altitude to zero gives the condition $K e^{-\beta h} = \tfrac12$, i.e. the peak occurs where the speed has fallen to $V = V_E e^{-1/2} = 0.6065\,V_E = 6.67\ \text{km/s}$. Substituting back: $$a_{max} = \frac{\beta V_E^{2}\sin\gamma}{2e}$$ Every vehicle-specific quantity has cancelled. The mass, drag coefficient and frontal area affect only the altitude at which the peak occurs, never its magnitude — which is why the question can be answered even though the vehicle mass is never stated.
5. Evaluate the maximum deceleration. With $\beta = 1.2\times10^{-4}\ \text{m}^{-1}$, $V_E = 11\,000\ \text{m}\,\text{s}^{-1}$ and $\sin 10^{\circ} = 0.17365$: $$a_{max} = \frac{1.2\times10^{-4} \times (11\,000)^{2} \times 0.17365}{2 \times 2.71828} = \frac{2521.4}{5.4366} = \boxed{464\ \text{m}\,\text{s}^{-2}}$$ which is $464/9.81 = 47.3\,g$ — survivable for instrumented hardware but far beyond human tolerance, and the reason crewed capsules enter at much shallower angles (Apollo entered at about 6.5° and pulled 6–7 g) or generate lift to stretch the deceleration out.
6. State where the peak occurs, and note the role of the given $C_D$ and $A$. The altitude of the peak follows from $Ke^{-\beta h} = \tfrac12$, i.e. $h = \ln(2K)/\beta$. This does need the mass; for an illustrative $m = 500\ \text{kg}$ vehicle with the stated $C_D A = 5.5\ \text{m}^{2}$, $$K = \frac{1.1 \times 5 \times 1.225}{2 \times 500 \times 1.2\times10^{-4} \times 0.17365} = 323.3 \quad\Longrightarrow\quad h = \frac{\ln 646.7}{1.2\times10^{-4}} = 53.9\ \text{km}$$ A heavier vehicle with the same $C_D A$ penetrates deeper before peaking, but experiences the same 464 m·s⁻². The drag data supplied in the question are thus needed only for this altitude, not for the deceleration the question asks for.
7. Part (c) — obtain the perigee and apogee radii. Orbital radii are measured from the centre of the Earth, so the stated altitudes must be increased by the Earth's radius: $$r_p = 6400 + 500 = 6900\ \text{km},\qquad r_a = 6400 + 1200 = 7600\ \text{km}$$ Since perigee and apogee are the two ends of the major axis, $$2a = r_p + r_a = 14\,500\ \text{km} \quad\Longrightarrow\quad a = 7250\ \text{km}$$
8. Find the eccentricity from the apsidal radii. Writing $r_p = a(1-e)$ and $r_a = a(1+e)$ and eliminating $a$: $$e = \frac{r_a - r_p}{r_a + r_p} = \frac{7600 - 6900}{14\,500} = \boxed{0.0483}$$ The orbit is only slightly elliptical — the focus lies 350 km from the geometric centre of an ellipse whose semi-major axis is 7250 km, as Figure 7.1 indicates (drawn exaggerated).
9. Apply vis-viva at apogee. The energy integral of the two-body problem gives the speed at any radius on an orbit of semi-major axis $a$: $$v^{2} = \mu\left(\frac{2}{r} - \frac{1}{a}\right), \qquad \mu = GM_E = 3.986\times10^{14}\ \text{m}^{3}\text{s}^{-2}$$ At apogee, $r = r_a = 7.600\times10^{6}\ \text{m}$: $$v_a^{2} = 3.986\times10^{14}\left(\frac{2}{7.600\times10^{6}} - \frac{1}{7.250\times10^{6}}\right) = 3.986\times10^{14} \times 1.2523\times10^{-7}$$ $$v_a = \sqrt{4.992\times10^{7}} = \boxed{7065\ \text{m}\,\text{s}^{-1}}\ (7.07\ \text{km/s})$$
10. Check the answer against conservation of angular momentum. The same relation at perigee gives $v_p = 7782\ \text{m}\,\text{s}^{-1}$, and since the position and velocity vectors are perpendicular at both apsides, angular momentum requires $r_p v_p = r_a v_a$: $$6.900\times10^{6} \times 7782 = 5.370\times10^{10} \approx 7.600\times10^{6} \times 7065 = 5.370\times10^{10}$$ The two products agree to four significant figures, confirming the pair of speeds. For completeness the orbital period is $T = 2\pi\sqrt{a^{3}/\mu} = 6144\ \text{s} = 102.4$ minutes, a typical low-Earth-orbit value. The satellite is slowest at apogee and fastest at perigee, exactly as Kepler's second law requires.