Level 5 — MasteryOrbital Mechanics & Astrodynamics

Orbital Mechanics & Astrodynamics

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Level 5 — Mastery (cross-domain: analysis + physics + computation) Time limit: 2 hours 30 minutes Total marks: 60

Constants (use where needed): Earth $\mu = GM_\oplus = 3.986004\times10^{5}\ \mathrm{km^3/s^2}$ , $R_\oplus = 6378\ \mathrm{km}$ , $J_2 = 1.08263\times10^{-3}$ .

Question 1 — Vis-viva, Kepler's equation, and a numerical propagator (24 marks)

A spacecraft is in an Earth orbit with perigee altitude $600\ \mathrm{km}$ and apogee altitude $35786\ \mathrm{km}$ (a GTO-like ellipse).

(a) From conservation of energy and angular momentum, derive the vis-viva equation $v^2=\mu\left(\dfrac{2}{r}-\dfrac{1}{a}\right)$ . Start from the specific orbital energy $\varepsilon=\tfrac12 v^2-\tfrac{\mu}{r}$ and show it equals $-\tfrac{\mu}{2a}$ by evaluating at perigee/apogee. (6)

(b) Compute the semi-major axis $a$ , eccentricity $e$ , orbital period $T$ , and the perigee speed $v_p$ for this orbit. (5)

(c) Derive Kepler's equation $M=E-e\sin E$ starting from Kepler's second law (areal velocity constant) and the geometry relating true anomaly $\nu$ to eccentric anomaly $E$ . State clearly the mean anomaly definition $M=n(t-t_p)$ . (6)

(d) For this orbit, find the true anomaly $\nu$ exactly $3\ \mathrm{hours}$ after perigee passage. Carry out at least two Newton–Raphson iterations for Kepler's equation (show iteration formula and each iterate), then convert $E\to\nu$ . (7)

Question 2 — Transfer optimization: Hohmann vs bi-elliptic, with plane change (20 marks)

Consider transfers between two coplanar circular Earth orbits: inner radius $r_1 = 7000\ \mathrm{km}$ , outer radius $r_2$ .

(a) Derive the total Hohmann $\Delta v$ as a function of $r_1,r_2,\mu$ , giving both burn magnitudes. (5)

(b) For a bi-elliptic transfer via an intermediate apoapsis radius $r_b\gg r_2$ , write the three $\Delta v$ contributions and state the limiting total as $r_b\to\infty$ . Show that the ratio $r_2/r_1$ above which bi-elliptic (in the limit) beats Hohmann is approximately $11.94$ . (8)

(c) Now suppose the outer orbit at $r_2 = 42164\ \mathrm{km}$ (GEO) also requires a plane change of $\Delta i = 28.5^\circ$ . Using the combined-maneuver formula $\Delta v = \sqrt{v_a^2 + v_b^2 - 2 v_a v_b \cos\Delta i},$ argue why performing the plane change at apoapsis of the transfer (rather than at the inner orbit) is far cheaper, and estimate the plane-change $\Delta v$ if done entirely at GEO circular speed vs at the transfer apoapsis speed. (7)

Question 3 — J₂ nodal precession and Sun-synchronous design (16 marks)

The $J_2$ oblateness perturbation causes secular drift of the right ascension of the ascending node (RAAN): $\dot\Omega = -\frac{3}{2}\,n\,J_2\left(\frac{R_\oplus}{p}\right)^2\cos i,\qquad p=a(1-e^2),\quad n=\sqrt{\mu/a^3}.$

(a) Explain physically the origin of the $\cos i$ dependence and the sign of $\dot\Omega$ for a prograde orbit. (4)

(b) A Sun-synchronous orbit requires $\dot\Omega = +360^\circ/365.25\ \mathrm{days}$ (to track the Sun). For a circular orbit at altitude $800\ \mathrm{km}$ , compute the required inclination $i$ . (8)

(c) State one operational advantage of an SSO for Earth-observation missions, tied directly to the property computed in (b). (4)

Answer keyMark scheme & solutions

Question 1

(a) Vis-viva derivation (6)

Energy conservation: $\varepsilon = \tfrac12 v^2 - \tfrac{\mu}{r}=$ const along orbit. (1)
Angular momentum $h=rv_\perp$ constant; at perigee & apogee $v=v_\perp$ (radial velocity zero). (1)
At perigee $r_p=a(1-e)$ , apogee $r_a=a(1+e)$ ; $h=r_pv_p=r_av_a$ . (1)
Evaluate $\varepsilon$ at perigee, use $v_p=h/r_p$ , and $r_p r_a=a^2(1-e^2)$ , $h^2=\mu a(1-e^2)$ : substituting gives $\varepsilon = \tfrac12\frac{h^2}{r_p^2}-\frac{\mu}{r_p} = -\frac{\mu}{2a}$ . (2)
Therefore $\tfrac12 v^2-\tfrac{\mu}{r}=-\tfrac{\mu}{2a}\Rightarrow v^2=\mu\!\left(\tfrac{2}{r}-\tfrac1a\right)$ . (1)

(b) Orbit parameters (5)

$r_p = 6378+600 = 6978\ \mathrm{km}$ ; $r_a = 6378+35786 = 42164\ \mathrm{km}$ . (1)
$a=\tfrac12(r_p+r_a)=\tfrac12(6978+42164)=24571\ \mathrm{km}$ . (1)
$e=\dfrac{r_a-r_p}{r_a+r_p}=\dfrac{35186}{49142}=0.71601$ . (1)
$T=2\pi\sqrt{a^3/\mu}=2\pi\sqrt{24571^3/3.986004\times10^5}=38318\ \mathrm{s}\approx 10.64\ \mathrm{h}$ . (1)
$v_p=\sqrt{\mu(2/r_p-1/a)}=\sqrt{3.986004\times10^5(2/6978-1/24571)}=10.16\ \mathrm{km/s}$ . (1)

(c) Kepler's equation derivation (6)

Mean anomaly $M=n(t-t_p)$ , $n=2\pi/T=\sqrt{\mu/a^3}$ ; grows linearly with time. (1)
Kepler II: area swept $\propto$ time, total area $\pi ab$ swept in $T$ ; so fraction of area = $M/2\pi$ . (1)
Introduce eccentric anomaly $E$ via the auxiliary circle radius $a$ ; ellipse point maps from circle point by vertical scaling $b/a$ . (1)
Area swept from perigee (focus) = (circular-sector area) − (triangle) scaled by $b/a$ : $\text{Area}=\tfrac{b}{a}\left(\tfrac12 a^2 E - \tfrac12 a^2 e\sin E\right)=\tfrac12 ab(E-e\sin E)$ . (1)
Equate to $\tfrac{M}{2\pi}\cdot\pi ab=\tfrac12 ab\,M$ . (1)
Cancel $\tfrac12 ab$ : $M=E-e\sin E$ . (1)

(d) Numerical propagation (7)

$n=2\pi/T = 2\pi/38318 = 1.63976\times10^{-4}\ \mathrm{rad/s}$ . (1)
$t=3\ \mathrm{h}=10800\ \mathrm{s}$ ; $M=n t=1.77094\ \mathrm{rad}$ . (1)
Newton–Raphson: $E_{k+1}=E_k-\dfrac{E_k-e\sin E_k-M}{1-e\cos E_k}$ , $e=0.71601$ . (1)
$E_0=M=1.77094$ : $f=E_0-e\sin E_0-M=-0.70168$ , $f'=1-e\cos E_0=1.14203$ ; $E_1=1.77094+0.61442=2.38536$ . (1)
$E_1=2.38536$ : $f=2.38536-0.71601\sin(2.38536)-1.77094=0.12081$ , $f'=1-0.71601\cos(2.38536)=1.52255$ ; $E_2=2.38536-0.07935=2.30601$ . (1)
$E_2$ : refine once more $\to E\approx 2.30240\ \mathrm{rad}$ (converged). (1)
Convert: $\tan\tfrac{\nu}{2}=\sqrt{\tfrac{1+e}{1-e}}\tan\tfrac{E}{2}=\sqrt{\tfrac{1.71601}{0.28399}}\tan(1.15120)=2.45846\times2.24=$ gives $\nu\approx 2.591\ \mathrm{rad}\approx 148.5^\circ$ . (1)

Question 2

(a) Hohmann (5)

Transfer ellipse: $a_t=(r_1+r_2)/2$ . (1)
Speeds: circular $v_1=\sqrt{\mu/r_1}$ , $v_2=\sqrt{\mu/r_2}$ ; transfer periapsis $v_{tp}=\sqrt{\mu(2/r_1-1/a_t)}$ , apoapsis $v_{ta}=\sqrt{\mu(2/r_2-1/a_t)}$ . (2)
$\Delta v_1=v_{tp}-v_1$ , $\Delta v_2=v_2-v_{ta}$ . (1)
$\Delta v_{tot}=\Delta v_1+\Delta v_2$ . (1)

(b) Bi-elliptic (8)

Burn 1 at $r_1$ raising apoapsis to $r_b$ : $\Delta v_1=\sqrt{\mu(2/r_1-2/(r_1+r_b))}-v_1$ . (1)
Burn 2 at $r_b$ raising periapsis $r_1\to r_2$ : $\Delta v_2=\sqrt{\mu(2/r_b-2/(r_b+r_2))}-\sqrt{\mu(2/r_b-2/(r_1+r_b))}$ . (1)
Burn 3 at $r_2$ lowering apoapsis $r_b\to r_2$ (deceleration): $\Delta v_3=v_2-\sqrt{\mu(2/r_2-2/(r_2+r_b))}$ . (1)
As $r_b\to\infty$ : burns approach parabolic; total $\to \sqrt{2}\,v_1(1-\sqrt{r_1/r_2})+\ldots$ giving the classic limit expression. (2)
Setting bi-elliptic-limit $\Delta v$ = Hohmann $\Delta v$ and solving numerically gives crossover ratio $R=r_2/r_1\approx 11.94$ . For $R<11.94$ Hohmann is always better; for $11.94<R<15.58$ bi-elliptic can win depending on $r_b$ ; above $15.58$ bi-elliptic wins for all sufficiently large $r_b$ . (2)
State: bi-elliptic pays with much longer flight time. (1)

(c) Combined plane change (7)

Plane-change cost $\propto v$ at which it is done: $\Delta v_{pc}=2v\sin(\Delta i/2)$ . (1)
At GEO circular speed $v_2=\sqrt{\mu/42164}=3.0747\ \mathrm{km/s}$ : $\Delta v_{pc,\text{GEO}}=2(3.0747)\sin(14.25^\circ)=1.513\ \mathrm{km/s}$ . (2)
At transfer apoapsis (GTO): $v_{ta}=\sqrt{\mu(2/42164-1/24571)}=1.597\ \mathrm{km/s}$ (using $a_t=24571$ ). $\Delta v_{pc,\text{apo}}=2(1.597)\sin(14.25^\circ)=0.786\ \mathrm{km/s}$ . (2)
Doing it at apoapsis (lower speed) roughly halves the cost; optimal practice combines the plane change with the circularization burn vectorially (law-of-cosines formula), cheaper still. (2)

Question 3

(a) Physical origin (4)

Oblateness adds an equatorial bulge; the extra mass exerts an out-of-plane torque component on the orbit. (2)
This torque causes the orbit plane to precess about Earth's spin axis; its magnitude scales with $\cos i$ (zero for polar $i=90^\circ$ , max at equatorial). Prograde orbit ( $i<90^\circ$ ): $\cos i>0\Rightarrow\dot\Omega<0$ (regression of nodes, westward). (2)

(b) SSO inclination (8)

$a=6378+800=7178\ \mathrm{km}$ ; circular $e=0$ , $p=a$ . (1)
$n=\sqrt{\mu/a^3}=\sqrt{3.986004\times10^5/7178^3}=1.03855\times10^{-3}\ \mathrm{rad/s}$ . (2)
Required $\dot\Omega=2\pi/(365.25\times86400)=1.99106\times10^{-7}\ \mathrm{rad/s}$ (positive). (2)
$\cos i = -\dfrac{\dot\Omega}{\tfrac32 n J_2 (R_\oplus/a)^2}$ . Compute denominator: $\tfrac32(1.03855\times10^{-3})(1.08263\times10^{-3})(6378/7178)^2=\tfrac32(1.03855\times10^{-3})(1.08263\times10^{-3})(0.78945)=1.33124\times10^{-6}$ . (2)
$\cos i=-1.99106\times10^{-7}/1.33124\times10^{-6}=-0.14957\Rightarrow i=98.60^\circ$ . (1)

(c) Operational advantage (4)

Because the orbit plane precesses at exactly $1$ revolution/year, it maintains a nearly constant local solar time at every equator crossing. (2)
Earth-observation targets are then imaged under consistent, repeatable illumination/shadow conditions, aiding change detection and calibration. (2)

[
  {"claim":"GTO a=24571 km, e≈0.71601","code":"rp=6978; ra=42164; a=(rp+ra)/2; e=(ra-rp)/(ra+rp); result = (abs(a-24571)<1) and (abs(float(e)-0.71601)<1e-4)"},
  {"claim":"Perigee speed ≈10.16 km/s","code":"mu=3.986004e5; rp=6978; a=24571; vp=sqrt(mu*(2/rp-1/a)); result = abs(float(vp)-10.16)<0.02"},
  {"claim":"Mean anomaly at 3h M≈1.771 rad","code":"import sympy as sp; mu=3.986004e5; a=24571; n=sp.sqrt(mu/a**3); M=n*10800; result = abs(float(M)-1.771)<0.005"},
  {"claim":"SSO inclination ≈98.6 deg","code":"import sympy as sp; mu=3.986004e5; Re=6378; J2=1.08263e-3; a=7178; n=sp.sqrt(mu/a**3); Odot=2*sp.pi/(365.25*86400); ci=-Odot/(sp.Rational(3,2)*n*J2*(Re/a)**2); i=sp.acos(ci)*180/sp.pi; result = abs(float(i)-98.6)<0.3"},
  {"claim":"GEO circular speed ≈3.075 km/s","code":"mu=3.986004e5; v2=sqrt(mu/42164); result = abs(float(v2)-3.0747)<0.01"}
]