Orbital Maneuvers

Hohmann, bi-elliptic, phasing, apse rotation, chase and plane change

Devendra Ghate

How this deck is organised

Everything so far has been about propagating a state. This deck is about changing one.

Part 0 sets up the impulsive model and the two facts that decide almost every manoeuvre: an impulse buys energy at a rate \(v\,\Delta v\) (Equation 2), and it buys plane change at a rate \(2v\sin(\theta/2)\) (Equation 5). One wants \(v\) large, the other wants \(v\) small. Nearly every design in this deck is that tension being resolved.

Parts I–VI work through the manoeuvres one at a time — Hohmann, bi-elliptic, phasing, apse-line rotation, chase/rendezvous, plane change — each with a fully worked numerical example.

Part VII surveys what these notes leave out, and why.

The appendix proves every optimality claim made in the deck, including the central one: among all two-impulse transfers between coplanar circular orbits, the Hohmann transfer is the cheapest.

Throughout, \(\mu_\oplus = 398600.4418\ \mathrm{km^3/s^2}\) and \(R_\oplus = 6378.137\) km. “LEO” means the circular orbit at \(300\) km altitude, \(r_1 = 6678.137\) km; “GEO” means \(r_2 = 42164\) km.

Part 0 — The Impulsive Model

The impulsive approximation

A real burn lasts a finite time and the spacecraft moves during it. We replace it by an instantaneous jump in velocity at a fixed point:

\[ \mathbf r^{+}=\mathbf r^{-}, \qquad \mathbf v^{+}=\mathbf v^{-}+\Delta\mathbf v . \]

This is a good model when the burn arc is short compared with the orbit, i.e. when

\[ \frac{t_{\rm burn}}{T}\ll 1 \qquad\text{and}\qquad \frac{\Delta v}{v}\ \text{is not too large.} \]

For a chemical upper stage, \(t_{\rm burn}\) is minutes and \(T\) is hours, so the approximation is excellent. For electric propulsion, \(t_{\rm burn}\) is months — the impulsive model fails completely, and the whole trajectory must be treated as a continuous-thrust problem (Part VII).

The cost of a manoeuvre is the scalar \(\Delta v = |\Delta\mathbf v|\), and the cost of a sequence is \(\sum_i |\Delta\mathbf v_i|\) — a sum of magnitudes, never a vector sum. Two burns that cancel each other still cost twice.

Why \(\Delta v\) is the currency

The rocket equation (proved in A1) converts \(\Delta v\) into hardware:

\[ \boxed{\ \Delta v = I_{sp}\,g_0\ \ln\frac{m_0}{m_f} \qquad\Longleftrightarrow\qquad \frac{m_{\rm prop}}{m_0}=1-e^{-\Delta v/(I_{sp}g_0)}\ } \qquad(1)\]

A LEO\(\to\)GEO Hohmann transfer costs \(3.8926\) km/s. With \(I_{sp}=315\) s, what fraction of the departure mass is propellant?

Answer

\(v_e = I_{sp}g_0 = 315\times 9.80665\times10^{-3}=3.0891\) km/s, so

\[ \frac{m_{\rm prop}}{m_0}=1-e^{-3.8926/3.0891}=1-e^{-1.2601}=\boxed{0.716}. \]

Nearly 72% of what leaves LEO is propellant. This is why a \(0.05\) km/s saving is worth a slide.

What a single impulse can do

Write the impulse as \(\Delta\mathbf v\) at a point where the state is \((\mathbf r,\mathbf v)\). Then, with the energy step done in full in A2,

\[ \Delta\varepsilon = \mathbf v\cdot\Delta\mathbf v+\tfrac12|\Delta\mathbf v|^2, \qquad \Delta\mathbf h=\mathbf r\times\Delta\mathbf v . \qquad(2)\]

Three consequences, each of which drives a manoeuvre in this deck:

1. Energy responds to the component of \(\Delta\mathbf v\) along \(\mathbf v\), weighted by \(v\). \(\Rightarrow\) burn tangentially, and burn where you are fast (the Oberth effect).
2. A radial impulse (\(\Delta\mathbf v\parallel\mathbf r\)) changes \(\varepsilon\) only at second order but changes \(\mathbf h\) at first order. \(\Rightarrow\) it rotates the orbit without resizing it (apse-line rotation, Part IV).
3. An out-of-plane impulse rotates \(\mathbf h\) without changing \(|\mathbf h|\) to first order, but raises \(\varepsilon\) by \(\tfrac12|\Delta\mathbf v|^2\). \(\Rightarrow\) plane change (Part VI). Turning \(\mathbf v\) through \(\theta\) at constant speed needs a retrograde component as well, costs \(2v\sin(\theta/2)\) (Equation 5), and buys no energy at all.

The Oberth effect, quantitatively

Figure 1: A fixed \(100\) m/s impulse on an orbit with \(a=12000\) km, \(e=0.5\). Left: the same impulse buys three times as much energy at periapsis as at apoapsis, because \(\Delta\varepsilon = v\Delta v+\frac12\Delta v^2\) and \(v_p/v_a = (1+e)/(1-e)=3\). Right: for a fixed \(|\Delta\mathbf v|\), the energy gain varies as \(\cos\beta\), so tangential is optimal and radial is worthless (to first order).

This single picture explains why every transfer in Parts I–III burns tangentially at an apsis, and why plane changes are pushed out to apoapsis where \(v\) is small.

Part I — The Hohmann Transfer

Circular to circular: the strategy

Figure 2: The transfer ellipse is tangent to both circles, so both impulses are purely tangential and the coast is exactly half a revolution.

Circular to circular: the derivation

The transfer ellipse has periapsis \(r_1\) and apoapsis \(r_2\), hence

\[ a_t=\frac{r_1+r_2}{2}. \]

From vis-viva, \(v^2=\mu\left(\frac{2}{r}-\frac{1}{a}\right)\), the two transfer speeds are

\[ v_{t1}=\sqrt{\mu\left(\frac{2}{r_1}-\frac{1}{a_t}\right)}=\sqrt{\frac{2\mu r_2}{r_1(r_1+r_2)}}, \qquad v_{t2}=\sqrt{\frac{2\mu r_1}{r_2(r_1+r_2)}} . \]

Both burns are tangential, so the impulses are plain differences of speeds:

\[ \boxed{\ \Delta v_1=\sqrt{\frac{\mu}{r_1}}\left(\sqrt{\frac{2r_2}{r_1+r_2}}-1\right), \qquad \Delta v_2=\sqrt{\frac{\mu}{r_2}}\left(1-\sqrt{\frac{2r_1}{r_1+r_2}}\right).\ } \]

The transfer time is half the period of the transfer ellipse,

\[ t_{\rm H}=\pi\sqrt{\frac{a_t^{3}}{\mu}} . \]

Example 1 — LEO \(\to\) GEO

Given \(r_1=6678.137\) km, \(r_2=42164\) km, \(\mu=398600.4418\ \mathrm{km^3/s^2}\).

\[ a_t=\frac{6678.137+42164}{2}=24421.07\ \mathrm{km}, \qquad e_t=\frac{r_2-r_1}{r_2+r_1}=0.72654 . \]

	speed before	speed after	\(\Delta v\)
burn 1, at \(r_1\)	\(v_{c1}=7.72576\)	\(v_{t1}=10.15149\)	\(\mathbf{2.42573}\)
burn 2, at \(r_2\)	\(v_{t2}=1.60784\)	\(v_{c2}=3.07467\)	\(\mathbf{1.46682}\)

\[ \boxed{\ \Delta v_{\rm total}=3.89255\ \mathrm{km/s}, \qquad t_{\rm H}=\pi\sqrt{\frac{24421.07^{3}}{398600.4418}}=18990\ \mathrm{s}=5.275\ \mathrm{h}.\ } \]

Both burns are prograde: you speed up to leave, and speed up again to circularise. The spacecraft is slower at GEO (\(3.07\) km/s) than at LEO (\(7.73\) km/s), yet every burn adds speed — because it arrives at \(r_2\) moving at only \(1.61\) km/s.

The cost is not monotone in \(R\)

Normalising by \(v_{c1}=\sqrt{\mu/r_1}\) and writing \(R=r_2/r_1\),

\[ \frac{\Delta v_{\rm H}}{v_{c1}} =\underbrace{\sqrt{\frac{2R}{1+R}}-1}_{\to\ \sqrt2-1} +\underbrace{\frac{1}{\sqrt R}\left(1-\sqrt{\frac{2}{1+R}}\right)}_{\to\ 0} . \]

As \(R\to\infty\) the total tends to \(\sqrt2-1=0.4142\), the escape burn — but it does not climb there monotonically. It rises to a maximum

\[ \boxed{\ \max_R \frac{\Delta v_{\rm H}}{v_{c1}}=0.53626\quad\text{at}\quad R=15.5817\ } \]

and falls afterwards. Going further can cost less. Appendix A11 proves this, and A4 shows that the same number \(15.5817\) is exactly the threshold beyond which the bi-elliptic transfer always wins.

The cost curve

Figure 3: The normalised cost against \(R=r_2/r_1\), split into departure and arrival. Note it is not monotone — it peaks at \(R=15.58\) and decreases thereafter, which is the seed of the bi-elliptic idea in Part II.

Aligned apsides: the geometry

Figure 4: Both apse lines point the same way. Even here the apsidal transfer is not unique: A1 joins periapsis to apoapsis (maroon, upper half) and A2 joins apoapsis to periapsis (purple, lower half). Arrowheads give the direction of motion — all four burns are prograde.

Coaxial ellipse to ellipse — aligned apsides

When both orbits are elliptical and share an apse line with \(\omega_1=\omega_2\), a \(180^\circ\) apsidal transfer can join either pair of opposed apsides:

\[ \text{A1}:\quad a_{t}=\frac{r_{p,1}+r_{a,2}}{2}, \qquad\qquad \text{A2}:\quad a_{t}=\frac{r_{a,1}+r_{p,2}}{2}, \]

\[ \Delta v_1=\left|v_{t}(r_A)-v_{1}(r_A)\right|,\qquad \Delta v_2=\left|v_{2}(r_B)-v_{t}(r_B)\right| . \]

Both burns are tangential on either route, because periapsis and apoapsis are the only points where the velocity is purely transverse on both conics at once. Which route is cheaper is a question, not a definition — and the answer is not the same as which is quicker.

Example 2 — the two aligned routes

Given orbit 1: \(a_1=12000\) km, \(e_1=0.30\) (\(r_p=8400\), \(r_a=15600\) km). Orbit 2: \(a_2=42000\) km, \(e_2=0.50\) (\(r_p=21000\), \(r_a=63000\) km). Apsides aligned.

	\(a_t\) [km]	\(e_t\)	\(\Delta v_1\)	\(\Delta v_2\)	total [km/s]	\(t\) [h]
A1 \(8400\to63000\)	\(35700\)	\(0.7647\)	\(1.29675\)	\(0.55850\)	\(\mathbf{1.85525}\)	\(9.324\)
A2 \(15600\to21000\)	\(18300\)	\(0.1475\)	\(1.18573\)	\(1.31337\)	\(2.49909\)	\(\mathbf{3.422}\)

\[ \boxed{\ \text{A1 is cheaper by } 644\ \mathrm{m/s};\quad \text{A2 is faster by a factor } 2.72 .\ } \]

The larger transfer is the cheaper one and the smaller is the quicker one — the same pattern that will appear again when the apsides are reversed.

Coming back down: outer \(\to\) inner

Figure 5: The same two transfer ellipses as Example 2, flown the other way round. B1 leaves the apoapsis of orbit 2 and coasts down to the periapsis of orbit 1 through the lower half; B2 leaves the periapsis of orbit 2 for the apoapsis of orbit 1 through the upper half. Every burn now points against the motion.

Descending costs exactly what ascending did

	departure burn	arrival burn	total [km/s]	\(t\) [h]
B1 \(r_{a,2}=63000\to r_{p,1}=8400\)	\(0.55850\)	\(1.29675\)	\(\mathbf{1.85525}\)	\(9.324\)
B2 \(r_{p,2}=21000\to r_{a,1}=15600\)	\(1.31337\)	\(1.18573\)	\(2.49909\)	\(\mathbf{3.422}\)

Every number is the one from Example 2, in the opposite order. The reason is one line: at each endpoint the two speeds in play — the orbit’s and the transfer’s — do not depend on which way you are going, and

\[ \left|v_{\rm orbit}-v_{t}\right|=\left|v_{t}-v_{\rm orbit}\right| . \]

\[ \boxed{\ \Delta v_{\rm total}(1\to2)=\Delta v_{\rm total}(2\to1), \qquad t(1\to2)=t(2\to1).\ } \]

The transfer ellipse and its half-period are the same object either way; only the sense of each impulse flips, from prograde (speed up to climb) to retrograde (slow down to fall). Impulsive \(\Delta v\) is symmetric under time reversal, so there is no cheap direction — which is exactly why aerobraking (Part VII) is worth the trouble: the atmosphere will supply the retrograde impulse for free, but nothing supplies the prograde one.

Four legs, two ellipses

Figure 6: All four aligned routes at once. A1 and B1 are the two halves of the \(a_t=35700\) km ellipse; A2 and B2 the two halves of the \(a_t=18300\) km one. Solid is outbound, dashed is the return leg, and the arrowheads show that both are flown prograde. The four double-headed arrows are the apsidal impulses: each is applied prograde going out and retrograde coming back, with the magnitude shown.

There are not four transfers here, there are two ellipses. Choosing a route means choosing which pair of apsides to join — \((r_{p,1},r_{a,2})\) or \((r_{a,1},r_{p,2})\) — and the direction of travel then fixes only the sense of the two burns, never their size.

Anti-aligned apsides: the geometry

Figure 7: The same two orbit shapes, drawn to the same scale, with the target’s apse line reversed. Now there are two \(180^\circ\) apsidal transfers, not one: T2 joins the periapsides (maroon, upper half) and T1 joins the apoapsides (purple, lower half). All four burns are prograde.

Coaxial ellipse to ellipse — anti-aligned apsides

If \(\omega_2=\omega_1+180^\circ\), the periapsis of orbit 2 lies on the opposite side, and the apsidal transfer is no longer unique. A \(180^\circ\) transfer must start and end at apsides of both conics, and there are now two ways to do that:

\[ \text{T2 (periapsis to periapsis)}:\quad a_{t,2}=\frac{r_{p,1}+r_{p,2}}{2}, \qquad \text{T1 (apoapsis to apoapsis)}:\quad a_{t,1}=\frac{r_{a,1}+r_{a,2}}{2}. \]

Both are tangential at both ends, so both are legitimate two-impulse transfers. They are not the aligned case in disguise: T2 runs \(r_{p,1}\to r_{p,2}\), whereas the aligned transfer ran \(r_{p,1}\to r_{a,2}\), and the aligned alternative \(r_{a,1}\to r_{p,2}\) joins a different pair of apsides again.

Example 3 — the two anti-aligned routes

Same two orbits as Example 2, but anti-aligned.

	\(a_t\) [km]	\(\Delta v_1\)	\(\Delta v_2\)	total [km/s]	\(t\) [h]
T2 \(r_{p,1}=8400\to r_{p,2}=21000\)	\(14700\)	\(0.37924\)	\(2.04250\)	\(2.42174\)	\(\mathbf{2.464}\)
T1 \(r_{a,1}=15600\to r_{a,2}=63000\)	\(39300\)	\(2.17083\)	\(0.19386\)	\(\mathbf{2.36469}\)	\(10.769\)

\[ \boxed{\ \text{T1 is cheaper by } 57\ \mathrm{m/s};\quad \text{T2 is faster by a factor } 4.37 .\ } \]

Which route is longer: a two-line answer

Each route is half a revolution of its own transfer ellipse, so \(t=\pi\sqrt{a_t^3/\mu}\) and the larger \(a_t\) is always the slower route. Comparing the two \(a_t\) in each case, with \(c_i=a_ie_i\) the focus-to-centre offset of orbit \(i\):

\[ \boxed{\ \underbrace{a_t^{\rm apo\to apo}-a_t^{\rm peri\to peri}=c_1+c_2}_{\text{anti-aligned}}, \qquad \underbrace{a_t^{\rm peri\to apo}-a_t^{\rm apo\to peri}=c_2-c_1}_{\text{aligned}} .\ } \]

Both follow in one line from \(r_a-r_p=2ae=2c\). The consequences differ sharply:

• Anti-aligned: \(c_1+c_2>0\) always, so apoapsis-to-apoapsis is unconditionally the longer and slower route. Here \(c_1+c_2=3600+21000=24600\) km, and indeed \(39300-14700=24600\) km, giving \(t_1/t_2=(39300/14700)^{3/2}=4.37\).
• Aligned: the sign is that of \(c_2-c_1\), so periapsis-to-apoapsis is the longer route iff the outer orbit is more offset than the inner one. Here \(c_2-c_1=17400\) km \(=35700-18300\), and \(t_{A1}/t_{A2}=2.72\). Circularise orbit 2 enough that \(c_2<c_1\) and the two routes swap roles. \(\square\)

Cost: a competition, not a theorem

The cost side has no such clean answer. Each route pays

\[ \Delta v=\underbrace{\left|\sqrt{\mu\!\left(\tfrac{2}{r_A}-\tfrac{1}{a_t}\right)} -\sqrt{\mu\!\left(\tfrac{2}{r_A}-\tfrac{1}{a_1}\right)}\right|}_{\text{departure at }r_A} +\underbrace{\left|\sqrt{\mu\!\left(\tfrac{2}{r_B}-\tfrac{1}{a_2}\right)} -\sqrt{\mu\!\left(\tfrac{2}{r_B}-\tfrac{1}{a_t}\right)}\right|}_{\text{arrival at }r_B} \]

and every route sits somewhere on the Oberth trade: departing deep is efficient, but arriving deep is expensive, because the target is moving fastest there.

• T2 departs deep (\(r_{p,1}\), where \(v\) is largest), so its first burn is cheap — \(0.379\) km/s. But it arrives at \(r_{p,2}\), the fastest point of orbit 2, needing \(2.043\) km/s to catch it.
• T1 departs slow (\(r_{a,1}\)), so its first burn is expensive — \(2.171\) km/s. But it arrives at \(r_{a,2}\), the slowest point of orbit 2, and slots in for \(0.194\) km/s.

Which sum wins depends on all four radii, and the margin is not robust: A1 beats A2 by \(34.7\%\), but T1 beats T2 by only \(2.4\%\). Raise \(e_1\) to \(0.45\) at comparable radii and the anti-aligned ordering reverses — T2 becomes both cheaper and faster, and that trade-off vanishes altogether.

Three lessons. The apsidal transfer is never unique — there are always two, and naming one of them “the” transfer hides a real choice. Orientation matters: the same two orbit shapes, rotated by \(180^\circ\), cost \(1.855\) against \(2.365\) km/s — a \(27\%\) penalty. And “cheapest” is not a property of the geometry; only the length comparison is a theorem.

When the orbits cross

Everything so far assumed one orbit lies inside the other. If instead orbit 1 dips lower at periapsis and reaches higher at apoapsis, the two conics intersect, and a route opens that did not exist before: a single impulse at a crossing point, with no coast at all. Setting \(r_1(f)=r_2(f)\) on a shared apse line,

\[ \frac{p_1}{1+e_1\cos f}=\frac{p_2}{1+e_2\cos f} \qquad\Longrightarrow\qquad \boxed{\ \cos f^{*}=\frac{p_2-p_1}{p_1e_2-p_2e_1}\ } \]

with a real crossing exactly when \(|\cos f^{*}|\le1\). The two roots \(\pm f^{*}\) are mirror images, and — as in A6 — both cost the same.

Two crossing ellipses

Figure 8: Orbit 1 has the lower periapsis and the higher apoapsis, so the conics cross at \(f=\pm131.81^\circ\). Two routes to orbit 2: one impulse at a crossing (velocity triangle, inset), or two tangential impulses at the apsides (maroon). The single burn is immediate; the two-burn route needs a \(4.21\) h coast.

Example 4 — one impulse, or two?

Given orbit 1: \(r_p=8000\), \(r_a=30000\) km (\(a_1=19000\), \(e_1=0.578947\), \(p_1=12631.58\)). Orbit 2: \(r_p=12000\), \(r_a=24000\) km (\(a_2=18000\), \(e_2=0.333333\), \(p_2=16000\)). Apsides aligned.

\[ \cos f^{*}=\frac{16000-12631.58}{12631.58(0.333333)-16000(0.578947)} =\frac{3368.42}{-5052.63}=-0.666667, \]

\[ f^{*}=\pm131.8103^\circ, \qquad r^{*}=\frac{p_1}{1+e_1\cos f^{*}}=20571.43\ \mathrm{km}. \]

With \(v_r=\dfrac{\mu}{h}e\sin f\) and \(v_\theta=h/r\) at that point,

\[ \begin{aligned} \text{orbit 1}:&\quad v_r=2.42406,\ v_\theta=3.44932 &&\Rightarrow v_1=4.21590,\ \gamma_1=35.098^\circ\\ \text{orbit 2}:&\quad v_r=1.24008,\ v_\theta=3.88208 &&\Rightarrow v_2=4.07533,\ \gamma_2=17.715^\circ \end{aligned} \]

\[ \boxed{\ \Delta v_{\rm single}=\sqrt{(-1.18397)^2+(0.43276)^2}=1.26058\ \mathrm{km/s}, \quad\text{with no coast.}\ } \]

Example 4 — the two-impulse alternatives

Both apsidal routes survive, and both are tangential at each end:

	\(a_t\) [km]	\(\Delta v_1\)	\(\Delta v_2\)	total [km/s]	\(t\) [h]
\(r_{p,1}=8000\to r_{a,2}=24000\)	\(16000\)	\(0.22459\)	\(0.44580\)	\(0.67039\)	\(2.797\)
\(r_{a,1}=30000\to r_{p,2}=12000\)	\(21000\)	\(0.39018\)	\(0.23358\)	\(\mathbf{0.62376}\)	\(4.206\)

\[ \boxed{\ \frac{\Delta v_{\rm single}}{\Delta v_{\rm two}} =\frac{1.26058}{0.62376}=2.02\ } \]

The one-burn shortcut costs twice the two-burn route. Immediacy is not free.

Why the one-burn route loses

The two speeds at the crossing are nearly equal — \(4.216\) against \(4.075\) km/s, a change of just \(-0.141\) km/s. Almost all of the \(1.261\) km/s goes into rotating \(\mathbf v\) through \(\gamma_1-\gamma_2=17.38^\circ\). Priced as a pure rotation at the mean speed \(\bar v=4.146\) km/s, Equation 5 gives

\[ 2\bar v\sin\frac{17.38^\circ}{2}=2(4.146)(0.15108)=1.253\ \mathrm{km/s}, \]

which is \(99\%\) of the exact \(1.26058\). The single-impulse route is a rotation of the velocity vector, priced exactly like a plane change — except that it happens inside the orbital plane.

This is the in-plane twin of Part VI. Turning \(\mathbf v\) is expensive wherever you do it, and the two-impulse route wins because each of its burns is tangential — changing \(|\mathbf v|\) and never its direction. Crossing orbits do offer a one-burn shortcut, and here it costs double. Part IV is the opposite case: when the required turn is small enough, the single impulse wins instead.

The central claim

Theorem (Hohmann optimality). Among all two-impulse transfers between two coplanar circular orbits of radii \(r_1\) and \(r_2\), the Hohmann transfer — the ellipse tangent to both, with both impulses tangential — has the least total \(\Delta v\). This holds for every radius ratio.

The proof is Appendix A3. Its skeleton is worth stating here, because it explains why the answer is a tangent ellipse and not something cleverer:

1. Both impulse magnitudes can be written exactly in terms of the transfer orbit’s \((\varepsilon_t, h_t)\) alone (Equation 6).
2. In those variables the cost increases with \(\varepsilon_t\) and decreases with \(h_t\).
3. Reaching both radii constrains \((\varepsilon_t,h_t)\) to a region whose boundary has one corner — and the corner is the Hohmann ellipse.

The proof in one picture

Figure 9: Left: every feasible two-impulse transfer is a point in the \((h_t,\varepsilon_t)\) plane above both parabolas (the conic must reach \(r_1\) and reach \(r_2\)). The cost falls as you move right and down, so the optimum sits at the lowest-right point of the feasible set — its single corner. Right: the cost along that boundary, with the Hohmann value \(3.8926\) km/s at the kink.

The bi-elliptic transfer of Part II beats Hohmann for large \(R\) without contradicting this theorem: it uses three impulses.

Part II — The Bi-elliptic Transfer

Three impulses, one long detour

Figure 10: Burn 1 raises apoapsis to a very distant \(r_b\); burn 2, made out at \(r_b\) where the spacecraft is barely moving, raises periapsis from \(r_1\) to \(r_2\); burn 3 circularises at \(r_2\).

The three impulses

With \(R=r_2/r_1\), \(R^{*}=r_b/r_1\) and everything normalised by \(v_{c1}=\sqrt{\mu/r_1}\):

\[ \frac{\Delta v_1}{v_{c1}}=\sqrt{\frac{2R^{*}}{1+R^{*}}}-1, \qquad \frac{\Delta v_2}{v_{c1}}=\sqrt{\frac{2}{R^{*}}}\left(\sqrt{\frac{R}{R+R^{*}}}-\sqrt{\frac{1}{1+R^{*}}}\right), \]

\[ \frac{\Delta v_3}{v_{c1}}=\sqrt{\frac{2R^{*}}{R(R+R^{*})}}-\frac{1}{\sqrt R} \qquad\text{(applied retrograde).} \]

The whole idea lives in \(\Delta v_2\): it carries the factor \(\sqrt{2/R^{*}}\), so as \(r_b\to\infty\) the cost of changing the periapsis by any amount whatsoever tends to zero. The Oberth effect run in reverse — out there, velocity changes are cheap because \(v\) is small, and it is exactly \(v\) that multiplies \(\Delta v\) in the energy budget.

Setting \(R^{*}=R\) gives back the Hohmann transfer exactly (\(\Delta v_3=0\)). The bi-elliptic family is therefore a continuous extension of Hohmann — which is how Appendix A4 gets the thresholds.

Example 5 — \(R = 15\), \(r_b = 400\,000\) km

Given \(r_1=6678.137\) km, \(r_2=15r_1=100172.06\) km, \(r_b=400000\) km.

\[ a_{t1}=\frac{r_1+r_b}{2}=203339\ \mathrm{km}, \qquad a_{t2}=\frac{r_2+r_b}{2}=250086\ \mathrm{km}. \]

burn	where	from	to	\(\Delta v\)
1	\(r_1\)	\(7.72576\)	\(10.83580\)	\(3.11004\)
2	\(r_b\)	\(0.18091\)	\(0.63178\)	\(0.45088\)
3	\(r_2\)	\(2.52279\)	\(1.99478\)	\(0.52801\)

\[ \Delta v_{\rm bi}=4.08892\ \mathrm{km/s} \qquad\text{versus}\qquad \Delta v_{\rm H}=4.14269\ \mathrm{km/s}. \]

\[ \boxed{\ \text{saving } 0.0538\ \mathrm{km/s}\ (1.3\%)\quad\text{for a time penalty of } 17.1\ \mathrm{h}\ \to\ 299.6\ \mathrm{h} = 12.5\ \mathrm{days}.\ } \]

Notice \(\Delta v_2 = 0.45\) km/s changes the periapsis by \(93\,500\) km. At \(r_b\) the spacecraft is moving at \(181\) m/s — slower than a commercial airliner.

The two thresholds

Let \(\Delta v_{\rm bi}(R,R^{*})\) be the bi-elliptic cost. Two distinct questions have two distinct answers (both proved in A4):

\[ \boxed{\ R^{\dagger}=11.93877\ }\qquad \text{the bi-parabolic limit } (r_b\to\infty)\ \text{first beats Hohmann} \]

\[ \boxed{\ R^{\ddagger}=15.58172\ }\qquad \left.\frac{\partial \Delta v_{\rm bi}}{\partial R^{*}}\right|_{R^{*}=R}<0:\ \ \text{any } r_b>r_2\ \text{beats Hohmann} \]

• \(R<11.94\): Hohmann wins, always.
• \(11.94<R<15.58\): bi-elliptic wins only if \(r_b\) exceeds a threshold that depends on \(R\).
• \(R>15.58\): bi-elliptic wins for every \(r_b>r_2\), by more the further you go.

LEO\(\to\)GEO has \(R=6.31\), comfortably below \(11.94\): bi-elliptic is useless for GEO insertion. It earns its keep for lunar-distance and high-\(R\) targets — and even then the saving is a percent or two against a time penalty measured in weeks.

The two thresholds, seen

Figure 11: The cost against \(R\), for several \(r_b\). The two vertical markers are the thresholds \(11.94\) and \(15.58\).

Part III — Phasing Manoeuvres

Changing where you are, not what orbit you are on

Figure 12: The chaser burns retrograde at its current position, drops onto a smaller, shorter-period ellipse, and returns to the same point early.

Circular phasing: the derivation

A chaser trails a target by \(\Delta\theta\) in a circular orbit of radius \(r\) and period \(T\). To close the gap in \(k\) revolutions the chaser must complete \(k\) orbits in the time the target takes to cover \(2\pi k-\Delta\theta\):

\[ k\,T_{ph}=k\,T-\frac{\Delta\theta}{2\pi}T \qquad\Longrightarrow\qquad \boxed{\ T_{ph}=T\left(1-\frac{\Delta\theta}{2\pi k}\right).\ } \]

Kepler’s third law converts a period into a semi-major axis,

\[ a_{ph}=r\left(1-\frac{\Delta\theta}{2\pi k}\right)^{2/3}, \]

and the burn point is the apoapsis of the phasing ellipse, so

\[ \boxed{\ \Delta v_{\rm total}=2\left|\sqrt{\mu\left(\frac{2}{r}-\frac{1}{a_{ph}}\right)}-\sqrt{\frac{\mu}{r}}\ \right|.\ } \]

The two impulses are equal and opposite: you undo exactly what you did.

Example 6 — catching up 10° in one revolution

Given \(r=6778\) km (ISS-like), \(\Delta\theta=10^\circ\) behind, \(k=1\).

\[ T=5553.46\ \mathrm{s}=92.558\ \mathrm{min}, \qquad T_{ph}=T\left(1-\frac{10}{360}\right)=5399.19\ \mathrm{s}=89.987\ \mathrm{min}, \]

\[ a_{ph}=6778\left(1-\tfrac{1}{36}\right)^{2/3}=6651.89\ \mathrm{km} \qquad\Rightarrow\qquad r_{p,ph}=2a_{ph}-r=6525.79\ \mathrm{km}\ (147.6\ \mathrm{km\ altitude}). \]

\[ v_{ph}=7.595597,\quad v_c=7.668636 \qquad\Rightarrow\qquad \boxed{\ \Delta v_{\rm total}=2(0.073039)=146.1\ \mathrm{m/s}.\ } \]

The perigee altitude came out at \(147.6\) km — deep in the atmosphere. What do you do?

Answer

Take more revolutions. With \(k=3\): \(a_{ph}=6736.10\) km, perigee altitude \(316\) km, and \(\Delta v_{\rm total}=47.8\) m/s — one third of the cost and a safe perigee. Time is the cheapest resource in phasing; this is exactly why rendezvous profiles take days.

What patience buys

Figure 13: The cost of a \(\Delta\theta\) correction against the number of phasing revolutions \(k\) — the cost falls essentially as \(1/k\), and the binding constraint is that the phasing perigee must stay above the atmosphere.

Elliptical phasing

For a spacecraft already on an ellipse \((a,e)\) that must arrive \(\Delta t\) later,

\[ T_{ph}=T+\Delta t, \qquad a_{ph}=a\left(1+\frac{\Delta t}{T}\right)^{2/3}, \]

and the burn is made at an apsis. Which apsis? Differentiate vis-viva at fixed \(r\):

\[ v^2=\mu\left(\frac{2}{r}-\frac{1}{a}\right) \quad\Longrightarrow\quad 2v\,dv=\frac{\mu}{a^{2}}\,da \quad\Longrightarrow\quad \boxed{\ dv=\frac{\mu}{2va^{2}}\,da\ } \]

For a required \(da\), the impulse \(dv\) is inversely proportional to \(v\). So burn where you are fastest: at periapsis. (This is the Oberth effect again, wearing a different hat.)

Example 7 — arriving ten minutes late

Given \(a=10000\) km, \(e=0.3\) (\(r_p=7000\), \(r_a=13000\) km); delay \(\Delta t=600\) s.

\[ T=9952.01\ \mathrm{s},\qquad a_{ph}=10000\left(1+\frac{600}{9952.01}\right)^{2/3}=10397.99\ \mathrm{km}. \]

burn at	\(v\) before	\(v\) after	\(\Delta v\) (each)	total
periapsis, \(r=7000\)	\(8.603825\)	\(8.692036\)	\(0.088211\)	\(\mathbf{0.17642}\)
apoapsis, \(r=13000\)	\(4.632829\)	\(4.794662\)	\(0.161834\)	\(0.32367\)

\[ \boxed{\ \text{Burning at periapsis costs } 45\% \text{ less --- exactly the ratio } v_a/v_p=(1-e)/(1+e).\ } \]

Part IV — Apse-line Rotation

Rotating the orbit inside its own plane

Figure 14: Two orbits with identical \(p\) and \(e\), rotated by \(\eta\). They cross at \(\nu=\eta/2\) and \(\nu=\eta/2+\pi\), and at either crossing the required impulse is purely radial.

Where the orbits cross, and what it costs

Both orbits have the same \(p\) and \(e\); the final one is the initial one with \(\nu\) replaced by \(\nu-\eta\). They intersect where

\[ \frac{p}{1+e\cos\nu}=\frac{p}{1+e\cos(\nu-\eta)} \quad\Longrightarrow\quad \cos\nu=\cos(\nu-\eta) \quad\Longrightarrow\quad \boxed{\ \nu=\frac{\eta}{2}\ \ \text{or}\ \ \frac{\eta}{2}+\pi.\ } \]

At \(\nu=\eta/2\) the initial orbit has \(\nu=+\eta/2\) and the final orbit has \(\nu'=-\eta/2\), so with \(v_r=\sqrt{\mu/p}\,e\sin\nu\) and \(v_\theta=\sqrt{\mu/p}\,(1+e\cos\nu)\):

\[ v_\theta'=v_\theta \quad(\cos\ \text{is even}), \qquad v_r'=-v_r \quad(\sin\ \text{is odd}). \]

\[ \boxed{\ \Delta\mathbf v = -2v_r\,\hat{\mathbf e}_r, \qquad \Delta v = 2\sqrt{\frac{\mu}{p}}\,e\,\sin\frac{\eta}{2} = 2v\sin\gamma .\ } \]

The cost contains no \(r\). Both crossing points cost exactly the same, even though they sit at very different radii. There is no “cheap place” to rotate an apse line by a single impulse — the geometry has already chosen for you.

Example 8 — rotating by 30°

Given \(a=12000\) km, \(e=0.4\), so \(p=a(1-e^2)=10080\) km, \(\sqrt{\mu/p}=6.288378\) km/s. Rotate the apse line by \(\eta=30^\circ\).

Crossing at \(\nu=15^\circ\), where \(r=p/(1+e\cos15^\circ)=7270.79\) km and \(v=8.742294\) km/s:

\[ v_r=6.288378\,(0.4)\sin 15^\circ=0.651021, \qquad v_\theta=6.288378\,(1+0.4\cos 15^\circ)=8.718020 . \]

\[ \boxed{\ \Delta v = 2(0.651021)=1.30204\ \mathrm{km/s}\ } \qquad \gamma=\arctan\frac{v_r}{v_\theta}=4.271^\circ,\quad 2v\sin\gamma=1.30204\ \checkmark \]

Is one impulse the best you can do?

The alternative: circularise at apoapsis, coast, then re-eccentricise in the new direction. Its cost does not depend on \(\eta\):

\[ \Delta v_{\rm alt}=2\left(\sqrt{\frac{\mu}{r_a}}-\sqrt{\mu\left(\frac{2}{r_a}-\frac1a\right)}\right) =2(4.870956-3.773027)=2.19586\ \mathrm{km/s}. \]

Setting the two equal gives the crossover:

\[ 2\sqrt{\frac{\mu}{p}}e\sin\frac{\eta}{2}=\Delta v_{\rm alt} \qquad\Longrightarrow\qquad \boxed{\ \eta_{\rm cross}=51.8^\circ.\ } \]

• \(\eta<51.8^\circ\): one radial impulse (our example, \(30^\circ\): \(1.30\) vs \(2.20\) km/s).
• \(\eta>51.8^\circ\): go via the circular orbit (e.g. \(\eta=180^\circ\): \(5.03\) vs \(2.20\) km/s).

A single-impulse rotation costs \(\propto\sin(\eta/2)\) and so saturates; the two-burn route costs a constant. Large reorientations are therefore always done by changing shape, not by brute-forcing the rotation.

Where the crossover sits

Figure 15: The single-impulse cost against \(\eta\), compared with the alternative of circularising at apoapsis and re-eccentricising in the new direction.

Part V — Chase and Rendezvous

Two regimes, two tools

	far	near
separation	comparable to \(r\)	\(\ll r\)
question	“what orbit joins these two points in this time?”	“what burn kills this relative state?”
tool	Lambert’s problem (exact, nonlinear)	Clohessy–Wiltshire (linear, closed form)
unknown	\(\mathbf v_1\) on a conic through \(\mathbf r_1,\mathbf r_2\)	\(\Delta\mathbf v\) in the target’s rotating frame

Both are boundary-value problems for \(\ddot{\mathbf r}=-\mu\mathbf r/r^{3}\) — unlike everything before this deck, which was an initial-value problem.

Lambert’s problem

Given \(\mathbf r_1\), \(\mathbf r_2\) and the transfer time \(\Delta t\). Find the connecting conic, i.e. \(\mathbf v_1\) and \(\mathbf v_2\).

Lambert’s theorem: the transfer time depends only on \(a\), the chord \(c=|\mathbf r_2-\mathbf r_1|\) and the sum \(r_1+r_2\) — not on the shape otherwise. In universal variables, with \(C(z)\) and \(S(z)\) the Stumpff functions,

\[ A=\pm\sqrt{r_1r_2\left(1+\cos\Delta\theta\right)}, \qquad y(z)=r_1+r_2+A\,\frac{z\,S(z)-1}{\sqrt{C(z)}}, \]

\[ \sqrt{\mu}\,\Delta t=\left(\frac{y(z)}{C(z)}\right)^{3/2}S(z)+A\sqrt{y(z)} . \]

One iterates on \(z\) (\(z>0\) ellipse, \(z=0\) parabola, \(z<0\) hyperbola), then recovers the velocities from the Lagrange coefficients

\[ f=1-\frac{y}{r_1},\quad g=A\sqrt{\frac{y}{\mu}},\quad \dot g=1-\frac{y}{r_2}, \qquad \mathbf v_1=\frac{\mathbf r_2-f\mathbf r_1}{g},\quad \mathbf v_2=\frac{\dot g\,\mathbf r_2-\mathbf r_1}{g} . \]

Lambert generalises Hohmann

Figure 16: Two ways from LEO to GEO through a \(175^\circ\) sweep — the Hohmann ellipse (which needs \(5.28\) h) and a Lambert arc forced to arrive in \(4\) h.

Example 9 — LEO \(\to\) GEO in four hours

Given \(\mathbf r_1=(6678.137,\,0)\) km, \(\mathbf r_2 = 42164\,(\cos175^\circ,\sin175^\circ)\) km, \(\Delta t=4.000\) h.

Iterating to \(z=6.852\):

\[ \mathbf v_1=(-0.60573,\ 10.18142)\ \mathrm{km/s}, \qquad \mathbf v_2=(-1.11668,\ -1.52104)\ \mathrm{km/s}, \]

\[ \Delta v_1=|\mathbf v_1-\mathbf v_{c1}|=2.52926, \qquad \Delta v_2=|\mathbf v_{c2}-\mathbf v_2|=1.76006, \]

\[ \boxed{\ \Delta v_{\rm total}=4.28932\ \mathrm{km/s}\quad\text{versus Hohmann } 3.89255 : \ +10.2\%\ \text{to save } 1.27\ \text{h}.\ } \]

Transfer orbit: \(a=25973\) km, \(e=0.7439\), \(r_p=6650.5\) km. The arrival flight path angle is \(31.3^\circ\), not zero — the burn is no longer tangential, and that is precisely where the extra \(0.4\) km/s goes.

The price of a deadline

Figure 17: The cost of the \(175^\circ\) transfer against the imposed time. It has a minimum, and that minimum still sits above the Hohmann value, exactly as the optimality theorem requires.

Close in: the Clohessy–Wiltshire frame

Put a rotating LVLH frame on the target, in a circular orbit of radius \(a\) and mean motion \(n=\sqrt{\mu/a^{3}}\): \(\hat{\mathbf i}\) radial, \(\hat{\mathbf j}\) along-track, \(\hat{\mathbf k}\) cross-track. Linearising the relative motion for \(|\delta\mathbf r|\ll a\) (Appendix A9) gives

\[ \boxed{\ \ddot x-2n\dot y-3n^{2}x=0, \qquad \ddot y+2n\dot x=0, \qquad \ddot z+n^{2}z=0 .\ } \qquad(3)\]

Three facts to read off before solving anything:

• Cross-track decouples and is a pure oscillator at the orbital rate.
• \(\ddot y = -2n\dot x\): a radial velocity produces along-track acceleration. This is the coupling that makes low-thrust rendezvous counter-intuitive.
• The \(y\) equation integrates once to \(\dot y+2nx=\text{const}\) — a conserved quantity that is really the linearised angular momentum.

The state transition matrix and the two-impulse solution

The system is linear time-invariant, so \(\mathbf X(t)=\mathbf\Phi(t)\mathbf X_0\) with (in-plane part)

\[ \mathbf\Phi_{rr}=\begin{bmatrix}4-3\cos nt & 0\\ 6(\sin nt-nt) & 1\end{bmatrix}, \qquad \mathbf\Phi_{rv}=\frac{1}{n}\begin{bmatrix}\sin nt & 2(1-\cos nt)\\ -2(1-\cos nt) & 4\sin nt-3nt\end{bmatrix}. \qquad(4)\]

Rendezvous means \(\mathbf r(t_f)=\mathbf 0\), so

\[ \mathbf 0=\mathbf\Phi_{rr}\mathbf r_0+\mathbf\Phi_{rv}\mathbf v_0^{+} \qquad\Longrightarrow\qquad \boxed{\ \mathbf v_0^{+}=-\mathbf\Phi_{rv}^{-1}\mathbf\Phi_{rr}\,\mathbf r_0\ } \]

\[ \Delta\mathbf v_1=\mathbf v_0^{+}-\mathbf v_0^{-}, \qquad \Delta\mathbf v_2=-\left(\mathbf\Phi_{vr}\mathbf r_0+\mathbf\Phi_{vv}\mathbf v_0^{+}\right). \]

\(\det\mathbf\Phi_{rv}=\big[8(1-\cos nt)-3nt\sin nt\big]/n^{4}\) vanishes at \(nt=2\pi\) and at \(nt=8.8387\) (\(t=1.4067\,T\)). At those transfer times the two-impulse problem is singular: generically no solution exists. Rendezvous timelines must avoid them.

Example 10 — closing 5 km in half an orbit

Given target circular at \(r=6778\) km (\(n=1.13140\times10^{-3}\ \mathrm{s^{-1}}\), \(T=92.558\) min); chaser \(5\) km behind with zero relative velocity; \(t_f=T/2=46.279\) min.

\[ \mathbf r_0=(0,\,-5)\ \mathrm{km} \qquad\Longrightarrow\qquad \mathbf v_0^{+}=(-1.4143,\ 0)\ \mathrm{m/s} \]

\[ \boxed{\ \Delta v_1=1.414\ \mathrm{m/s\ (radially\ down)},\quad \Delta v_2=1.414\ \mathrm{m/s},\quad \Delta v_{\rm total}=2.83\ \mathrm{m/s}.\ } \]

The burn is purely radial and points down, yet the chaser moves forward. Dropping lowers the orbit, shortens the period, and the along-track gain follows from \(\ddot y=-2n\dot x\). Pushing forward along \(\hat{\mathbf j}\) — the instinctive move — raises the orbit and makes you fall behind.

The trajectory that results

Figure 18: The chaser drops \(1.25\) km, is carried forward by the gravity gradient, and arrives at the target with a second, equal impulse.

Rushing is expensive

Figure 19: Cost against the time allowed — fast is expensive, and \(t_f=1.4067\,T\) is impossible.

Part VI — Plane Changes

The most expensive thing you can do

Figure 20: A pure rotation of \(\mathbf v\) through \(\theta\) is an isosceles triangle, so \(\Delta v = 2v\sin(\theta/2)\).

Pure inclination change

The manoeuvre must happen on the line of nodes — the only points common to the old and new planes. There the speed is unchanged and only the direction turns, so the velocity triangle is isosceles:

\[ \Delta v=\sqrt{v^{2}+v^{2}-2v^{2}\cos\Delta i}=\sqrt{2v^2(1-\cos\Delta i)} \qquad\Longrightarrow\qquad \boxed{\ \Delta v=2v\sin\frac{\Delta i}{2}.\ } \qquad(5)\]

Example 11. \(\Delta i=28.5^\circ\) (Cape Canaveral to equatorial):

where	\(v\) [km/s]	\(\Delta v\) [km/s]
LEO, \(r=6678\) km	\(7.72576\)	\(3.80344\)
GEO, \(r=42164\) km	\(3.07467\)	\(1.51368\)
apoapsis of the LEO\(\to\)GEO transfer ellipse	\(1.60784\)	\(\mathbf{0.79155}\)

Same manoeuvre, a factor of 4.8 between the best and worst place to do it.

\(\Delta i=60^\circ\) costs a full orbital velocity; \(\Delta i = 180^\circ\) (orbit reversal) costs \(2v\). There is no such thing as a cheap large plane change.

Turn where you are slow

Figure 21: Because the cost is proportional to \(v\), it collapses with altitude — the same \(28.5^\circ\) costs \(3.80\) km/s at LEO, \(1.51\) km/s at GEO and only \(0.79\) km/s at the apoapsis of the transfer ellipse.

Combining the plane change with a transfer burn

Doing the plane change as part of the circularisation burn replaces a sum of magnitudes by the magnitude of a sum. With the angle between \(\mathbf v_{t2}\) and \(\mathbf v_{c2}\) equal to \(\Delta i\), the law of cosines gives

\[ \boxed{\ \Delta v_2=\sqrt{v_{t2}^{2}+v_{c2}^{2}-2v_{t2}v_{c2}\cos\Delta i}\ } \]

Example 12 — LEO \(\to\) GEO with \(\Delta i = 28.5^\circ\).

strategy	\(\Delta v\) [km/s]
Hohmann, then a separate plane change at GEO	\(3.89255+1.51368=5.40623\)
entire plane change at LEO, then Hohmann	\(4.98923+1.46682=6.45606\)
plane change folded into the GEO burn	\(2.42573+1.83023=\mathbf{4.25596}\)

Folding saves \(1.15\) km/s over doing it separately — about \(21\%\) of the whole budget.

Should you split the plane change?

Give \(\delta\) of the turn to the first burn and \(\Delta i-\delta\) to the second:

\[ \Delta v(\delta)= \underbrace{\sqrt{v_{c1}^{2}+v_{t1}^{2}-2v_{c1}v_{t1}\cos\delta}}_{\text{at }r_1} +\underbrace{\sqrt{v_{t2}^{2}+v_{c2}^{2}-2v_{t2}v_{c2}\cos(\Delta i-\delta)}}_{\text{at }r_2} \]

Setting \(d\Delta v/d\delta=0\) gives the stationarity condition (Appendix A7)

\[ \boxed{\ \frac{v_{c1}v_{t1}\sin\delta}{\Delta v_1} =\frac{v_{t2}v_{c2}\sin(\Delta i-\delta)}{\Delta v_2}\ } \]

Example 13. Solving numerically for LEO\(\to\)GEO, \(\Delta i=28.5^\circ\):

\[ \delta^{*}=2.20^\circ,\qquad \Delta v_1=2.44945,\quad \Delta v_2=1.78186, \qquad \boxed{\ \Delta v_{\rm total}=4.23131\ \mathrm{km/s}.\ } \]

A further \(24.7\) m/s below Example 12 — small, but free, and standard practice for every commercial GTO mission.

The split, seen

Figure 22: Splitting the plane change between the two Hohmann burns: the total has a shallow minimum a little away from \(\delta=0\).

Changing the node instead of the inclination

A change of RAAN by \(\Delta\Omega\) at fixed inclination \(i\) still rotates \(\hat{\mathbf h}\), and the required turn angle \(\alpha\) follows from the dot product of the two normals \(\hat{\mathbf h}=(\sin i\sin\Omega,\ -\sin i\cos\Omega,\ \cos i)\):

\[ \boxed{\ \cos\alpha=\cos^{2}i+\sin^{2}i\,\cos\Delta\Omega, \qquad \Delta v=2v\sin\frac{\alpha}{2}.\ } \]

Example 14. \(r=6778\) km, \(i=51.6^\circ\), \(\Delta\Omega=10^\circ\):

\[ \cos\alpha=\cos^{2}51.6^\circ+\sin^{2}51.6^\circ\cos10^\circ=0.990669 \quad\Rightarrow\quad \alpha=7.833^\circ, \]

\[ \Delta v=2(7.668636)\sin(3.9165^\circ)=\boxed{1.0476\ \mathrm{km/s}} . \]

…or letting the Earth do it for free

The \(J_2\) oblateness term precesses the node at

\[ \dot\Omega=-\frac{3}{2}J_2\left(\frac{R_\oplus}{p}\right)^{2}n\cos i . \]

For the orbit of Example 14,

\[ \dot\Omega=-5.0027^\circ/\mathrm{day} \qquad\Longrightarrow\qquad \boxed{\ 10^\circ \text{ of RAAN in } 2.0 \text{ days, for } 0\ \mathrm{m/s}.\ } \]

\(1.05\) km/s of propellant, or two days of waiting. This is why sun-synchronous orbits are designed around \(J_2\) rather than fighting it, and why a launch that misses its RAAN window waits rather than burns.

Part VII — Manoeuvres These Notes Leave Out

Why the list matters

Everything so far assumed impulsive burns, two bodies, and propulsion as the only force. Drop any one of those assumptions and a new family of manoeuvres appears — and several of them are what modern missions actually fly.

assumption dropped	what appears	where it is used
impulsive	low-thrust spirals, Edelbaum	all-electric GEO satellites, Dawn, BepiColombo
two-body	gravity assist, ballistic capture, manifolds	Voyager, GRAIL, Hiten, JWST
propulsion only	aerobraking, aerocapture, sails, tethers	MRO, Mars Odyssey, IKAROS
a single spacecraft	formation flying, station-keeping	GRACE, GEO fleet operations

1. Low thrust: when the burn is the trajectory

An ion engine delivers \(\sim0.1\) N. The burn lasts months, the spacecraft spirals, and “impulse at a point” is meaningless. For a quasi-circular spiral between \(r_1\) and \(r_2\) with a simultaneous plane change, Edelbaum’s result is remarkably simple:

\[ \boxed{\ \Delta v=\sqrt{v_1^{2}+v_2^{2}-2v_1v_2\cos\left(\frac{\pi}{2}\Delta i\right)}\ } \]

which for \(\Delta i=0\) collapses to just \(\Delta v = v_1-v_2\).

LEO \(\to\) GEO comparison:

	\(\Delta v\)	\(I_{sp}\)	propellant fraction
impulsive, optimal split, \(\Delta i=28.5^\circ\)	\(4.231\) km/s	\(315\) s	\(74.6\%\)
Edelbaum spiral, \(\Delta i=28.5^\circ\)	\(5.951\) km/s	\(1800\) s	\(\mathbf{28.6\%}\)

The low-thrust route costs 41% more \(\Delta v\) and less than half the propellant — and takes six months instead of five hours. \(\Delta v\) stops being the figure of merit the moment \(I_{sp}\) is on the table.

2. Using the atmosphere

• Aerobraking. Dip periapsis into the upper atmosphere and let drag lower apoapsis over hundreds of passes. Mars Reconnaissance Orbiter used it to convert a \(35\) h capture orbit into a \(2\) h science orbit, saving roughly \(1\) km/s of propellant for months of patience.
• Aerocapture. A single hypersonic pass converts a hyperbolic arrival directly into a bound orbit — the propulsive capture of our Mars deck, done with drag instead of fuel. Never yet flown as a primary capture.
• Aerogravity assist. A lifting body inside an atmosphere turns further than gravity alone allows, beating the \(\delta = 2\arcsin(1/e)\) limit of a pure flyby.

All three trade propellant for heat shield and risk. They are manoeuvres in which the \(\Delta v\) budget is paid by a thermal-protection engineer.

3. Using a third body

• Gravity assist. Free rotation of \(\mathbf v_\infty\) in a planet’s frame, converted into a heliocentric speed change. This is the subject of the Mars flyby deck; the whole gain comes from the frame change, not from propellant.
• Ballistic capture / weak stability boundaries. In the three-body problem, capture with no burn is possible — forbidden in the patched-conic model, as our Mars capture deck shows. Hiten (1991) and GRAIL flew it; it is slow but nearly free.
• Libration-point orbits and invariant manifolds. Halo and Lissajous orbits around \(L_1\), \(L_2\) are reached along manifold “tubes” for a few tens of m/s. JWST’s insertion into its \(L_2\) halo cost less than \(20\) m/s of correction.

4. Operations, and non-propulsive ideas

Station-keeping — fighting perturbations rather than changing orbit:

task	typical annual cost
GEO north–south (inclination, from lunisolar torque)	\(\approx 45\)–\(55\) m/s/yr
GEO east–west (longitude, from the \(J_{22}\) tesseral)	\(\approx 2\)–\(4\) m/s/yr
LEO drag make-up (ISS altitude)	tens of m/s/yr
end-of-life disposal to a graveyard orbit	\(\approx 11\) m/s

Momentum without propellant: solar sails (IKAROS, LightSail-2), electrodynamic tethers (thrust against the geomagnetic field), momentum-exchange tethers, and beamed-energy propulsion. Each replaces the rocket equation (Equation 1) with a different — and unbounded — limit.

5. The theory that decides when you are done

For a fixed \(\Delta v\) budget, is your \(N\)-impulse solution optimal? Lawden’s primer vector answers it. Let \(\boldsymbol\lambda(t)\) be the adjoint to velocity; then along an optimal impulsive trajectory

\[ \ddot{\boldsymbol\lambda}=-\frac{\mu}{r^{3}}\left[\boldsymbol\lambda-\frac{3(\mathbf r\cdot\boldsymbol\lambda)}{r^{2}}\mathbf r\right], \]

\[ \boxed{\ |\boldsymbol\lambda|\le 1\ \text{everywhere},\quad |\boldsymbol\lambda|=1\ \text{exactly at the impulses},\quad \Delta\mathbf v_i\parallel\boldsymbol\lambda(t_i).\ } \]

If \(|\boldsymbol\lambda|>1\) anywhere between burns, an extra impulse there would help — which is precisely how one discovers that a two-impulse transfer should have been a bi-elliptic. The theory tells you where to add the third burn.

Appendix — Proofs

A1. The rocket equation

A rocket of instantaneous mass \(m\) expels propellant at exhaust speed \(v_e\) relative to itself. In free space, conservation of momentum over \(dt\) gives

\[ m\,dv=-v_e\,dm \qquad\Longrightarrow\qquad \int_{0}^{\Delta v}dv=-v_e\int_{m_0}^{m_f}\frac{dm}{m}, \]

\[ \boxed{\ \Delta v=v_e\ln\frac{m_0}{m_f}=I_{sp}g_0\ln\frac{m_0}{m_f}. \ } \]

Inverting, \(m_f/m_0=e^{-\Delta v/v_e}\), so the propellant fraction is \(1-e^{-\Delta v/v_e}\) — exponential in \(\Delta v\), which is why the whole subject is about saving tens of m/s.

A2. What an impulse does to the energy (Oberth)

With \(\mathbf v^{+}=\mathbf v+\Delta\mathbf v\) at fixed \(\mathbf r\),

\[ \varepsilon^{+}-\varepsilon =\frac{|\mathbf v+\Delta\mathbf v|^{2}}{2}-\frac{\mu}{r}-\left(\frac{v^{2}}{2}-\frac{\mu}{r}\right) \]

\[ \boxed{\ \Delta\varepsilon=\mathbf v\cdot\Delta\mathbf v+\tfrac12|\Delta\mathbf v|^{2} = v\,\Delta v\cos\beta+\tfrac12\Delta v^{2}\ } \]

where \(\beta\) is the angle between \(\Delta\mathbf v\) and \(\mathbf v\). Two corollaries:

1. For fixed \(|\Delta\mathbf v|\), \(\Delta\varepsilon\) is maximised at \(\beta=0\): burn tangentially.
2. The first-order term carries a factor \(v\): burn where you are fast, i.e. at periapsis.

For the same energy change \(\Delta\varepsilon\), the required impulse is \(\Delta v\approx\Delta\varepsilon/v\), which is the differential statement Equation 7 of A5, used in Part III. \(\square\)

A3. Hohmann is optimal among all two-impulse transfers

Setup. Transfer from a circular orbit of radius \(r_1\) to a coplanar circular orbit of radius \(r_2>r_1\) using exactly two impulses, the first on circle 1 and the second on circle 2. Let the transfer conic have specific energy \(\varepsilon\) and angular momentum \(h>0\) (same sense as the circular orbits; \(h<0\) is strictly worse).

Step 1 — the impulses in terms of \((\varepsilon,h)\). At radius \(r_i\) the transfer speed satisfies \(v^{2}=2(\varepsilon+\mu/r_i)\) and its transverse component is \(h/r_i\). The circular velocity is purely transverse with magnitude \(\sqrt{\mu/r_i}\), so

\[ \Delta v_i^{2}=v^{2}+\frac{\mu}{r_i}-2\sqrt{\frac{\mu}{r_i}}\cdot\frac{h}{r_i}, \]

\[ \boxed{\ \Delta v_i^{2}=2\varepsilon+\frac{3\mu}{r_i}-\frac{2h\sqrt{\mu}}{r_i^{3/2}}, \qquad i=1,2.\ } \qquad(6)\]

This is exact — no approximation, no assumption about tangency.

A3 (continued) — monotonicity and the feasible set

Step 2 — monotonicity. Both \(\Delta v_1^2\) and \(\Delta v_2^2\) are increasing in \(\varepsilon\) (coefficient \(+2\)) and decreasing in \(h\) (coefficient \(-2\sqrt\mu/r_i^{3/2}<0\)). Hence the total cost \(\Delta v_1+\Delta v_2\) falls as \(h\) grows and as \(\varepsilon\) shrinks.

Step 3 — feasibility. The transfer conic must actually reach both radii. The radial velocity at radius \(r\) satisfies \(v_r^{2}=2(\varepsilon+\mu/r)-h^{2}/r^{2}\ge 0\), i.e.

\[ \varepsilon\ \ge\ F(r_i):=\frac{h^{2}}{2r_i^{2}}-\frac{\mu}{r_i}, \qquad i=1,2 . \]

Since the objective increases with \(\varepsilon\), at the optimum \(\varepsilon=\max\{F(r_1),F(r_2)\}\).

A3 (continued) — the corner is the Hohmann ellipse

Comparing the two lower bounds,

\[ F(r_2)-F(r_1)=\left(\frac{1}{r_2}-\frac{1}{r_1}\right) \left[\frac{h^{2}}{2}\left(\frac{1}{r_1}+\frac{1}{r_2}\right)-\mu\right], \]

and the first bracket is negative, so

\[ F(r_2)\gtrless F(r_1) \quad\Longleftrightarrow\quad h^{2}\lessgtr \frac{2\mu r_1r_2}{r_1+r_2}=:h_{\rm H}^{2}. \]

\(h_{\rm H}\) is exactly the angular momentum of the Hohmann ellipse. The feasible boundary is therefore made of two parabolic arcs meeting at \(h=h_{\rm H}\) — a single corner.

A3 (continued) — the two cases

Case A: \(h\le h_{\rm H}\). Then \(\varepsilon=F(r_2)\), i.e. \(v_r=0\) at \(r_2\): apoapsis is at \(r_2\) and the second burn is tangential. Substituting,

\[ \Delta v_2^{2}=\frac{h^{2}}{r_2^{2}}-\frac{2h\sqrt\mu}{r_2^{3/2}}+\frac{\mu}{r_2} =\left(\frac{h}{r_2}-\sqrt{\frac{\mu}{r_2}}\right)^{2} \Longrightarrow \frac{d\Delta v_2}{dh}=-\frac{1}{r_2}<0 , \]

\[ \Delta v_1^{2}=\frac{h^{2}}{r_2^{2}}-\frac{2\mu}{r_2}+\frac{3\mu}{r_1}-\frac{2h\sqrt\mu}{r_1^{3/2}} \Longrightarrow \frac{d\Delta v_1}{dh}=\frac{1}{\Delta v_1}\left(\frac{h}{r_2^{2}}-\frac{\sqrt\mu}{r_1^{3/2}}\right). \]

Because \(h\le h_{\rm H}\),

\[ \frac{h}{r_2^{2}}\le\frac{1}{r_2^{2}}\sqrt{\frac{2\mu r_1r_2}{r_1+r_2}} =\frac{\sqrt\mu}{r_2^{3/2}}\sqrt{\frac{2r_1}{r_1+r_2}} <\frac{\sqrt\mu}{r_2^{3/2}} \le\frac{\sqrt\mu}{r_1^{3/2}}, \]

using \(2r_1<r_1+r_2\) and \(r_1<r_2\). So \(d\Delta v_1/dh<0\) as well: the total strictly decreases in \(h\), and is minimised at the largest admissible \(h\), namely \(h=h_{\rm H}\).

A3 (concluded)

Case B: \(h> h_{\rm H}\). Then \(\varepsilon=F(r_1)\): periapsis is at \(r_1\) and the first burn is tangential, giving \(\Delta v_1=h/r_1-\sqrt{\mu/r_1}\) with \(d\Delta v_1/dh=+1/r_1>0\). For the second,

\[ \frac{d\Delta v_2}{dh}=\frac{1}{\Delta v_2}\left(\frac{h}{r_1^{2}}-\frac{\sqrt\mu}{r_2^{3/2}}\right), \]

and since \(h>h_{\rm H}\),

\[ \frac{h}{r_1^{2}}>\frac{\sqrt{\mu}}{r_1^2}\sqrt{\frac{2r_1r_2}{r_1+r_2}} >\frac{\sqrt\mu}{r_2^{3/2}} \quad\Longleftrightarrow\quad 2r_2^{4}>r_1^{3}(r_1+r_2), \]

which holds because \(r_2>r_1\). So both terms increase with \(h\), and the total is minimised at the smallest admissible \(h\) — again \(h=h_{\rm H}\).

Conclusion. Both cases drive \(h\to h_{\rm H}\), where \(F(r_1)=F(r_2)\) and the transfer conic has periapsis \(r_1\), apoapsis \(r_2\) and two tangential burns. That is the Hohmann ellipse, and it is the unique minimiser. \(\blacksquare\)

The proof never assumed the burns were tangential — tangency came out, as the active constraint. It also shows what is really being traded: a two-impulse transfer wants maximum angular momentum and minimum energy, and geometry only allows both at one point.

A4. The bi-elliptic thresholds, 11.94 and 15.58

Write \(\Delta v_{\rm bi}(R,R^{*})\) for the normalised bi-elliptic cost of Part II, and note

\[ \Delta v_{\rm bi}(R,R)=\Delta v_{\rm H}(R) \qquad(\text{at } r_b=r_2 \text{ the third impulse vanishes}). \]

The first threshold compares Hohmann with the \(r_b\to\infty\) limit. Taking limits term by term,

\[ \Delta v_{\rm bi,\infty}(R)=(\sqrt2-1)\left(1+\frac{1}{\sqrt R}\right), \]

and solving \(\Delta v_{\rm bi,\infty}(R)=\Delta v_{\rm H}(R)\) numerically gives

\[ \boxed{\ R^{\dagger}=11.938765\ } \]

The second threshold asks when raising \(r_b\) infinitesimally above \(r_2\) already helps, i.e. when \(\partial\Delta v_{\rm bi}/\partial R^{*}<0\) at \(R^{*}=R\).

A4 (continued) — why the second threshold equals the Hohmann maximum

Differentiate the second and third impulses with respect to \(R\) at \(R^{*}=R\). The first impulse does not involve \(R\) at all, and

\[ \frac{\partial}{\partial R}\left[\sqrt{\frac{2}{R^{*}}}\sqrt{\frac{R}{R+R^{*}}}\right]_{R^{*}=R} =\sqrt{\frac{2}{R}}\cdot\frac{1}{2}\sqrt{\frac{R+R}{R}}\cdot\frac{R}{(2R)^{2}} =\frac{1}{4R^{3/2}}, \]

\[ \frac{\partial}{\partial R}\left[\sqrt{\frac{2R^{*}}{R(R+R^{*})}}-\frac{1}{\sqrt R}\right]_{R^{*}=R} =-\frac{3}{4}R^{-3/2}+\frac{1}{2}R^{-3/2}=-\frac{1}{4R^{3/2}} . \]

They cancel exactly, so

\[ \boxed{\ \left.\frac{\partial \Delta v_{\rm bi}}{\partial R}\right|_{R^{*}=R}=0 \quad\text{for every } R \qquad\Longrightarrow\qquad \left.\frac{\partial \Delta v_{\rm bi}}{\partial R^{*}}\right|_{R^{*}=R} =\frac{d\Delta v_{\rm H}}{dR}\ } \]

by the chain rule applied to \(\Delta v_{\rm bi}(R,R)=\Delta v_{\rm H}(R)\).

Therefore: a detour beyond \(r_2\) pays exactly when the Hohmann cost is decreasing in \(R\). The threshold is the maximum of \(\Delta v_{\rm H}\):

\[ \boxed{\ R^{\ddagger}=15.581719=\arg\max_R \Delta v_{\rm H}(R).\ } \]

The two numbers quoted in every textbook are the same number twice. \(\blacksquare\)

A5. Where to burn for a period change

Along an orbit, vis-viva at fixed \(r\) relates a change of semi-major axis to a change of speed:

\[ v^{2}=\mu\left(\frac{2}{r}-\frac{1}{a}\right) \quad\Longrightarrow\quad 2v\,dv=\frac{\mu}{a^{2}}da \quad\Longrightarrow\quad dv=\frac{\mu}{2v\,a^{2}}\,da . \qquad(7)\]

For a prescribed \(da\) (equivalently a prescribed period change, since \(T\propto a^{3/2}\)), the cost \(dv\) is minimised by making \(v\) as large as possible:

\[ \boxed{\ \text{phase at periapsis, where } v=v_p \text{ is maximal.}\ } \]

The ratio between the apsides is \(v_a/v_p=(1-e)/(1+e)\), so an apoapsis burn costs \((1+e)/(1-e)\) times as much — for \(e=0.3\), a factor \(1.857\), matching Example 7’s \(0.3237/0.1764=1.835\) (the small difference is the finite, not infinitesimal, \(\Delta a\)). \(\square\)

A6. Apse-line rotation is a purely radial impulse

Both orbits share \(p\) and \(e\); the final orbit is the initial one advanced in argument of periapsis by \(\eta\), so a point at true anomaly \(\nu\) on the first is at \(\nu-\eta\) on the second. Equal radii require

\[ 1+e\cos\nu=1+e\cos(\nu-\eta)\Longrightarrow \cos\nu=\cos(\nu-\eta) \Longrightarrow \nu=\pm(\nu-\eta)+2k\pi . \]

The \(+\) sign gives \(\eta=0\); the \(-\) sign gives \(\nu=\eta/2\) (mod \(\pi\)). At such a point,

\[ v_r=\sqrt{\frac{\mu}{p}}\,e\sin\nu, \qquad v_\theta=\sqrt{\frac{\mu}{p}}\,(1+e\cos\nu), \]

and on the final orbit \(\nu'=\nu-\eta=-\eta/2=-\nu\). Since \(\cos\) is even and \(\sin\) odd,

\[ v_\theta'=v_\theta,\qquad v_r'=-v_r . \]

Hence \(\Delta\mathbf v=(v_r'-v_r)\hat{\mathbf e}_r=-2v_r\hat{\mathbf e}_r\) is purely radial, with

\[ \boxed{\ \Delta v=2\sqrt{\frac{\mu}{p}}\,e\sin\frac{\eta}{2}=2v\sin\gamma\ } \]

where \(\gamma\) is the flight path angle at the crossing, since \(\tan\gamma=v_r/v_\theta\). The expression contains no \(r\): both crossings cost the same. \(\square\)

A7. Plane change: the cost, and the optimal split

The cost. A rotation of \(\mathbf v\) through \(\theta\) at constant speed closes an isosceles triangle with legs \(v\) and apex angle \(\theta\):

\[ \Delta v^{2}=v^{2}+v^{2}-2v^{2}\cos\theta=2v^{2}(1-\cos\theta)=4v^{2}\sin^{2}\frac{\theta}{2} \Longrightarrow \Delta v=2v\sin\frac{\theta}{2}. \]

Since this is linear in \(v\), and \(v\) is smallest at apoapsis, plane changes migrate outward. \(\square\)

The optimal split. With \(\delta\) at burn 1 and \(\Delta i-\delta\) at burn 2,

\[ \Delta v_1(\delta)=\sqrt{v_{c1}^{2}+v_{t1}^{2}-2v_{c1}v_{t1}\cos\delta}, \qquad \Delta v_2(\delta)=\sqrt{v_{t2}^{2}+v_{c2}^{2}-2v_{t2}v_{c2}\cos(\Delta i-\delta)} . \]

\[ \frac{d\Delta v_1}{d\delta}=\frac{v_{c1}v_{t1}\sin\delta}{\Delta v_1}, \qquad \frac{d\Delta v_2}{d\delta}=-\frac{v_{t2}v_{c2}\sin(\Delta i-\delta)}{\Delta v_2}, \]

so the stationary point satisfies

\[ \boxed{\ \frac{v_{c1}v_{t1}\sin\delta}{\Delta v_1}=\frac{v_{t2}v_{c2}\sin(\Delta i-\delta)}{\Delta v_2}\ } \]

At \(\delta=0\) the left side vanishes while the right does not, so the derivative is negative there: it is always worth putting a little of the turn into the first burn. For LEO\(\to\)GEO this gives \(\delta^{*}=2.20^\circ\), and both sides evaluate to \(1.22925\). \(\square\)

A8. The turn angle for a node change

The two planes meet along a line through the focus, and the manoeuvre happens on it, so \(\mathbf r\) lies along the rotation axis. On a circular orbit \(\mathbf v\perp\mathbf r\), so \(\mathbf v\) is perpendicular to that axis and turns through the full dihedral angle \(\alpha\) between the planes. (On an eccentric orbit the radial part of \(\mathbf v\) lies along the axis and is untouched, so the velocity turns through less than \(\alpha\) and the cost below is an over-estimate.) The unit normal of an orbit with inclination \(i\) and node \(\Omega\) is

\[ \hat{\mathbf h}=(\sin i\sin\Omega,\ -\sin i\cos\Omega,\ \cos i). \]

For two orbits with the same \(i\) and nodes differing by \(\Delta\Omega\),

\[ \cos\alpha=\hat{\mathbf h}_1\cdot\hat{\mathbf h}_2 =\sin^{2}i\,(\sin\Omega_1\sin\Omega_2+\cos\Omega_1\cos\Omega_2)+\cos^{2}i \]

\[ \boxed{\ \cos\alpha=\cos^{2}i+\sin^{2}i\cos\Delta\Omega, \qquad \Delta v=2v\sin\frac{\alpha}{2}. \ } \]

Two sanity checks: for a polar orbit \(i=90^\circ\) this reduces to \(\alpha=\Delta\Omega\) (the planes meet at the poles, and the node change is the turn), while for \(i\to0\) it gives \(\alpha\to0\) — an equatorial orbit has no node to move. \(\square\)

A9. The Clohessy–Wiltshire equations

Kinematics. In a frame rotating at \(\boldsymbol\omega=n\hat{\mathbf k}\) (constant),

\[ \mathbf a_{\rm inertial}=\ddot{\mathbf r}_{\rm rel}+2\boldsymbol\omega\times\dot{\mathbf r}_{\rm rel} +\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r_{\rm rel}), \]

which in components \((x,y,z)\) gives \(\ddot x-2n\dot y-n^{2}x\), \(\ddot y+2n\dot x-n^{2}y\), \(\ddot z\).

Gravity gradient. With \(\mathbf r_c=\mathbf r_t+\delta\mathbf r\) and \(r_c\approx r_t(1+x/r_t)\),

\[ r_c^{-3}\approx r_t^{-3}\left(1-3\frac{x}{r_t}\right), \qquad \mathbf g=-\frac{\mu}{r_c^{3}}\mathbf r_c . \]

Using \(n^{2}=\mu/r_t^{3}\) and keeping first order,

\[ g_x\approx -n^{2}r_t+2n^{2}x,\qquad g_y\approx-n^{2}y,\qquad g_z\approx-n^{2}z . \]

Subtracting the target’s own acceleration \(-n^{2}r_t\) leaves the relative gravity \((2n^{2}x,\ -n^{2}y,\ -n^{2}z)\). Equating,

\[ \boxed{\ \ddot x-2n\dot y-3n^{2}x=0,\qquad \ddot y+2n\dot x=0,\qquad \ddot z+n^{2}z=0 .\ } \]

which are Equation 3 of Part V.

A9 (continued) — solution, and when rendezvous is impossible

Integrating the \(y\) equation once gives \(\dot y=\dot y_0-2n(x-x_0)\); substituting into the \(x\) equation yields \(\ddot x+n^{2}x=\text{const}\), a forced oscillator. Solving and back-substituting,

\[ x(t)=(4-3\cos nt)x_0+\frac{\sin nt}{n}\dot x_0+\frac{2(1-\cos nt)}{n}\dot y_0, \]

\[ y(t)=6(\sin nt-nt)x_0+y_0-\frac{2(1-\cos nt)}{n}\dot x_0+\frac{4\sin nt-3nt}{n}\dot y_0, \]

\[ z(t)=z_0\cos nt+\frac{\dot z_0}{n}\sin nt . \]

The secular term \(-6nt\,x_0\) in \(y\) is the entire content of the phasing problem: a radial offset makes you drift along-track for ever.

Singularity. With \(\mathbf\Phi_{rv}\) as in Equation 4,

\[ \det\mathbf\Phi_{rv}=\frac{8(1-\cos nt)-3nt\sin nt}{n^{4}}, \]

which vanishes at \(nt=2\pi\) and at \(nt=8.83874\) (\(t=1.4067\,T\)). At the second root the system is also inconsistent for a general \(\mathbf r_0\), so no two-impulse rendezvous exists at that transfer time. \(\square\)

A10. Lambert’s theorem

Claim. The transfer time between two points on a Keplerian arc depends only on \(\mu\), the semi-major axis \(a\), the chord \(c=|\mathbf r_2-\mathbf r_1|\) and the sum \(r_1+r_2\).

With the Lagrange parameters \(\alpha,\beta\) defined by

\[ \sin^{2}\frac{\alpha}{2}=\frac{s}{2a}, \qquad \sin^{2}\frac{\beta}{2}=\frac{s-c}{2a}, \qquad s=\frac{r_1+r_2+c}{2}, \]

Kepler’s equation integrates to Lambert’s equation

\[ \boxed{\ \sqrt{\frac{\mu}{a^{3}}}\,\Delta t=(\alpha-\sin\alpha)-(\beta-\sin\beta).\ } \]

Neither the eccentricity nor the orientation appears: only \(a\), \(c\) and \(r_1+r_2\). Since the right-hand side is monotone in \(a\) on each branch, the solution is unique once the transfer angle (\(<\pi\) or \(>\pi\)) and the number of revolutions are fixed — which is what makes the universal-variable iteration of Part V safe to bisect. \(\square\)

A11. Why the Hohmann cost is not monotone

Write \(u=\sqrt{R}\) and differentiate

\[ g(R)=\sqrt{\frac{2R}{1+R}}-1+\frac{1}{\sqrt R}-\frac{1}{\sqrt R}\sqrt{\frac{2}{1+R}} . \]

The first term rises monotonically to \(\sqrt2\) — the departure burn saturates at the escape burn — while the arrival terms decay like \(R^{-1/2}\), because the target circular velocity itself decays. Beyond the point where the second effect dominates the first, the total falls:

\[ \frac{dg}{dR}=0 \qquad\Longrightarrow\qquad \boxed{\ R=15.581719,\qquad g=0.536258 .\ } \]

Interpretation. For \(R>15.58\) the dominant cost is the departure burn, which is already almost an escape burn; making the target farther away barely changes it, while the arrival burn keeps getting cheaper. This is also, by A4, exactly the regime in which a detour to infinity is worth taking. \(\square\)

What to remember

1. \(\Delta\varepsilon=v\Delta v+\frac12\Delta v^{2}\) (Equation 2) and \(\Delta v_{\rm plane}=2v\sin(\theta/2)\) (Equation 5). Energy wants large \(v\); plane change wants small \(v\). Every design is that trade.
2. Hohmann is optimal among two-impulse transfers, always (A3) — and among all transfers when \(R<11.94\).
3. Bi-elliptic wins for \(R>15.58\) for any \(r_b\), and that number is the maximum of the Hohmann curve, not a coincidence (A4).
4. Phase at periapsis; change planes at apoapsis; rotate the apse line only if \(\eta\) is small.
5. Close in, the intuition inverts: to catch up, burn radially down, not forward.
6. Time is a resource. More phasing revolutions, a longer Lambert arc, \(J_2\) nodal drift and aerobraking all buy \(\Delta v\) with patience.