Worked examples — Patched conic method — interplanetary trajectory design

This page is a drill ground. The parent note built the machinery: SOI radii, Hohmann ellipses, the vis-viva speed formula, and hyperbolic escape with the Oberth effect baked in. Here we push every one of those tools through every kind of input it can meet — inner vs outer target, zero excess speed, the degenerate "same orbit" case, a real launch-window word problem, and an exam twist.

Everything below reuses only symbols the parent earned:

• $\mu$ = $G$ times the central body's mass (Sun-frame ${\mu }_{\odot }$, Earth-frame ${\mu }_E$) — "how strong the pull is."
• $r$ = distance from the central body's centre. $a$ = semi-major axis (half the long width of an ellipse).
• $v_c=\sqrt{\mu /r}$ = circular-orbit speed. $v_{\infty }$ = leftover speed relative to a planet far from it.

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The scenario matrix

Every problem this topic throws is one cell of this grid. The worked examples afterwards each tag the cell(s) they cover, so by the end no cell is left un-shown.

Cell	What makes it different	Covered by
A. Inner → outer ($r_2>r_1$)	speed up to climb out; $v_{t,1}>v_{c,1}$	Ex 1
B. Outer → inner ($r_2<r_1$)	slow down to fall in; $v_{t,1}<v_{c,1}$, sign flip	Ex 2
C. Zero / degenerate ($r_2=r_1$)	limiting case, $v_{\infty }\to 0$	Ex 3
D. Escape burn from a well	$\Delta v$ with Oberth, deep vs shallow parking orbit	Ex 4
E. Limiting $v_{\infty }\to 0$	burn reduces to pure escape speed	Ex 4 (part c)
F. Arrival side (capture)	$v_{\infty }$ into target SOI, capture burn	Ex 5
G. SOI size / sanity	is the "tiny SOI" assumption fair?	Ex 6
H. Real-world word problem	timing a launch, phase angle	Ex 7
I. Exam twist	given $\Delta v$ budget, solve backwards for $v_{\infty }$	Ex 8

Constants used throughout (SI): ${\mu }_{\odot }=1.327\times 10^{20}$, ${\mu }_E=3.986\times 10^{14}$, ${\mu }_{Mars}=4.283\times 10^{13}$, $r_{Earth}=1.496\times 10^{11}$, $r_{Mars}=2.279\times 10^{11}$, $r_{Venus}=1.082\times 10^{11}$, $R_E=6.378\times 10^6$ (all metres, m³/s²).

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Ex 1 — Inner → outer (Cell A)

Forecast: Mars is farther out, so to reach it the spacecraft must be faster than Earth at departure. Guess: $v_{\infty }$ a few km/s, positive.

1. $a_t=\frac{r_1+r_2}{2}=\frac{1.496+2.279}{2}\times 10^{11}=1.888\times 10^{11}$ m. Why this step? The cheapest (Hohmann) ellipse just kisses both circular orbits: perihelion at $r_1$, aphelion at $r_2$. Its half-width is the average.
2. $v_{c,1}=\sqrt{{\mu }_{\odot }/r_1}=\sqrt{1.327\times 10^{20}/1.496\times 10^{11}}=2.978\times 10^4$ m/s. Why this step? This is Earth's own speed around the Sun — the baseline we compare against.
3. $v_{t,1}=\sqrt{{\mu }_{\odot }\left(\frac{2}{r_1}-\frac{1}{a_t}\right)}=3.279\times 10^4$ m/s. Why this step? Vis-viva gives the transfer-ellipse speed $v_t$ at its perihelion, which sits exactly at Earth's orbit.
4. $v_{\infty }^{dep}=v_{t,1}-v_{c,1}=3.279-2.978=0.301\times 10^4\approx 3.01$ km/s (positive). Why this step? The difference is the extra Sun-frame speed the departure hyperbola must inject. Positive ⇒ we speed up. See the red gap in the figure.

Recall Verify

Units: $\sqrt{({\text{m}}^3/{\text{s}}^2)/\text{m}}=\sqrt{{\text{m}}^2/{\text{s}}^2}=$ m/s ✓. Sanity: $v_{t,1}>v_{c,1}$ because at perihelion an ellipse moves fastest, and this ellipse reaches out to Mars (Kepler's 2nd law). Positive $v_{\infty }$ confirms "speed up for outer targets."

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Ex 2 — Outer → inner (Cell B: the sign flip)

Forecast: Venus is closer to the Sun. To fall inward the spacecraft must slow down relative to Earth. Guess: $v_{\infty }$ points opposite Earth's motion — but its magnitude is what a burn must supply.

1. $a_t=\frac{1.496+1.082}{2}\times 10^{11}=1.289\times 10^{11}$ m. Why? Now Earth's orbit is the aphelion of the transfer ellipse (the far point), Venus the perihelion.
2. $v_{c,1}=2.978\times 10^4$ m/s (same Earth baseline as Ex 1). Why? Departure planet is still Earth.
3. $v_{t,1}=\sqrt{{\mu }_{\odot }\left(\frac{2}{r_1}-\frac{1}{a_t}\right)}=2.734\times 10^4$ m/s. Why? Vis-viva gives the transfer-ellipse speed $v_t$ at Earth's orbit, which is now the ellipse's slowest point (aphelion). So $v_{t,1}<v_{c,1}$.
4. $v_{\infty }^{dep}=v_{t,1}-v_{c,1}=2.734-2.978=-0.244\times 10^4=-2.44$ km/s. Why? Negative sign = the transfer speed is less than Earth's. Physically the excess velocity vector points behind Earth's motion. The burn magnitude a rocket must deliver is $|v_{\infty }|=2.44$ km/s — retrograde relative to Earth's orbital velocity.

Recall Verify

Units m/s ✓. Sanity: outer target (Ex 1) gave $+3.01$, inner target gives $-2.44$ — opposite signs, exactly the expected slow-down. Both are only a few km/s vs the $\sim$30 km/s orbital speed, so a Hohmann is a small nudge on a big orbit ✓.

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Ex 3 — Degenerate case $r_2=r_1$ (Cell C)

Forecast: Same orbit ⇒ same speed ⇒ nothing to change. Guess $v_{\infty }=0$.

1. $a_t=\frac{r_1+r_1}{2}=r_1$. Why? When both radii are equal the "ellipse" degenerates to the circle itself — semi-major axis equals the radius.
2. $v_{t,1}=\sqrt{{\mu }_{\odot }\left(\frac{2}{r_1}-\frac{1}{r_1}\right)}=\sqrt{{\mu }_{\odot }/r_1}=v_{c,1}$. Why? With $a_t=r_1$ vis-viva collapses to the circular-speed formula. The transfer speed $v_t$ is the circular speed.
3. $v_{\infty }^{dep}=v_{t,1}-v_{c,1}=0$. Why? No radius change ⇒ no energy change ⇒ zero excess. The limiting case checks out and confirms the formulas don't blow up.

Recall Verify

$2/r_1-1/r_1=1/r_1$ exactly, so $v_{t,1}=v_{c,1}$ identically ⇒ $v_{\infty }=0$. No units issue (it's zero) ✓.

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Ex 4 — Escape burn with Oberth, deep vs shallow (Cells D & E)

Forecast: Burning deep in the well (small $r_p$, fast local speed) should be cheaper per unit $v_{\infty }$ — the Oberth effect. Guess (a) needs less $\Delta v$ than (b), even though both add the same $v_{\infty }$.

Part (a) — deep (LEO):

1. $v_{circ}=\sqrt{{\mu }_E/r_p}=\sqrt{3.986\times 10^{14}/6.578\times 10^6}=7.784$ km/s. Why? The speed you already have while circling.
2. $v_p=\sqrt{v_{\infty }^2+2{\mu }_E/r_p}=\sqrt{3010^2+2(3.986\times 10^{14})/6.578\times 10^6}=1.145\times 10^4$ m/s $=11.45$ km/s. Why? On a hyperbola energy adds: escape energy ($2{\mu }_E/r_p$) plus leftover $v_{\infty }^2$. Here $v_p$ is the perigee speed defined above.
3. $\Delta v_a=v_p-v_{circ}=11.45-7.784=3.67$ km/s. Why? The burn is the jump from circular to hyperbolic-perigee speed.

Part (b) — shallow (high orbit): 4. $v_{circ}=\sqrt{{\mu }_E/r_p}=\sqrt{3.986\times 10^{14}/4.0\times 10^7}=3.157$ km/s. Why? Same circular-speed formula $v_{circ}=\sqrt{{\mu }_E/r_p}$ — just evaluated at the larger radius, so it comes out slower (things orbit slower higher up). 5. $v_p=\sqrt{v_{\infty }^2+2{\mu }_E/r_p}=\sqrt{3010^2+2(3.986\times 10^{14})/4.0\times 10^7}=5.38\times 10^3$ m/s $=5.38$ km/s. Why? Same hyperbolic-perigee energy formula $v_p=\sqrt{v_{\infty }^2+2{\mu }_E/r_p}$ — escape energy plus leftover $v_{\infty }^2$, now at the higher $r_p$ where the escape term $2{\mu }_E/r_p$ is smaller. 6. $\Delta v_b=5.38-3.157=2.23$ km/s. Why compare? The departure burn itself is smaller here ($2.23<3.67$) — but that ignores the cost to reach this high orbit, computed next.

Completing the net-cost argument: to raise the parking orbit from LEO ($r_{LEO}=6.578\times 10^6$ m) to $r_{high}=4.0\times 10^7$ m first costs its own $\Delta v$. A Hohmann raise between those two circles has $a_{raise}=\frac{6.578\times 10^6+4.0\times 10^7}{2}=2.329\times 10^7\text{ m}.$

• Burn 1 (leave LEO): transfer perigee speed $v_{peri}=\sqrt{{\mu }_E(2/r_{LEO}-1/a_{raise})}=\sqrt{3.986\times 10^{14}(2/6.578\times 10^6-1/2.329\times 10^7)}=9.10\times 10^3=9.10$ km/s, so $\Delta v_1=v_{peri}-v_{circ,LEO}=9.10-7.784=1.32$ km/s. Why this step? Leaving the low circle onto the raising ellipse is itself a vis-viva speed jump at its perigee ($r_{LEO}$); the burn is that jump above the LEO circular speed.
• Burn 2 (circularise high): transfer apogee speed $v_{apo}=\sqrt{{\mu }_E(2/r_{high}-1/a_{raise})}=\sqrt{3.986\times 10^{14}(2/4.0\times 10^7-1/2.329\times 10^7)}=1.496\times 10^3=1.50$ km/s, so $\Delta v_2=v_{circ,high}-v_{apo}=3.157-1.50=1.66$ km/s. Why this step? Arriving at the high circle the ellipse is slower than the circle there (apogee is the ellipse's slowest point), so a prograde burn tops the speed up to circular.
• Raising cost $\approx 1.32+1.66=2.98$ km/s.
• Total shallow path: $2.98+2.23=5.21$ km/s, far more than the direct deep burn $3.67$ km/s.

Why this matters: once you add back the cost to climb up, the shallow route is much worse ($5.21$ vs $3.67$ km/s). The genuine Oberth saving comes from burning deep when you are already there, not from first flying to a higher orbit.

Oberth efficiency comparison — how much $v_{\infty }$ each burn buys per km/s spent:

Part (c) — limiting $v_{\infty }\to 0$: 7. Set $v_{\infty }=0$ in (a): $v_p\to \sqrt{2{\mu }_E/r_p}=\sqrt{2}{v}_{circ}=11.01$ km/s, the escape speed. Why? Zero leftover speed means "just barely escape." The hyperbola degenerates to a parabola, and $\Delta v=\sqrt{2}{v}_{circ}-v_{circ}=0.414v_{circ}=3.22$ km/s. Why it matters: the general formula reduces to the textbook escape burn — the limit behaves.

Recall Verify

Units m/s ✓. Deep-burn Oberth check: at LEO, $3.67$ km/s buys $v_{\infty }=3.01$; the full climb-and-go shallow route totals $\approx 5.21$ km/s — much worse. Limit: with $v_{\infty }=0$, $v_p=\sqrt{2}{v}_{circ}$ exactly ✓.

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Ex 5 — Arrival side / capture (Cell F)

Forecast: Symmetric to departure but backwards: we shed speed to fall from a hyperbola into a bound circle. Guess $\Delta v$ of order 1–2 km/s.

1. $v_p=\sqrt{v_{\infty }^2+2{\mu }_{Mars}/r_p}=\sqrt{2650^2+2(4.283\times 10^{13})/3.6\times 10^6}=5.55\times 10^3$ m/s $=5.55$ km/s. Why? Same hyperbolic-perigee energy formula — $v_p$ is the incoming hyperbola's fastest (closest) point.
2. $v_{circ}=\sqrt{{\mu }_{Mars}/r_p}=\sqrt{4.283\times 10^{13}/3.6\times 10^6}=3.449$ km/s. Why? The speed of the circular orbit we want to end up on.
3. $\Delta v_{capture}=v_p-v_{circ}=5.55-3.449=2.10$ km/s (a braking burn, retrograde). Why? We remove exactly enough kinetic energy to convert the open hyperbola into a closed circle at perigee.

Recall Verify

Units m/s ✓. $v_p>v_{circ}$ (hyperbola always faster than a circle at the same radius) so a positive braking magnitude — physical ✓. Order 2 km/s matches typical Mars-orbit-insertion budgets.

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Ex 6 — SOI size sanity check (Cell G)

Forecast: The patch method assumes the SOI is negligibly small compared to the orbit, so we can pretend the craft leaves "at Earth's position." Guess: ratio well under 1%.

1. $r_{SOI}=a_p{\left(\frac{m_E}{m_{\odot }}\right)}^{2/5}=1.496\times 10^{11}\times (3.003\times 10^{-6}{)}^{0.4}$. Why? This is the crossover radius the parent derived by setting the two perturbation ratios equal.
2. $(3.003\times 10^{-6}{)}^{0.4}=6.17\times 10^{-3}$, so $r_{SOI}=1.496\times 10^{11}\times 6.17\times 10^{-3}=9.24\times 10^8$ m. Why? Raising a small number to the $2/5$ power softens it — the SOI is bigger than $m_E/m_{\odot }$ alone would suggest but still small.
3. Ratio $r_{SOI}/a_p=6.17\times 10^{-3}\approx 0.6\%$. Why it matters: well under 1%. So treating the craft as leaving from Earth's exact orbital position introduces sub-percent error — the "tiny SOI" assumption is justified.

Recall Verify

Units m ✓. Known Earth SOI $\approx 9.2\times 10^5$ km $=9.2\times 10^8$ m — matches step 2 to two figures ✓. Ratio dimensionless, $0.6\%\ll 1$ ✓.

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Ex 7 — Real-world word problem: catching Mars (Cell H)

Forecast: Mars is slower than the transfer's arrival point rotates, so at launch Mars must be ahead of the arrival spot, but not a full 180°. Guess: Mars leads Earth by roughly 40–50°.

1. Mars's period from Kepler's 3rd: $T_{Mars}=2\pi \sqrt{r_2^3/{\mu }_{\odot }}=2\pi \sqrt{(2.279\times 10^{11}{)}^3/1.327\times 10^{20}}=5.94\times 10^7$ s $=687$ days. Why? We need Mars's angular speed to know where it drifts during flight.
2. Angle Mars sweeps in 259 days: ${\theta }_{Mars}=360°\times \frac{259}{687}=135.7°$. Why? Angular speed $\times$ time = angle covered.
3. The probe arrives at a point $180°$ around from Earth (Hohmann = half an ellipse). Why? Aphelion of the transfer sits diametrically opposite the launch point.
4. Required lead angle = $180°-{\theta }_{Mars}=180°-135.7°=44.3°$. Why? Mars only needs to cover the remaining arc to reach the meeting point exactly when the probe does. So at launch Mars must be 44.3° ahead of Earth in its orbit — a launch window condition.

Recall Verify

$360\times 259/687=135.7°$ ✓. Lead $=180-135.7=44.3°$, in the forecast 40–50° band ✓. The textbook Earth→Mars phase angle is $\approx 44°$ ✓.

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Ex 8 — Exam twist: solve backwards (Cell I)

Forecast: We normally go $v_{\infty }\to \Delta v$. Here we invert: given $\Delta v$, back out $v_{\infty }$, then back out the transfer ellipse. Expect $v_{\infty }$ slightly under 3 km/s (from Ex 4a, $3.67$ km/s gave $3.01$).

1. $v_p=v_{circ}+\Delta v=7.784+3.60=11.384$ km/s. Why? Reverse of $\Delta v=v_p-v_{circ}$: add the burn back onto the circular speed to recover the perigee speed $v_p$.
2. $v_{\infty }=\sqrt{v_p^2-2{\mu }_E/r_p}=\sqrt{11384^2-2(3.986\times 10^{14})/6.578\times 10^6}=2.86\times 10^3$ m/s $=2.86$ km/s. Why? Invert the hyperbolic energy relation $v_p^2=v_{\infty }^2+2{\mu }_E/r_p$ to strip off the escape term.
3. Heliocentric transfer speed at Earth: $v_{t,1}=v_{c,1}+v_{\infty }=29.78+2.86=32.64$ km/s. Why? Add Earth's orbital velocity (patch condition) — the Sun-frame transfer-ellipse speed $v_t$ at perihelion.
4. Back out $a_t$ from vis-viva: $v_{t,1}^2={\mu }_{\odot }\left(\frac{2}{r_1}-\frac{1}{a_t}\right)\Rightarrow \frac{1}{a_t}=\frac{2}{r_1}-\frac{v_{t,1}^2}{{\mu }_{\odot }}$. Numerically $\frac{2}{r_1}=1.3369\times 10^{-11}$, $\frac{v_{t,1}^2}{{\mu }_{\odot }}=\frac{(32640{)}^2}{1.327\times 10^{20}}=8.029\times 10^{-12}$, so $\frac{1}{a_t}=5.340\times 10^{-12}$, $a_t=1.873\times 10^{11}$ m. Why? This is the algebraic inverse of Ex 1 step 3 — same equation, different unknown.
5. Aphelion reached: $r_{aph}=2a_t-r_1=2(1.873\times 10^{11})-1.496\times 10^{11}=2.250\times 10^{11}$ m. Why? An ellipse's aphelion is $2a_t$ minus the perihelion $r_1$. This is just short of Mars ($2.279\times 10^{11}$), telling us $3.60$ km/s from LEO nearly — but not quite — reaches Mars's orbit.

Recall Verify

Consistency: $2.86$ km/s ($v_{\infty }$ from $3.60$ budget) is just below the $3.01$ km/s that needed $3.67$ — monotonic ✓. Aphelion $2.250\times 10^{11}$ m sits just inside Mars's $2.279\times 10^{11}$ m, matching "3.60 km/s is slightly short of a true Mars Hohmann" ✓.

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Recall Feynman: explain the whole page to a 12-year-old

• What did $r_1$ and $r_2$ mean? ::: The Sun-distance of where you leave from and where you're going.
• Why does an outer target need speeding up but an inner target slowing down? ::: To climb outward you need more speed; to fall inward you shed speed.
• Why is burning deep in a gravity well efficient? ::: You're already moving fast there, so the same burn adds more usable energy (Oberth).