Patched conic method — interplanetary trajectory design: Question bank

Before we start, the symbols this page leans on the whole way down. Every one is earned here before it appears in a trap.

Definition The vocabulary and symbols you must own

Patch — one two-body conic (a single simple orbit shape) that we stitch to the next.
$μ$ ("mew"), the gravitational parameter — a shorthand for "mass of the central body times the gravity constant $G$ ": $μ = G m$ . It is the only thing about a body that matters for orbits around it, so we bundle $G$ and $m$ into one number. Its subscript tells you which body is central: $μ_{⊙} = G m_{⊙}$ (the Sun, symbol $⊙$ ), $μ_{E} = G m_{E}$ (Earth).
$r_{p}$ , perigee radius — the closest distance of the departure hyperbola to Earth's centre; in practice the radius of the circular parking orbit you launch from.
$v_{c i r c}$ , circular (parking-orbit) speed — the speed of a circular orbit at radius $r_{p}$ : $v_{c i r c} = μ_{E} / r_{p}$ . It is the speed you already have before the burn.
$v_{esc}$ , escape speed — the speed at $r_{p}$ that just barely reaches infinity with nothing left over: $v_{esc} = 2 μ_{E} / r_{p}$ . Note the factor $2$ : escaping costs twice the square of circular speed.
$v_{\infty}$ ("vee-infinity") — the leftover speed a spacecraft still has after it has fully escaped a planet, measured relative to that planet, at the edge of its Sphere of influence. Related to the others by energy bookkeeping: $v_{p}^{2} = v_{esc}^{2} + v_{\infty}^{2}$ .
Heliocentric — "Sun-centered"; a speed or orbit measured with the Sun as the fixed reference.

The one picture that makes every "frame" trap obvious — the patch stitch and the vector addition at the boundary:

Figure — Patched conic method — interplanetary trajectory design

Look at the SOI circle: the hyperbola's velocity $v_{\infty}$ is drawn relative to Earth, and Earth's own orbital velocity $v_{c, 1}$ is the second arrow. The heliocentric transfer speed is their tip-to-tail sum, not an arithmetic addition — keep this figure in mind for every "Spot the error" below.

True or false — justify

Each answer must be a reason, not a verdict.

The patched conic method gives the exact Earth→Mars trajectory.

False. It stitches idealized two-body conics and pretends only one body pulls at a time; the real field is many-body and continuous, so this is a great first guess you later refine numerically.

Inside Earth's Sphere of influence, the Sun's gravity is switched off.

False. The Sun still pulls; we merely choose to ignore it as a small perturbation because Earth dominates the motion there. The SOI is a modelling boundary, not a physical shield.

The sphere of influence is where Earth's and the Sun's gravitational forces on the craft are equal.

False. It is where treating Earth vs the Sun as the primary body is equally good — a ratio of primary-to-perturbing accelerations, not equal forces. Equal-force point is much closer to Earth.

Across a patch boundary the spacecraft's velocity is continuous.

True, but only after adding the planet's orbital velocity. Position is continuous outright; the hyperbola's

v_{\infty}

(planet-relative) must be vector-added to Earth's orbital speed to match the heliocentric ellipse.

A Hohmann transfer is exactly half of an ellipse.

True. Perihelion touches the inner orbit, aphelion the outer, so the craft coasts through exactly

18 0^{\circ}

— hence transfer time

t = π a_{t}^{3} / μ_{⊙}

(derived in the box below).

For an outbound Earth→Mars trip you speed up relative to the Sun.

True. Mars is farther out, so the transfer ellipse needs a higher perihelion speed than Earth's circular speed; the burn adds heliocentric speed (

v_{t, 1} > v_{c, 1}

The departure burn

Δ v

equals

v_{\infty}

False.

v_{\infty}

is measured at infinity where the craft is slow; you burn at perigee where it is fast, and energy (not speed) is conserved on the coast.

Δ v = v_{\infty}^{2} + 2 μ_{E} / r_{p} - μ_{E} / r_{p}

, usually smaller than

v_{\infty}

The departure conic is always a hyperbola, never a parabola or ellipse.

True for an interplanetary trip. You need

v_{\infty} > 0

to reach another planet's SOI, and positive excess energy means a hyperbola;

v_{\infty} = 0

would be a parabola (barely escapes), and negative would be a bound ellipse.

The Vis-viva equation works for the hyperbolic departure leg too.

True. Vis-viva

v = μ (2/ r - 1/ a)

holds for any conic; for a hyperbola the semi-major axis

a

is negative, which keeps

v^{2}

positive even at infinity.

Formula Sidebar 1 — where the

$2/5$ power really comes from The parent's $r_{S O I} = a_{p} (m_{p} / m_{⊙})^{2/5}$ deserves an honest sketch, because " $\propto r^{3}$ vs $\propto 1/ r^{2}$ " was too glib. Here is the clean version, using $μ = G m$ .

Near the planet at distance $r$ from it and $\approx a_{p}$ from the Sun, we compare how badly the second body disturbs the clean two-body motion of the first. A disturbance ratio is (perturbing acceleration) / (primary acceleration).

Planet as primary: primary pull $= \frac{μ _{p}}{r ^{2}}$ . The Sun's disturbing effect is not its full pull but the tidal difference across the small distance $r$ , which scales as $\frac{μ _{⊙} r}{a _{p}^{3}}$ (a pull that changes by $\sim \frac{d}{d a _{p}} (\frac{μ _{⊙}}{a _{p}^{2}}) \cdot r$ ). So $A_{A} = \frac{μ _{⊙} r / a _{p}^{3}}{μ _{p} / r ^{2}} = \frac{m _{⊙}}{m _{p}} \frac{r ^{3}}{a _{p}^{3}} (\propto r^{3}) .$

Sun as primary: primary pull $= \frac{μ _{⊙}}{a _{p}^{2}}$ . Now the planet's full pull $\frac{μ _{p}}{r ^{2}}$ is the disturbance (the craft is close to it), so $A_{B} = \frac{μ _{p} / r ^{2}}{μ _{⊙} / a _{p}^{2}} = \frac{m _{p}}{m _{⊙}} \frac{a _{p}^{2}}{r ^{2}} (\propto 1/ r^{2}) .$

The two descriptions are equally good where $A_{A} = A_{B}$ : $\frac{m _{⊙}}{m _{p}} \frac{r ^{3}}{a _{p}^{3}} = \frac{m _{p}}{m _{⊙}} \frac{a _{p}^{2}}{r ^{2}} \Rightarrow r^{5} = a_{p}^{5} (\frac{m _{p}}{m _{⊙}})^{2} \Rightarrow r = a_{p} (\frac{m _{p}}{m _{⊙}})^{2/5} .$ The squared mass ratio appears because one side carries $m_{⊙} / m_{p}$ and the other $m_{p} / m_{⊙}$ ; multiplying across to isolate $r^{5}$ stacks the ratio twice. The fifth root of that square is the $2/5$ . This is why it is $2/5$ and not $1/2$ : the primary/perturbing accelerations scale as $r^{3}$ and $1/ r^{2}$ , and $3 - (- 2) = 5$ is the power on $r$ .

Spot the error

Each line contains a wrong claim; the reveal names why it fails.

"Since

v_{\infty} = 2.94

km/s and the parking speed is

7.78

km/s, just add them:

Δ v = 2.94

km/s."

Wrong because

v_{\infty}

lives at infinity, not at perigee; you must first find perigee speed

v_{p} = v_{\infty}^{2} + v_{esc}^{2}

with

v_{esc} = 2 μ_{E} / r_{p}

, then

Δ v = v_{p} - v_{c i r c}

. The escape term is what makes

v_{\infty}

"cheap" to buy deep in the well (Oberth effect).

v_{\infty}

is the spacecraft's speed around the Sun after leaving Earth."

Wrong reference frame —

v_{\infty}

is measured relative to Earth. The heliocentric transfer speed is

v_{c, 1} \pm v_{\infty}

, obtained by vector-adding Earth's orbital velocity (see the stitch figure at the top).

"The SOI radius

r_{S O I} = a_{p} (m_{p} / m_{⊙})^{2/5}

scales with

m_{p}

, so Jupiter's SOI is enormous mainly because it's massive."

Both factors matter, but the

2/5

power weakens the mass dependence, while the linear

a_{p}

(Jupiter is far from the Sun) contributes strongly; distance is doing a lot of the work.

"To go from Earth to Mars we should burn to reach Mars's current position."

Wrong — Mars moves during the ~259-day coast. You aim at where Mars will be at arrival, which is why launch windows exist.

"Because a hyperbola is an escape orbit, its total energy is zero."

Zero energy is a parabola. A hyperbola has strictly positive specific energy

ε = v_{\infty}^{2} /2 > 0

; that leftover positive energy is exactly the excess speed at infinity.

"On the transfer ellipse the craft's speed is constant."

Kepler's second law forbids it — the craft sweeps equal areas in equal times, so it is fastest at perihelion (near Earth) and slowest at aphelion (near Mars).

"The parking orbit radius

r_{p}

doesn't affect the departure

Δ v

It does: a smaller

r_{p}

means a higher

v_{esc} = 2 μ_{E} / r_{p}

, so the burn happens deeper and cheaper — the Oberth effect rewards low perigees.

Formula Sidebar 2 — why the transfer time is

$t = π a_{t}^{3} / μ_{⊙}$ This formula is asserted all over the parent page; here is why the $π$ and the $3/2$ power appear.

Step 1 — full period first. Kepler's third law for any orbit of semi-major axis $a$ around a body of parameter $μ$ says the full period is $T = 2 π \frac{a ^{3}}{μ} .$ Where the $3/2$ comes from: an orbit is a balance of gravity ( $\propto 1/ a^{2}$ ) and the speed needed to stay on it; working the geometry through, the time to go around scales as $a^{3/2}$ — bigger orbits are slower and longer, and $3/2$ is exactly how those two effects compound. The $2 π$ is the "one full lap" factor, just like the circumference of a circle carries a $2 π$ .

Step 2 — a Hohmann transfer is half a lap. The transfer ellipse touches Earth at perihelion and Mars at aphelion, so the craft coasts through exactly $18 0^{\circ}$ — half the ellipse. Half the period: $t = \frac{1}{2} T = \frac{1}{2} \cdot 2 π \frac{a _{t}^{3}}{μ _{⊙}} = π \frac{a _{t}^{3}}{μ _{⊙}} .$ The $2 π$ halves to a lone $π$ — that is precisely why a single $π$ (not $2 π$ ) sits in the transfer-time formula. The figure below shows the half-ellipse and the $18 0^{\circ}$ sweep:

Why questions

Answer the mechanism, not just the outcome.

Why does the SOI derivation give a

2/5

power rather than, say,

1/2

Because the planet-as-primary disturbance ratio scales as

r^{3}

and the Sun-as-primary ratio as

1/ r^{2}

; setting them equal gives

r^{5} \propto a_{p}^{5} (m_{p} / m_{⊙})^{2}

, and the fifth root of the squared mass ratio is the

2/5

(full sketch in Sidebar 1).

Why do velocities "add" at the patch boundary but not simply?

Position is shared, but the hyperbola's velocity is planet-relative while the ellipse's is Sun-relative; you convert between frames by vector-adding the planet's orbital velocity, so magnitudes don't add arithmetically unless the vectors are aligned.

Why is burning deep in a gravity well (small

r_{p}

) more efficient?

At small

r_{p}

the craft moves fast, and kinetic energy

\propto v^{2}

; a fixed

Δ v

added to a large

v

raises

v^{2}

(and thus

v_{\infty}^{2}

) more than the same

Δ v

added to a slow craft — that's the Oberth effect.

Why does the Hohmann transfer orbit use exactly two tangential burns?

Tangential burns change speed without wasting energy on direction changes, and touching both circular orbits (perihelion at

r_{1}

, aphelion at

r_{2}

) is the minimum-energy two-impulse connection between coplanar circles.

Why can we treat the craft as "leaving the SOI at Earth's position"?

Because

r_{S O I}

is tiny compared to Earth's Sun distance

a_{p}

(~1000× smaller), so on the heliocentric scale the whole SOI collapses to a point at Earth's location.

Why is the transfer time fixed once

r_{1}

and

r_{2}

are chosen?

Time depends only on

a_{t} = (r_{1} + r_{2}) /2

through

t = π a_{t}^{3} / μ_{⊙}

(Kepler's third law); the geometry alone determines the coast duration.

Edge cases

The cases the topic invites but rarely shows.

What if

v_{\infty} = 0

at departure?

The departure conic becomes a parabola (marginal escape); the craft just barely leaves Earth's SOI with zero relative speed, so it can only drift to bodies sharing Earth's orbit — no real transfer.

What if the target planet's orbit is inside Earth's (e.g. Venus)?

The transfer ellipse's aphelion is at Earth and perihelion at Venus, so you must slow down relative to the Sun (

v_{t, 1} < v_{c, 1}

); the departure burn is retrograde, and

v_{\infty} = ∣ v_{t, 1} - v_{c, 1} ∣

still uses the magnitude.

What if the two planet orbits are not coplanar?

The basic Hohmann patch fails — a plane-change component is needed, adding

Δ v

; the pure patched conic assumes coplanar circular orbits as an idealization.

What happens to

r_{S O I}

m_{p} \to 0

(a tiny asteroid)?

r_{S O I} = a_{p} (m_{p} / m_{⊙})^{2/5} \to 0

, meaning the body has essentially no region where it dominates — the Sun controls the motion everywhere around it.

What happens to

r_{S O I}

a_{p} \to 0

(a body very close to the Sun)?

The SOI shrinks linearly to zero because the Sun's overwhelming nearby pull leaves no region where the small body can dominate.

What if

r_{1} = r_{2}

(same orbit)?

Then

a_{t} = r_{1}

, the "transfer ellipse" degenerates into the circular orbit itself,

v_{t, 1} = v_{c, 1}

, and

v_{\infty} = 0

— no burn and no transfer are needed.

At Mars arrival, is

v_{\infty}

the same value as at Earth departure?

No — arrival

v_{\infty}

is the planet-relative speed at Mars, computed from

∣ v_{t, 2} - v_{c, 2} ∣

at aphelion; it captures how much faster/slower the ellipse moves than Mars, and drives the arrival hyperbola.

Recall One-line self-test

Name the single fact that fixes every one of Errors 1–3 on the parent page. ::: $v_{\infty}$ is measured relative to the planet, at infinity — remember that and you fix the "just add speeds," the "heliocentric = $v_{\infty}$ ," and the frame-mixing mistakes at once.