Spheres of Influence and the Patched-Conic Model

Laplace's sphere, the Hill sphere, the three-phase trajectory and the size of the error

Apollo 8, 23 December 1968

About fifty-five hours after launch, more than 300,000 km from Earth, Apollo 8 crosses an invisible boundary. Nothing happens on board. On the ground, the flight dynamics officers change the centre of their calculations from the Earth to the Moon: from here on, the spacecraft's trajectory is computed as a hyperbola about the Moon, with the Earth's pull treated as a small perturbation. The boundary is the Moon's sphere of influence. It is not a physical object, and nothing in the equations of motion changes there. It is a bookkeeping decision about which body to treat as primary — and it is the idea that made interplanetary navigation computable with pencils, slide rules and the computers of the 1960s.

Watch first · presented by Shoaib Aziz

The numbers in this film are the labs' own: Earth's sphere of influence is 924,647 km (Jupiter's 48,219,601 km; the Moon's, relative to the Earth, 66,183 km) and its Hill sphere about 1.5 million km; Laplace's boundary found numerically is 924,657 km along Earth's orbit but 801,924 km towards the Sun, a factor of 0.867; and the three-body trial of an Earth–Mars Hohmann departure (v∞ = 2.945 km/s) leaves Earth's sphere after 3.16 days still moving at 3.08 km/s, has a = 1.2641 AU and e = 0.2088 after 40 days (patched conic: 1.2618 AU, 0.2075), and passes 640,445 km beyond Mars's orbit at aphelion.

The true problem is an

n

-body problem: Sun, planets and spacecraft pulling on each other, with no closed-form solution. The patched-conic approximation replaces it with a chain of two-body problems:

Departure — a hyperbola about the Earth, from the parking orbit out to Earth's sphere of influence.
Cruise — an ellipse about the Sun, from Earth's sphere to the target's. Its endpoints are the planet centres, because at heliocentric scale each sphere is a point. This is the Lambert arc of the unit on Lambert's problem.
Arrival — a hyperbola about the target, from its sphere of influence in to periapsis.

The conics are joined (patched) by the frame-change rule

\vec v_{\text{helio}} = \vec v_{\text{planet}} + \vec v_\infty

At the patch point the spacecraft's heliocentric velocity is the planet's velocity plus the hyperbolic excess velocity. The sphere is a point to the Sun and infinitely large to the planet — a deliberate inconsistency that turns out to cost surprisingly little.

💡Laplace's sphere of influence

Inside some region round a planet of mass

m

(orbiting a Sun of mass

M

at distance

a

) it is better to use the planet as the primary and the Sun as a perturbation; outside it the reverse. Laplace's criterion compares the ratio of perturbation to primary acceleration in the two descriptions:

planet-centred: primary $\mu_p/d^2$ ; perturbation = the Sun's tidal acceleration (its pull on the spacecraft minus its pull on the planet);
Sun-centred: primary $\mu_\odot/r^2$ ; perturbation = the planet's pull on the spacecraft minus its pull on the Sun.

The boundary is where the two ratios are equal. Expanding both to leading order in

d/a

gives Laplace's result

r_{SOI} = a\left(\frac{m}{M}\right)^{2/5}

Earth: 924,600 km. Mars: 577,200 km. Jupiter: 48.2 million km — almost a third of an astronomical unit. The Moon relative to Earth: 66,200 km.

💡Not quite a sphere

Laplace's full result is a surface,

r(\theta) = a(m/M)^{2/5}(1 + 3\cos^2\theta)^{-1/10}

, with

\theta

measured from the Sun–planet line. Across the line (

\theta = 90°

) it equals the formula; along the line it is smaller by

4^{-1/10} = 0.871

. In the second lab you find the boundary numerically: 924,657 km along-track for Earth, but 801,924 km toward the Sun — a ratio of 0.867. Everyone uses the sphere anyway; the approximation it supports is cruder than the shape.

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Body	r_SOI (km)	Hill radius (km)	r_SOI / body radius
Mercury	112,400	220,700	46
Venus	616,300	1,011,000	102
Earth	924,600	1,497,000	145
Mars	577,200	1,084,000	170
Jupiter	48,220,000	53,150,000	674
Saturn	54,810,000	65,470,000	909
Uranus	51,840,000	70,190,000	2,028
Neptune	86,780,000	116,200,000	3,504
Moon (w.r.t. Earth)	66,200	61,500	38

⚠Sphere of influence versus [Hill sphere](https://en.wikipedia.org/wiki/Hill_sphere)

The Hill sphere,

r_H = a\,(m/3M)^{1/3}

, answers a different question: inside roughly a third to a half of it, a satellite can orbit the planet stably for a long time. It is always larger than Laplace's sphere for the planets, and it comes from the restricted three-body problem (Lagrange points L1 and L2 sit near its edge). Use the sphere of influence to decide where to switch primaries in a patched-conic calculation; use the Hill sphere to ask whether a moon or a captured asteroid can survive.

How long does the planet own the spacecraft? Leaving a 200 km orbit on a Mars-bound hyperbola (

v_\infty = 2.9

km/s), it takes about 3.2 days to reach Earth's sphere of influence; on a Jupiter-bound one (8.8 km/s), 1.2 days. Compare with a cruise of 6–30 months: the planetocentric phases are short, which is why the heliocentric arc can treat each sphere as a point.

But the planet does not let go at the boundary. At

r_{SOI}

the hyperbola's speed is

\sqrt{v_\infty^2 + 2\mu/r_{SOI}}

, not

v_\infty

. For the Mars departure that is 3.088 km/s against 2.945 — the patched conic ignores the last 0.143 km/s of Earth's pull.

📐Worked Example: putting the approximation on trial

Integrate the real three-body motion — Sun, Earth and spacecraft — for a departure from a 200 km orbit onto an Earth–Mars Hohmann hyperbola (

v_\infty = 2.945

km/s), and compare with the patched-conic prediction.

The spacecraft crosses Earth's sphere of influence after 3.16 days, moving at 3.082 km/s relative to Earth (two-body hyperbola: 3.088).
Forty days later its heliocentric orbit has $a = 1.2641$ AU and $e = 0.2088$ ; the patched conic predicts the Hohmann values 1.2618 AU and 0.2075.
Aphelion lands at 1.5280 AU instead of Mars's 1.5237 AU: 640,000 km too far — larger than Mars's own sphere of influence.

The patched conic is right to about 0.2% in the orbital elements and wrong by hundreds of thousands of kilometres at the target. Both statements matter: it is an excellent design tool and an unusable navigation tool.

Figure 10.2

The three conics of an interplanetary transfer, and the two patch points where the description changes hands.

🕰Where patched conics fail

The approximation is excellent when the spacecraft moves fast relative to the planet (large

v_\infty

) and spends little time near the boundary. It fails for slow, low-energy trajectories that linger near the edge of a sphere of influence: lunar transfers via the Sun–Earth Lagrange points (Hiten, 1991; GRAIL, 2011), ballistic capture at Mars, and Lagrange-point missions such as JWST. Those need the three-body problem — the subject of the unit on the three-body problem and Lagrange points — not a chain of conics.

↓But there is a problem

Three conics, two patches, one frame-change rule. The middle conic is solved. The first one is not. The Lambert solution of the unit on Lambert's problem handed us a vector

\vec v_{\infty}

— a magnitude and a direction in space. The spacecraft is sitting in a circular orbit 200 km above Cape Canaveral, moving at 7.8 km/s in a direction set by the launch azimuth. Where in that orbit must the burn happen, how big must it be, and is the orbit even in the right plane?