Exercises — Sphere of influence — radius derivation

Aisha is correct: the Laplace SOI carries the ==$2/5$ power== of the mass ratio, the fingerprint of a tidal perturbation. Ben's $1/2$ power is the force-equality point $r_{eq}=R{q}^{1/2}$ — where the planet's raw pull equals the Sun's raw pull. It is a different (smaller) radius and is not the SOI. Mnemonic: "Tides are two-fifths."

• $R$ = distance between the two large bodies (e.g. planet ↔ Sun).
• $m$ = mass of the smaller body — the one whose bubble we compute.
• $M$ = mass of the larger central body.
• Always $m\ll M$, so $q=m/M<1$ and $q^{2/5}<1$, making $r_{SOI}<R$ as it must be (the bubble sits well inside the separation).

Step 1 — the mass ratio: $q=\frac{5.97\times 10^{24}}{1.989\times 10^{30}}=3.00\times 10^{-6}$. Step 2 — raise to $2/5$ using logs (recall: log turns the power into a multiply). ${\log }_{10}q=-5.523$; times $0.4$ gives $-2.209$; so $q^{0.4}=10^{-2.209}=6.18\times 10^{-3}$. Step 3 — multiply by $R$: $r_{SOI}=1.496\times 10^{11}\times 6.18\times 10^{-3}\approx 9.2\times 10^8\text{ m}\approx 924,000\text{ km}.$ That's about 145 Earth radii.

$q=6.42\times 10^{23}/1.989\times 10^{30}=3.228\times 10^{-7}$. ${\log }_{10}q=-6.491$; $\times 0.4=-2.596$; $q^{0.4}=10^{-2.596}=2.535\times 10^{-3}$. $r_{SOI}=2.279\times 10^{11}\times 2.535\times 10^{-3}\approx 5.78\times 10^8\text{ m}\approx 578,000\text{ km}.$ Smaller than Earth's despite Mars being farther out, because Mars is much lighter — the mass ratio wins over the distance.

Force-equality uses the $1/2$ power: $r_{eq}=R\sqrt{q}=2.279\times 10^{11}\times \sqrt{3.228\times 10^{-7}}.$ $\sqrt{3.228\times 10^{-7}}=5.681\times 10^{-4}$, so $r_{eq}=2.279\times 10^{11}\times 5.681\times 10^{-4}=1.295\times 10^8$ m ≈ 129,500 km. Ratio: $r_{SOI}/r_{eq}=5.78\times 10^8/1.295\times 10^8\approx 4.46$. The SOI is about 4.5× larger than the force-balance point — because the tidal criterion ($2/5$ power on the small $q$) beats the raw-force criterion ($1/2$ power).

$r_{SOI}\propto q^{2/5}\propto m^{2/5}$. Doubling $m$ multiplies the radius by $2^{2/5}=2^{0.4}=\mathrm{1.320.}$ So the SOI grows by about 32%, not 100%. The $2/5$ power tames the mass change — a big lever on mass gives a modest lever on the bubble. This softness is the same tidal $2/5$ we pictured at the top.

$(1/2{)}^{1/5}=2^{-0.2}=0.8706$. Applying it to L2.1's 924,000 km: $924,000\times 0.8706\approx 804,000\text{ km}.$ A ~13% reduction. It comes from the factor 2 in the tidal derivative $\frac{d}{dR}\left(\frac{GM}{R^2}\right)=-\frac{2GM}{R^3}$. Being close to 1, convention drops it.

We are patching between an Earth-centred orbit and a Moon-centred orbit, so the SOI uses the Moon–Earth pair. $q=7.35\times 10^{22}/5.97\times 10^{24}=0.01231$. ${\log }_{10}q=-1.9097$; $\times 0.4=-0.7639$; $q^{0.4}=10^{-0.7639}=0.1722$. $r_{SOI}=3.84\times 10^8\times 0.1722\approx 6.61\times 10^7\text{ m}\approx 66,000\text{ km}.$ A spacecraft closer than ~66,000 km to the Moon is best treated as orbiting the Moon; farther, as orbiting the Earth. This connects directly to the Patched Conic Approximation and to planning a Gravity Assist / Flyby.

Rearrange $r_{SOI}=R{q}^{2/5}$ for $q$: $q={\left(\frac{r_{SOI}}{R}\right)}^{5/2}.$ Why the $5/2$ power? Undoing $q^{2/5}$ means raising to the reciprocal $5/2$. $R=5\times 1.496\times 10^{11}=7.48\times 10^{11}$ m. Ratio $r_{SOI}/R=4.82\times 10^{10}/7.48\times 10^{11}=0.06444$. $q=0.06444^{2.5}$. ${\log }_{10}0.06444=-1.1908$; $\times 2.5=-2.977$; $q=10^{-2.977}=1.055\times 10^{-3}$. $m=qM=1.055\times 10^{-3}\times 1.989\times 10^{30}\approx 2.10\times 10^{27}\text{ kg}.$ That's roughly a Jupiter-mass planet.

The nested-sphere figure below makes this visual. The Moon sits 384,000 km from Earth; its bubble extends ±66,000 km, so it spans roughly 318,000 – 450,000 km from Earth. Earth's own SOI reaches 924,000 km. Since $450,000<924,000$, the entire Moon SOI is nested inside Earth's SOI. Meaning: a craft near the Moon is inside two nested bubbles. We patch conics in stages — Sun→Earth→Moon — a hierarchy the Restricted Three-Body Problem and Hill Sphere treatments make precise. The Patched Conic Approximation handles each handoff as a fresh Two-Body Problem.

Hill: $q/3=3.00\times 10^{-6}/3=1.00\times 10^{-6}$; cube root $=(10^{-6}{)}^{1/3}=10^{-2}=0.01$. $r_H=1.496\times 10^{11}\times 0.01=1.496\times 10^9\text{ m}\approx 1.50\times 10^6\text{ km}.$ The Hill sphere (~1.5 million km) is larger than the SOI (~924,000 km). They answer different questions: the SOI asks "where do I switch which body I orbit for trajectory accuracy?" (a $2/5$ tidal-ratio law), while the Hill Sphere asks "where can a moon stay gravitationally bound against the Sun?" (a $1/3$ centrifugal-balance law in the Restricted Three-Body Problem). Different physics ⇒ different exponent ⇒ different radius. The figure below stacks the three radii to scale.

Inside the SOI, Tidal Forces from the Sun are, by construction, a negligible perturbation relative to Mars's pull — so Mars's gravity alone bends the probe on a clean two-body hyperbola. Outside, Mars's pull is the negligible perturbation and the Sun dominates, so the probe follows a heliocentric conic. The SOI boundary is exactly the surface where these two "which-is-the-perturbation" verdicts flip (Laplace's equal-ratio condition). Patching the two conics at that surface — matching position and velocity — gives the flyby's net velocity change with minimal error. That is the whole point of the Patched Conic Approximation.

Forward: $r_{SOI}=R{q}^{2/5}$. ::: Compute $q=m/M$ first, then $q^{2/5}$, then $\times R$. Invert for mass ratio: $q=(r_{SOI}/R{)}^{5/2}$. ::: Reciprocal power $5/2$ undoes $2/5$. Mass $\times 2\Rightarrow$ SOI $\times$? ::: $2^{2/5}\approx 1.32$ (32% bigger). Force-equality radius? ::: $r_{eq}=R{q}^{1/2}$ — a raw force balance, smaller than the SOI. SOI vs Hill for Earth: ::: SOI ≈ 0.92 Mkm, Hill ≈ 1.5 Mkm — different exponents, different questions. When does $r_{SOI}=R{q}^{2/5}$ break? ::: When $q\to 1$ or the orbit is very eccentric — use the full three-body treatment.