Sphere of influence — radius derivation: Worked examples

Before we start, one reminder of what each letter means — never use a symbol you haven't re-anchored:

The scenario matrix

Here is every "cell" — every distinct kind of case this one formula can be asked about. Each worked example below is tagged with the cell it lands in.

Cell	What makes it different	Example
C1 — Standard planet	ordinary $m / M$ , plug and chug	Ex 1 (Earth)
C2 — Nested / smaller pair	roles swap: moon around planet	Ex 2 (Moon)
C3 — The trap	force-balance vs Laplace SOI	Ex 3 (compare radii)
C4 — Large mass ratio	a heavy secondary (Jupiter)	Ex 4 (Jupiter)
C5 — Limit $m \to 0$	tiny asteroid, SOI $\to 0$	Ex 5 (Ceres)
C6 — Limit $R$ large	same body, farther out	Ex 6 (scaling)
C7 — Degenerate $m / M \to 1$	formula assumption $m ≪ M$ fails	Ex 7 (breakdown)
C8 — Exam twist	given $r_{S O I}$ , find $m$ (invert)	Ex 8 (inverse)

We will need a couple of arithmetic tools. Let me anchor them before using them.

Worked examples

Ex 1 — Standard planet (Cell C1)

Worked example Earth's sphere of influence

Given $m_{\oplus} = 5.97 \times 1 0^{24} kg$ , $M_{⊙} = 1.989 \times 1 0^{30} kg$ , and $R = 1.496 \times 1 0^{11} m$ (one AU), find $r_{S O I}$ .

Forecast: Guess an order of magnitude first. The mass ratio is about $3 \times 1 0^{- 6}$ . Raising that to $2/5$ makes it bigger (fractional powers of small numbers move them toward 1). Do you expect Earth's bubble to be thousands, hundreds of thousands, or millions of km wide?

Step 1 — Form the dimensionless ratio. $\frac{m}{M} = \frac{5.97 \times 1 0 ^{24}}{1.989 \times 1 0 ^{30}} = 3.00 \times 1 0^{- 6}$ Why this step? Only the pure number gets the power. If we accidentally divided in metres or mixed $R$ in here, the exponent would land on the wrong thing.

Step 2 — Apply the $2/5 = 0.4$ power using logs. $lo g_{10} (3.00 \times 1 0^{- 6}) = lo g_{10} (3.00) + (- 6) = 0.477 - 6 = - 5.523$ Multiply by $0.4$ : $- 5.523 \times 0.4 = - 2.209$ . Undo the log: $(\frac{m}{M})^{0.4} = 1 0^{- 2.209} = 6.18 \times 1 0^{- 3}$ Why this step? The log turns the awkward $0.4$ -power into a plain multiplication, then $1 0^{(\cdot)}$ brings us back.

Step 3 — Multiply by $R$ (the length carrier). $r_{S O I} = 1.496 \times 1 0^{11} \times 6.18 \times 1 0^{- 3} = 9.24 \times 1 0^{8} m \approx 924, 000 km$ Why this step? $R$ enters to the first power, so units stay in metres — the only length in the whole calculation.

Verify: Units: $m \times (dimensionless) = m$ ✔. Magnitude: $\approx 145$ Earth radii — matches the standard textbook value $\approx 0.925$ million km ✔. It's in the "hundreds of thousands of km" band — check your forecast.

Ex 2 — Roles swap: moon around planet (Cell C2)

Worked example The Moon's SOI relative to Earth

Now the pair we patch between is Moon↔Earth. So the smaller body is the Moon: $m = 7.35 \times 1 0^{22} kg$ , the central body is Earth: $M = 5.97 \times 1 0^{24} kg$ , and $R = 3.84 \times 1 0^{8} m$ (Earth–Moon distance).

Forecast: The mass ratio here ( $\sim 0.012$ ) is thousands of times bigger than Earth-around-Sun. Bigger ratio → bigger $(m / M)^{0.4}$ . But $R$ is far smaller. Which wins?

Step 1 — Ratio, always smaller over larger. $\frac{m}{M} = \frac{7.35 \times 1 0 ^{22}}{5.97 \times 1 0 ^{24}} = 0.01231$ Why this step? SOI is always drawn around the lighter body using its mass over the heavier central body. Get this backwards and you'd draw a bubble bigger than the whole system.

Step 2 — Apply the $0.4$ power. $lo g_{10} (0.01231) = - 1.910, - 1.910 \times 0.4 = - 0.764$ $(m / M)^{0.4} = 1 0^{- 0.764} = 0.172$

Step 3 — Scale by $R$ . $r_{S O I} = 3.84 \times 1 0^{8} \times 0.172 = 6.61 \times 1 0^{7} m \approx 66, 000 km$

Verify: Units metres ✔. About $1/6$ of the Earth–Moon distance — a spacecraft passing within $\sim 66, 000$ km of the Moon should be modelled as orbiting the Moon. This is the standard lunar SOI value ✔. See Patched Conic Approximation for when you flip the conic at this boundary.

Ex 3 — The trap: force-balance vs Laplace SOI (Cell C3)

Worked example Where forces are equal, and why that's the WRONG boundary

For Earth, compare the Laplace SOI (Ex 1) with the naive equal-force point, where the Sun's pull and Earth's pull cancel. To avoid clashing with $r_{S O I}$ (the SOI boundary), let $d$ be the variable distance from Earth to the balance point we are solving for.

Forecast: The parent note warns: "if you ever get a , you did a force balance." Predict whether the force-balance radius is bigger or smaller than the true SOI.

Step 1 — Set the two pulls equal. Equal-force means Earth's pull at distance $d$ from Earth equals the Sun's pull at distance $R - d$ from the Sun. Using Newton's gravity ( $G$ = the gravitational constant defined above): $\frac{G m}{d ^{2}} = \frac{GM}{( R - d ) ^{2}}$ Why this step? This is the "obvious" definition of dominance — where the planet literally pulls as hard as the Sun. We're building it precisely so we can watch it fail. Note $G$ sits on both sides.

Step 2 — Solve. First, $G$ cancels from both sides (the SOI never depends on it). Since $d ≪ R$ , approximate $R - d \approx R$ : $\frac{m}{d ^{2}} = \frac{M}{R ^{2}} \Rightarrow d = R \frac{m}{M}$ Why this step? The square root ( $1/2$ power) is the fingerprint of a raw force balance — contrast the $2/5$ power of the true SOI.

Step 3 — Number. $3.00 \times 1 0^{- 6} = 1.732 \times 1 0^{- 3}$ $d = 1.496 \times 1 0^{11} \times 1.732 \times 1 0^{- 3} = 2.59 \times 1 0^{8} m \approx 259, 000 km$

Step 4 — Compare. SOI $\approx 924, 000$ km $>$ force-balance $\approx 259, 000$ km. Ratio $\approx 3.6$ . Why this step? This quantifies the parent's warning: the tidal criterion gives a bubble almost 4× larger than the "forces equal" guess.

Verify: $924, 000/259, 000 = 3.57$ . Purely symbolically the ratio is $(m / M)^{2/5} / (m / M)^{1/2} = (m / M)^{- 1/10} = (3 \times 1 0^{- 6})^{- 0.1} \approx 3.5$ ✔. The SOI is larger — check your forecast. See Tidal Forces for why the differential term matters more than the raw force.

The figure below draws both radii to scale around Earth, so you can see that the orange SOI circle swallows the plum force-balance circle almost four times over — the whole point of this "trap" example.

Figure — Sphere of influence — radius derivation

The orange ring is the Laplace SOI ( $924, 000$ km); the dashed plum ring is the force-balance point ( $259, 000$ km). The arrow to the left points toward the Sun. Notice how much room the true SOI covers — a spacecraft well outside the plum ring is still firmly inside Earth's sphere of influence.

Ex 4 — Heavy secondary (Cell C4)

Worked example Jupiter's SOI — the biggest planetary bubble

$m_{J} = 1.898 \times 1 0^{27} kg$ , $M_{⊙} = 1.989 \times 1 0^{30} kg$ , $R = 7.785 \times 1 0^{11} m$ (5.2 AU).

Forecast: Jupiter is $\sim 318 \times$ heavier than Earth AND farther out. Both push the SOI up. Guess: is Jupiter's bubble bigger than the Earth–Sun distance ( $1.5 \times 1 0^{11}$ m)?

Step 1 — Ratio. $\frac{m}{M} = \frac{1.898 \times 1 0 ^{27}}{1.989 \times 1 0 ^{30}} = 9.54 \times 1 0^{- 4}$ Why this step? We again isolate the pure dimensionless mass ratio, because only this number is allowed to carry the $2/5$ power. Jupiter's ratio ( $\sim 1 0^{- 3}$ ) is $\sim 300 \times$ Earth's, so we already expect a much larger bubble.

Step 2 — Power $0.4$ . $lo g_{10} (9.54 \times 1 0^{- 4}) = - 3.020, \times 0.4 = - 1.208$ $(m / M)^{0.4} = 1 0^{- 1.208} = 0.0619$ Why this step? The log turns the $0.4$ -power into a simple multiply, exactly as in Ex 1 — the same tool applied to a heavier body so we can compare fairly.

Step 3 — Scale. $r_{S O I} = 7.785 \times 1 0^{11} \times 0.0619 = 4.82 \times 1 0^{10} m \approx 48.2 million km$ Why this step? $R$ enters to the first power and carries the length units; Jupiter's larger $R$ compounds its larger mass ratio, producing the Solar System's biggest planetary SOI.

Verify: Units metres ✔. Standard Jupiter SOI is $\approx 0.322$ AU $= 4.82 \times 1 0^{10}$ m ✔. It's the largest of all planets — the double boost from heavy $m$ and large $R$ paid off, though it's still smaller than 1 AU (your forecast: no). Jupiter's huge SOI is why it dominates gravity assists.

Ex 5 — Limit $m \to 0$ (Cell C5)

Worked example A tiny asteroid: does the SOI shrink to nothing?

Ceres (a dwarf planet, small): $m = 9.4 \times 1 0^{20} kg$ , $M_{⊙} = 1.989 \times 1 0^{30} kg$ , $R = 4.14 \times 1 0^{11} m$ (2.77 AU).

Forecast: As $m \to 0$ , the formula says $r_{S O I} \to R \cdot 0^{2/5} = 0$ . So the bubble should be tiny. But $2/5$ is a gentle power — it doesn't crush the number as fast as you'd think. Predict: kilometres, thousands of km, or hundreds of thousands?

Step 1 — Ratio. $\frac{m}{M} = \frac{9.4 \times 1 0 ^{20}}{1.989 \times 1 0 ^{30}} = 4.73 \times 1 0^{- 10}$ Why this step? This is a very small number — the whole point of the limiting case.

Step 2 — Power. $lo g_{10} (4.73 \times 1 0^{- 10}) = - 9.325, \times 0.4 = - 3.730$ $(m / M)^{0.4} = 1 0^{- 3.730} = 1.86 \times 1 0^{- 4}$ Why this step? Notice: even a ratio of $\sim 1 0^{- 10}$ only drops to $\sim 1 0^{- 4}$ after the $0.4$ power. The exponent softens extreme smallness — that's the tidal "gentle" effect at work.

Step 3 — Scale. $r_{S O I} = 4.14 \times 1 0^{11} \times 1.86 \times 1 0^{- 4} = 7.71 \times 1 0^{7} m \approx 77, 000 km$

Verify: Units metres ✔. Ceres's radius is only $\sim 470$ km, yet its SOI is $\sim 77, 000$ km — over 160 body-radii. The limit $m \to 0$ does give $r \to 0$ , but slowly: the $2/5$ power means small bodies still have surprisingly large bubbles. Forecast: tens of thousands of km ✔.

Ex 6 — Limit: same body, larger $R$ (Cell C6)

Worked example What if Earth were twice as far from the Sun?

Keep Earth's mass, but set $R^{'} = 2 R = 2.992 \times 1 0^{11} m$ . Find the new SOI without redoing the mass part.

Forecast: The mass ratio is unchanged, so $(m / M)^{0.4}$ is the same $6.18 \times 1 0^{- 3}$ from Ex 1. Only $R$ doubled — and $R$ appears to the first power. Predict the new SOI relative to $924, 000$ km.

Step 1 — Isolate the $R$ -dependence. $\frac{r _{S O I}^{'}}{r _{S O I}} = \frac{R ^{'} ( m / M ) ^{0.4}}{R ( m / M ) ^{0.4}} = \frac{R ^{'}}{R} = 2$ Why this step? Because $R$ is linear, the SOI scales directly with distance — no exponent surprises. This is a clean sanity lever.

Step 2 — Number. $r_{S O I}^{'} = 2 \times 9.24 \times 1 0^{8} = 1.848 \times 1 0^{9} m \approx 1, 848, 000 km$

Verify: Direct recompute: $R^{'} (m / M)^{0.4} = 2.992 \times 1 0^{11} \times 6.18 \times 1 0^{- 3} = 1.85 \times 1 0^{9}$ m ✔. Exactly double — confirming the linear $R$ -dependence and your forecast. (Physically, farther from the Sun the Sun's tide weakens, so a planet holds sway over a larger region.)

Ex 7 — Degenerate case $m / M \to 1$ (Cell C7)

Worked example When the formula BREAKS: two equal masses

Suppose someone plugs $m = M$ (a binary star, or a planet as heavy as its star) into $r_{S O I} = R (m / M)^{2/5}$ . What comes out, and why is it nonsense?

Forecast: With $m / M = 1$ , the formula gives $r_{S O I} = R \cdot 1^{2/5} = R$ . So the "bubble" would reach all the way to the other body. Sensible?

Step 1 — Plug in. $r_{S O I} = R (\frac{M}{M})^{2/5} = R \times 1 = R$ Why this step? We deliberately violate the derivation's assumption $m ≪ M$ to see it fail — this is the degenerate cell.

Step 2 — Spot the contradiction. The derivation used $r ≪ R$ to write $F_{B} = GM / R^{2}$ (craft sits at $\approx R$ from the Sun). But here $r = R$ , so $r ≪ R$ is false. The approximation collapses — the answer $r = R$ literally puts the SOI edge on the other body. Why this step? Every formula carries hidden assumptions; a degenerate input exposes them.

Step 3 — What to use instead. When masses are comparable, there is no clean spherical SOI; you need the Hill Sphere (which stays finite for $m / M \to 1$ ) or the full Restricted Three-Body Problem with its Lagrange points.

Verify: Symbolically $lim_{m \to M} R (m / M)^{2/5} = R$ ✔ — and $R$ is exactly the invalid boundary, confirming the formula must not be used near $m / M \approx 1$ . The formula is only trustworthy when $m / M ≲ 1 0^{- 2}$ or so (Moon–Earth at $0.012$ is roughly the practical edge).

Ex 8 — Exam twist: invert the formula (Cell C8)

Worked example Given the SOI, find the planet's mass

A newly found exoplanet orbits a Sun-like star ( $M = 1.989 \times 1 0^{30} kg$ ) at $R = 3.0 \times 1 0^{11} m$ , and its measured SOI radius is $r_{S O I} = 1.5 \times 1 0^{9} m$ . Find the planet's mass $m$ .

Forecast: We must undo the $2/5$ power. Undoing a $2/5$ power means raising to the reciprocal power $5/2 = 2.5$ . Expect that step to amplify errors — so keep digits.

Step 1 — Rearrange the formula for the ratio. Start from $r_{S O I} = R (m / M)^{2/5}$ : $\frac{r _{S O I}}{R} = (\frac{m}{M})^{2/5}$ Why this step? Isolate the power-carrying term so we can take its inverse power.

Step 2 — Raise both sides to $5/2$ . $\frac{m}{M} = (\frac{r _{S O I}}{R})^{5/2}$ Why this step? $(x^{2/5})^{5/2} = x^{1}$ — raising to $5/2$ exactly cancels the $2/5$ power, freeing $m / M$ .

Step 3 — Numbers. $\frac{r _{S O I}}{R} = \frac{1.5 \times 1 0 ^{9}}{3.0 \times 1 0 ^{11}} = 5.0 \times 1 0^{- 3}$ $(5.0 \times 1 0^{- 3})^{2.5} : lo g_{10} (5.0 \times 1 0^{- 3}) = - 2.301, \times 2.5 = - 5.753$ $\frac{m}{M} = 1 0^{- 5.753} = 1.766 \times 1 0^{- 6}$ Why this step? We apply the log tool once more — this time multiplying by $2.5$ instead of $0.4$ — to evaluate the $5/2$ power cleanly.

Step 4 — Multiply by the star mass. $m = 1.766 \times 1 0^{- 6} \times 1.989 \times 1 0^{30} = 3.51 \times 1 0^{24} kg$ Why this step? The ratio $m / M$ is dimensionless; multiplying by the known star mass $M$ restores kilograms and gives the planet's actual mass.

Verify: Plug back forward: $(1.766 \times 1 0^{- 6})^{0.4} = 1 0^{- 2.301} = 5.0 \times 1 0^{- 3}$ , times $R = 3.0 \times 1 0^{11}$ gives $1.5 \times 1 0^{9}$ m — the given SOI ✔. The mass $3.51 \times 1 0^{24}$ kg is a bit under Earth's — a plausible rocky planet ✔.

Recall Quick self-test across the matrix

Earth SOI value? ::: $\approx 9.24 \times 1 0^{8}$ m $\approx 924, 000$ km. Moon SOI value? ::: $\approx 6.6 \times 1 0^{7}$ m $\approx 66, 000$ km. Which is bigger for Earth — SOI or force-balance point, and by roughly what factor? ::: SOI, by $\approx 3.6 \times$ ( $924, 000$ vs $259, 000$ km). If $R$ doubles and $m$ is unchanged, SOI does what? ::: Doubles (linear in $R$ ). Why does $m = M$ break the formula? ::: It gives $r = R$ , violating the $r ≪ R$ assumption used in the derivation; use the Hill Sphere instead. To find $m$ from a measured SOI, what power do you use? ::: Raise $(r_{S O I} / R)$ to the $5/2$ power (undoes the $2/5$ ).

Connections

Sphere of influence — radius derivation — the parent derivation these examples exercise.
Patched Conic Approximation — where you actually switch two-body models at the SOI edge.
Hill Sphere — the alternative bubble that survives the $m \to M$ degenerate case.
Restricted Three-Body Problem — the exact framework when the SOI picture is too crude.
Gravity Assist / Flyby — Jupiter's giant SOI (Ex 4) makes it the prime flyby target.