Question bank — Sphere of influence — radius derivation

False. The SOI (Laplace) balances the ratio perturbation/main force in two viewpoints, giving a $2/5$ power; raw force-equality gives a $1/2$ power and a smaller radius (Earth: ~260,000 km vs the SOI's ~924,000 km). :::

False. Near Earth the Sun still pulls the craft roughly as hard as it pulls Earth itself. What is negligible is only the tidal difference of that pull across the small distance $r$ — the part that actually perturbs a planet-centred orbit. :::

False (approximately spherical). The real balance surface bulges toward and away from the Sun because the tidal term depends on direction; "sphere" is a convenient idealisation, which is why we write $r_{SOI}\approx$ not $=$. :::

True. $r_{SOI}\propto (m/M{)}^{2/5}$, so increasing the planet mass $m$ increases the radius — bigger boss, bigger bubble. :::

False. Because of the $2/5$ power, doubling $m$ multiplies the radius by only $2^{2/5}\approx 1.32$ — a 32% increase, not 100%. :::

True. $r_{SOI}\propto R$ (to the first power), so an outer planet at larger $R$ has a proportionally larger bubble — Jupiter's SOI is enormous partly because $R$ is large. :::

True. Keeping it would shrink $r_{SOI}$ by about 13%; dropping it inflates the value slightly, which is the standard convention. :::

False. Every force term was written per unit mass (acceleration), so the craft's own mass cancels everywhere — the bubble is a property of the planet–Sun pair only. :::

True. The tidal perturbation came from $\frac{d}{dR}(1/R^2)\propto 1/R^3$; that extra power of $R$ is exactly what turns a  ​ balance into the $2/5$ Laplace result. See Tidal Forces. :::

Dimensionally broken. Only the dimensionless ratio $m/M$ is raised to $2/5$; the length $R$ must stay to the first power so the answer has units of metres: $r_{SOI}=R(m/M{)}^{2/5}$. :::

Wrong pairing. The SOI is defined for the pair you patch between. For the Moon's SOI relative to Earth, $m=$ Moon and $M=$ Earth — the local primary, not the Sun. :::

Wrong. Craft and planet both fall toward the Sun together, so the common pull does nothing to the relative orbit. Only the difference across $r$, the tidal term $2GMr/R^3$, perturbs it. :::

A  ​ is the signature of a force balance, not a Laplace SOI. A correct SOI derivation must yield a $2/5$ power; a $1/2$ power means you equated raw forces. :::

No. Inside the SOI the ratio $(P/F{)}_A$ is smaller than $(P/F{)}_B$, meaning the planet-centred picture is the cleaner one — the switch is about which two-body approximation is less wrong, not about which raw force is bigger. :::

No. In the Sun-centred view the planet is the perturber, and the planet acts directly on the craft — its full $Gm/r^2$ pull, not a tidal difference. Only the Sun (the shared distant body) enters as a tidal term. :::

A perturbation matters only relative to the main force holding the orbit together. A big disturbance around a tightly-bound orbit is harmless; a small one around a weakly-bound orbit is fatal — so the meaningful quantity is perturbation ÷ main. :::

The full three-body problem has no clean closed-form solution. The SOI lets us stitch two exact two-body orbits together — the Patched Conic Approximation — swapping the central body as the craft crosses the boundary. :::

Because the Sun's disturbing influence in the planet frame is only a gentle tidal effect ($\propto r/R^3$), the planet stays "in charge" much farther out than the crude force-equality point suggests — the criterion is more forgiving. :::

It encodes that the Sun enters as a derivative (tidal) rather than a direct force. The exponent is a fingerprint: seeing $2/5$ tells you the tidal reasoning was used correctly. :::

The boundary is fuzzy and non-spherical, and the switch of central body is instantaneous in the model but gradual in reality. During a Gravity Assist / Flyby the patched-conic seam introduces a small error that real missions correct with tracking. :::

$r_{SOI}=R(m/M{)}^{2/5}\to 0$. A massless body has no bubble — there is nothing to dominate, so its "planet-centred" region shrinks to a point. :::

The derivation assumed $m\ll M$ and $r\ll R$; those approximations break. For comparable masses you need the full Restricted Three-Body Problem and the Hill Sphere instead, which handles the Lagrange-point geometry properly. :::

No, though they're cousins. The Hill Sphere comes from balancing gravity and centrifugal effects in the rotating frame and scales as $(m/3M{)}^{1/3}$; the Laplace SOI uses the $(m/M{)}^{2/5}$ perturbation-ratio criterion. Different exponents, different sizes. :::

Neither is better — that's the definition: both approximations are equally imperfect there. In practice you pick one side by convention and accept that the seam is the region of largest patched-conic error. :::

The true tidal term is directional, so the balance surface is stretched along the planet–Sun line and squeezed perpendicular to it. The single number $r_{SOI}$ is an averaged, idealised radius — reality is a lobed surface. :::