Tidal Forces and Tidal Radii

Gravity is universal, but what disrupts a satellite is not the mean pull of its host — that merely carries the center of mass along its orbit — it is the variation of that pull across the satellite's body. A star on the near side of a globular cluster is tugged slightly harder toward the Galactic center than a star on the far side; in the frame falling with the cluster, that residual is the tidal force. To first order in the ratio of satellite size to orbital distance it is linear in position, stretching the satellite along the line to the host and squeezing it transversely — the same field that raises two ocean bulges on the Earth and has locked the Moon's rotation. The competition between this disruptive shear and the satellite's own binding self-gravity fixes a critical scale: the tidal radius $r_t$ . Inside $r_t$ self-gravity wins and stars stay bound; outside, the tide dominates and stars leak away through the two Lagrange points on the satellite–host axis, streaming into leading and trailing tidal tails. In the rotating frame of a circular orbit the same boundary is the Jacobi (or Hill) surface bounding the Roche lobe of the restricted three-body problem. The exact coefficient depends on the setup: circular versus eccentric orbit, a point-mass host versus an extended halo (which softens the tide through the local slope of the enclosed mass), and a rigid solid versus a self-gravitating fluid. All are one estimate wearing different coefficients — set the tidal stretch equal to the satellite's surface gravity. This physics truncates clusters, sculpts dwarf galaxies, draws the Sagittarius and Palomar 5 streams that weigh the Milky Way, and, through the same tensor, torques protogalaxies in the cosmic web.

Key takeaways

A tidal force is the differential gravitational pull across an extended body: the near side is pulled harder than the far side, giving a stretch along the line to the perturber and a compression transverse to it.
To leading order the tidal acceleration is the gradient of the external field, $\sim (GM/D^3)\,r$ , growing with the perturber mass $M$ and the body size $r$ and falling as the cube of the separation $D$ .
A self-gravitating satellite survives only inside its tidal (Jacobi) radius $r_t$ , where self-gravity balances the tide; matter beyond $r_t$ is stripped through $L_1$ and $L_2$ into tidal tails.
For a satellite of mass $m$ on a circular orbit of radius $D$ about a host of enclosed mass $M(D)$ , $r_t \approx D\,[m/(kM)]^{1/3}$ with $k=3$ for a point mass, $k=2$ for a flat rotation curve.
The Roche limit is the same balance for a strengthless fluid body: it disrupts inside $d \approx 2.44\,R_M\,(\rho_M/\rho_m)^{1/3}$ of its primary — the physics that sizes globular clusters, dwarf spheroidals, tidal streams, and planetary rings.
Key idea: the tidal force is the gradient of the external gravitational field across a body; balancing it against the body's self-gravity yields the tidal (Jacobi/Hill/Roche) radius $r_t \approx D\,[m/(kM)]^{1/3}$ , beyond which the satellite is stripped.

Where this lives today

Tidal radii are the working currency of near-field cosmology. Wide surveys — Gaia and now the Vera Rubin Observatory's LSST — map cold stellar streams like Palomar 5 and the Sagittarius stream that are tidal tails unwound along their progenitors' orbits; a stream's length, width, and density gaps constrain the Milky Way's mass profile, the shape of its dark halo, and the dark subhalos whose flybys carve gaps into it. King models with a tidal-radius cutoff remain the standard fit to globular-cluster surface-brightness profiles, and the King-radius-to- $r_t$ ratio diagnoses whether a cluster is tidally limited. In the dwarf-galaxy regime, comparing measured half-light radii with nominal tidal radii is a primary argument that the classical dwarf spheroidals are dark-matter dominated. And the extreme case — a star torn apart on a plunging orbit into a supermassive black hole — is now an observed transient class: tidal disruption events flare as the debris returns and accretes, turning Roche's fluid-body calculation into a live probe of dormant black holes.

The physical picture

Stand at the center of a satellite in free fall toward a host. The host's gravity there is cancelled by the acceleration of the frame, so you feel nothing — the equivalence principle at work. But a point on the near side is closer and pulled harder, one on the far side pulled less, and both are dragged away from the center along the line to the host, while points off to the side are pulled inward along converging field lines. What survives the cancellation is a stretch along the radial axis and a squeeze perpendicular to it — the traceless quadrupole, its radial eigenvalue $-2GM/D^3$ exactly twice the transverse $+GM/D^3$ . The first figure makes that pattern concrete.

Figure

The differential field across a satellite falling toward its host: the uniform pull cancels at the center, leaving a radial stretch of −2GM/D³ along the line to the host and a transverse squeeze of +GM/D³ — the traceless quadrupole that opens the Lagrange points L₁ and L₂ at the tidal radius r_t

Now let the satellite be a swarm of stars, and think of $r_t$ as a tug-of-war radius. The satellite's own gravity pulls a test star inward with $Gm/r^2$ ; the host's tidal shear, boosted by the orbit's centrifugal term, pulls it outward with $\sim 3GMr/D^3$ . Where these balance, the star sits on a knife's edge at the inner Lagrange point $L_1$ (or the outer $L_2$ ) — the saddle of the effective potential. Stars inside are bound; stars pushed past the saddle slide into the leading and trailing tails, carrying slightly less and slightly more orbital energy than the progenitor. The second figure runs this through with real numbers.

Figure

Worked tidal radius of a Milky Way globular cluster: mass m = 2×10⁵ M⊙ on a circular orbit at D = 8 kpc in a flat rotation curve v_c = 220 km/s gives r_t = D (m/2M)^⅓ ≈ 83 pc, and stripped stars escape through L₁ and L₂ into leading and trailing tidal tails

The core equations

In the frame that falls with the satellite's center of mass, the uniform part of the host's pull cancels and only the first non-trivial Taylor term survives. That residual is linear in position and is packaged in one symmetric object — the tidal tensor, the Hessian of the host potential at the satellite. Everything downstream specializes this tensor.

\mathbf{g}_{\rm tid}(\mathbf{r}) = -\,(\mathbf{r}\cdot\nabla)\,\mathbf{g}_{\rm ext}\big|_0 = -\,T_{ij}\,r_j,\qquad T_{ij}=\frac{\partial^2 \Phi_{\rm ext}}{\partial x_i\,\partial x_j}\bigg|_0 \qquad (1)

where: $\mathbf{g}_{\rm tid}$ — tidal acceleration in the satellite's rest frame; $\mathbf{r}$ — position relative to the satellite center; $T_{ij}$ — tidal tensor, the Hessian of the external potential; $\Phi_{\rm ext}$ — host potential at the satellite. Reading: the tide is a linear map $r_j\mapsto -T_{ij}r_j$ , so a spherical body deforms into an ellipsoid aligned with the tensor's eigenvectors. For a point-mass host the tensor is diagonal in the radial-and-transverse basis. Differentiating the Keplerian potential $\Phi=-GM/D$ twice gives a radial eigenvalue exactly twice the transverse ones and of opposite sign.

T_{ij} = \mathrm{diag}\!\left(-\frac{2GM}{D^3},\; \frac{GM}{D^3},\; \frac{GM}{D^3}\right) \qquad (2)

where: $M$ — host point mass; $D$ — separation to the host; $G$ — gravitational constant. Reading: the trace is zero — Laplace's equation in vacuum — so a pure tide stretches and squeezes but never compresses volume; that changes once the satellite sits inside the host's own matter. The crudest estimate equates the outward tidal stretch at radius $r$ to the satellite's inward surface gravity. Ignoring the orbit gives the naive coefficient two.

\frac{2GM}{D^3}\,r_t \;=\; \frac{Gm}{r_t^2}\;\;\Longrightarrow\;\; r_t = D\left(\frac{m}{2M}\right)^{1/3} \qquad (3)

where: $m$ — satellite mass interior to $r_t$ ; $r_t$ — tidal radius. Reading: the cube-root makes $r_t$ insensitive to the mass ratio — a factor of a thousand in $m/M$ moves it by only ten. Put the satellite on a circular orbit and move to the co-rotating frame: the centrifugal term adds one more unit of outward push, turning the two into a three. This is the Hill (or Jacobi) radius.

r_t = D\left(\frac{m}{3M}\right)^{1/3} \qquad (4)

where: $r_t$ — Hill/Jacobi radius; $m,M$ — satellite and host masses, $m\ll M$ ; $D$ — orbital separation. Reading: the "3" is $2$ (tidal stretch) $+\,1$ (centrifugal); it becomes a variable coefficient once the host is extended (Derivation 3). The rotating frame that produces the factor three has its own conserved quantity: the effective potential adds a centrifugal well to the two gravitational potentials, and its value at $L_1$ sets the surface separating bound from escaping orbits.

\Phi_{\rm eff}(\mathbf{r}) = \Phi_M + \Phi_m - \tfrac{1}{2}\Omega^2\!\left(x^2+y^2\right),\qquad \Omega^2 = \frac{GM}{D^3} \qquad (5)

where: $\Phi_M,\Phi_m$ — host and satellite potentials; $\Omega$ — orbital angular frequency; the last term is the centrifugal potential. Reading: $E_J=\tfrac12 v^2+\Phi_{\rm eff}$ is the Jacobi integral, the only conserved quantity of the circular restricted three-body problem (the Coriolis force does no work); its zero-velocity surfaces bound the accessible region. Real cluster orbits are eccentric, and stripping bites hardest at perigalacticon where the satellite plunges deepest. King's 1962 formula evaluates the tide at that closest approach.

r_t = a(1-e)\left[\frac{m}{M\,(3+e)}\right]^{1/3} \qquad (6)

where: $a$ — orbital semi-major axis; $e$ — eccentricity; $a(1-e)$ — perigalactic distance. Reading: as $e\to0$ this returns the circular Hill radius; larger $e$ shrinks the perigalacticon and stiffens the tide, truncating eccentric clusters more tightly. Finally, the same balance for a strengthless fluid body is the Roche limit, best written through the density ratio so the satellite's own size drops out.

d_{\rm Roche} = 2.44\,R_M\left(\frac{\rho_M}{\rho_m}\right)^{1/3} \qquad (7)

where: $R_M$ — radius of the primary; $\rho_M,\rho_m$ — mean densities. Reading: the rigid-body balance gives the smaller coefficient $2^{1/3}\approx1.26$ ; a fluid deforms into an ellipsoid that weakens its self-gravity, pushing disruption out to the classical $2.44$ .

Derivation AThe tidal radius from the co-rotating force balance

The goal is to pin down the coefficient in $r_t = D[m/(kM)]^{1/3}$ for the case that dominates galactic dynamics — a satellite on a circular orbit — by writing the full radial force balance in the frame that rotates with the orbit, where the naive tidal estimate of equation (3) is corrected by the centrifugal term and lands on the Hill value $k=3$ . We work along the satellite–host line with the $x$ -axis pointing away from the host, and locate the inner Lagrange point $L_1$ where a test star feels no net radial force.

Step 1. Collect the three radial accelerations acting on a test star displaced by $x$ from the satellite center: the tidal stretch $+2GMx/D^3$ pushing outward, the centrifugal term $+\Omega^2 x$ also pushing outward, and the satellite's own gravity $-Gm/x^2$ pulling inward. Why: in the rotating frame the tide and the centrifugal support are the two destabilizing terms, and only the satellite's self-gravity opposes them.

Step 2. Fix $\Omega$ from the circular-orbit condition. For a circular orbit about a point mass, Kepler's third law gives $\Omega^2 = GM/D^3$ , identical to the coefficient of the tidal stretch, so the two outward terms combine into $3GMx/D^3$ . Why: it is this equality — centrifugal frequency equal to the tidal frequency — that converts the naive "2" into the Hill "3".

Step 3. Set the net radial force to zero at $x=r_t$ : $\,3GM\,r_t/D^3 = Gm/r_t^2$ . This is exactly the location of $L_1$ , the saddle of $\Phi_{\rm eff}$ through which stars first escape. Why: the vanishing of the effective radial force marks the boundary between orbits that fall back toward the satellite and orbits that run off down the potential.

Step 4. Solve the cube: $r_t^3 = (m/3M)\,D^3$ , hence $r_t = D\,(m/3M)^{1/3}$ . Dropping the rotation ( $\Omega\to0$ ) recovers the naive $k=2$ . Why: the coefficient is not a universal constant but a bookkeeping of which support terms are present.

Go deeper · Derivation B · 1 minThe Roche limit from self-gravity balance

Derivation BThe Roche limit from self-gravity balance

The goal is to derive the distance at which a strengthless self-gravitating body is torn apart by its primary's tide, expressed through the two bodies' mean densities so the satellite's own radius cancels — recovering both the rigid coefficient $1.26$ and the fluid $2.44$ that separate an intact moon from a ring of debris. We compare the outward tidal stretch at the satellite's surface with the inward self-gravity there, then substitute masses for densities.

Step 1. Write the surface tidal stretch on a satellite of radius $r$ at separation $d$ : $a_{\rm tid} = 2GM\,r/d^3$ , evaluated on the near and far faces where the differential pull is largest. Why: the near/far surface points are where the tide most opposes the body's self-gravity, so they are the first to be lifted off.

Step 2. Write the inward self-gravity at that same surface, $a_{\rm self} = Gm/r^2$ , and set the two equal for the marginal case, $2GM\,r/d^3 = Gm/r^2$ . Why: beyond this separation the tide exceeds the surface gravity everywhere on the sub-host face and the body can no longer hold its outer layers.

Step 3. Replace masses by mean densities, $M=\tfrac43\pi R_M^3\rho_M$ and $m=\tfrac43\pi r^3\rho_m$ , and cancel the common $\tfrac43\pi G$ . The satellite radius $r$ drops out entirely, leaving $d^3 = 2\,R_M^3\,(\rho_M/\rho_m)$ . Why: disruption depends on how compact the two bodies are, not on their absolute sizes — a key simplification that makes the Roche limit a property of densities alone.

Step 4. Take the cube root: $d_{\rm rigid} = R_M(2\rho_M/\rho_m)^{1/3}\approx1.26\,R_M(\rho_M/\rho_m)^{1/3}$ . A fluid body does not stay spherical — it stretches into a prolate ellipsoid whose elongated tips have weaker self-gravity and feel a stronger tide, so it comes apart farther out, at the classical $2.44\,R_M(\rho_M/\rho_m)^{1/3}$ . Why: the coefficient encodes the body's rheology; the same density scaling governs rigid rubble piles, fluid moons, and everything between.

Go deeper · Derivation C · 1 minThe tidal tensor, its trace, and extended hosts

Derivation CThe tidal tensor, its trace, and extended hosts

The goal is to show why the point-mass coefficient "3" is not universal but is set by the host's mass distribution, by computing the tidal tensor of an extended spherical host and reading off the effective radial coefficient as a function of the local logarithmic slope of the enclosed mass — the correction that makes a cluster's tidal radius larger in a diffuse halo than around a point mass of equal enclosed weight. The trace, fixed by Poisson's equation, is the diagnostic.

Step 1. Start from $T_{ij}=\partial_i\partial_j\Phi_{\rm ext}$ and take its trace: $\mathrm{Tr}\,T = \nabla^2\Phi_{\rm ext} = 4\pi G\rho(D)$ by Poisson's equation. In vacuum $\rho=0$ and the tensor is traceless (eigenvalues $-2,+1,+1$ in units of $GM/D^3$ ); inside matter the trace is positive. Why: the trace measures the local density the satellite is embedded in, which softens the pure vacuum tide.

Step 2. For a spherical host the radial eigenvalue is $T_{rr}=\dfrac{\mathrm{d}}{\mathrm{d}D}\!\left(\dfrac{GM(D)}{D^2}\right)=\dfrac{GM'(D)}{D^2}-\dfrac{2GM(D)}{D^3}$ , where $M(D)$ is the enclosed mass and $M'=4\pi D^2\rho(D)$ . Why: only the mass interior to the orbit contributes to the mean pull, so the tide now depends on how fast that enclosed mass grows outward.

Step 3. Form the effective radial tidal coefficient in the rotating frame, $-T_{rr}+\Omega^2$ with $\Omega^2=GM(D)/D^3$ , giving $-T_{rr}+\Omega^2 = \dfrac{GM(D)}{D^3}\!\left(3-\dfrac{\mathrm{d}\ln M}{\mathrm{d}\ln D}\right)$ . This bracket replaces the bare "3" of the point-mass Hill radius. Why: the point-mass "3" was $2$ (stretch) $+1$ (centrifugal); the enclosed-mass gradient subtracts from it because interior matter you orbit outside of no longer tugs differentially the way a point source does.

Step 4. Evaluate the limiting hosts. A point mass has $\mathrm{d}\ln M/\mathrm{d}\ln D=0$ , giving coefficient $3$ ; a singular isothermal sphere with a flat rotation curve has $M\propto D$ so the slope is $1$ , giving coefficient $2$ . Since $r_t\propto(\text{coeff})^{-1/3}$ , the isothermal tidal radius is larger by $(3/2)^{1/3}\approx1.14$ . Why: a cluster in a realistic dark halo is stripped less aggressively than the point-mass estimate would suggest, a $\sim14\%$ effect that matters when reading cluster edges as mass probes.

Limiting cases

The coefficient in $r_t = D\,[m/(kM)]^{1/3}$ is set by the geometry of the encounter:

Case	Result	Meaning
Point-mass host, circular orbit (Hill/Jacobi)	$r_t = D\left(\dfrac{m}{3M}\right)^{1/3}$	The classic Hill radius; the factor $3$ is $2$ tidal stretch $+\,1$ centrifugal in the rotating frame.
Extended host, circular orbit	$r_t = D\left[\dfrac{m}{M(D)\,(3 - \mathrm{d}\ln M/\mathrm{d}\ln D)}\right]^{1/3}$	For a flat rotation curve ( $M\propto D$ ) the coefficient drops from $3$ to $2$ , so tidal radii grow by $\approx14\%$ .
Eccentric orbit (King 1962)	$r_t = a(1-e)\left[\dfrac{m}{M\,(3+e)}\right]^{1/3}$	Stripping is set at perigalacticon $a(1-e)$ ; the $(3+e)$ stiffens the tide toward radial orbits.
Fluid / rubble body (Roche limit)	$d = 2.44\,R_M\left(\dfrac{\rho_M}{\rho_m}\right)^{1/3}$	A strengthless body distorts and disrupts; the $2.44$ (rigid $1.26$ ) comes from the equilibrium ellipsoidal figure.

Worked examples

1 · The Hill radius of the Earth (confidence builder)

How far from the Earth does the Sun's tide overwhelm the Earth's gravity, and is the Moon safely bound? Take $m_\oplus=5.97\times10^{24}$ kg, $M_\odot=1.99\times10^{30}$ kg, and $D=1\,\mathrm{AU}=1.50\times10^{11}$ m.

Step 1. Use the circular Hill radius, equation (4): $r_t = D\,(m_\oplus/3M_\odot)^{1/3}$ . Form the mass ratio first: $m_\oplus/(3M_\odot) = 5.97\times10^{24}/(5.97\times10^{30}) = 1.00\times10^{-6}$ .

Step 2. Take the cube root: $(1.00\times10^{-6})^{1/3}=1.00\times10^{-2}$ , one hundredth of an AU: $r_t = 1.50\times10^{11}\times1.00\times10^{-2} = \mathbf{1.50\times10^{9}\ m \approx 1.5\times10^{6}\ km}$ .

Step 3. Compare with the Moon at $3.84\times10^{5}$ km. The ratio $3.84\times10^5 / 1.50\times10^6 = 0.26$ puts the Moon at about a quarter of the Hill radius — comfortably bound against solar stripping. The Earth's Hill sphere is ≈ 1.5 million km, and the Moon sits at 0.26 r_t, safely inside.

2 · The tidal radius of a Milky Way globular cluster (fully worked)

A globular cluster of mass $m=2\times10^{5}\,M_\odot$ orbits at $D=8$ kpc where the rotation curve is flat at $v_c=220$ km/s. What is its tidal radius? A flat rotation curve is isothermal, so the coefficient is $2$ (Derivation 3), not the point-mass $3$ .

Step 1. Get the enclosed Galactic mass from the circular speed: $M(D)=v_c^2 D/G$ . With $v_c=2.20\times10^5$ m/s and $D=8\,\mathrm{kpc}=2.47\times10^{20}$ m, this is $1.79\times10^{41}$ kg $\approx \mathbf{9.0\times10^{10}\,M_\odot}$ .

Step 2. Form the isothermal balance $r_t = D\,[m/(2M(D))]^{1/3}$ . The mass ratio $m/(2M) = 1.11\times10^{-6}$ has cube root $1.04\times10^{-2}$ .

Step 3. Multiply by the orbital radius: $r_t = 8\,\mathrm{kpc}\times1.04\times10^{-2} = \mathbf{83\ pc}$ . The point-mass coefficient $3$ would have given $83\times(2/3)^{1/3}=73$ pc — a $12\%$ underestimate that shows why the host profile matters.

Step 4. Interpret. A King model fit should show a truncation near $r_t\approx83$ pc; stars found beyond it are in the act of being stripped into tails. The cluster is tidally truncated at r_t ≈ 83 pc, set by the flat-rotation-curve coefficient 2.

3 · The Roche limit of Saturn's rings

Why do Saturn's rings stop where they do? Model the ring particles as strengthless icy bodies of mean density $\rho_m\approx900$ kg m⁻³ orbiting Saturn ( $\rho_M=687$ kg m⁻³, equatorial radius $R_M=6.03\times10^4$ km).

Step 1. Use the fluid Roche limit, equation (7): $d = 2.44\,R_M(\rho_M/\rho_m)^{1/3}$ . The density ratio $\rho_M/\rho_m = 687/900 = 0.763$ has cube root $0.914$ .

Step 2. Assemble: $d = 2.44\times(6.03\times10^4)\times0.914 = \mathbf{1.34\times10^{5}\ km}$ .

Step 3. Compare with the observed rings, whose bright A-ring outer edge lies near $1.37\times10^5$ km. The Roche limit falls just outside the main ring system: inside it, tides prevent particles from accreting into a moon, so material stays spread as rings, while Saturn's regular moons all orbit beyond it. The rings occupy the region inside the ≈ 1.34×10⁵ km Roche limit, exactly where a moon cannot gravitationally assemble.

Physics of the portrait

Tidal disruption is one of the oldest problems in celestial mechanics, and its master scale carries the name of the man who first computed it — a quiet French astronomer who worked out, decades before anyone had a use for it, how close a fluid moon may approach its planet before it is torn apart.

Origins · 1 minÉdouard Roche and how this was discovered

ÉR

Édouard Roche

1820–1883 · celestial mechanics and the tidal disruption of satellites

Édouard Albert Roche spent his entire career at Montpellier, far from the great observatories, working on the figures of rotating and tidally distorted fluid bodies. In an 1849 memoir on the shape of a fluid mass under the attraction of a distant point, he asked when a self-gravitating satellite can no longer hold itself together against its primary's tide, and derived the critical distance — about $2.44$ planetary radii for equal densities — now universally called the Roche limit. He had no direct application in mind; the result sat as a curiosity of mathematical astronomy. Its vindication came repeatedly: Saturn's rings lie just inside Saturn's Roche limit, where accretion into a moon is forbidden, and in 1994 comet Shoemaker–Levy 9 was strung into a chain of fragments inside Jupiter's. The same balance, transplanted from fluids to stellar systems, became the tidal radius of star clusters (King, building on the restricted three-body work of Lagrange, Jacobi, and Hill), the truncation radius of dwarf galaxies, and the launching scale of the tidal streams that now weigh the Milky Way. A single geometric insight about where self-gravity yields to the tide turned out to organize phenomena from planetary rings to galactic halos.

History

1772

— Lagrange locates the five equilibrium points of the restricted three-body problem

1836

— Jacobi identifies the conserved integral of motion in the rotating frame

1849

— Roche fixes the critical distance for tidal breakup of a fluid moon

1878

— Hill develops the zero-velocity surfaces and the lunar Hill sphere

1962

— King derives the tidal radius of a star cluster at perigalacticon

1994

— Shoemaker–Levy 9, split inside Jupiter's Roche limit, strikes the planet

Going further

Texts. Carroll & Ostlie, An Introduction to Modern Astrophysics, 2nd ed., §§19, 25 (tidal forces, Roche limit, clusters) · Binney & Tremaine, Galactic Dynamics, 2nd ed., §8.3 (tidal fields), §8.4 (tidal radius), §3.1 (restricted three-body) · Mo, van den Bosch & White, Galaxy Formation and Evolution, §12 (tidal stripping and streams) · King, I. R. 1962, AJ 67, 471 (tidal radius of clusters) · Dodelson, Modern Cosmology (tidal field of the cosmic web). Every result follows from first principles: the tidal tensor is the Taylor expansion of the host potential (eq. 1), with point-mass eigenvalues $-2,+1,+1$ in units of $GM/D^3$ (eq. 2, trace vanishes); the Hill radius comes from the co-rotating balance with $3=2+1$ ; the extended-host coefficient $3-\mathrm{d}\ln M/\mathrm{d}\ln D$ follows from the Poisson trace, giving $3$ for a point mass and $2$ for a flat rotation curve (note: a common mis-statement gives the isothermal coefficient as $1$ rather than $2$ ); the Roche coefficients are $1.26$ (rigid) and $2.44$ (fluid) from the density-ratio balance; King's formula is from AJ 67, 471. Key anchor numbers: Earth's Hill radius $1.5\times10^6$ km; a $2\times10^5\,M_\odot$ cluster at 8 kpc truncated at $r_t\approx83$ pc; Saturn's Roche limit $\approx1.34\times10^5$ km, just inside the A-ring edge near $1.37\times10^5$ km.

Prerequisites. Gravitational Two-Body Problem · Galactic Potential Models · Orbits in Spherical Potentials · Virial Theorem in Stellar Systems. Leads to. Gravitational N-Body Problem · Collisionless Boltzmann Equation · Two-Body Relaxation Time · Distribution Function and Phase Space.

Ask the tutor: ask me to walk through why a cluster's tidal radius grows when you swap a point mass for a realistic dark halo, or to launch a tidal tail step by step from the Lagrange points.

Why doesn't the host's total gravity appear in the tidal radius, only its gradient?

Because in the frame that falls with the satellite's center of mass, the uniform part of the host field is cancelled exactly by the center-of-mass acceleration — the equivalence principle. A body sitting in a perfectly uniform gravitational field feels no internal stress at all, no matter how strong the field. What survives is the variation of the field across the body, the first Taylor term $-(\mathbf{r}\cdot\nabla)\mathbf{g}$ , which is the Hessian $T_{ij}$ of the potential. That differential field is what stretches and squeezes, so only the second derivative of the potential, never its value or its first derivative, enters the tidal radius.

Why is the coefficient 3 for a point mass but 2 for an isothermal halo?

The point-mass "3" is really $2+1$ : two units from the radial tidal stretch $2GM/D^3$ and one from the centrifugal term $\Omega^2=GM/D^3$ of the circular orbit. When the host is extended, the mean pull at radius $D$ comes only from the enclosed mass $M(D)$ , and the tide is softened by how fast that mass grows outward — the coefficient becomes $3-\mathrm{d}\ln M/\mathrm{d}\ln D$ . A singular isothermal sphere with a flat rotation curve has $M\propto D$ , so the slope is $1$ and the coefficient falls to $2$ . Since $r_t\propto(\text{coeff})^{-1/3}$ , the isothermal tidal radius is about $14\%$ larger than the point-mass estimate for the same enclosed mass.

What is the difference between the tidal radius, the Hill radius, the Jacobi radius, and the Roche limit?

They are four names for facets of one balance. The tidal radius is the generic scale where a satellite's self-gravity equals the host's tide. The Hill radius is that scale for a small body on a circular orbit about a point mass, $D(m/3M)^{1/3}$ , familiar from planetary dynamics. The Jacobi radius is the same boundary described through the zero-velocity surface of the Jacobi integral in the rotating frame — identical number, three-body language. The Roche limit applies the balance to a strengthless fluid body and quotes it as a distance in primary radii through the density ratio, $2.44\,R_M(\rho_M/\rho_m)^{1/3}$ . Stellar systems use "tidal/Hill/Jacobi"; fluids and rubble piles use "Roche".

Why does a fluid moon disrupt farther out than a rigid one (2.44 versus 1.26)?

A rigid body holds its spherical shape and resists the tide until the stretch at its surface simply exceeds its surface gravity — that happens at $1.26\,R_M(\rho_M/\rho_m)^{1/3}$ . A fluid body cannot resist deformation: as it approaches, the tide draws it into a prolate ellipsoid pointed at the primary. The elongated tips stick out farther, where the tide is stronger and the body's own self-gravity is weaker, so the fluid comes apart at a larger separation, $2.44\,R_M(\rho_M/\rho_m)^{1/3}$ . The coefficient is a statement about the body's rheology — rigid bodies survive closer in, fluids and loose rubble piles are stripped sooner.

Do tidal tails trace the satellite's orbit?

Almost, but not exactly, and that near-miss is what makes them useful. Stars stripped through the inner Lagrange point $L_1$ lose a little orbital energy and angular momentum and fall onto slightly smaller, faster orbits, forming the leading tail; stars leaving through the outer point $L_2$ gain a little and drift onto larger, slower orbits, forming the trailing tail. Each tail lies close to the progenitor's orbit but is offset by roughly the tidal radius and spread by the small energy kick, so a stream is a slightly smeared orbit. Because the orbit is set by the Galactic potential, fitting the stream's track and velocities measures the Milky Way's mass distribution and halo shape.

Why are dwarf spheroidals thought to be dark-matter dominated?

Because they have not been torn apart. A dwarf spheroidal's stars show a large internal velocity dispersion, which by the virial theorem implies a large enclosed mass. If that mass were only the visible stars, the galaxy's mean density would be low, its tidal radius small, and the object should be actively disrupting in the Milky Way's tide — yet most classical dwarfs are round, bound, and show stars well within their nominal tidal radii. The resolution is a massive dark-matter halo: it raises the mean density and hence the tidal radius, letting the dwarf survive. The exceptions, like the Sagittarius dwarf whose stars trail into a giant stream, are precisely the ones caught mid-disruption.