Space environment — LEO radiation (SAA, Van Allen), atomic oxygen, MMOD debris: Visual walkthrough

This is the visual companion to the parent topic.

Step 1 — A moving charge, and the one arrow that pushes it

WHAT. Picture a single tiny charged speck — a proton — flying through space. First we fix a coordinate frame so left/right, up/down and in/out of the page all have precise meaning: let $\overset{x}{^}$ point right, $\overset{y}{^}$ point up, and $\overset{z}{^}$ point out of the page toward you (this is the standard right-handed frame). Now draw the velocity as an arrow $v = v \overset{x}{^}$ (pointing right) and switch on a magnetic field $B = B \overset{y}{^}$ (pointing up).

WHY the cross product and not ordinary multiplication. A magnetic field does not simply push a charge forward or backward. Experiment shows the force is sideways — perpendicular to both the motion and the field. The one mathematical tool that eats two arrows and spits out a third arrow at right angles to both is the cross product $\times$ . That is exactly why the Lorentz law uses it:

$F = q (v \times B)$

$q$ — the charge (a number; positive for a proton). It scales the whole force.
$v$ — velocity arrow: how fast and which way the particle moves.
$B$ — magnetic field arrow: how strong the field is and which way it points.
$v \times B$ — a new arrow perpendicular to both, whose length is $v B sin θ$ ( $θ$ = angle between $v$ and $B$ ). This is what "sideways" means precisely.

Working the direction out with the right-hand rule. Point your right hand's fingers along $v$ (right, $+ \overset{x}{^}$ ) and curl them toward $B$ (up, $+ \overset{y}{^}$ ); your thumb points out of the page ( $+ \overset{z}{^}$ ). In symbols $\overset{x}{^} \times \overset{y}{^} = + \overset{z}{^}$ , so $v \times B = v B \overset{z}{^}$ . For our positive proton ( $q > 0$ ) the force therefore points out of the page, toward you.

PICTURE. In the figure the teal arrow ( $v$ ) points right, the plum arrow ( $B$ ) points up, and the orange symbol — a dot inside a circle — marks the force $F$ popping straight out of the page toward you, perpendicular to the plane the two arrows span.

Figure — Space environment — LEO radiation (SAA, Van Allen), atomic oxygen, MMOD debris

Step 2 — Decomposing the velocity so we can talk about "the part that bends"

WHAT. Before we can even write a clean force balance, we must name the two pieces of the velocity, because — as Step 1's cross product already hinted — the field only grabs part of the motion. Break the velocity arrow $v$ into two perpendicular pieces measured against the field line:

$v_{⊥}$ — the component of $v$ across (perpendicular to) $B$ .
$v_{∥}$ — the component of $v$ along (parallel to) $B$ .

By the Pythagoras of these two right-angle pieces, $v^{2} = v_{⊥}^{2} + v_{∥}^{2}$ .

WHY split now — why the cross product forces this on us. The length of $v \times B$ is $v B sin θ$ , and $v sin θ$ is exactly $v_{⊥}$ , the across-the-field part. The along-the-field part contributes $sin 0° = 0$ : it feels no force and simply coasts. So the magnetic force only ever acts on $v_{⊥}$ . To write any honest equation, $v_{⊥}$ must be a defined symbol first — which is why we decompose here, before the circle.

PICTURE. The teal arrow $v$ splits into an orange leg $v_{⊥}$ (across the field line) and a teal leg $v_{∥}$ (along it); the right angle between them and the field line is marked, so you can see $v^{2} = v_{⊥}^{2} + v_{∥}^{2}$ as a right triangle.

Step 3 — A force that is always sideways makes a circle

WHAT. Focus on $v_{⊥}$ alone (defined in Step 2). Its force is always at 90° to it. So the force never speeds it up or slows it down — it only bends the direction. Bend the direction the same amount, over and over, and the across-motion closes into a circle.

WHY a circle and not a curve that runs away. Think of a ball on a string swung overhead. The string pulls inward (toward your hand) at 90° to the ball's motion; the ball goes in a circle. The magnetic force plays the exact role of that string. This is why we can borrow the idea of centripetal force — the inward pull needed to hold circular motion:

$F_{centripetal} = \frac{m v _{⊥}^{2}}{r}$

$m$ — mass of the particle (heavier → harder to bend).
$v_{⊥}$ — the across-the-field speed defined in Step 2 (the part that actually gets bent).
$r$ — radius of the circle it settles into.
The fraction says: faster or heavier particle → bigger circle; tighter turn needs more force.

PICTURE. Watch the orange force arrow always aiming at the centre while the teal velocity arrow rides the rim — the two stay locked at 90° all the way round the circle.

Step 4 — Setting the two forces equal earns the gyroradius $r_{L}$

WHAT. Two descriptions of the same inward force must agree. The magnetic force supplies the centripetal force. Set them equal. Because a force magnitude is never negative, we use the magnitude of the charge, written $∣ q ∣$ (the size of $q$ , ignoring its plus/minus sign).

WHY set them equal. "Supplies" is the key word: the magnetic push is the thing curving the path, so its size must equal the size the circle demands. Equal cause, equal effect.

$magnetic push ∣ q ∣ v_{⊥} B = circle demands \frac{m v _{⊥}^{2}}{r}$

Now solve for $r$ . Cancel one $v_{⊥}$ from both sides:

$∣ q ∣ B = \frac{m v _{⊥}}{r} ⟹ r_{L} = \frac{m v _{⊥}}{∣ q ∣ B}$

$r_{L}$ — the Larmor radius (gyroradius): the size of the little circle the particle traces around a field line.
$∣ q ∣$ — the magnitude of the charge. Using $∣ q ∣$ (not $q$ ) guarantees $r_{L}$ comes out positive, as a radius must be; the sign of $q$ only decides which way the particle circles, not how big the circle is.
Bigger $m$ or $v_{⊥}$ → bigger circle (harder to turn).
Bigger $∣ q ∣$ or $B$ → smaller, tighter circle (stronger grip).

This is exactly the formula the parent note stated (with $∣ q ∣$ made explicit) — now it is ours.

PICTURE. The plum circle is labelled with its radius $r_{L}$ ; the orange arrow shows the magnetic push $∣ q ∣ v_{⊥} B$ pointing in, the teal arrow shows the circle's demand $m v_{⊥}^{2} / r$ , and the equation beneath reads them off as equal.

Step 5 — Adding $v_{∥}$ back: the spiral is born

WHAT. Step 3–4 only used $v_{⊥}$ . Now put the coasting piece $v_{∥}$ (from Step 2) back in. It feels no force, so it just glides steadily along the field line while the across-part keeps circling.

WHY this gives a helix. Circle (from $v_{⊥}$ ) + steady glide along the line (from $v_{∥}$ ) = a helix. Using the pitch angle $α$ (the tilt of $v$ from the field line), the two pieces are

$v_{⊥} = v sin α v_{∥} = v cos α$

$α$ — the pitch angle: angle between the velocity arrow and the field line.
$v_{⊥} = v sin α$ — the circling part. At $α = 90°$ this is the whole speed.
$v_{∥} = v cos α$ — the glide-along part. At $α = 0°$ this is the whole speed.

That corkscrew is the first of the belt's three motions.

PICTURE. The plum spiral winds around the dashed field line; the orange arrow $v_{⊥} = v sin α$ drives the circling and the teal arrow $v_{∥} = v cos α$ drives the steady climb along $B$ .

Step 6 — Why the particle bounces between the poles (the magnetic mirror)

WHAT. Earth's field is not uniform — it is a dipole, so field lines crowd together (get stronger) near the magnetic poles and spread out (get weaker) near the equator. A particle spiralling toward a pole runs into a stronger $B$ .

WHY $μ$ stays constant — the visual argument. Define the magnetic moment of the gyrating particle:

$μ = \frac{m v _{⊥}^{2}}{2 B}$

$m v_{⊥}^{2} /2$ — the kinetic energy tied up in the circling motion.
$B$ — the local field strength where the particle currently sits.
$μ$ — this ratio; physically it is the magnetic moment of the tiny current loop the gyration makes.

Here is why $μ$ barely changes. A gyrating charge is a tiny current loop of area $A = π r_{L}^{2}$ . The magnetic flux threading that loop is $Φ = B A = B π r_{L}^{2}$ . Substitute $r_{L} = m v_{⊥} / (∣ q ∣ B)$ :

$Φ = B π (\frac{m v _{⊥}}{∣ q ∣ B})^{2} = \frac{π m ^{2} v _{⊥}^{2}}{∣ q ∣ ^{2} B} = \frac{2 π m}{∣ q ∣ ^{2}} μ .$

So $Φ$ and $μ$ are the same quantity up to the fixed constant $2 π m /∣ q ∣^{2}$ . A basic result of slowly-varying fields (Faraday's law applied to a loop the field can't change faster than one orbit) is that the flux through the gyro-loop is conserved — the loop simply shrinks to keep the same number of field lines through it. Since $Φ$ is fixed, $μ$ is fixed. This holds only when the field changes slowly: barely over one loop-width $r_{L}$ in space, barely over one gyration in time — which Earth's dipole comfortably satisfies. This is what "adiabatic invariant" means.

The quantitative reflection. Two conservation laws now act together:

$μ = \frac{m v _{⊥}^{2}}{2 B} = const, v^{2} = v_{⊥}^{2} + v_{∥}^{2} = const$

The second holds because the magnetic force does no work (Step 3), so the total speed $v$ never changes. From the first, $v_{⊥}^{2} = \frac{2 μ}{m} B$ — so as the particle climbs into stronger $B$ , $v_{⊥}^{2}$ grows in exact proportion to $B$ . Substitute into the second:

$v_{∥}^{2} = v^{2} - \frac{2 μ}{m} B .$

$v_{∥}$ shrinks as $B$ rises, and hits zero exactly where $\frac{2 μ}{m} B = v^{2}$ .

Anchor it to the equator, where the particle has pitch angle $α_{eq}$ and field $B_{eq}$ . There $v_{⊥}^{2} = v^{2} sin^{2} α_{eq}$ , so $\frac{2 μ}{m} = \frac{v ^{2} sin ^{2} α _{eq}}{B _{eq}}$ . The turning point ( $v_{∥} = 0$ ) needs $\frac{2 μ}{m} B_{mirror} = v^{2}$ , giving the clean mirror condition:

$\frac{B _{mirror}}{B _{eq}} = \frac{1}{sin ^{2} α _{eq}}$

$B_{mirror}$ — field strength at the reflection point.
$α_{eq}$ — pitch angle back at the equator.
Read it: a particle with a small equatorial pitch angle needs a large field ratio to turn around — it must climb deep toward the pole.

The particle bounces pole-to-pole like a bead between two mirrors — the magnetic mirror. This is the second motion.

PICTURE. Converging plum field lines make a funnel; the spiral tightens (its $r_{L}$ shrinks, keeping flux $Φ$ constant — the shrinking loop is drawn to scale) as it climbs, $v_{∥}$ dwindles to zero at the marked point $B_{mirror} = B_{eq} / sin^{2} α_{eq}$ , and the path turns around.

Step 7 — Edge cases: what happens at the extreme pitch angles

WHAT. We must cover every particle, not just the tidy ones. The pitch angle $α$ decides everyone's fate.

WHY these cases matter. They are the boundaries of "trapped." Miss them and you'd wrongly claim all particles are safely bottled.

$α = 90°$ (pure perpendicular). All speed is $v_{⊥}$ ; $v_{∥} = 0$ . From the mirror condition, $B_{mirror} / B_{eq} = 1/ sin^{2} 90° = 1$ — it mirrors right at the equator and barely drifts along the line: the most deeply trapped case.
$α = 0°$ (pure parallel). All speed is $v_{∥}$ ; $v_{⊥} = 0$ , so $r_{L} = 0$ — no circle at all, and $μ = 0$ . The mirror condition demands an infinite field ratio, which never happens; the particle streams straight down the field line, dives into the atmosphere near the pole, and is lost. This is the heart of the loss cone: any particle with too small a pitch angle mirrors only below the atmosphere and escapes.
$B \to 0$ (field vanishes, e.g. far from the dipole). Then $r_{L} = \frac{m v _{⊥}}{∣ q ∣ B} \to \infty$ : the "circle" becomes a straight line. No field, no trapping. This is one reason the belts end — where the field is too weak, particles fly free.
$q = 0$ (a neutral particle). Force $= 0$ . It ignores the field entirely and travels in a straight line. Only charged particles are trapped — which is why the belts are made of protons and electrons, not neutrons or dust.

PICTURE. Four mini-panels: the fat equatorial circle ( $90°$ ), the straight dive into the loss cone ( $0°$ ), the straightening path as $B \to 0$ , and the indifferent straight shot of a neutral particle.

Step 8 — The third motion: gradient and curvature drift build a full ring

WHAT. Now watch the centre of the little gyro-circle, not the particle. Two features of the dipole field push that centre sideways a tiny bit each loop, and those tiny steps add up until the centre has walked all the way around Earth.

WHY the centre steps — gradient drift, made concrete. Follow one gyro-circle. Because $B$ is stronger on the Earth-facing side of that loop and weaker on the far side, the local radius $r_{L} = \frac{m v _{⊥}}{∣ q ∣ B}$ is smaller on the strong-field arc and larger on the weak-field arc. The particle therefore sweeps a tight half-loop on the strong side and a wide half-loop on the weak side. A circle stitched from a tight arc and a wide arc cannot close — after one full gyration the particle lands a little to the side of where it started. That leftover displacement is the drift step. Carrying the geometry through gives the gradient-drift velocity

$v_{\nabla B} = \frac{m v _{⊥}^{2}}{2 ∣ q ∣ B ^{2}} \frac{B \times \nabla B}{B}$

$\nabla B$ — the arrow pointing the way $B$ increases (toward Earth here).
$B \times \nabla B$ — the cross product that sends the step sideways, perpendicular to both the field and the direction it strengthens.
$\propto 1/∣ q ∣$ with the sign of $q$ carried inside the cross product: protons drift one way (westward), electrons the opposite way (eastward). Reverse the charge and you reverse the whole drift direction — that opposite motion of the two species is literally an electric current, the belt's ring current.
Bigger $v_{⊥}$ (fatter loop) → bigger side-step, since the tight/wide arcs differ more.

WHY a second drift — curvature drift. In a dipole the field lines are also curved, not straight. A particle gliding along a curved line (via $v_{∥}$ ) feels a centrifugal push outward from the curve; that push, crossed with $B$ , produces a second sideways step in the same sense as the gradient drift:

$v_{curv} = \frac{m v _{∥}^{2}}{∣ q ∣ B ^{2}} \frac{R _{c} \times B}{R _{c}^{2}}$

$R_{c}$ — the radius-of-curvature arrow of the field line (how sharply it bends).
$v_{∥}$ — the glide speed (Step 5); a fast glider along a sharp bend drifts more.
Same $1/∣ q ∣$ sign rule → protons and electrons again drift oppositely, adding to the gradient drift rather than fighting it.

Together, gradient + curvature drift make each particle's guiding centre circle the whole planet.

A number so you feel the accumulation. A belt proton has $r_{L} \sim 100 km$ and its drift speed is only a tiny fraction of its gyration speed, so each single gyration nudges the centre sideways by far less than one gyroradius. But there are millions of gyrations per pass, and the steps all point the same way, so they add coherently: a typical inner-belt proton completes a full loop around Earth in minutes to hours. Slow per turn, unstoppable in aggregate — exactly how a clock's second hand crawls yet still laps the dial.

PICTURE. Top-down view above Earth: on the Earth-facing arc each little loop is drawn tight, on the far arc it is drawn wide; the mismatch nudges each successive loop's centre one small step sideways (the steps are marked with short arrows), and stacking ~a-dozen loops traces the belt's full ring — the inset zooms two consecutive loops to show the single sideways step between their centres.

The one-picture summary

Three motions, one particle: a fast tight gyration around the field line (Steps 3–5), a slow bounce between the two magnetic poles set by $B_{mirror} / B_{eq} = 1/ sin^{2} α_{eq}$ (Step 6), and a very slow drift all the way around Earth from the uneven, curved field (Step 8). Layer them and you get the doughnut of the Van Allen belts. The single formula $r_{L} = \frac{m v _{⊥}}{∣ q ∣ B}$ is the seed of all of it: it sets the circle size, it explains the mirror (bigger $B$ → smaller $r_{L}$ , more $v_{⊥}$ , $v_{∥}$ to zero), and it explains the drift (uneven $B$ → uneven arc → sideways step).

Recall Feynman retelling — tell it back in plain words

Fix your axes first: right is $\overset{x}{^}$ , up is $\overset{y}{^}$ , out of the page is $\overset{z}{^}$ . A proton zips right through a field pointing up; the right-hand rule says $v \times B$ points out of the page, so a positive charge is shoved toward you (an electron, into the page). Because that cross product only grabs the part of the velocity across the field, first split $v$ into $v_{⊥}$ (across, gets bent) and $v_{∥}$ (along, coasts). The forever-sideways shove curls $v_{⊥}$ into a circle, and matching "magnetic shove = circle's demand" gives its size, $r_{L} = m v_{⊥} / (∣ q ∣ B)$ — we use $∣ q ∣$ so a radius stays positive. Add the coasting $v_{∥}$ back and circle + glide = a corkscrew. Head toward a pole and the field crowds up; the flux through the gyro-loop is pinned, so as the loop shrinks $v_{⊥}$ climbs, and since total speed is fixed the forward glide drains to zero right where $B_{mirror} = B_{eq} / sin^{2} α_{eq}$ — the particle bounces. Meanwhile the field is stronger on Earth's side of each little loop, so the loop is tight on one side and wide on the other and can't close: its centre steps sideways a hair each turn (gradient drift), and the curved field lines add a second sideways nudge (curvature drift) in the same direction — protons walking west, electrons east. Millions of coherent little steps carry each particle right around the planet in minutes to hours. Gyrate, bounce, drift — smear that over months and you've painted a glowing doughnut. Escape clauses: point straight down the line (loss cone) and you dive into the air and die; carry no charge, or find no field, and the trap simply lets you go.

Recall Quick self-check

Why is the magnetic force always perpendicular to velocity? ::: Because it comes from the cross product $q v \times B$ , which by definition points at 90° to both $v$ and $B$ . With $v$ right and $B$ up, which way does the force on a proton point? ::: Out of the page ( $+ \overset{z}{^}$ ), by the right-hand rule $\overset{x}{^} \times \overset{y}{^} = \overset{z}{^}$ ; for an electron it flips into the page. Why does the gyroradius formula use $∣ q ∣$ ? ::: A radius must be positive; $∣ q ∣$ drops the sign of the charge, which only decides the direction of circling, not the size. What conserved quantity forces the bounce, and why is it conserved? ::: The magnetic moment $μ = m v_{⊥}^{2} / (2 B)$ ; it is fixed because the magnetic flux through the gyro-loop is conserved when the field changes slowly over one orbit. Where does the particle mirror? ::: Where $B_{mirror} = B_{eq} / sin^{2} α_{eq}$ , i.e. where $v_{∥}$ reaches zero. Why are the belts ring-shaped, and which way does each species drift? ::: Gradient and curvature drift step each loop's centre sideways every turn, walking the particle around Earth into a torus; protons drift west, electrons east (opposite, because of the sign of $q$ ).

See also: Van Allen Probes mission · Magnetic field modeling · Single-event effects · Spacecraft materials selection