This page assumes you have seen nothing. We build every letter, symbol and picture the parent note leans on, in the order you need them. When you can answer the checklist at the bottom, go read the parent derivation.
The little number is a subscript — just a name tag, not multiplication. M2 reads "mass number two", not "M times 2".
We also use m (small letter) for the test mass — the satellite or dust grain so tiny it pulls on nobody. Big letters pull; the small letter just goes along for the ride.
We line the two big bodies up along the horizontal line y=0 (the x-axis). Everything happens in this one flat plane, so two numbers are enough.
How to read figure s01: the big lavender dot on the left is M1; the smaller coral dot on the right is M2; the little mint dot up and to the right is our test mass m at some (x,y). The butter dot at the crossing of the dashed lines is the origin (0,0). Notice the mint dot is off the axis (its y=0) — that off-axis position is exactly what we need for L4/L5 later. The two double-headed arrows at the bottom show that M1 is only a short distance μ from the origin, while M2 is a long distance (1−μ) away.
Two bodies do not orbit each other like one is nailed down. They both circle a shared pivot — the barycenter (centre of mass). Picture two kids on a see-saw: the heavy one sits close to the middle, the light one far out, so it balances.
First we need names for where each body is. Let x1 be the x-coordinate of body M1 and x2 the x-coordinate of body M2 (both lie on the axis, so their y=0).
To describe "how lopsided" the pair is with ONE number, define the mass ratio:
The parent puts M1 at x1=−μ and M2 at x2=+(1−μ). Check the balance rule with scaled masses M1=1−μ, M2=μ: (1−μ)(−μ)+μ(1−μ)=0 ✓. In figure s01 the heavy body sits a short distance μ left of the origin, the light body a long distance (1−μ) right — exactly the see-saw rule.
We will constantly need "how far is the test mass from each big body?" Give those two distances names first:
Now, how do we compute a straight-line distance from coordinates? First define coordinate differences:
How to read figure s02: the lavender dot is M1, the mint dot is the test mass m. The coral horizontal leg is Δx=x+μ (the test mass's x minus M1's x=−μ), the butter vertical leg is Δy=y, and the slate slanted line is the distance r1 itself — the hypotenuse. The little square at the corner marks the right angle that makes Pythagoras apply.
So the two distances the parent writes are just Pythagoras from m to each body:
r1=(x+μ)2+y2(to M1),r2=(x−(1−μ))2+y2(to M2).
The (x+μ) is "how far right m is from M1", because M1 sits at x1=−μ and subtracting a negative adds.
We choose to ride along with the spinning pair — a rotating reference frame. In this frame the two big bodies stop moving (they're frozen), which is the whole trick that makes "parking spots" possible.
How to read figure s03: the slate "×" at the centre is the spin axis; the lavender and coral dots are M1 and M2, frozen because we spin with them. The mint arrows all point straight outward from the axis, and they get longer the farther out they start — that pictures ω2ρ growing with ρ. The faint grey curl reminds you the whole frame is turning at rate ω. This outward push is the third force that, together with the two gravities, must cancel at a Lagrange point.
Both these forces are built properly in Coriolis and Centrifugal Forces (Rotating Frames); the whole spinning-frame setup is the Restricted Three-Body Problem.
The parent picks units where D=1 and G(M1+M2)=1, so ω2=1, i.e. ω=1. This is just a smart choice of rulers and clocks so the equations shed clutter — nothing physical changes. See Kepler's Third Law.
