Two-body problem — equations of motion, reduction to one-body: Visual walkthrough

We build up to this single boxed result: $\ddot{r} = - \frac{μ}{r ^{2}} \hat{r}, μ = G (m_{1} + m_{2}) .$ But first — what is even a $r$ ? What is a dot on top of it? Let's start from nothing.

Step 0 — The alphabet (earn every symbol first)

Why do we need vectors and not plain numbers? Because gravity has a direction — it points from one body toward the other. A plain number can't say "which way"; an arrow can. That is the whole reason we use vectors here and not ordinary algebra.

Figure — Two-body problem — equations of motion, reduction to one-body

Look at the figure: two dots, one origin, two arrows. Everything else on this page is built from exactly these ingredients.

Step 1 — The relative arrow $r$

WHAT we did: we defined a single arrow that connects the two bodies.

WHY: gravity depends only on the gap between the bodies, not on where the pair sits in space. If both dots slide 5 metres to the right, the force is unchanged. So the physics lives entirely inside $r$ . Tracking $r$ throws away nothing that matters.

PICTURE: in the figure, $r$ is the pale-yellow arrow. Its length $r$ is the separation; its direction is "from 1 toward 2." We also draw the unit arrow $\hat{r} = r / r$ — same direction, but length exactly 1. Dividing an arrow by its own length just shrinks/stretches it to length 1 while keeping the direction; that is what "unit vector" means, and we use it to say direction only, no length.

Step 2 — Newton's law on each dot (with the correct sign)

Read the left equation piece by piece:

$m_{1} \ddot{r}_{1}$ = (mass of body 1) $\times$ (its acceleration) = the net force on body 1 (that's just $F = ma$ ).
$\frac{G m _{1} m _{2}}{r ^{2}}$ = the strength of gravity: $G$ is Newton's constant, the two masses multiply, and $1/ r^{2}$ says the pull weakens with the square of the distance.
$\hat{r}$ = direction: body 1 is pulled toward body 2, which is the $+ \hat{r}$ direction (recall $\hat{r}$ points from 1 to 2). Hence the $+$ .
On the right equation the sign flips to $- \hat{r}$ because body 2 is pulled back toward body 1 — the opposite direction.

WHAT: two applications of $F = ma$ , one per body.

WHY the signs: each body falls toward the other. The strengths are identical ( $G m_{1} m_{2} / r^{2}$ appears in both) — that is Newton's third law, action equals reaction, already baked in. If you got the signs backwards you'd have the bodies flying apart, which is wrong for gravity.

PICTURE: two force arrows pointing at each other, equal length, opposite direction.

Step 3 — Add the equations: the center drifts freely

WHAT: we literally added the two boxed equations from Step 2.

WHY: the right-hand sides are equal and opposite, so they annihilate. The left side is the second derivative of $m_{1} r_{1} + m_{2} r_{2}$ . Define the center of mass: $R = \frac{m _{1} r _{1} + m _{2} r _{2}}{m _{1} + m _{2}},$ the mass-weighted average position — the balance point of the pair. Then $\ddot{R} = 0$ .

$\ddot{R} = 0$ means zero acceleration, so $R$ moves in a straight line at constant speed (or stays put). No external force acts on the isolated pair, so their balance point can't accelerate.

PICTURE: the two dots wheel around, but the marked balance point glides on a perfectly straight track. It carries no orbital information — so we hop into a frame riding along with it and forget it. That deletes 3 of our 6 unknowns.

Step 4 — Subtract the equations: the relative motion

First divide each Step-2 law by its own mass (so we isolate acceleration): $\ddot{r}_{1} = + \frac{G m _{2}}{r ^{2}} \hat{r}, \ddot{r}_{2} = - \frac{G m _{1}}{r ^{2}} \hat{r} .$ Now subtract the first from the second. Because $r = r_{2} - r_{1}$ , differentiating twice gives $\ddot{r} = \ddot{r}_{2} - \ddot{r}_{1}$ :

Term by term in the final result:

$\ddot{r}$ = acceleration of the relative arrow — how the gap-vector speeds up or turns.
$G (m_{1} + m_{2})$ = both masses appear, because both dots accelerate. Their two accelerations add up in the relative motion.
$1/ r^{2}$ and $\hat{r}$ = same inverse-square strength and same "toward the center" direction.
The minus sign = the relative arrow is pulled to shrink (bodies attract), always pointing inward.

WHAT: we combined the two motions into one equation for one arrow.

WHY this is the whole point: this equation has exactly the form of one particle falling toward a fixed center. We have converted a 2-body chase into a 1-body Kepler problem.

PICTURE: forget the two dots; draw a single "ghost" particle at the tip of $r$ , always yanked toward a fixed origin. That ghost traces the orbit.

Step 5 — The reduced mass (so energy comes out right)

We can also write the relative law as an honest $F = ma$ for one particle. The real force magnitude is $F = G m_{1} m_{2} / r^{2}$ . What mass $μ_{red}$ makes $μ_{red} \ddot{r} = F$ true?

$μ_{red} \ddot{r} = \frac{m _{1} m _{2}}{m _{1} + m _{2}} (- \frac{G ( m _{1} + m _{2} )}{r ^{2}} \hat{r}) = - \frac{G m _{1} m _{2}}{r ^{2}} \hat{r} = F . ✓$

WHAT / WHY: two different repackagings of the same motion — one uses $μ$ (strength) with a bare acceleration, one uses $μ_{red}$ (a real effective mass) with the real force. The second makes kinetic energy split cleanly: $T = \frac{1}{2} M \dot{R}^{2} + \frac{1}{2} μ_{red} \dot{r}^{2}, M = m_{1} + m_{2},$ "energy of the drifting center" $+$ "energy of the internal orbit," with no cross-term.

PICTURE: one effective particle of mass $μ_{red}$ swinging on the end of the force $F$ .

Step 6 — The degenerate & edge cases (never leave a gap)

The figure shows all three: heavy-light (tiny wobble), equal (twin circles), and the collapse cone where $r \to 0$ .

Recall Check the coefficients that recover the dots

$r_{1} = R - \frac{m _{2}}{M} r$ and $r_{2} = R + \frac{m _{1}}{M} r$ . The heavier body has the smaller coefficient, so it stays closer to the COM. Why do these work? ::: Subtract them: $r_{2} - r_{1} = \frac{m _{1} + m _{2}}{M} r = r$ ✓. Mass-weight and add: the $r$ -terms cancel, leaving $M R$ ✓.

The one-picture summary

Left of the arrow: two dots, two arrows, two coupled equations, 6 unknowns. Right of the arrow: one drifting balance point $R$ (straight, boring) $+$ one ghost particle on $r$ obeying $\ddot{r} = - μ \hat{r} / r^{2}$ . Two-body mess → one Kepler problem.

Recall Feynman: the whole walkthrough in plain words

Two kids spin on ice holding a rope. Instead of watching both kids (hard — everything moves), watch two simpler things. First, the middle of the rope: it just slides along a straight line, dead boring, because nothing outside is pushing the pair — that's the center of mass, and we throw it away. Second, the rope itself, the arrow from one kid to the other: it stretches and swings, and it holds all the spinning information. When we write down Newton's push for each kid and add them, the two pushes cancel (that's why the middle drifts free). When we subtract them, the two pushes pile up, so the rope feels a pull set by both kids' weights together — that's the $m_{1} + m_{2}$ . Result: instead of two chasing kids, we have one imaginary bead sliding on the rope, pulled toward an invisible center. That single bead is the whole orbit. Heavy kid barely budges, light kid swings wide — which is exactly why a planet looks fixed while its moon races around.

Connections

Parent topic — the full statement and worked examples.
Kepler's laws — solving $\ddot{r} = - μ \hat{r} / r^{2}$ gives conic orbits.
Conservation of angular momentum (central force) — why the ghost particle stays in a plane.
Vis-viva equation — the energy of the reduced one-body system.
Center of mass frame — the "ride-along" frame we used in Step 3.
Reduced mass in molecular vibrations — the same $μ_{red}$ trick for two atoms on a spring.
Three-body problem — what breaks when a third dot is added.