Two-body problem — equations of motion, reduction to one-body: Worked examples

Before anything, two words we will lean on constantly:

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

Look at the figure: the arrow $r$ runs from body 1 to body 2, and the amber dot (COM) always sits on the line between them, closer to the heavier mass. Everything below is just this picture under different weights.

The scenario matrix

Every case orbital-reduction can throw at you falls into one of these cells. Each cell is claimed by at least one worked example.

Cell	What varies	Degenerate / limiting?	Covered by
A — dominant mass ( $m_{2} ≪ m_{1}$ )	huge mass ratio	limit $μ \to G m_{1}$	Ex 1
B — equal masses ( $m_{1} = m_{2}$ )	symmetric	COM exactly midway	Ex 2
C — general unequal masses	finite ratio	none	Ex 3
D — reduced mass vs gravitational $μ$	two different $μ$ 's	keep them apart	Ex 4
E — energy split (COM + internal)	velocities	cross-terms vanish	Ex 5
F — $m_{2} \to 0$ (test particle)	zero mass input	$μ_{red} \to 0$ , COM $\to m_{1}$	Ex 6
G — real-world word problem	Sun–Jupiter wobble	Sun does move	Ex 7
H — exam twist: recover both orbits	inversion + signs	heavier body closer	Ex 8

We deliberately cover the two ends — the test-particle limit (Cell F, a mass so small it's basically zero) and the equal-mass case (Cell B) — because these are where the "planet looks fixed" intuition either holds perfectly or fails completely.

The examples

Worked example Ex 1 — Cell A: huge mass ratio (low Earth satellite)

$m_{1} = m_{\oplus} = 5.97 \times 1 0^{24} kg$ , satellite $m_{2} = 1000 kg$ , $G = 6.674 \times 1 0^{- 11} m^{3} kg^{- 1} s^{- 2}$ . Find the exact $μ = G (m_{1} + m_{2})$ and show the satellite mass is invisible.

Forecast: guess — will the answer differ from the textbook $G M_{\oplus}$ in any digit you'd write down? (Guess yes/no, then check.)

Add the masses: $m_{1} + m_{2} = 5.97 \times 1 0^{24} + 1 0^{3} = 5.970 000 \dots \times 1 0^{24} kg$ . Why this step? $μ$ uses the sum (mnemonic: "SUM for the force"), never one mass alone.
The added $1 0^{3}$ sits $1 0^{21}$ places below the leading digit — it cannot change any figure we'd quote. Why this step? This is the limit $m_{2} ≪ m_{1} \Rightarrow m_{1} + m_{2} \approx m_{1}$ made concrete.
$μ = 6.674 \times 1 0^{- 11} \times 5.97 \times 1 0^{24} = 3.984 \times 1 0^{14} m^{3} / s^{2}$ . Why this step? This is the "standard gravitational parameter" $G M_{\oplus}$ satellites are designed around.

Verify: units $[m^{3} kg^{- 1} s^{- 2}] [kg] = m^{3} s^{- 2}$ ✓. Numeric $μ \approx 3.98 \times 1 0^{14}$ , matching the accepted $3.986 \times 1 0^{14}$ (small offset only from rounded $G, m_{\oplus}$ ).

Worked example Ex 2 — Cell B: equal-mass binary stars

$m_{1} = m_{2} = m = 2.0 \times 1 0^{30} kg$ , separation $r = 1.0 \times 1 0^{11} m$ . Find $μ$ , $μ_{red}$ , the COM location, and each star's orbit radius.

Forecast: where is the COM — nearer star 1, star 2, or exactly between? Guess before reading.

$μ = G (m_{1} + m_{2}) = G (2 m) = 6.674 \times 1 0^{- 11} \times 4.0 \times 1 0^{30} = 2.6696 \times 1 0^{20} m^{3} / s^{2}$ . Why this step? Force parameter always uses the total mass.
$μ_{red} = \frac{m \cdot m}{2 m} = \frac{m}{2} = 1.0 \times 1 0^{30} kg$ . Why this step? "product over sum" collapses to half the mass when equal.
COM: $r_{1} = R - \frac{m _{2}}{M} r = R - \frac{1}{2} r$ , so star 1 sits $r /2$ from COM. By symmetry star 2 too. COM is exactly midway. Why this step? Equal coefficients $m_{2} / M = m_{1} / M = \frac{1}{2}$ force the balance point to the middle.
Each orbit radius $= r /2 = 5.0 \times 1 0^{10} m$ . Why this step? The orbit radius of a body is exactly its distance from the COM — and step 3 showed each body sits $\frac{m _{2}}{M} r = \frac{1}{2} r$ from the COM. Each star circles that fixed point, so the circle's radius is that same offset.

Verify: the two orbit radii sum to $r /2 + r /2 = r = 1 0^{11}$ ✓ (they must add to the separation). $μ_{red} = 1 0^{30} = m /2$ ✓.

Worked example Ex 3 — Cell C: general unequal masses (Earth–Moon)

$m_{1} = m_{\oplus} = 5.97 \times 1 0^{24} kg$ , $m_{2} = m_{moon} = 7.35 \times 1 0^{22} kg$ , separation $r = 3.84 \times 1 0^{8} m$ . Find the COM distance from Earth's center.

Forecast: is the COM inside the Earth (radius $6.37 \times 1 0^{6}$ m) or outside it? Guess.

Earth's distance from COM: $∣ r_{1} - R ∣ = \frac{m _{2}}{M} r$ . Why this step? From the undo formula, the offset of body 1 from the COM is $\frac{m _{2}}{M} r$ .
$M = 5.97 \times 1 0^{24} + 7.35 \times 1 0^{22} = 6.0435 \times 1 0^{24} kg$ . Why this step? The offset formula divides by the total mass $M$ , so we must form $M = m_{1} + m_{2}$ before we can build any mass fraction — the fraction $m_{2} / M$ literally asks "what share of the whole is body 2?"
$\frac{m _{2}}{M} = \frac{7.35 \times 1 0 ^{22}}{6.0435 \times 1 0 ^{24}} = 0.012162$ . Why this step? We isolate the dimensionless mass fraction first (a pure number $< 1$ ) so that multiplying by the separation next gives a length; it also tells us at a glance how small the offset will be ( $\approx 1.2%$ of $r$ ).
Distance $= 0.012162 \times 3.84 \times 1 0^{8} = 4.670 \times 1 0^{6} m$ . Why this step? Multiply the mass fraction by the separation to get the physical offset.

Verify: $4.67 \times 1 0^{6} m < 6.37 \times 1 0^{6} m$ — the COM (barycenter) lies inside the Earth ✓, a famous fact. Units: (dimensionless)×m = m ✓.

Worked example Ex 4 — Cell D: the two

$μ$ 's must not be confused Same Earth–Moon numbers as Ex 3. Compute both $μ = G (m_{1} + m_{2})$ (a volume-per-time-squared) and $μ_{red}$ (a mass) and confirm their units differ.

Forecast: which of the two is a mass, and which is not? Say it out loud.

$μ = G M = 6.674 \times 1 0^{- 11} \times 6.0435 \times 1 0^{24} = 4.033 \times 1 0^{14} m^{3} / s^{2}$ . Why this step? This drives the shape/speed of the orbit (it appears in $\ddot{r} = - μ r / r^{3}$ ).
$μ_{red} = \frac{m _{1} m _{2}}{M} = \frac{5.97 \times 1 0 ^{24} \times 7.35 \times 1 0 ^{22}}{6.0435 \times 1 0 ^{24}} = 7.261 \times 1 0^{22} kg$ . Why this step? This is the effective mass carrying kinetic energy of the internal motion.
Note $μ_{red} < m_{2}$ (smaller than the lighter body). Always true. Why this step? $μ_{red} = \frac{m _{2}}{1 + m _{2} / m _{1}} < m_{2}$ , a useful sanity bound.

Verify: $μ \approx 4.03 \times 1 0^{14} m^{3} / s^{2}$ (units of $G \cdot$ mass), $μ_{red} \approx 7.26 \times 1 0^{22} kg$ (a mass, close to but below $m_{2} = 7.35 \times 1 0^{22}$ ) ✓. See Reduced mass in molecular vibrations for the identical $μ_{red}$ in diatomics.

Worked example Ex 5 — Cell E: kinetic energy splits cleanly

Let == $T$ denote the total kinetic energy== of the two bodies, $T = \frac{1}{2} m_{1} \dot{r}_{1}^{2} + \frac{1}{2} m_{2} \dot{r}_{2}^{2}$ (each term is the familiar $\frac{1}{2} m v^{2}$ ). Given $m_{1} = 3 kg$ at velocity $\dot{r}_{1} = (2, 0) m/s$ , $m_{2} = 1 kg$ at $\dot{r}_{2} = (0, 4) m/s$ , verify $T = \frac{1}{2} M \dot{R}^{2} + \frac{1}{2} μ_{red} \dot{r}^{2}$ .

Forecast: guess whether the two ways of computing $T$ give the same number (they must).

Direct: $T = \frac{1}{2} (3) (2^{2}) + \frac{1}{2} (1) (4^{2}) = 6 + 8 = 14 J$ . Why this step? Raw definition of $T$ , no tricks — our ground truth.
$M = 4$ , $\dot{R} = \frac{3 ( 2 , 0 ) + 1 ( 0 , 4 )}{4} = \frac{( 6 , 4 )}{4} = (1.5, 1.0)$ , so $\dot{R}^{2} = 2.25 + 1.0 = 3.25$ . Why this step? COM velocity is the mass-weighted average.
COM term: $\frac{1}{2} (4) (3.25) = 6.5 J$ . Why this step? The whole system's mass $M$ drifts bodily at velocity $\dot{R}$ ; the kinetic energy of that bulk drift is $\frac{1}{2} M \dot{R}^{2}$ , the ordinary $\frac{1}{2} m v^{2}$ applied to the total mass moving as one lump.
$\dot{r} = \dot{r}_{2} - \dot{r}_{1} = (0, 4) - (2, 0) = (- 2, 4)$ , $\dot{r}^{2} = 4 + 16 = 20$ . $μ_{red} = \frac{3 \cdot 1}{4} = 0.75$ . Why this step? The internal motion is described by the single relative-velocity vector $\dot{r}$ (how fast body 2 moves as seen from body 1), and its energy behaves like one particle of the effective mass $μ_{red}$ — so we need both pieces before assembling $\frac{1}{2} μ_{red} \dot{r}^{2}$ .
Internal term: $\frac{1}{2} (0.75) (20) = 7.5 J$ . Why this step? This is the orbit's energy — the only part vis-viva cares about.

Verify: $6.5 + 7.5 = 14 J$ = direct answer ✓. The cross-terms cancel exactly because $R, r$ are independent coordinates.

Worked example Ex 6 — Cell F: the degenerate test-particle limit

$m_{2} \to 0$ Take $m_{1} = M_{⊙} = 1.99 \times 1 0^{30} kg$ and let $m_{2} \to 0$ (a dust grain). What happens to $μ$ , $μ_{red}$ , and the COM?

Forecast: which quantity collapses to zero — the gravitational $μ$ or the reduced mass? Guess before computing.

$μ = G (m_{1} + m_{2}) \to G m_{1}$ (finite, nonzero). Why this step? The orbit's shape is set entirely by the big mass — the grain orbits normally.
$μ_{red} = \frac{m _{1} m _{2}}{m _{1} + m _{2}} \to \frac{m _{1} \cdot 0}{m _{1}} = 0$ . Why this step? A zero-mass particle carries zero internal kinetic energy — energy split degenerates.
COM: $R = \frac{m _{1} r _{1} + 0}{m _{1}} = r_{1}$ → the COM sits on the big body. Why this step? With no mass on the other side, the balance point is the heavy body itself; the "planet fixed at the focus" picture becomes exact.

Verify: numerically, at $m_{2} = 1 kg$ : $μ_{red} = \frac{1.99 \times 1 0 ^{30}}{1.99 \times 1 0 ^{30} + 1} \approx 1.0000 kg \to$ tracks $m_{2}$ , heading to $0$ as $m_{2} \to 0$ ✓. $μ \to G m_{1} = 1.328 \times 1 0^{20}$ , unchanged ✓.

Worked example Ex 7 — Cell G: real-world word problem (does the Sun move?)

Jupiter $m_{2} = 1.898 \times 1 0^{27} kg$ , Sun $m_{1} = 1.989 \times 1 0^{30} kg$ , mean separation $r = 7.78 \times 1 0^{11} m$ . How far does the Sun sit from the Sun–Jupiter barycenter, and is that inside or outside the Sun (radius $6.96 \times 1 0^{8}$ m)?

Forecast: guess — barycenter inside the Sun, or hovering in empty space above its surface?

Sun's offset from COM $= \frac{m _{2}}{M} r$ , $M = 1.989 \times 1 0^{30} + 1.898 \times 1 0^{27} = 1.9909 \times 1 0^{30} kg$ . Why this step? Same undo-formula offset as Ex 3, now with the giant planet.
$\frac{m _{2}}{M} = \frac{1.898 \times 1 0 ^{27}}{1.9909 \times 1 0 ^{30}} = 9.533 \times 1 0^{- 4}$ . Why this step? The Sun's wobble is driven by Jupiter's share of the total mass, so we form the planet-to-total ratio $m_{2} / M$ first — a pure number — then scale it by $r$ ; forming it separately also reveals the wobble is only $\approx 0.095%$ of the orbit size.
Offset $= 9.533 \times 1 0^{- 4} \times 7.78 \times 1 0^{11} = 7.417 \times 1 0^{8} m$ . Why this step? This is the physical wobble radius of the Sun.

Verify: $7.42 \times 1 0^{8} m > 6.96 \times 1 0^{8} m$ — the barycenter lies just outside the Sun's surface ✓. So the Sun genuinely orbits a point in space — the wobble astronomers use to detect exoplanets. This is Cell A's "planet fixed" intuition failing by a hair.

Worked example Ex 8 — Cell H: exam twist — recover both orbit radii from a ratio

A binary has mass ratio $m_{1} / m_{2} = 4$ and separation $r = 1.0 \times 1 0^{10} m$ . Without knowing individual masses, find each body's distance from the COM and confirm the heavier body is closer.

Forecast: which body orbits the smaller circle — the heavy one or the light one? Guess.

Let $m_{2} = 1$ unit $\Rightarrow m_{1} = 4$ , $M = 5$ . Why this step? Only the ratio enters the offsets, so we can pick convenient units.
Body 1 (heavy) distance $= \frac{m _{2}}{M} r = \frac{1}{5} \times 1 0^{10} = 2.0 \times 1 0^{9} m$ . Why this step? Body 1's offset carries the other mass fraction $m_{2} / M$ .
Body 2 (light) distance $= \frac{m _{1}}{M} r = \frac{4}{5} \times 1 0^{10} = 8.0 \times 1 0^{9} m$ . Why this step? Each body's offset is weighted by the other body's mass fraction (from the undo formulas: $r_{1}$ uses $m_{2} / M$ , $r_{2}$ uses $m_{1} / M$ ), so we swap the fractions when moving from body 1 to body 2 — this is exactly why the heavier partner pins the COM close and the lighter one swings wide.
Ratio of radii $= \frac{8 \times 1 0 ^{9}}{2 \times 1 0 ^{9}} = 4 = m_{1} / m_{2}$ : the light body swings 4× wider. Why this step? Distances from COM go inversely as the masses — heavier = closer.

Verify: radii sum $2 \times 1 0^{9} + 8 \times 1 0^{9} = 1.0 \times 1 0^{10} = r$ ✓. Heavy-body radius ( $2 \times 1 0^{9}$ ) < light-body radius ( $8 \times 1 0^{9}$ ) ✓, confirming "heavier stays closer to the COM."

Recall Which cell was which?

Match the scenario to its lesson. Big mass ratio ::: Cell A — $μ \approx G m_{1}$ , satellite mass invisible (Ex 1). Equal masses ::: Cell B — COM midway, each orbits at $r /2$ (Ex 2). One mass $\to 0$ ::: Cell F — $μ_{red} \to 0$ , COM sits on the big body (Ex 6). Sun wobbles ::: Cell G — barycenter just outside the Sun's surface (Ex 7). Distances from COM ::: go inversely as the masses — heavier is closer (Ex 8).

Connections

Parent topic — the derivation these examples exercise.
Center of mass frame — the coordinate trick behind every offset formula here.
Vis-viva equation — uses the internal energy term isolated in Ex 5.
Kepler's laws — the orbit shapes the reduced equation produces.
Conservation of angular momentum (central force) — why each orbit stays planar.
Reduced mass in molecular vibrations — same $μ_{red}$ from Ex 4 in diatomics.
Three-body problem — where Cell-style clean reduction stops working.