Two-body problem — equations of motion, reduction to one-body: Question bank

Before we start, one reminder so no symbol here is unearned. We have two point masses: call the first mass $m_{1}$ (say the heavier body, e.g. Earth or a star) and the second $m_{2}$ (the lighter one, e.g. a satellite or a companion star). Both are just numbers in kilograms.

$r_{1}, r_{2}$ = position vectors of the two masses $m_{1}$ and $m_{2}$ .
$r = r_{2} - r_{1}$ = relative position (arrow from body 1 to body 2), $r = ∣ r ∣$ .
Center of mass (COM) = the mass-weighted average point $R = \frac{m _{1} r _{1} + m _{2} r _{2}}{M}$ , with total mass $M = m_{1} + m_{2}$ . From here on we abbreviate "center of mass" as COM.
$μ = G (m_{1} + m_{2})$ = gravitational parameter (a sum, units $m^{3} / s^{2}$ ).
$μ_{red} = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$ = reduced mass (a product-over-sum, units kg). To avoid clashing with the gravitational parameter $μ$ , this page always writes the reduced mass as $μ_{red}$ — never a bare $μ$ .
A dot means "rate of change in time": $\dot{r}$ = velocity, $\ddot{r}$ = acceleration.

The figure below fixes the geometry of these objects — keep it in view while you work the items.

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

And this second figure shows the two individual orbits and the free drift of the COM, so the "scaled copies" language later has a picture behind it.

True or false — justify

The center of mass (COM) of an isolated two-body system always moves in a straight line at constant speed.

True. Adding the two Newton equations, the internal gravity cancels (

m_{1} \ddot{r}_{1} + m_{2} \ddot{r}_{2} = 0

), so

\ddot{R} = 0

— zero acceleration means straight, uniform drift.

The gravitational parameter for a satellite around Earth is exactly

G M_{\oplus}

False. It is exactly

μ = G (m_{\oplus} + m_{sat})

;

G M_{\oplus}

is an excellent approximation only because

m_{sat} ≪ m_{\oplus}

. For comparable masses it would be wrong.

The reduced equation

\ddot{r} = - μ r / r^{3}

describes a body sitting at the actual center of mass.

False. It describes a fictitious particle at the tip of

r

orbiting a fixed center that produces parameter

μ

. Neither real body sits there; the COM is a different point entirely.

m_{2} \to 0

, the relative motion parameter

μ

becomes

G m_{1}

True.

μ = G (m_{1} + m_{2}) \to G m_{1}

, recovering the "test particle around a fixed mass" limit — the light body has no back-reaction.

For two equal masses, the reduced mass equals half of one mass.

True.

μ_{red} = \frac{m \cdot m}{m + m} = \frac{m}{2}

. The product-over-sum shrinks the effective mass toward the smaller body.

The heavier body orbits farther from the center of mass than the lighter one.

False. The heavier body sits closer:

r_{1} - R = - \frac{m _{2}}{M} r

, so its distance scales with the other mass. Big mass, small swing.

The two symbols

μ

and

μ_{red}

have the same units.

False.

μ = G (m_{1} + m_{2})

has units

m^{3} / s^{2}

;

μ_{red}

is a mass in kg. Same-looking letters, totally different objects — which is exactly why we keep the subscript.

The kinetic energy of the system is

\frac{1}{2} μ_{red} \dot{r}^{2}

once we ignore the drifting COM.

True.

T = \frac{1}{2} M \dot{R}^{2} + \frac{1}{2} μ_{red} \dot{r}^{2}

; dropping the COM term leaves the internal (orbital) energy carried by

μ_{red}

The reduced-mass one-body reduction works for any central force between the two bodies, not just an inverse-square attraction.

True. The derivation only used "force directed along

\hat{r}

, equal and opposite" (Newton's third law). Any central force

F (r) \hat{r}

reduces the same way — inverse-square is just the special gravitational case; the identical algebra gives the reduced mass in Reduced mass in molecular vibrations for a spring-like force.

Once reduced, a three-body system also collapses to one clean one-body equation.

False. The trick relies on internal forces cancelling pairwise with only one relative vector. With three mutual pulls the coordinates no longer decouple — see Three-body problem.

Spot the error

"Since Earth barely moves, only the satellite accelerates, so

μ = G m_{sat}

Two errors: Earth does accelerate (tiny but nonzero), and

μ

uses the sum

G (m_{\oplus} + m_{sat}) \approx G M_{\oplus}

. Using the satellite's mass alone is backwards.

"The center of mass is pulled inward by gravity, so it accelerates toward the orbit."

Internal forces cannot move the COM — they cancel in the sum. Only an external force could accelerate

R

; here there is none, so

\ddot{R} = 0

r = r_{1} - r_{2}

points from body 2 to body 1, but that doesn't change the physics."

The definition fixes the sign of every downstream formula. With

r = r_{2} - r_{1}

the force on the reduced particle is attractive (

- \hat{r}

); flip it and you must flip all the reconstruction coefficients too, or you'll place bodies on the wrong sides.

"In

μ_{red} \ddot{r} = F

the force is

- \frac{G ( m _{1} + m _{2} )}{r ^{2}} \hat{r}

No — the physical force is

F = - \frac{G m _{1} m _{2}}{r ^{2}} \hat{r}

. The extra

(m_{1} + m_{2})

belongs to

\ddot{r} = - μ r / r^{3}

; multiplying by

μ_{red}

cancels it back down to

G m_{1} m_{2}

"Because both masses appear in

μ

, doubling both masses leaves the orbit period unchanged."

Wrong.

μ = G (m_{1} + m_{2})

doubles, so the relative acceleration doubles and the orbit period shortens (

T \propto μ^{- 1/2}

at fixed size, via Kepler's laws).

"The relative orbit and each body's individual orbit have the same size."

They are scaled copies. Body 1's orbit has radius

\frac{m _{2}}{M}

times the relative orbit, body 2's has

\frac{m _{1}}{M}

times — both smaller than

r

's trace unless the other mass is zero.

Why questions

Why do we bother subtracting the two Newton equations instead of solving them directly?

Subtracting collapses the two coupled acceleration equations into one equation for

r

alone, turning a 6-variable mess into a solvable one-body problem.

Why does adding the equations give the center-of-mass motion, while subtracting gives the relative (orbital) motion?

Add → the equal-and-opposite forces cancel → force-free center-of-mass motion (uniform drift). Subtract → the forces reinforce with a factor

(m_{1} + m_{2})

→ the relative motion (the actual orbit). The two combinations decouple the system into center-of-mass motion plus relative motion.

Why must the reduced mass be smaller than either individual mass?

μ_{red} = \frac{m _{1} m _{2}}{m _{1} + m _{2}} < min (m_{1}, m_{2})

always, because dividing the product by the (larger) sum shrinks it. Physically, sharing the acceleration between two movable bodies makes the pair "respond" like something lighter than either.

Why is splitting energy into center-of-mass part + relative part legitimate?

Because

R

and

r

are independent coordinates whose cross-terms vanish, the kinetic energy decouples:

\frac{1}{2} M \dot{R}^{2} + \frac{1}{2} μ_{red} \dot{r}^{2}

. That lets us solve the orbit while ignoring the center-of-mass drift entirely — the key idea behind the Center of mass frame.

Why does the reduced problem conserve angular momentum, keeping the orbit in a plane?

The force lies along

\hat{r}

(central), so it exerts no torque about the center — see Conservation of angular momentum (central force). Constant angular momentum vector pins the motion to a single plane.

Why does the vis-viva relation use

μ = G (m_{1} + m_{2})

and not just

GM

Vis-viva equation is derived from the reduced equation of motion, whose parameter is the full sum

μ

. Only when one mass dominates does it simplify to the familiar

GM

form.

Edge cases

What happens to the reduction when

m_{2} = 0

exactly?

μ_{red} = 0

and

μ = G m_{1}

: the reduced equation still holds, but body 2 is a massless test particle that traces the relative orbit exactly while body 1 stays fixed at the (now coincident) COM.

What if the two masses are equal — where does each body's orbit centre sit?

The COM lies exactly midway, and each star orbits it on a circle of radius

r /2

— mirror-image scaled copies of the relative orbit of size

r

What if the initial relative velocity is zero (bodies momentarily at rest)?

The relative angular momentum is zero, so

r

has no sideways motion — the bodies fall straight toward each other along the line joining them (a degenerate radial "orbit"), while the COM keeps its own drift.

What if

r \to 0

(bodies collide)?

The force

\propto 1/ r^{2}

blows up — the point-mass model breaks down. Real bodies have finite size, so before

r = 0

they touch; the idealized equation only holds while they stay separated.

What happens to

μ_{red}

as one mass grows without bound (

m_{1} \to \infty

μ_{red} = \frac{m _{1} m _{2}}{m _{1} + m _{2}} \to m_{2}

. The reduced mass approaches the lighter body's mass, matching the intuition that a very heavy partner acts like a fixed anchor.

Can this two-body reduction be applied inside a three-body system by treating a close pair as one object?

Only approximately, and only if the third body is far away. The tight pair can be modelled as a single mass at its COM, but true reduction fails once the third pull is comparable — the hallmark of the Three-body problem.

Connections

Parent: full derivation
Kepler's laws · Vis-viva equation · Conservation of angular momentum (central force)
Center of mass frame · Reduced mass in molecular vibrations · Three-body problem