Centre-of-Mass and Relative Motion in Coupled Systems

Recommended resources

The freely available resources this lesson will direct you to.

Welcome back. Last lesson, you wrote the coupled equations for two bodies that attract one another. Each acceleration depends on the same separation, yet neither body is generally stationary. This lesson asks whether a better choice of coordinates can separate the motion of the system as a whole from the motion of the bodies relative to one another.

We will derive that change of coordinates, show why the centre of mass moves uniformly, and obtain a single gravitational equation for the separation vector. This reduction will let you understand and eventually test a two-body simulation without pretending that the heavier body is fixed.

Two ways to describe the same pair of bodies

Keep the inertial-frame positions $r_{1}$ and $r_{2}$ from last lesson, and define the relative position from body 1 to body 2 by

d = r_{2} - r_{1}, d = ∥ d ∥.

Now let $M = m_{1} + m_{2}$ and define the centre-of-mass position

R = \frac{m _{1} r _{1} + m _{2} r _{2}}{M} .

The mass-weighted average places $R$ between the bodies, nearer the more massive one. Both $R$ and $d$ are vectors, so we have replaced two three-dimensional position vectors with two different three-dimensional position vectors. No information has been discarded.

That sign convention is worth checking whenever you consult another derivation. The physics does not depend on which direction you choose for the relative vector, but every subsequent equation must use your choice consistently.

For a short visual overview, watch the coordinate change in the following video. Its presenter uses $r_{1} - r_{2}$ for the relative vector—the reverse of our $d$ —and states rather than works through part of the derivation. We will supply those steps below.

Project V Two-Body Problem (Part A) Lecture 58 Numerical Methods for Engineers

Watch “Project V Two-Body Problem (Part A)” by Jeffrey Chasnov for a visual account of why centre-of-mass and relative coordinates simplify the problem.

Watch the new coordinates for the two vector definitions, then the reduced equations for the results of the transformation. Focus on which equation describes overall drift and which still describes gravitational motion; remember that the video’s relative vector points opposite to ours.

To make the coordinate change useful, we must also be able to reverse it. Substitute $r_{2} = r_{1} + d$ into the definition of $R$ :

R = \frac{m _{1} r _{1} + m _{2} ( r _{1} + d )}{M} = r_{1} + \frac{m _{2}}{M} d .

Solving for the two original positions gives

r_{1} = R - \frac{m _{2}}{M} d, r_{2} = R + \frac{m _{1}}{M} d .

These formulas have a physical interpretation. Relative to the barycentre, the bodies lie on opposite sides of it, at distances inversely weighted by their masses. If $m_{1} = 4 m_{2}$ , body 1 is one-fifth of the separation from the barycentre, while body 2 is four-fifths away.

Because the masses are constant, differentiating also recovers the individual velocities:

v_{1} = R ˙ - \frac{m _{2}}{M} d ˙, v_{2} = R ˙ + \frac{m _{1}}{M} d ˙ .

Thus positions and velocities can be translated between the two descriptions.

The centre of mass does not feel internal gravity

Last lesson’s equations, in our chosen sign convention, were

r ¨_{1} = G m_{2} \frac{d}{d ^{3}}, r ¨_{2} = - G m_{1} \frac{d}{d ^{3}}, d > 0.

Differentiate the definition of $R$ twice and substitute:

R ¨ = \frac{m _{1} r ¨ _{1} + m _{2} r ¨ _{2}}{M} = \frac{1}{M} (G m_{1} m_{2} \frac{d}{d ^{3}} - G m_{1} m_{2} \frac{d}{d ^{3}}) = 0 .

The cancellation is Newton’s third law expressed in centre-of-mass coordinates: gravity transfers momentum between the two bodies but cannot accelerate their isolated system as a whole. Consequently,

R (t) = R_{0} + V_{0} (t - t_{0}), V_{0} = R ˙ (t_{0}) = \frac{m _{1} v _{1} ( t _{0} ) + m _{2} v _{2} ( t _{0} )}{M} .

“Moves uniformly” includes the special case $V_{0} = 0$ , when the barycentre stays put. It does not say that either body moves uniformly.

The coordinate system whose origin travels with $R (t)$ is a barycentric rest frame. Since its origin has constant velocity and its axes need not rotate, it is still an inertial frame. Choosing it sets $R = 0$ , but that is a convenient choice of frame, not a new force or a claim that one body is fixed.

The relative motion is a one-body problem

The acceleration of the separation vector is the difference between the individual accelerations:

d ¨ = r ¨_{2} - r ¨_{1} = - G m_{1} \frac{d}{d ^{3}} - G m_{2} \frac{d}{d ^{3}} .

Therefore,

d ¨ = - G (m_{1} + m_{2}) \frac{d}{∥ d ∥ ^{3}} .

The original coupled gravitational equations have separated into two equations with different jobs: $R ¨ = 0$ gives the barycentre’s straight-line motion, while the boxed equation determines how the separation changes. Notice the sum of the masses. Both bodies respond to gravity, so both contribute to the acceleration of their separation. Replacing $m_{1} + m_{2}$ with $m_{1}$ would silently return to an approximation in which body 1 hardly moves.

You can also write the relative equation in Newton’s familiar force-equals-mass form. Define the reduced mass

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

Multiplying the relative equation by $μ_{red}$ gives

μ_{red} d ¨ = - G m_{1} m_{2} \frac{d}{d ^{3}} .

Mathematically, this is the motion of one effective particle of mass $μ_{red}$ , at position $d$ , under an attractive central force. The effective particle is a device for representing relative motion; it is not a third physical body, and the origin of this relative-coordinate equation is not necessarily the physical barycentre.

For a compact written derivation to compare with ours, use the following notes. They call the bodies $M$ and $m$ , use the relative vector from $M$ to $m$ , and show both the centre-of-mass reconstruction and the reduced-mass equation.

Central Force Motion: Kepler's Laws

Read the opening two-body reduction in the MIT OpenCourseWare notes to check how the original positions are recovered and how subtracting the acceleration equations produces a single relative equation.

On p. 2, find the section “the two-body problem and Kepler’s problem.” Read the reduction, from its opening description through the paragraph following equation (6). Stop before the paragraph beginning “Alternatively.” Pay particular attention to equation (4), which reconstructs both bodies’ positions, and equations (5)–(6), which distinguish the total mass in the acceleration equation from the reduced mass in the force equation.

A useful boundary on this result: one-body does not mean one-dimensional. At this stage $d$ remains a vector in three-dimensional space. The simplification is that its gravitational equation can be solved without simultaneously solving an equation for $R$ . A later lesson will use angular momentum to establish when the relative trajectory lies in a plane.

Reconstructing a moving two-body system

Suppose you know the four initial vectors from last lesson. Compute the new initial data as

R_{0} = \frac{m _{1} r _{1} ( t _{0} ) + m _{2} r _{2} ( t _{0} )}{M}, V_{0} = \frac{m _{1} v _{1} ( t _{0} ) + m _{2} v _{2} ( t _{0} )}{M},

d_{0} = r_{2} (t_{0}) - r_{1} (t_{0}), u_{0} = v_{2} (t_{0}) - v_{1} (t_{0}) .

The first pair determines $R (t)$ explicitly. The second pair supplies initial conditions for $d ¨ = - GM d / d^{3}$ . Once $d (t)$ is known, recover the physical positions using

r_{1} (t) = R (t) - \frac{m _{2}}{M} d (t), r_{2} (t) = R (t) + \frac{m _{1}}{M} d (t) .

This reconstruction matters for simulation: a plot of $d (t)$ shows the relative orbit, whereas plots of $r_{1} (t)$ and $r_{2} (t)$ show the two bodies in the chosen inertial frame. If the initial total momentum is nonzero, their barycentre drifts even while their motion about it may look orbital.

For instance, with mass ratio $m_{1} : m_{2} = 4 : 1$ and $R = 0$ , a separation $d = (3, 0)$ gives $r_{1} = (- 0.6, 0)$ and $r_{2} = (2.4, 0)$ . Add any constant velocity $V_{0}$ to the barycentre, and both positions acquire the same straight-line drift $V_{0} (t - t_{0})$ ; their separation is unchanged. This is why a simulation can show two bodies travelling across the screen without any external force.

Key takeaways

Centre-of-mass and relative coordinates reorganise, rather than approximate, the isolated two-body model. Internal gravitational forces cancel from the centre-of-mass equation, so $R$ moves at constant velocity. Subtracting the two acceleration equations gives a single relative-motion equation with gravitational coefficient $G (m_{1} + m_{2})$ , or equivalently a one-body force equation with reduced mass $μ_{red} = m_{1} m_{2} / (m_{1} + m_{2})$ . The inverse coordinate formulas then recover both physical trajectories.

Next, we will examine angular momentum. It will give another constraint on the relative motion and explain why an isolated two-body orbit lies in a plane.