Two-Body Reduction and Relative Motion

Two-Body Reduction and Relative Motion

Many mechanics problems begin with two interacting particles:

m1r1	= F12,	(1)
m2r2	= F21.	(2)

If the interaction obeys Newton’s third law,

(3)

then the six scalar coordinates contained in r₁ and r₂ can be reorganized into two much more useful vectors:

• the center of mass coordinate R,
• the relative coordinate r.

Define

(4)

and

(5)

For an isolated two-body system, these coordinates separate the dynamics into

(6)

and

(7)

where

(8)

is the reduced mass.

Thus the internal two-body motion is dynamically equivalent to a single effective particle of mass μ moving in the relative coordinate.

1 Why change coordinates?

The original coordinates r₁ and r₂ mix two distinct motions:

• translation of the pair as a whole,
• motion of the particles relative to one another.

The center of mass and relative coordinates separate these roles.

The center of mass answers:

(9)

The relative coordinate answers:

(10)

For isolated systems, the first problem is trivial uniform motion. The second contains the actual interaction dynamics.

2 Definitions

Let

(11)

be the total mass.

The center of mass coordinate is

(12)

The relative coordinate is chosen as

(13)

Thus r points from particle 2 toward particle 1.

The choice of sign is conventional. If the opposite definition is used, all corresponding force and angular-momentum signs must be changed consistently.

3 Inverse coordinate transformation

Starting from

MR	= m1r1 + m2r2,	(14)
r	= r1 − r2,	(15)

solve for the original positions.

From

(16)

substitute into the center of mass equation:

MR	= m1(r + r2) + m2r2	(17)
	= m1r + Mr2.	(18)

Therefore

(19)

Then

(20)

These equations reconstruct the individual particle positions from R and r.

4 Distances from the center of mass

Define positions relative to the center of mass:

ρ1	= r1 − R,	(21)
ρ2	= r2 − R.	(22)

Using the inverse transformation,

ρ1 = r,		(23)
ρ2 = −r.		(24)

Their magnitudes satisfy

(25)

The more massive particle stays closer to the center of mass.

5 Velocity transformation

Differentiate the inverse position relations:

v1 = V + r,		(26)
v2 = V −r,		(27)

where

(28)

The relative velocity is

(29)

Thus r is the velocity of particle 1 relative to particle 2.

6 Total momentum

The total linear momentum is

P	= m1v1 + m2v2	(30)
	= m1 + m2.	(31)

The relative-velocity terms cancel:

(32)

Therefore the center of mass motion carries the total linear momentum.

7 The center of mass frame

In the center of mass frame,

(33)

Then

v1′	= r,	(34)
v2′	= −r.	(35)

The corresponding momenta are

p1′	= m1v1′ = r,	(36)
p2′	= m2v2′ = −r.	(37)

Define the reduced mass

(38)

Then

p1′ = μr,		(39)
p2′ = −μr.		(40)

This motivates the definition of the relative momentum

(41)

8 Reduced mass

The reduced mass is

(42)

It also satisfies

(43)

Several useful limits are:

8.1 Equal masses

If

(44)

then

(45)

8.2 One mass much larger

If

(46)

then

(47)

The lighter particle behaves approximately as though it were moving around a fixed heavy center.

8.3 Reduced mass is smaller than either mass

For finite positive masses,

(48)

9 Kinetic-energy decomposition

The total kinetic energy is

(49)

Substitute

v1	= V + r,	(50)
v2	= V −r.	(51)

Then

K

= m12 + m22.

(52)

The cross terms cancel. The result is

(53)

Thus

(54)

The two terms are

KCM = MV 2,		(55)
Krel = μ2.		(56)

10 Why the cross terms cancel

The cross terms from the two kinetic energies are

(57)

They are equal and opposite.

Therefore

(58)

This cancellation is the energy analogue of the cancellation of the relative-motion terms in total linear momentum.

11 Equations of motion without external forces

Let F₁₂ be the force on particle 1 due to particle 2.

Then

m1r1	= F12,	(59)
m2r2	= −F12.	(60)

Add the two equations:

(61)

Since

(62)

we obtain

(63)

Thus

(64)

12 Relative equation of motion

Subtract the particle accelerations:

r	= r1 −r2	(65)
	= + .	(66)

Factor:

(67)

Using

(68)

we obtain

(69)

This is the essential two-body reduction.

Two interacting particles have become one effective particle of mass μ moving in the relative coordinate.

13 External forces

If external forces also act,

m1r1	= F12 + F1,ext,	(70)
m2r2	= −F12 + F2,ext.	(71)

Adding gives

(72)

Subtracting gives

(73)

Thus external forces can affect both:

• center of mass translation,
• relative motion.

If both particles experience the same external acceleration,

(74)

then the external field does not directly alter the relative equation.

A spatially uniform gravitational field is an important example.

14 Potential energy depending only on separation

Suppose the internal interaction is conservative and depends only on

(75)

Then

(76)

The total mechanical energy is

(77)

Using the kinetic-energy decomposition,

(78)

Define

(79)

and

(80)

For an isolated conservative system, both parts are independently constant because the center of mass and relative motions separate.

15 Angular-momentum decomposition for two particles

The total angular momentum about an origin is

(81)

For a two-particle system,

(82)

Substitute

ρ1	= r,	v1′	= r,	(83)
ρ2	= −r,	v2′	= −r.	(84)

After collecting terms,

(85)

Therefore the relative motion has the angular momentum of an effective particle of mass μ.

Define

(86)

16 Central interaction

If the internal force is central,

(87)

then

(88)

The relative angular momentum is therefore conserved:

(89)

In planar polar coordinates,

(90)

The relative orbit therefore obeys exactly the same central-force structure developed in M04-08, with the replacement

(91)

17 Relative effective potential

For a conservative central interaction,

(92)

Define

(93)

Then

(94)

Thus all of the turning-point, circular-orbit, and stability methods from M04-08 apply directly to the relative coordinate.

18 Newtonian gravitational two-body problem

For two point masses,

(95)

The relative equation is

(96)

Use

(97)

Divide by μ:

(98)

Equivalently,

(99)

The relative coordinate therefore moves exactly like a test particle in a Kepler problem with gravitational parameter

(100)

This is one of the most useful results in celestial mechanics.

19 Gravitational relative energy

The gravitational potential energy is

(101)

Thus

(102)

Because

(103)

we can divide by μ and define the specific relative energy

(104)

This is the familiar specific orbital energy written for the relative orbit.

20 Circular gravitational two-body motion

For a circular relative orbit of constant separation r,

(105)

The relative equation gives

(106)

Therefore

(107)

Since

(108)

we obtain

(109)

The orbital period is

(110)

so

(111)

For the two-body problem, the mass entering the gravitational parameter is the total mass m₁ + m₂.

21 Barycentric radii and speeds

For circular motion about the center of mass,

ρ1	= r,	(112)
ρ2	= r.	(113)

Both particles have the same angular speed ω.

Their speeds are

v1′	= ωρ1 = vrel,	(114)
v2′	= ωρ2 = vrel.	(115)

Thus

(116)

in magnitude, consistent with zero total momentum in the center of mass frame.

22 Example 1: equal masses

Let

(117)

Then

(118)

and

(119)

The center of mass lies halfway between the particles:

ρ1	= r,	(120)
ρ2	= −r.	(121)

In a circular gravitational orbit of separation r,

(122)

Each body moves on a circle of radius

(123)

about the barycenter.

23 Example 2: large mass-ratio limit

Let

(124)

Then

(125)

and

μ	=	(126)
	= m1	(127)
	≈ 0.990m1.	(128)

Thus

(129)

The center of mass lies close to the heavy body:

ρ2	= r	(130)
	= r.	(131)

The familiar “light particle orbiting a fixed heavy body” model is therefore the large mass-ratio limit of the exact two-body problem.

24 Mapping the relative orbit back to the individual orbits

Suppose the relative coordinate traces a curve

(132)

The particle motions about the center of mass are

ρ1(t) = r(t),		(133)
ρ2(t) = −r(t).		(134)

Thus both individual barycentric trajectories have the same shape as the relative orbit, scaled by different factors and located on opposite sides of the center of mass.

25 Example 3: reconstructing positions

Let

m1	= 2 kg,	(135)
m2	= 3 kg,	(136)

so

(137)

At one instant, suppose

(138)

and

(139)

Then

r1	= R + r	(140)
	= (4ex + ey) + (3ex − 1.2ey)	(141)
	= (7ex − 0.2ey) m.	(142)

Similarly,

r2	= R −r	(143)
	= (4ex + ey) − (2ex − 0.8ey)	(144)
	= (2ex + 1.8ey) m.	(145)

The difference is

(146)

as required.

26 Example 4: relative kinetic energy

Two particles have

m1	= 2 kg,	(147)
m2	= 6 kg,	(148)

and relative speed

(149)

The reduced mass is

μ	=	(150)
	= 1.5 kg.	(151)

Therefore

(152)

Thus

(153)

27 Connection to collision mechanics

The same reduced mass appeared in one-dimensional collision energy:

(154)

This is not a coincidence.

Collision mechanics and orbital mechanics both involve the same decomposition:

• center of mass motion,
• relative motion.

In a collision, the relative kinetic energy can be redistributed or dissipated.

In a conservative central-force problem, the relative energy evolves between kinetic and potential forms while remaining constant.

28 Connection to scattering

In the center of mass frame,

(155)

A two-body scattering event can therefore be analyzed as the deflection of the relative momentum

(156)

by an interaction potential U(r).

This reduction is the starting point for classical scattering theory.

29 Common mistakes

1. Confusing the center of mass coordinate R with the relative coordinate r.
2. Forgetting which direction the chosen relative vector points.
3. Using m₁ + m₂ instead of the reduced mass in the relative kinetic energy.
4. Using either physical mass directly in μr = F₁₂.
5. Forgetting that r = v₁ − v₂.
6. Assuming the center of mass is fixed in every inertial frame. It is fixed only in the center of mass frame for an isolated system.
7. Forgetting that the more massive particle lies closer to the center of mass.
8. Treating the relative orbit as identical in scale to either barycentric orbit.
9. Forgetting that G(m₁ + m₂), not merely Gm₂, governs the exact relative gravitational motion.
10. Using the single-particle angular momentum mr²𝜃 instead of μr²𝜃 for genuine two-body relative motion.
11. Mixing laboratory-frame total energy with center of mass-frame relative energy.
12. Assuming uniform external gravity alters the relative motion. A perfectly uniform gravitational field accelerates both masses equally.
13. Forgetting that tidal or otherwise nonuniform external forces can alter relative motion.

30 Practice exercises

1. Starting from R = (m₁r₁ + m₂r₂)∕M and r = r₁ − r₂, derive the inverse coordinate transformation.
2. Show that r = v₁ − v₂.
3. Derive P = MV_CM using the transformed velocities.
4. Show that in the center of mass frame,

   (157)

5. Derive the identity 1∕μ = 1∕m₁ + 1∕m₂.
6. Prove the kinetic-energy decomposition

   (158)

7. Derive MR = 0 for an isolated two-body system.
8. Derive the reduced relative equation μr = F₁₂.
9. Derive the external-force relative equation

   (159)

10. Derive L_CM = μr ×r.
11. For a central interaction, show that ℓ = μr²𝜃 is conserved.
12. Derive the relative effective potential for a conservative central force.
13. For Newtonian gravity, derive

    (160)

14. Derive the circular two-body result

    (161)

15. Show how a known relative orbit r(t) maps into the two individual barycentric orbits.

31 Summary

Define

(162)

with

(163)

The inverse transformation is

r1 = R + r,		(164)
r2 = R −r.		(165)

The reduced mass is

(166)

For an isolated two-body system,

MR = 0,		(167)
μr = F12.		(168)

The kinetic energy separates as

(169)

For a central interaction,

(170)

is conserved.

For Newtonian gravity,

(171)

The two-body problem has therefore been reduced to center of mass translation plus a one-particle relative-motion problem.

References

References