Mechanics

Celestial Mechanics

Two-Body Problem

The two-body problem considers two rigid point masses in mutual
orbit about each other. To determine the motion of these bodies,
first find the vector equations of motion. Given two bodies with
masses and , let be the vector from the center of mass to and be the vector from the center of mass to . From the definition of center of
mass

(1)

Define

(2)

(3)

where is the so-called reduced
mass. The vector displacement from to is

(4)

The distance between the two mutually orbiting bodies is

(5)

Equations (1)-(5) lead to the identities

(6)

(7)

so

(8)

(9)

In terms of r,

(10)

(11)

The differential equations of motion are then

(12)

(13)

Plugging (10)-(11) into (12)-(13),

(14)

(15)

or

(16)

The problem therefore reduces to the one-body problem of a
massless test particle about a mass M. When (16) is solved
for r, and are found from equations (12)-(13). Equation
(16) is a second-order ordinary
differential equation, so there are six constants of motion. Three of them
correspond to the vector constant known as angular
momentum. Define the angular momentum per unit mass

(17)

Upon taking the derivative, it therefore follows that

(18)

Therefore, h is conserved. Since h is
perpendicular to the plane determined by the position vector
r and the velocity vector , the motion is confined to a plane. Two
parameters are needed to specify the orientation of the plane (whose
normal is parallel to h) and one to describe the magnitude of
h. So, switch to polar
coordinates in the plane of motion, where is the angle from an arbitrary reference
direction

(19)

(20)

Since h is constant, (20) is equivalent to Kepler's
third law of equal areas in equal times

(21)

(22)

Differentiating (20), the following expression is obtained

(23)

Since , it follows that

(24)

Now, find in polar
coordinates and plug into (16),

(25)

(26)

Express in terms of h using (20)

(27)

Equation (27) is a second-order differential equation, and
therefore contains two constants of integration. One of these is the
orbital energy. Equation (27) can be solved for r(t)
by multiplying both sides by

(28)

and noting that this is equal to

(29)

(30)

Define orbital energy as the sum of kinetic
energy and gravitational
potential energy,

(31)

then

(32)

so (30) gives

(33)

A second way to show (33) plugs (16) into

(34)

to obtain

(35)

so E is also a conserved quantity. This determined
one of the constants for the second-order differential equation
(28). Conservation
of energy can be used to obtain the velocity at any point in the
orbit. The velocity at any point in the orbit may be derived from
the energy equation.

We thus have described seven constants of motion: the scalars
a, e, E, , and the vector h. However, a,
e, and are related and E may be evaluated at
the periapsis
of the particle's orbit, where to relate a and E.

(36)

Therefore, only five of the six parameters are known. Note
that (36) implies that energy for bound orbits (with a < 0) and
energy for unbound orbits (with a > 0).
The last conserved quantity may be found starting with expression
(16)

(37)

(38)

but

(39)

so

(40)

Now, observe that

(41)

(42)

Therefore, (40) can be written as

(43)

(44)

Define a vector

(45)

known as the Laplace-Runge-Lenz
vector. Although it has not been shown here, it contains the one
additional constant of motion. However, since it is a vector,
A must be related to other constants by two equations. One is
given by the orthogonality of A and h,

(46)

The other may be found by observing that

(47)

(48)

where the constant A is related to the eccentricity
by

(49)

In order to express , note that the angular
velocity may be simply expressed using

(50)

(51)

and plugging in for , and the vis-viva
equation for

(52)

Using the relationship

(53)

where p is the semilatus
rectum, this becomes

(54)

A closed-form time-independent solution to the equation of motion
uses the following substitution

(55)

(56)

Plugging this in,

(57)

(58)

The solution is

(59)

(60)

where B and are arbitrary constants. Now, define the
eccentricity and focal parameter by

(61)

(62)

Equation (60) is then identifiable as the polar equation
for a conic section, so

(63)

where a is the semimajor
axis , e is the eccentricity
and is the argument
of pericenter.

Therefore, there are 10 constants of motion known: the scalars
a, e, E, , and the vectors A and h.
However, these quantities are related by the four relations

(64)

(65)

(66)

(67)

so there are six constants of motion in all.

The velocity may be found from the energy equation,

(68)

(69)

This equation is known as the vis-viva
equation.

To obtain closed-form solutions for r(t) and , rewrite (32) as

(70)

and rearrange

(71)

(72)

t(r) is therefore given by

(73)

Let

(74)

(75)

(76)

(77)

(78)

(79)

then

(80)

Similarly, it is possible to solve for by brute force as follows. From (20)

(81)

so

(82)

Defining

(83)

then

(84)

The integrals of motion have different solutions depending on the
value of the eccentricity
e, as summarized in the following table.

eccentricity

orbit type

0

circular
orbit

elliptical
orbit

1

parabolic
orbit

e > 1

hyperbolic
orbit

Circular
Orbit, Elliptical
Orbit, Hyperbolic
Orbit, Kepler
Problem, Many-Body
Problem, n-Body
Problem, Orbit,
Parabolic
Orbit, Relativistic
Two-Body Problem, Restricted
Three-Body Problem, Three-Body
Problem

References

Basdevant, J.-L. and Dalibard, J. "The Two-Body Problem." §9.1 in
The
Quantum Mechanics Solver: How to Apply Quantum Theory to Modern
Physics. Berlin: Springer-Verlag, p. 61, 2000.

© Eric W. Weisstein