San José State University

applet-magic.com
Thayer Watkins
Silicon Valley
USA

 The Reduction of a Two-Body Central Force Problem to a One-Body Problem

## The Basic Problem

Let R1 and R2 be the three dimensional position vectors of two bodies of masses m1 and m2, respectively. (Symbols in red denote three dimensional vectors.)

The force between the two bodies is directed along the vector r=R1-R2 and its magnitude is a function of the distance between them |r|. The equations of motion for the two bodies are:

#### m1(d²R1/dt²) = - F = - f(|r|)(r/|r|) |   m2(d²R2/dt²) = +F = + f(|r|)(r/|r|)

where (r/|r|) denotes a unit vector in the direction of r.

## Center of Mass of the System

Let the position vector of the center of mass be denoted as R where R is given by

#### (m1+m2)R = m1R1 + m2R2

Adding the two equations of motion together shows that

#### (m1+m2)d²R/dt² = m1d²R1/dt² + m2d²R2/dt² = -F + F = 0because the force between the two bodies is equal in magnitude and oppositely directed.

Thus d²R/dt² = 0 and hence if dR/dt = 0 initially then the center of mass does not move.

## Movement Within the System

Let r1= R1-R and r2= R2-R. Then

And likewise

#### r2 = - (m1/(m1+m2)) r

This of course also means that

#### r2 = -(m1/m2)r1and m1r1 + m2r2 = 0

which could have been derived directly.

Thus r1, r2 and r are proportional to each other. Any one of them could be used for determining the internal motion of the system.

The equation of motion for r is given by

#### d²r/dt² = d²R1/dt² - d²R2/dt² = -(1/m1)F - (1/m2)F = -((1/m1 + 1/m2)F

This is equivalent to a body of mass ((1/m1+1/m2)-1=m1m2/(m1+m2) moving in a force field of f(|r|). The quantity m1m2/(m1+m2) is called the reduced mass mμ of the system. For m1=m2=m the reduced mass has the counterintuitive value of m/2. Another phrasing of the concept of reduced mass is that it is one half of the harmonic mean of the two masses.

If r1 is used as the variable of analysis then since r1 = (m2/(m1+m2))r

Thus

#### m1d²r1/dt² = -Fand likewise m2d²r2/dt² = +F

In contrast to the case with the use of r and the counterintuitive concept of reduced mass mμ the equations for r1 and r2 involve the appropriate masses, m1 and m2, respectively.

Note that the full equations of motion in terms of r1 and r2 are

#### m1d²r1/dt² = - f((1+m2/m1)|r1|)(r1/|r1|) and likewise m2d²r2/dt² = + f((1+m1/m2)|r2|)(r2/|r2|)

That is to say, in the force formula |r| must be replaced by (1+m2/m1)|r1| or (1+m1/m2)|r2|.

When a solution is found for r1(t) the solution for r2 is just −(m1/m2)r1(t).

## Summary :

In the reduction of a two-body problem to a one-body problem there are three approaches; i.e.,

• Retain r, the distance between the bodies, as the distance variable but utilize the reduced mass in the dynamic equation.
• Utilize the mass of the first body m1 but replace r by (1+m2/m1)r1
where r1 is the distance of the first body to the center of mass of the system.
• Utilize the mass of second body m2 but replace r by (1+m1/m2)r2
where r2 is the distance of the second body to the center of mass of the system.

Of course, the designation of first and second body is completely arbitrary.

## Illustrations of the Two-Body Systems

The operative equation for examining two-body systems is:

#### r2 = -(m1/m2)r1

If the Earth were the only satellite of the Sun and the Earth were moving in an elliptical orbit about the center of mass of the Sun-Earth system then the Sun would also be moving in an elliptical orbit about that same center of mass. The scales of the orbits would be in proportion to the ratio of the masses. The Sun is about 333 thousand times as massive as the Earth. Thus if the average radius of the Earth's orbit is 93 million miles (150 million km) then the radius of the Sun's counter orbit is about 280 miles (450 km). The center of mass of the Sun-Earth system is well within the body of the Sun.

When the orbits are elliptical the orbit is with respect to one of the foci of the ellipse. The counter orbit of the other body is with respect ot the opposite foci for its ellipse. This has to be the case so both bodies reach their apogees (maximum distances) and perigees (minimum distaces) at the same time.

For the Sun-Jupiter system the magnitude of the Sun's counter orbit is not so insignificant. The mass of the Sun is only about 1050 times as large as that of Jupiter. The radius of Jupiter's orbit around the center of mass of the two bodies is about 484 million miles so the radius of the Sun's counter orbit around their center of mass would be about 462 thousand miles. This is about 28 thousand miles above the surface of the Sun.

For the Earth-Moon system the statistics are that the ratio of the mass of the moon to that of the Earth is 0.0123 and therefore since the radius of the Moon's orbit is about 240 thousand miles (384 thousand km) the radius of the orbit that the Earth would move in with respect to the Earth-Moon center of mass if there were no other bodies in the solar system is about three thousand miles (4700 km). Since the radius of the Earth'radius is approximately four thousand miles the center of mass of the system is about a thousand miles below the surface of the Earth. Because the Earth is rotating that center of mass is not at a fixed location with respect to the Earth's features. In other words, the Earth rotates through that center of mass so it is under different points of the Earth's surface at different times of the day.

## Two-Body Systems with Masses Distributed Over Spherical Shells

For inverse distance squared attraction the effect of the distributed mass is the same as if the masses of the bodies were concentrated at its center of masses providing there is no overlap of their volumes. Thus the dynamics for the case of separated bodies would be the same as for point-masses bodies. However if one body is in the form of a shell and the other body is located within that shell there is no net attractive force on the body which is within the shell. This leads to more complicated dynamics. See the Distributed Two-Body Problem.