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 The General Characteristics of the Trajectories of Particles Moving in a Central Force Field

The astronomer Johannes Kepler found laws followed by the orbits of the planets of the solar system. Isaac Newton explained those laws as a result of the gravitation attraction between the Sun and the planets that obeyed an inverse distance squared formula. Some of Kepler's laws however are the result of a more general force formula. This material is to derive those properties of the trajectories of particles moving in a force field that depend only on the force acting centrally.

If the force on a particle depends only upon the distance r from the center of the force field then there is spherical symmentry to the system and angular momentum is conserved. Let r and v denote the radial vector and velocity vector, respectively, of the particle. (In the following vectors are depicted in red.) If m is the mass of the particle then its momentum p is equal to mv. The angular momentum M of the particle with respect to the center of the force field is given by

Since

#### r×M = r×(r×p) = 0and v×M = v×(r×p) = 0

r and p must always be perpendicular to M. If M0 then r and p must lie in the plane whose normals are parallel to M and passes through the center of the force field.

This means the analysis can be restricted to the plane. The origin of the coordinate system may be taken to be at the center of the force field. The trajectory of the particle can be represented by the radial vectors and these are completely described by their magnitude r and an angle θ.

Let r^ and θ^ be the unit vectors in the increasing r and θ directions, respectively. The velocity of the particle is then

#### v = (dr/dt)r^ + r(dθ/dt)θ^

Since r^ and θ^ are perpendicular the squared magnitude of velocity is

#### (dr/dt)² + r²(dθ/dt)²

This means that the kinetic energy K of the particle is

#### K = ½m[(dr/dt)² + r²(dθ/dt)²]

If the potential energy of the particle is denoted as V(r) then the total energy E of the particle is

## Angular Momentum and the Area Swept by the Radial Vector Along the Trajectory

The angular momentum of the particle with respect to the center of the force field is mr²(dθ/dt) and this remains constant; i.e.,

#### d(mr²(dθ/dt))/dt = 0

Since the mass m is constant this means that r²(dθ/dt) is also constant.

Now consider the area swept out by the radial vector as the particle moves, as shown below

Thus the rate at which area is swept by the radial vector is equal to r²(dθ/dt) which was found to be constant as a result of the conservation of angular moment. Therefore the rate at which area is swept out is everywhere constant along the trajectory of the particle as a result of the conservation of its angular momentum moving in a central force field. This is Kepler's Second Law of planetary motion.

## Obtaining the Trajectory Through Quadrature (Integration)

The angular momnetum of the particle with respect to the center of the force field is constant. Let this constant be denoted as L. Thus

#### Since E = K + V = ½m[(dr/dt)² + r²(dθ/dt)²] + V(r) this can be reduced to E = K + V = ½m[(dr/dt)² + r²(L/(mr²))²] + V(r) or, equivalently E = ½m(dr/dt)² + ½L²/(mr²) + V(r)

This latter equation can be solved for (dr/dt); i.e.,

#### (dr/dt) = {(2/m)[E - V(r) - ½L²/(mr²)]}½

Since (dθ/dt) = L/(mr²) this means (dt/dθ) = mr²/L. Therefore

#### (dr/dθ) = (dr/dt)(dt/dθ) = {(2/m)[E - V(r) - ½q²/(mr²)]}½(mr²/q) which can be represented as dθ = dr(mr²/L)/{(2/m)[E - V(r) - ½L²/(mr²)]}½

Given the formula for V(r) the above equation may be integrated to give θ in terms of r and hence, in principle, r in terms of θ. This would be the equation of the orbit of the particle.

The integration can be carried out only over the range in which the expression in the denominator is positive. The limits of integration are given as the solutions to the equation

#### E - V(r) - ½L²/(mr²) = 0

These solutions would be the distances for the perihelion and aphelion points of the orbit.

The coordinate system can be oriented so that θ=0 corresponds to the smallest root of the above equation. When V(r)=−G/r the above equation is a quadratic equation in (1/r) and is easily solved analytically. When

#### V(r)=−∫r∞(He-λz/z²)dz

the equation is transcendental and has to be solved numerically.

From the largest solution for r, say rP, from the above equation other characteristics of the orbit may be obtained. For example,

#### ωP = (dθ/dt)P = L/(mr²)

This means that the kinetic energy KP may be computed

#### KP = ½mr²ω² or KP = E − V(rP)

It might have been noted that E−V is just the kinetic energy K. At the maximum and minimum r, dr/dt is equal to 0 so K=½mr²(dθ/dt)². Thus the equation to solved reduces to

#### ½mr²ω² − ½L²/(mr²) = 0 or, equivalently m²r4ω² = L²

Although angular momentum L is equal to mr²ω this does not lead to a solution. This is because the above equation is equivalent to:

#### L² = L²

(To be continued.)