San José State University

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Thayer Watkins
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The Dynamics of a Particle
Moving in a Spherically
Symmetric Potential Field

Consider a particle of mass m moving in a potential V(r). The particle will follow a planar orbit determined by the plane containing its velocity vector and its position vector. The radial force on the particle is −V'(r). The net radial force on the particle is then the so-called centrifugal force less the potential gradient force; i.e.,

Fr = mv²/r − V'(r)

where v is the tangential velocity. Let u be the radial velocity (dr/dt), Then

m(du/dt) = mv²/r − V'(r)

The angular momentum of the particle is mr²(dθ/dt). Angular momentum L is conserved so

L = mr²(dθ/dt) = mvr

Thus

v = L/(mr)
and hence
m(du/dt) = m(L/(mr))²/r − V'(r)
= L²/(mr³) − V'(r)
or, equivalently
m(d²r/dt²) = L²/(mr³) − V'(r)

If both sides of this equation are multiplied by (dr/dt) the result is equivalent to

½m(dr/dt)² = −½L²(d(1/r²)/dt) − dV(r)/dt

Integrating from 0 to t gives

½mu(t)² − ½mu(0)² = ½L²[1/r(t) − 1/r(0)] − [V(r(t) − V(r(0)]
which can be rearranged as
½mu(t)² − ½L²/r(t) + V(r(t)) = ½mu(0)² − ½L²/r(0) + V(r(0)]

Thus ½mu(t)² − ½L²/r(t) + V(r(t)) is constant over time. Let its value be denoted as E. Then

½mu(t)² − ½L²/r(t) + V(r(t)) = E
and hence
u(t) = (dr/dt) = [2E + L²/r(t) − 2V(r(t))]½/m½

Therefore

dr/[(E + ½L²/r(t) − V(r(t))] = dt/(2m)½

The quantity [E−V(r(t))] represents kinetic energy as a function of location. Let this be denoted as K(r(t)). Thus

dr/(K(r(t)) + ½L²/r(t)) = dt/(2m)½

The integration of this expression from 0 to t gives a function F(r(t)) = t/(2m)½. The inversion of this function gives r(t)=G(t/(2m)½).

Having r(t) then gives

v(t) = L/(mr) = r(dθ/dt)
and hence
(dθ/dt) = L/(mr²)

Integration then gives θ(t). Thus the trajectory (r(t), θ(t)) is determined. Furthermore the velocity vector (u(t), v(t)) is determined and likewise w(t)=[u(t)²+v(t)²]½.


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