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Moving in a Spherically Symmetric Potential Field |
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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.,
where v is the tangential velocity. Let u be the radial velocity (dr/dt), Then
The angular momentum of the particle is mr²(dθ/dt). Angular momentum L is conserved so
Thus
If both sides of this equation are multiplied by (dr/dt) the result is equivalent to
Integrating from 0 to t gives
Thus ½mu(t)² − ½L²/r(t) + V(r(t)) is constant over time. Let its value be denoted as E. Then
Therefore
The quantity [E−V(r(t))] represents kinetic energy as a function of location. Let this be denoted as K(r(t)). Thus
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
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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