# PlanetPhysics/Motion in Central Force Field

Let us consider a body with mass $m$ in a gravitational force field exerted by the origin and directed always from the body towards the origin.\, Set the plane through the origin and the velocity vector ${\vec {v}}$ of the body.\, Apparently, the body is forced to move constantly in this plane, i.e. there is a question of a planar motion.\, We want to derive the trajectory of the body.
Equip the plane of the motion with a polar coordinate system $r,\,\varphi$ and denote the position vector of the body by ${\vec {r}}$.\, Then the velocity vector is
   ${\begin{matrix}{\vec {v}}\;=\;{\frac {d{\vec {r}}}{dt}}\;=\;{\frac {d}{dt}}(r{\vec {r}}^{\,0})\;=\;{\frac {dr}{dt}}{\vec {r}}^{\,0}+r{\frac {d\varphi }{dt}}{\vec {s}}^{\,0},\end{matrix}}$
where ${\vec {r}}^{\,0}$ and ${\vec {s}}^{\,0}$ are the unit vectors in the direction of ${\vec {r}}$ and of ${\vec {r}}$ rotated 90 degrees anticlockwise (${\vec {r}}^{\,0}={\vec {i}}\cos \varphi +{\vec {j}}\sin \varphi$,\, whence\, </math>\frac{\vec{r}^{\,0}}{dt} = (-\vec{i}\sin\varphi+\vec{j}\cos\varphi)\frac{d\varphi}{dt} = \frac{d\varphi}{dt}\vec{s}^{\,0}$).\,Thusthe[[../KineticEnergy/|kineticenergy]]ofthebodyis<math>E_{k}\;=\;{\frac {1}{2}}m\left|{\frac {d{\vec {r}}}{dt}}\right|^{2}\;=\;{\frac {1}{2}}m\left(\!\left({\frac {dr}{dt}}\right)^{2}\!+\!\left(r{\frac {d\varphi }{dt}}\right)^{2}\right)\!.$ Because the gravitational force on the body is exerted along the position vector, its moment is 0 and therefore the angular momentum ${\vec {L}}\;=\;{\vec {r}}\!\times \!m{\frac {d{\vec {r}}}{dt}}\;=\;mr^{2}{\frac {d\varphi }{dt}}{\vec {r}}^{\,0}\!\times \!{\vec {s}}^{\,0}$ of the body is constant; thus its magnitude is a constant, $mr^{2}{\frac {d\varphi }{dt}}\;=\;G,$ whence
   ${\begin{matrix}{\frac {d\varphi }{dt}}\;=\;{\frac {G}{mr^{2}}}.\end{matrix}}$
The central force\, ${\vec {F}}:=-{\frac {k}{r^{2}}}{\vec {r}}^{\,0}$\, (where $k$ is a constant) has the scalar potential \, $U(r)=-{\frac {k}{r}}$.\, Thus the total energy\, $E=E_{k}\!+\!U(r)$ of the body, which is constant, may be written $E\;=\;{\frac {1}{2}}m\!\left({\frac {dr}{dt}}\right)^{2}\!+{\frac {1}{2}}mr^{2}\!\left({\frac {G}{mr^{2}}}\right)^{2}\!-\!{\frac {k}{r}}\;=\;{\frac {m}{2}}\!\left({\frac {dr}{dt}}\right)^{2}\!+{\frac {G^{2}}{2mr^{2}}}\!-\!{\frac {k}{r}}.$ This equation may be revised to $\left({\frac {dr}{dt}}\right)^{2}\!+{\frac {G^{2}}{m^{2}r^{2}}}-{\frac {2k}{mr}}+{\frac {k^{2}}{G^{2}}}\;=\;{\frac {2E}{m}}+{\frac {k^{2}}{G^{2}}},$ i.e. $\left({\frac {dr}{dt}}\right)^{2}\!+\left({\frac {k}{G}}-{\frac {G}{mr}}\right)^{2}\;=\;q^{2}$ where $q\;:=\;{\sqrt {{\frac {2}{m}}\left(\!E\!+\!{\frac {mk^{2}}{2G^{2}}}\right)}}$ is a constant.\, We introduce still an auxiliary angle $\psi$ such that
   ${\begin{matrix}{\frac {k}{G}}-{\frac {G}{mr}}\;=\;q\cos \psi ,\quad {\frac {dr}{dt}}\;=\;q\sin \psi .\end{matrix}}$
Differentiation of the first of these equations implies ${\frac {G}{mr^{2}}}\cdot {\frac {dr}{dt}}\;=\;-q\sin \psi {\frac {d\psi }{dt}}\;=\;-{\frac {dr}{dt}}\cdot {\frac {d\psi }{dt}},$ whence, by (2), ${\frac {d\psi }{dt}}\;=\;-{\frac {G}{mr^{2}}}\;=\;-{\frac {d\varphi }{dt}}.$ This means that\, $\psi =C\!-\!\varphi$, where the constant $C$ is determined by the initial conditions.\, We can then solve $r$ from the first of the equations (3), obtaining
   ${\begin{matrix}r\;=\;{\frac {G^{2}}{km\left(1-{\frac {Gq}{k}}\cos(C-\varphi )\right)}}\;=\;{\frac {p}{1-\varepsilon \cos(\varphi -C)}},\end{matrix}}$
where $p\;:=\;{\frac {G^{2}}{km}},\quad \varepsilon \;:=\;{\frac {Gq}{k}}.$\\
The result (4) shows that the trajectory of the body in the gravitational field of one point-like sink is always a conic section whose focus contains the sink causing the field.\\
As for the type of the conic, the most interesting one is an ellipse.\, It occurs when\, $\varepsilon <1$.\, This condition is easily seen to be equivalent with a negative total energy $E$ of the body.\\
One can say that any planet revolves around the Sun along an ellipse having the Sun in one of its foci --- this is Kepler's first law .