11.5: Differential Orbit Equation

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The differential orbit equation relates the shape of the orbital motion, in plane polar coordinates, to the radial dependence of the two-body central force. A Binet coordinate transformation, which depends on the functional form of $\mathbf{F}(\mathbf{r}),$ can simplify the differential orbit equation. For the inverse-square law force, the best Binet transformed variable is $u$ which is defined to be

\[u\equiv \frac{1}{r}\]

Inserting the transformed variable $u$ into equation $(11.4.2)$ gives

\[\dot{\psi}=\frac{lu^{2}}{\mu }\]

From the definition of the new variable

\[\frac{dr}{dt}=-u^{-2}\frac{du}{dt}=-u^{-2}\frac{du}{d\psi }\dot{\psi}=-\frac{ l}{\mu }\frac{du}{d\psi }\]

Differentiating again gives

\[\frac{d^{2}r}{dt^{2}}=-\frac{l}{\mu }\frac{d}{dt}\left( \frac{du}{d\psi } \right) =-\left( \frac{lu}{\mu }\right) ^{2}\frac{d^{2}u}{d\psi ^{2}}\]

Substituting these into Lagrange’s radial equation of motion gives

\[\frac{d^{2}u}{d\psi ^{2}}+u=-\frac{\mu }{l^{2}}\frac{1}{u^{2}}F(\frac{1}{u}) \label{11.39}\]

Binet’s differential orbit equation directly relates $\psi$ and $r$ which determines the overall shape of the orbit trajectory. This shape is crucial for understanding the orbital motion of two bodies interacting via a two-body central force. Note that for the special case of an inverse square-law force, that is where $F(\frac{1}{u})=ku^{2}$, then the right-hand side of Equation \ref{11.39} equals a constant $-\frac{\mu k}{l^{2}}$ since the orbital angular momentum is a conserved quantity.

Example $\PageIndex{1}$: Central force leading to a circular orbit $r = 2R\cos \theta$

9.5.1.PNG — Figure $\PageIndex{1}$: Circular trajectory passing through the origin of the central force.

Binet’s differential orbit equation can be used to derive the central potential that leads to the assumed circular trajectory of $r=2R\cos \theta$ where $R$ is the radius of the circular orbit. Note that this circular orbit passes through the origin of the central force when $r=2R\cos \theta =0$

Inserting this trajectory into Binet’s differential orbit Equation \ref{11.39} gives

\[\frac{1}{2R}\frac{d^{2}\left( \cos \theta \right) ^{-1}}{d\theta ^{2}}+\frac{ 1}{2R}\left( \cos \theta \right) ^{-1}=-\frac{\mu }{l^{2}}4R^{2}\left( \cos \theta \right) ^{2}F(\frac{1}{u}) \tag{$\alpha $}\]

Note that the differential is given by

\[\frac{d^{2}\left( \cos \theta \right) ^{-1}}{d\theta ^{2}}=\frac{d}{d\theta } \left( \frac{\sin \theta }{\cos ^{3}\theta }\right) =\frac{2\sin ^{2}\theta }{\cos ^{3}\theta }+\frac{1}{\cos \theta }\notag\]

Inserting this differential into equation $\alpha$ gives \[\frac{2\sin ^{2}\theta }{\cos ^{3}\theta }+\frac{1}{\cos \theta }+\frac{1}{ \cos \theta }=\frac{2}{\cos ^{3}\theta }=-\frac{\mu }{l^{2}}8R^{3}\left( \cos \theta \right) ^{2}F(\frac{1}{u})\notag\]

Thus the radial dependence of the required central force is

\[F=-\frac{l^{2}}{8R^{3}\mu }\frac{2}{\cos ^{5}\theta }=-\frac{8R^{2}l^{2}}{ \mu }\frac{1}{r^{5}}=-\frac{k}{r^{5}}\notag\]

This corresponds to an attractive central force that depends to the fifth power on the inverse radius $\mathit{r}$. Note that this example is unrealistic since the assumed orbit implies that the potential and kinetic energies are infinite when $r\rightarrow 0$ at $\theta \rightarrow \frac{\pi }{2}$.