Differential equations,
non-linear oscillations
and celestial mechanics

The gravity inverse square law was deducted from the Third Kepler law: only then for circular orbits squares of periods are proportional to cubes of radii. This was observed by Isaac Newton and Robert Hooke but while Hooke announced this observation, Newton was working towards justifying all Kepler laws for arbitrary orbits.

We consider a massive star and a massless planet but actually this works for two objects of arbitrary masses (but not for three!)

Central forces and conservation laws

Consider first motion in the cenral field \begin{equation} \mathbf{r}'' = - V'( r) \mathbf{r}^0 \label{eq-1} \end{equation} where $r=|\mathbf{r}|$ is the distance to the origin, $\mathbf{r}^0=r^{-1}\mathbf{r}$ and $V( r)$ is the potential. Then there are two conservation laws: \begin{gather} \frac{1}{2}\mathbf{r}'^2+V( r)=E, \label{eq-2}\\ \mathbf{r}'\times \mathbf{r} =\mathbf{M}; \label{eq-3} \end{gather} (\ref{eq-2}) is an energy conservation law and (\ref{eq-3}) is an angular momentum conservation law.

Indeed, $2rr'=(r^2)'=(\mathbf{r}\cdot\mathbf{r})'= 2\mathbf{r}\cdot \mathbf{r}'$ and therefore $r'=\mathbf{r}^0\cdot\mathbf{r}'$; then \begin{equation*} E' = \mathbf{r}'\cdot \mathbf{r}''+ V'( r) r' = \mathbf{r}'\cdot \bigl(\mathbf{r}'' + V'( r) \mathbf{r}^0\bigr)=0 \end{equation*} due to (\ref{eq-1}) and also \begin{equation*} \mathbf{M}'= \mathbf{r}''\times \mathbf{r}+ \mathbf{r}'\times \mathbf{r}'=0 \end{equation*} where the second term vanishes due to properties of the cross-product and the first term is $-V'( r)\mathbf{r}^0\times \mathbf{r}$ due to (\ref{eq-1}) and it vanishes due to properties of the cross-product.

Then $\mathbf{r}$ is orthogonal to $\mathbf{M}$ and all the movement is in the plane orthogonal to $\mathbf{M}$.

Remark. Sure it is possible that $\mathbf{M}=0$ but then one can prove easily that all the movement is along a single line (ray actually). This is not interesting.

Using polar coordinates

Consider in the plane polar coordinates $(r,\theta)$. Then one can express $E$ and \begin{equation} E= \frac{1}{2}r'^2 + r^2\theta'^2+ V( r), \label{eq-4} \end{equation} and $\mathbf{M}=M\mathbf{k}$ where $\mathbf{k}$ is a unit vector orthogonal to the plane and \begin{equation} M= \theta ' r^2. \label{eq-5} \end{equation}

Exercise. Prove these equalities.

Remark. Preservation of $M$ is the Second Kepler Law as $\theta' r^2=S'$ where $S=S(t)$ is an area swept from some moment $t_0$ to $t$.

Finding $\theta'= Mr^{-2}$ from (\ref{eq-5}) and plugging into (\ref{eq-4}) we find that \begin{equation} E= \frac{1}{2}r'^2 + W( r), \text{ with } W( r):=\frac{1}{2}M^2 r^{-2}+ V( r). \label{eq-6} \end{equation} We call $W( r)$ reduced potential. Differentiating (\ref{eq-6}) we find $r'(r'' +W'( r))=0$ and therefore $r$ is described by non-linear oscillation equation \begin{equation} r'' = -W'( r). \label{eq-7} \end{equation}

Due to (\ref{eq-6}) \begin{equation} r'= \pm \sqrt{2\bigl(E-W( r)\bigr) } \label{eq-8} \end{equation} and $r$ oscillates beween two roots $r_1$ and $r_2$ of equation \begin{equation} W( r)=E \label{eq-9} \end{equation} at least as graph of $W( r)$ looks like on the picture below and $E<0$ (for $E\ge 0$ movement will be unbounded).

Since (\ref{eq-8}) implies $dt= \pm \frac{dr}{\sqrt{2\bigl(E-W( r)\bigr) }}$ we conclude that period is \begin{equation} T= 2\int_{r_1}^{r_2} \frac{dr}{\sqrt{2\bigl(E-W( r)\bigr) }} \label{eq-10} \end{equation} where we multiply by $2$ because movement goes from $r_1$ to $r_2$ and then back.

Also $d\theta =\theta' dt= Mr^{-2}dt$ and therefore \begin{equation} d\theta = \pm \frac{Mr^{-2}dr}{\sqrt{2\bigl(E-W( r)\bigr) }}. \label{eq-11} \end{equation}

Using gravity law

So far we have not used that $V( r)=-\frac{m}{r}$ (and $V'=\frac{m}{r^2}$) but now we use it: \begin{equation} d\theta = \pm \frac{Mr^{-2}dr}{\sqrt{ \bigl(2E+2mr^{-1} - M^2r^{-2}\bigr) }} \label{eq-12} \end{equation} and \begin{multline*} \theta = -\int \frac{Mr^{-2}dr}{\sqrt{ \bigl(2E+2mr^{-1} - M^2r^{-2}\bigr) }} \int \frac{dz}{\sqrt{ \bigl(E+2mM^{-1}z -z^2\bigr) }}= \\[3pt] =\arccos ((z- c_1)c_2^{-1}) \end{multline*} with $z=Mr^{-1}$, $c_2^2=2E+m^2M^{-2}$, $c_1=mM^{-1}$ (it must be positive) where we just include constant in the definition of $\theta$ and therefore $z=c_1+c_2\cos (\theta)$, \begin{equation} r= \frac{M}{c_1+c_2\cos(\theta)}=\frac{1}{d_1+d_2\cos(\theta)}. \label{eq-13} \end{equation}

Exercise. Prove that (\ref{eq-13}) describes an ellipse for $d_1>d_2$, hyperbola for $d_1<d_2$ and parabola for $d_1=d_2$; in all cases origin is one of the focuses (or the focus in the case of parabola). It is The First Kepler Law.

Remark. For any law different from $V\sim - r^{-1}$ (and $V\sim r^2$) most of the orbits are not periodic.

Calculating period

So (\ref{eq-10}) becomes \begin{equation*} T= 2\int_{r_1}^{r_2} \frac{dr}{\sqrt{\bigl(2E+ 2mM^{-1}r^{-1} -M^2r^{-2} \bigr) }}. \end{equation*} So \begin{equation*} T= 2\int_{r_1}^{r_2} \frac{r dr}{\sqrt{\bigl(-2E (r_2-r)(r-r_1) \bigr) }} \end{equation*} where we multiplied numerator and denominator by $r$ and used that denominator vanishes for $r=r_1$ and for $r=r_2$. Then \begin{equation*} T= 2\int_{r_1}^{r_2} \frac{r dr}{\sqrt{\bigl(-2E (d^2 -(r-a)^2) \bigr) }} \end{equation*} where $r_1+r_2=2a$ where $a$ is a large semiaxis of the ellipse and $r_2-r_1=2d$, $d=M/\sqrt{-2E}$, $-Ea=m$. Making substitution $r=a+x$ we get \begin{equation*} T= 2\int_{-d}^{d} \frac{(x+a)dx}{\sqrt{\bigl(-2E (d^2 -x^2) \bigr) }}=2\pi a/\sqrt{-2E}= \sqrt{2}\pi a^{\frac{3}{2}}m^{-\frac{1}{2}}. \end{equation*} So for any two orbits $T_1^2/T_2^2=a_1^3/a_2^3$ which is the Third Kepler Law.

Remark. Sure we compare orbits in the same stellar system as $m$ is proportional to the mass of the star.

Remark. $r_1$ is called a perihelion, $r_2$ is called an aphelion.

orbits