Kepler Problem

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The Kepler problem is a two-body problem that describes planetary motion. According to Newton's gravitational law the potential is of the form \begin{align} U(r)=-\frac{\alpha}{r}. \end{align} For the two-body problem we have effective potential \begin{align} U_{eff}(r)=-\frac{\alpha}{r} + \frac{L^2}{2m r^2} \end{align}

Kepler.png

The potantial has its minimum at $r_{min}$

\begin{align} \frac{d}{dr} \left( -\frac{\alpha}{r} + \frac{L^2}{2m r^2} \right) =0\\ \frac{\alpha}{r_{min}^2} =\frac{L^2}{m r_{min}^3}\\ r_{min}=\frac{L^2}{m \alpha} \end{align} The potential at $r_{min}$ is \begin{align} (U_{eff})_{min}=-\frac{\alpha^2m}{L^2} + \frac{\alpha^2 m}{2 L^2}=-\frac{\alpha^2 m}{2 L^2} \end{align}


Turning Points

From the above graph, we immediately see that for $E>0$ motion is infinite and for $E<0$ motion is finite.

Kepler2.png

For $E>0$ there is just one turning point for $U_{eff}=E$ \begin{align} U_{eff}(r)=-\frac{\alpha}{r_t} + \frac{L^2}{2m r_t^2}=E\\ E r_t^2+\alpha r_t - \frac{L^2}{2m}=0 \end{align} \begin{align} r_t=\frac{-\alpha + \sqrt{\alpha^2+\frac{2EL^2}{m}}}{2E} \end{align} For $E<0$ there are two turning points \begin{align} r_{t \pm}=\frac{-\alpha \pm \sqrt{\alpha^2+\frac{2EL^2}{m}}}{-2|E|}\\ r_{t \pm}=\frac{\alpha \pm \sqrt{\alpha^2+\frac{2EL^2}{m}}}{2|E|} \end{align}


Equations of Motion

\begin{align} L= \frac{m \cdot \mathbf{r}^2}{2}-\frac{\alpha}{r_t} + \frac{L^2}{2m r_t^2} \end{align} From the article on Central Forces we know \begin{align} \varphi= \int \frac{\frac{L}{r^2}dr}{\sqrt{2m[E-U(r)]-\frac{M^2}{r^2}}}+const. \end{align} which for $U=-\frac{\alpha}{r}$ is \begin{align} \varphi= \int \frac{\frac{L}{r^2}dr}{\sqrt{2m[E+\frac{\alpha}{r}]-\frac{L^2}{r^2}}}+const.=\int \frac{dr}{r^2\sqrt{\frac{2mE}{L^2}+\frac{2m\alpha}{rL^2}-\frac{1}{r^2}}}+const. \end{align} Now we make the transformation $u=1/r$ with $du=-1/r^2 \,dr$ \begin{align} \varphi=-\int \frac{du}{\sqrt{\frac{2mE}{L^2}+\frac{2m\alpha}{L^2}u-u^2}}+const. \end{align} Integrals of this type have the solution \begin{align} \int \frac{dx}{\sqrt{a + bx +cx^2}}=\frac{1}{\sqrt{-c}} \arccos \left( -\frac{b + 2cx}{\sqrt{b^2-4ac}} \right) \end{align} We compare terms and identify \begin{align} a=\frac{2mE}{L^2}\\ b=\frac{2m\alpha}{L^2}\\ c=-1 \end{align} and get the solution

\begin{align} \varphi-\varphi_0&=\arccos \left( -\frac{\frac{2m\alpha}{L^2}- 2u}{\sqrt{\frac{4m^2\alpha^2}{L^4}+\frac{8mE}{L^2}}} \right)\\ &=\arccos \left( \frac{\frac{1}{r}-\frac{m\alpha}{L^2}}{\sqrt{\frac{m^2\alpha^2}{L^4} \left(1+\frac{2mEL^2}{\alpha^2} \right)}} \right)\\ &=\arccos \left( \frac{\frac{1}{r}-\frac{m\alpha}{L^2}}{\frac{m\alpha}{L^2}\sqrt{\left(1+\frac{2mEL^2}{\alpha^2} \right)}} \right)\\ &=\arccos \left(\frac{\frac{L^2}{m\alpha} \left( \frac{1}{r}-1 \right)}{\sqrt{\left(1+\frac{2mEL^2}{\alpha^2} \right)}} \right)\\ \end{align} We define the so called parameter \begin{align} p=\frac{L^2}{m \alpha} \end{align} and the eccentricity \begin{align} e=\sqrt{1+\frac{2EL^2}{m\alpha^2}} \end{align} With this notation and choosing $\varphi_0=0$, we can rewrite $\varphi$ in compact form \begin{align} \cos \varphi&=\frac{\frac{p}{r}-1}{e}\\ \end{align}

\begin{align} e \cos \varphi +1&=\frac{p}{r}\\ \end{align} For $\varphi_0=0$ we see from this equation that for $\varphi =0$ the trajectory is closest to the center.

Kepler3.png

From geometry it is known that the major axis $a$ and minor axis $b$ are \begin{align} \tag{1} a&=\frac{p}{1-e^2}=\frac{\alpha}{2|E|}\\ b&=\frac{p}{\sqrt{1-e^2}}=\frac{L}{\sqrt{2m|E|}}\\ \end{align}

Orbital Period

We use the fact \begin{align} L=2\cdot m \cdot \dot{f} \tag{2} \end{align} derived in the Central Forces article and the fact that the area of the ellipse is $\pi a b$. Integrating (2) for one period from $0$ to $T$ we have \begin{align} TL=2\cdot m \cdot f=2\cdot m \cdot \pi \cdot a \cdot b \end{align} Now we plug in equation (1) for $b$ \begin{align} TL=2\cdot m \cdot \pi \cdot a \cdot \frac{L}{\sqrt{2m|E|}}\\ T=2\cdot \sqrt{m} \cdot \pi \cdot a \cdot \frac{1}{\sqrt{2|E|}}\\ \end{align} and use the equation for $a$ in (1) to get \begin{align} T=2\cdot \sqrt{m} \cdot \pi \cdot a \cdot \frac{1}{\sqrt{\alpha /a}}\\ \end{align} and find that the orbital period is \begin{align} T=2\pi \cdot a^{3/2} \cdot \sqrt{\frac{m}{\alpha}}\\ \end{align} which leads us to Keplers third law \begin{align} \frac{T}{a^{3/2}}=2\pi \cdot \sqrt{\frac{m}{\alpha}}=const.\\ \end{align}