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## Finding the Shape of the Orbit (Kepler's 1st Law) Lecture (25 minutes)

Central Forces Notes Section 10

- First we note that the equations of motion cannot be solved explicitly for $r(t)$ and $\phi(t)$ in the general case and that this system of equations can and often is solved using numerical approximations.
- Then we go on to solve for the shape of the orbit $r(\phi)$ using the fact that the angular momentum is conserved. Thus $\frac{d}{dt} = \frac{d\phi}{dt} \frac{d}{d\phi}=\frac{l}{\mu r^2}\frac{d}{d\phi}$ and $\frac{d^2}{dt^2} = \frac{l}{\mu r^2}\frac{d}{d\phi} \left(\frac{l}{\mu r^2}\frac{d}{d\phi} \right)$. Which leads to $$\frac{dr}{dt} =\frac{l}{\mu r^2}\frac{dr}{d\phi}= -\frac{l}{\mu}\frac{d(r^{-1})}{d\phi} = -\frac{l}{\mu}\frac{du}{d\phi}$$ It is useful when introducing students to the change of variables to $u=r^{-1}$ to have them work backwards (right to left) through the last equation above to see explicitly how this change of variables simplifies the equations. This then leads to $$-\frac{l^2}{\mu^2}u^2\frac{d^2 u}{d\phi^2} -\frac{l^2}{\mu^2}u^3= f\left(\frac{1}{u}\right)$$ which yields $$\frac{d^2 u}{d\phi^2} + u = -\frac {\mu}{l} \frac{1}{u^2} f{\left(\frac{1}{u}\right)}$$.
- It is helpful to students to point out that this equation should look familiar since it is just an inhomogeneous version of the harmonic oscillator equation.
- To proceed, we substitute in the gravitational force function ${\bf{f}}(r) = -\frac{k}{r^2} {\bf\hat{r}}$ and solve the equation for both the homogeneous and inhomogeneous solutions to get $$u = \frac{1}{r} = C \cos{(\phi+\delta)} + \frac{\mu}{l^2} k$$ which reduces to $$r=\frac{\frac{l^2}{\mu k}}{1+C' \cos{(\phi+\delta)}}$$ which is just the equation for a conic section.
- Student confusion may arise if they mix up $u$ and $\mu$ in these equations.