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\)
Section 21.12 Shape of the Orbit
In
Section 21.10 , we reduced the central force problem to a pair of uncoupled ordinary differential equations for the variables
\(r\) and
\(\phi\) as functions of time, i.e.
(21.10.5) and
(21.10.6) .
If we are only interested in the shape of the orbit, we can do something simpler than solving the equations of motion for
\(r\) and
\(\phi\) as functions of
\(t\text{;}\) we can solve for the shape of the orbit, i.e. instead of using the variable
\(t\) as a parameter in
(21.10.5) and
(21.10.6) , we will use the variable
\(\phi\) and solve for
\(r(\phi)\text{,}\) the polar equation for the shape of the orbit.
To do this, we need to change the time derivatives into \(\phi\) derivatives:
\begin{equation}
\frac{d}{dt}
= \frac{d\phi}{dt}\frac{d}{d\phi}
= \dot\phi\frac{d}{d\phi}
= \frac{\ell}{\mu r^2}\frac{d}{d\phi}\tag{21.12.1}
\end{equation}
It turns out that the differential equation which we obtain will be much easier to solve if we also change independent variable from \(r\) to
\begin{equation}
u=r^{-1}\tag{21.12.2}
\end{equation}
(I don’t know how to motivate this clever guess.) Therefore,
\begin{equation}
\frac{dr}{dt}
= \frac{\ell}{\mu r^2}\frac{dr}{d\phi}
= -\frac{\ell}{\mu}\frac{dr^{-1}}{d\phi}
= -\frac{\ell}{\mu}\frac{du}{d\phi}\tag{21.12.3}
\end{equation}
(To verify the second equality, work from right to left.) Then the second derivative is given by
\begin{equation}
\frac{d^2r}{dt^2}
= \frac{d}{dt}\frac{dr}{dt}
= \frac{\ell}{\mu}u^2\frac{d}{d\phi}
\left(-\frac{\ell}{\mu}\frac{du}{d\phi}\right)
= -\frac{\ell^2}{\mu^2}u^2\frac{d^2u}{d\phi^2}\tag{21.12.4}
\end{equation}
\begin{equation}
\frac{d^2u}{d\phi^2} + u
= -\frac{\mu}{\ell^2}\frac{1}{u^2} \>f\!\left(\frac1u\right)\tag{21.12.5}
\end{equation}
For the special case of inverse square forces
\(f(r)=-k/r^2\) (spherical gravitational and electric sources), it turns out that the right-hand side of
(21.12.5) is constant so that the equation is particularly easy to solve. First, solve the homogeneous equation (with
\(f(r)=0\) ), which is just the harmonic oscillator equation with general solution
\begin{equation}
u_h = A \cos(\phi + \delta)\tag{21.12.6}
\end{equation}
Add to this any particular solution of the inhomogeneous equation (with \(f(r)=-k/r^2\) ). By inspection, such a solution is just
\begin{equation}
u_p = \frac{\mu k}{\ell^2}\tag{21.12.7}
\end{equation}
so that the general solution of
(21.12.5) for an inverse square force is
\begin{equation}
r^{-1} = u = u_h +
u_p = A\cos(\phi+\delta) + \frac{\mu k}{\ell^2}\tag{21.12.8}
\end{equation}
Then solving for
\(r\) in
(21.12.8) we obtain
\begin{equation}
r
= \frac{1}{\frac{\mu k}{\ell^2} + A\cos(\phi+\delta)}
= \frac{\frac{\ell^2}{\mu k}}{1+A'\cos(\phi+\delta)}\tag{21.12.9}
\end{equation}
This is the equation for a conic section, expressed in polar coordinates. You can explore how the graph of this equation depends on the various parameters in
Section 21.11 .
