Equation 1.10.1.
\begin{equation*}
\difftwo{r}{t}-\frac{h^2}{r^3} = -\frac{GM}{r^2}
\end{equation*}
Section 1.10 Optional — Planetary Motion
We now return to the claim, made in §1.9 on central forces, that if \(\vr(t)\) obeys Newton’s inverse square law
\begin{equation*}
\difftwo{\vr}{t} = -\frac{GM}{r^3}\vr = -\frac{GM}{r^2}\hat\vr
\end{equation*}
then the curve obeys Kepler’s laws
- \(\vr(t)\) runs over an ellipse having one focus at the origin and
- \(\vr(t)\) sweeps out equal areas in equal times and
- the square of the period is proportional to the cube of the major axis of the ellipse.
We just showed, in §1.9, that the fact that \(-\frac{GM}{r^3}\vr\) is parallel to \(\vr\) implies that \(\vr(t)\) lies in a plane through the origin and sweeps out equal area in equal times. We now verify the remaining Kepler laws.
We start by just rewriting Newton’s laws above in polar coordinates. We saw in Lemma 1.8.2.c, that if we write \(\vr(t) = r(t)\,\hat\vr(t)\text{,}\) then
\begin{align*}
\difftwo{\vr}{t} &=
\left(\difftwo{r}{t}-r\ \left(\diff{\theta}{t}\right)^2\right) \hat\vr
+\left(r\ \difftwo{\theta}{t} + 2 \diff{r}{t}\ \diff{\theta}{t}\right)\hat\vth\\
&= -\frac{GM}{r^3}\vr
= -\frac{GM}{r^2}\hat\vr
\end{align*}
The \(\hat\vr\) and \(\hat\vth\) components of this equation are
\begin{align*}
\difftwo{r}{t}-r\ \left(\diff{\theta}{t}\right)^2 &= -\frac{GM}{r^2}\\
r\ \difftwo{\theta}{t} + 2 \diff{r}{t}\ \diff{\theta}{t} &= 0
\end{align*}
The second of these two equations only tells us that
\begin{equation*}
\diff{\ }{t}\left\{r^2\,\diff{\theta}{t}\right\}
= r\left\{
r\ \difftwo{\theta}{t} + 2 \diff{r}{t}\ \diff{\theta}{t}\right\} = 0
\implies
r^2\,\diff{\theta}{t} = h,\quad\text{a constant}
\end{equation*}
which we already knew. Substituting \(\diff{\theta}{t} =\frac{h}{r^2}\) into the first equation gives
This equations contains a lot of \(\frac{1}{r}\)’s. So let’s set \(u=\frac{1}{r}\text{.}\) Furthermore, for the first of Kepler’s laws, we really want \(r\) as a function of \(\theta\) rather than \(t\text{.}\) So let’s make \(u\) a function of \(\theta\) and write
\begin{equation*}
r(t) = \frac{1}{u(\theta(t))}
\end{equation*}
Then
\begin{align*}
\diff{r}{t}(t) &= -\frac{1}{u^2}
\diff{u}{\theta}\big(\theta(t)\big)\diff{\theta}{t}(t)
= - h\diff{u}{\theta}\big(\theta(t)\big)\qquad
\text{since } \diff{\theta}{t} = \frac{h}{r^2}=hu^2\\
\difftwo{r}{t}(t)
&= -h\difftwo{u}{\theta}\big(\theta(t)\big)\diff{\theta}{t}(t)
= -h^2 u\big(\theta(t)\big)^2 \difftwo{u}{\theta}\big(\theta(t)\big)
\end{align*}
and our equation becomes
Equation 1.10.2.
\begin{equation*}
-h^2u^2\difftwo{u}{\theta} -h^2u^3= -GM u^2\qquad\text{or}\qquad
\difftwo{u}{\theta} + u = \frac{GM}{h^2}
\end{equation*}
This is a second order, linear, ordinary differential equation with constant coefficients. Recall that the general solution of such an equation is the sum of a “particular solution” (i.e. any one solution, which in this case we can take to be the constant function \(\frac{GM}{h^2}\)) plus the general solution of the homogeneous equation \(u'+u=0\text{,}\) which one often writes as
1
See Appendix A.9.
\begin{equation*}
A\cos\theta +B\sin\theta
\end{equation*}
with \(A\) and \(B\) arbitrary constants. In this particular application it is more convenient to write the solution in a different, standard but less commonly used, form. Namely, we can use the triangle
to write \(A= C\cos\alpha\) and \(B=C\sin\alpha\) so that the general solution of the homogeneous equation \(u'+u=0\) becomes
\begin{equation*}
C\cos\alpha\cos\theta +C\sin\alpha\sin\theta
=C\cos(\theta-\alpha)
\end{equation*}
\begin{equation*}
u(\theta) = \frac{GM}{h^2} + C\cos(\theta-\alpha)
\end{equation*}
and the general solution to 1.10.1 is
\begin{equation*}
r(t) = \frac{1}{\frac{GM}{h^2} + C\cos(\theta(t)-\alpha)}
\end{equation*}
The angle \(\alpha\) just shifts the zero point of our coordinate \(\theta\text{.}\) By rotating our coordinate system by \(\alpha\text{,}\) we can arrange that \(\alpha=0\) and then
\begin{equation*}
r(t) = \frac{1}{\frac{GM}{h^2} + C\cos(\theta(t))}
= \frac{\ell}{1+\veps\cos\theta}
\qquad\text{with}\quad
\ell=\frac{h^2}{GM},\
\veps=\frac{Ch^2}{GM}
\end{equation*}
As we saw in Example 1.8.4, this is exactly the equation of a conic section with eccentricity \(\veps\text{.}\)
That leaves only the last of Kepler’s laws, relating the period to the semi-major axis. As we are talking about planets, whose orbits remain bounded, our conic section must be a circle or ellipse, rather than a parabola or hyperbola. Looking back at Example 1.8.5, we see that the semi-major and semi-minor axes of our ellipse are
\begin{gather*}
a=\frac{\ell}{1-\veps^2}
\qquad
b=\frac{\ell}{\sqrt{1-\veps^2}}
\end{gather*}
The period \(T\) of our orbit is just the length of time it takes the radius vector \(\vr(t)\) to sweep out the area of the ellipse, which is \(\pi ab\text{.}\) As the rate at which the radius vector is sweeping out area is \(\frac{1}{2} r^2\diff{\theta}{t} = \frac{h}{2}\text{,}\) we have
2
You probably computed the area of an ellipse in first year calculus. If not, you should be able to do it now in a few lines.
\begin{gather*}
T^2 = \Big(\frac{\pi a b}{h/2}\Big)^2
=\frac{4\pi^2 a^2 b^2}{h^2}
=\frac{4\pi^2 a^2 b^2}{GM\ell}
=\frac{4\pi^2}{GM}\ a^3\qquad
\text{since }\ell=\frac{b^2}{a}
\end{gather*}
