What is the "associated scalar equation" of equations of motion? In an essay I am reading on celestial mechanics the equations of motion for a 2 body problem is given as:
$$\mathbf{r}''=\nabla(\frac{\mu}{r})=-\frac{\mu \mathbf{r}}{r^3}$$
Fine.  Then it says the "associated scalar equation" is:
$$r''=-\frac{\mu}{r^2}+\frac{c^2}{r^3}$$
I've never heard of such a thing.  Can someone please explain what the "associated scalar equation" of an equation of motion is.  If it is just the equation of motion in scalar form, then why does that extra term $\frac{c^2}{r^3}$ appear?
Oh, $\mu$ is the mass constant.  It's not clear from the essay what $c^2$ is.  It might be the speed of light squared, or perhaps a constant of integration.
EDIT: The essay in question can be found here.  The equations in question are found on page 5.
 A: The "associated scalar equation" is just the formula for the time evolution of the scalar magnitude of the displacement, $r$, rather than all its vector components. It really only makes sense to write such an equation if the right-hand side can be expressed in terms of $r$ only, and not $\mathbf{r}$. Then you can use it to analyze the evolution of $r$ in simple scalar terms, without worrying about vector quantities.
To see where it comes from, first note the scalar $r$ can be written $r = \sqrt{\mathbf{r} \cdot \mathbf{r}}$. Then
$$ r' = \frac{1}{2} (\mathbf{r} \cdot \mathbf{r})^{-1/2} (\mathbf{r} \cdot \mathbf{r}' + \mathbf{r}' \cdot \mathbf{r}) = \frac{\mathbf{r}'\cdot\mathbf{r}}{r}. $$
Continuing with the next derivative, we find
\begin{align}
r'' & = \frac{1}{r^2} \left((\mathbf{r}'' \cdot \mathbf{r} + \mathbf{r}' \cdot \mathbf{r}') r - (\mathbf{r}' \cdot \mathbf{r}) r'\right) \\
& = \frac{1}{r^2} \left(\left(-\frac{\mu}{r^3} \mathbf{r} \cdot \mathbf{r} + \mathbf{r}' \cdot \mathbf{r}'\right) r - \frac{(\mathbf{r}'\cdot\mathbf{r})^2}{r}\right),
\end{align}
where we use the formula we found for $r'$ as well as $\mathbf{r}'' = -\mu \mathbf{r} / r^3$. Recalling $\mathbf{r} \cdot \mathbf{r} = r^2$, we can write
$$ r'' = -\frac{\mu}{r^2} + \frac{1}{r^3} \left((\mathbf{r}' \cdot \mathbf{r}') (\mathbf{r} \cdot \mathbf{r}) - (\mathbf{r}' \cdot \mathbf{r})^2\right), $$
which is the same form as the given associated scalar equation.
It remains to show that the parenthesized expression is constant. Recognizing and then manipulating some triple products yields
\begin{align}
r'' & = -\frac{\mu}{r^2} - \frac{1}{r^3} \mathbf{r} \cdot (\mathbf{r}' \times (\mathbf{r}' \times \mathbf{r})) \\
& = -\frac{\mu}{r^2} - \frac{1}{r^3} (\mathbf{r} \times \mathbf{r}') \cdot (\mathbf{r}' \times \mathbf{r}).
\end{align}
But $\mathbf{r}' \times \mathbf{r}$ is just the specific relative angular momentum $\mathbf{h}$, which is conserved in the two-body problem. Thus we recover the given formula with the constant $c^2 = \mathbf{h} \cdot \mathbf{h}$.
A: My answer is the same as Chris', but formulated in a different way (it's essentially the same as this wiki article):
In polar coordinates, the position vector is
$$
\mathbf{r} = r\,(\cos\varphi,\sin\varphi) = r\,\mathbf{\hat{r}},
$$
with $\mathbf{\hat{r}}$ the radial unit vector. The velocity is then
$$
\mathbf{v} = \dot{r}\,(\cos\varphi,\sin\varphi) + r\dot{\varphi}\,(-\sin\varphi,\cos\varphi) = \dot{r}\,\mathbf{\hat{r}} + r\dot{\varphi}\,\boldsymbol{\hat{\varphi}},
$$
with $\boldsymbol{\hat{\varphi}}$ the azimuthal unit vector. The acceleration is
$$
\mathbf{a} = (\ddot{r} - r\dot{\varphi}^2)\,\mathbf{\hat{r}} + (2\dot{r}\dot{\varphi} + r\ddot{\varphi})\,\boldsymbol{\hat{\varphi}}.
$$
But since gravity is a radial force, the azimuthal accerelation must be zero, so that
$$
2\dot{r}\dot{\varphi} + r\ddot{\varphi} = \frac{1}{r}\frac{\text{d}}{\text{d}t}(r^2\dot{\varphi}) = 0.
$$
In other words, the specific relative angular momentum $h=r^2\dot{\varphi}$ is a constant. Therefore, the (radial) acceleration becomes
$$
a_r = \ddot{r} - r\dot{\varphi}^2 = \ddot{r} - \frac{h^2}{r^3} = -\frac{\mu}{r^2},
$$
which gives the desired result.
