Applying characteristic distance and time 
Q: Consider the 1 -dimensional motion of a particle in the potential $V(x)=g|x|^{\alpha}$. Let $t$ and $d$ denote the characteristic time and distance, respectively, for this motion. Show by dimensional analysis that
$$
t \propto d^{1-\frac{\alpha}{2}}
$$
Apply this result to (a) the harmonic oscillator, (b) a particle in free fall on the surface of the earth, and (c) a planet falling radially toward a star.

I've been able to solve the first part fairly simply, deriving it using dimensional analysis as the problem requests.
\begin{array}{r}
g|x|^{\alpha}=M^{2} \frac{\left[d^{2}\right]}{\left[t^{2}\right]} \\
t^{2}=m^{2} * \frac{\left[d^{2}\right]}{g \cdot|x|^{\alpha}} \\
t^{2} \propto d^{2-\alpha} \\
t \propto d^{1-\frac{\alpha}{2}}
\end{array}
However, I don't really understand how to apply this interesting result to the harmonic oscillator, let alone parts b and c. Does this result imply that the unit of time is always proportional to a specific distance? I feel as if though I am missing the significance of this result.
 A: 
the unit of time is always proportional to a specific distance

The relevant timescale scales with the relevant lengthscale to some power, yes.
For a hamonic oscillator, $V\propto x^2$ so $\alpha=2$ and $t \propto d^0 = 1$. I.e. for a harmonic oscillator, the relevant timescale (i.e. the period) does not depend on space, i.e. it is not a function of position, but a constant. You can easily check this against the periods for a spring (only depends on spring constant $k$ and mass $m$) and pendulum (length $\ell$ and acceleration $g$).
For a particle in free fall on the surface of the Earth, $V\propto mgx$ so $\alpha=1$ and and $t\propto d^{1/2}$. From the suvat equation, the time for a projectile to reach and fall from a height $h$ scales as $\sqrt{h}$ so it checks out.
Finally, for planetary motion you would use Newton's formula so $V\propto 1/x$, so $\alpha=-1$ and hence $t\propto d^{3/2}$. Kepler's third law says that the orbital period of planets in an elliptical orbit of semi-major axis $a$ goes as $T\propto \sqrt{a^3} \propto a^{3/2}$. So, again, it checks out.
