Q:

Non-dimensionalization and perturbative expansion

I need to expand an equation, of the form
$$\dot{r} = \gamma(a,\mu) F_1 + g(\mu,\ell,h,R) F_2$$
in powers of $\epsilon = a/\ell$.
So I thoughts I would non-dimensionalize it first. 
I know that $$\gamma(a,\mu) = 1/6\pi a \mu,$$ with $\mu$ a dynamic viscosity, $a$ a length scale.   $g$ is a complicated function of $\mu$ and $\ell,h,R$ which are also length scales.  $F_{1,2}$ are forces, $\dot{r}$ is a velocity. 
I would like to measure lengths in units of $\ell$, and times in units of $\Omega^{-1}$, where $\Omega$ is a frequency.  
It is not clear to me how can I get to a position where I can perform an expansion in $\epsilon$. 

A:

You need to measure quantities in units of $\ell,\Omega$, and μ, so, then, viscosity units instead of mass--which has units of $\mu\ell/\Omega$. 
The left-hand-side has units of $\ell \Omega$, which must also agree with the units of the right-hand side. 
The forces F have units of $\mu \Omega \ell^2$; and $g$ has the same units as $\gamma$, namely $1/\ell \mu$.
As a consequence, in these units, the right-hand side reads, for dimensionless $f\equiv F/   \mu \Omega \ell^2  $, 
$$\ell \Omega\left (\frac{f_1}{6\pi \epsilon } + g\mu \ell ~f_2 \right ).     $$
The parenthesis now consists of dimensionless quantities.  As discussed, the dependence of the otherwise complicated function g on $\mu$ is simple, given its over-all dimension,
$$ g(\mu,\ell,h,R)=\tilde g (\ell,h,R)  /\mu ,$$
where $\tilde g$ still has units of $1/\ell$, which can be trivially scaled out of all of its length variables, as indicated below.  
Defining $\tilde h\equiv h/\ell$ and $\tilde R\equiv R/\ell$, 
you end up with a dimensionless r.h.-side parenthesis,
$$ \left  (\frac{f_1}{6\pi \epsilon } + \tilde g(1,\tilde h, \tilde R)  ~f_2 \right ).    $$
You might experiment with simple notional example functions to appreciate the logic of the scaling. 
If, indeed, you need to expand in terms of ε and not 1/ε, the form of the equation is unforgiving, and multiplying the  equation by  ε merely dictates excessive dominance of the of the first term to zeroth order, leading to its vanishing.  But this non-dimensionalization is unique, given your desiderata.  Perhaps you wish to modify them by, e.g., measuring lengths in units of a, instead.  

Edit on alternate length units as per comment  : If you wanted, instead, to use a instead of $\ell$ as your unit length, $\epsilon$ would disappear from the  first term, and would only live on the second term.  The second term might be proportional to $\epsilon$,  depending on the explicit form of g. Recall g must have dimensions of 1/μ×length; so, e.g., for a purely hypothetical toy original
$$
g= \frac{ 1}{\mu \ell} ( 1+ (h/R)^2 + h/\ell +...)
$$
the corresponding nondimensionalized function multiplying the nondimensionalized second force $F_2$ would be 
$$
a\mu g =\epsilon (1+(h/R)^2 + \epsilon (h/a)+...) . 
$$
and then expansion in $\epsilon$ would be regular... But without knowledge of specifics, one really cannot say much... 

