The most common case where this comes up is when you're dealing with a problem where it's helpful to linearize using a Taylor expansion. For example, a decaying exponential
$$
e^{-t/t_0} = 1
- \frac{t}{t_0}
+ \frac12\left(\frac{t}{t_0}\right)^2
- \frac1{3!}\left(\frac{t}{t_0}\right)^3
+ \cdots
$$
or a smallish logarithm
$$
\ln(1+x) = x - \frac{x^2}{2}
+ \frac{x^3}{3} - \frac{x^4}{4}
+ \cdots
$$
or the sine of an angle
$$
\sin\theta = \theta - \frac1{3!}\theta^3 + \frac{1}{5!} \theta^5
- \frac{1}{7!} \theta^7
+ \cdots
$$
or an actual binomial, like the acceleration due to gravity near Earth's surface
$$
g(h) = \frac{GM_\oplus}{(R_\oplus + h)^2}
= \frac{GM_\oplus}{R_\oplus^2} \left(
1 - 2 \frac{h}{R_\oplus}
+ \frac{(-2)(-3)}{2!} \left( \frac{h}{R_\oplus} \right)^2
+ \frac{(-2)(-3)(-4)}{3!} \left( \frac{h}{R_\oplus} \right)^3
+ \cdots
\right)
$$
Now, when we compute a physically interesting quantity, like
$$
g\approx 9.806\,65\,\rm m/s^2 \approx 9.81\, m/s^2 \approx 9.8\,m/s^2 \approx 10\,m/s^2,
$$
we have a built-in way to indicate the precision that we intend: we truncate the "insignificant" digits from the end of the decimal representation. It's pretty common to limit a result to two or three significant figures, which corresponds to an implied precision of somewhere under 1%.
In the Taylor expansions, there is always some dimensionless parameter which is raised to a different power in every term ($h/R_\oplus$ in the last example). If that dimensionless parameter is smaller than 0.1, then each term in the Taylor expansion corresponds, roughly, to a single significant digit in the final result. And if the dimensionless parameter is smaller than 0.01, you can expect to get your three-ish significant digits from the first two terms in the series! This is why you hear people, when pressed, say that "$\gg$" means something like "different by a factor of ten or more."
For your specific example: we have data on the moon's orbit from lunar laser ranging that tells us the orbit is described by general relativity to a precision of about a centimeter. A one-centimeter shift in the orbit of the moon would change the moon-earth gravitational potential energy by
$$
\delta U
= GM_\oplus M_\text{moon}
\left(
\frac{1}{d} - \frac{1}{d+1\,\rm cm}
\right)
= \frac{GM_\oplus M_\text{moon}}{d}
\left(
\frac{1\,\rm cm}{d}
+ \cdots
%\mathcal O\left(
%\frac{1\,\rm cm}{d}
%\right)^2
\right)
$$
where $d$ is the average Earth-Moon distance. I think your numerical intuition should confirm that $d\gg 1\,\rm cm$.
In jabirali's answer he shows that the electrostatic and gravitational interactions would have roughly equal strength if the earth-moon system had $\sqrt{Qq}\approx 10^{14}\,\rm C$. By specifying how well we know the earth-moon distance we can put up a better limit:
\begin{align}
\sqrt{Qq} &\lesssim 10^{14}\,\rm C \frac{1\,cm}{384,400\,km}
\\&\lesssim 10^3\,\rm C
\end{align}
The feebleness of the gravitational force is pretty well-discussed, but I'm actually surprised at how small that is.