Let me chip in from a physicists vantage... It's all about orders of magnitude. Are the observed effects so different in order that your simplified model is sufficiently accurate? In this case, sufficient depends on how closely you need to look and how feasible it is to even look deeper.
You'll rarely if ever see the $\ll$ sign or its counterpart unless the numbers are at least an order of magnitude apart. I love avid19's answer on this question, as the part on control is very intriguing. I can see that in many contexts, but I don't find that the notion of control is universally applied, in that we cannot choose the numbers that we measure from reality. We can control our required levels of tolerance in modelling reality, though.
Physically, then, there is a benchmark, and that is "Does the model work?" We approximate because the real world is a mess of particle interactions. Additionally, many effects that we observe are non-linear, that is, not too friendly to effectively compute. The harmonic oscillator is a great example.
To add to this quandry, let me reference Schroeder's book An Introduction to Thermal Physics. As a preface, this discussion is in regards to systems involving many, many elements, such as modelling atoms in a gas. You're either going to love or hate this excerpt:
There are three kinds of numbers that commonly occur in statistical mechanics: small numbers, large numbers, and very large numbers...
You already know how to manipulate small numbers.
Large numbers are much larger than small numbers... The most important property of large numbers is that you can add a small number to a large number without changing it. For example, $10^{23} + 23 = 10^{23}$. The only exception is when you plan to eventually subtract off the same large number: $10^{23} +42 - 10^{23} = 42$.
Very large numbers are even larger than large numbers... [they] have the amazing property that you can multiply them by large numbers without changing them. For instance, $10^{10^{23}} \times 10^{23} = 10^{10^{23} +23}= 10^{10^{23}}$.