In this answer to a question on this site, the gravitational potential of the Earth is expanded as $$U(r) \approx U(r_0) + \left.\frac {dU(r)}{dr}\right|_{r=r_0}(r-r_0),$$ keeping only the linear term. Why is this series cut off here, with the rest of the terms ignored?

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    $\begingroup$ It's an approximation, and the answer explicitly says so. What is your question about that? $\endgroup$ – ACuriousMind Jan 20 '16 at 15:30

Because it's an excellent approximation. The gravitational potential from a spherical mass $M$ of radius $R$ is, to second order \begin{align} U(r) & = -\frac{GM}{r} \\ & = U(R)+\frac{dU}{dr}(r-R)+\frac12\frac{d^2U}{dr^2}(r-R)^2 +O\left(\left(\frac{r-R}{R}\right)^3\right) \\ & = -\frac{GM}{R}+\frac{GM}{R^2}(r-R)-\frac{GM}{R^3}(r-R)^2 +O\left(\left(\frac{r-R}{R}\right)^3\right) \\ & = -\frac{GM}{R}+\frac{GM}{R^2}(r-R)\left[1-\frac{r-R}{R}\right] +O\left(\left(\frac{r-R}{R}\right)^3\right). \end{align} For points near the surface of the Earth at $R=6300\:\mathrm{km}$, with $h=r-R$ on the order of, say $|h|\lesssim 1\:\mathrm{km}$ (unreasonably tall!), the quadratic term is about $$\frac{h}{R}\lesssim10^{-4}.$$ Each further term in the series scales in importance with respect to the previous one by that same factor.

In practice, the angular variation of the potential - the change in $g$ from one location to another caused by the local distribution of rocks and other masses - is in the order of 1%, so it completely trumps the vertical variation, even for huge changes in height.

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