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Working through some questions in a text book and came across one which I did not quite understand how to find the solution. I looked up the solution in the manual and was unable to understand how they came to their approximation.

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  • $\begingroup$ A general rule when looking for approximations like this, is that it's a lot easier if it's in the form $(1+\epsilon)$, so factor things out like I did in my answer, to get it into this form. $\endgroup$ – CDCM Jan 6 '18 at 5:07
  • $\begingroup$ Ahh I had just copied across a minus sign wrong, so my answer was off by a factor of $-1$. I've edited the answer now. $\endgroup$ – CDCM Jan 6 '18 at 5:41
  • $\begingroup$ As far as getting the approximation, like I say, pick your choice. The Taylor series works too, $\frac{d}{dx}\left( 1+x\right)^n=n(1+x)^{n-1}$, which around $x=0$ gives a Taylor series of $1+nx$ to first order. $\endgroup$ – CDCM Jan 6 '18 at 5:44
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So, the answer is suggestively written to suggest they've factored out $\frac{1}{R^2_{ME}}$, so an initial step is to see what that gives you. Since $(R_{ME}+R_E)^2 = \left(R_{ME}(1+\frac{R_E}{R_{ME}})\right)^2 = R_{ME}^2\left(1+\frac{R_E}{R_{ME}}\right)^2$ $$F_1 - F_0 = \frac{GM_m m}{R^2_{ME}}\left(\frac{1}{(1-\frac{R_E}{R_M})^2} -\frac{1}{(1+\frac{R_E}{R_M})^2}\right).$$ Now since $R_E/R_{ME}$ is small, you can expand using your favourite method. e.g binomial expansion $(1+x)^n\approx 1+nx $ for small x. Here we have negative $n$, but it still works.

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