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Using the equation; $F_{r}=12\frac{U_{0}}{R_{0}}\left[\left(\frac{R_{0}}{R_{0}+x}\right)^{13}-\left(\frac{R_{0}}{R_{0}+x}\right)^{7}\right]$ I must apply the binomial theroem to; $(\frac{1}{R_{0}+x})^{13}$ And $(\frac{1}{R_{0}+x})^7$,

In order to show that $F_{r}\approx -\left(\frac{72U_{0}}{R_{0}^{2}}\right)x$

Also, x is very small comapared to $R_{0}$ so, $\left| \frac{x}{R_{0}}\right|<1$

What Ive done so far is, $$\frac{1}{R^{13}_{0}}.\frac{1}{1+(\frac{x}{R_{0}})^{13}}$$ $=\frac{1}{R^{13}_{0}}\left(1+(\frac{x}{R_{0}})\right)^{-13}\approx \frac{1}{R^{13}_{0}}\left(1+(-13)\frac{x}{R_{0}}\right)=\frac{1}{R^{13}_{0}}\left(1-13\frac{x}{R_{0}}\right)$

So, does? $F_{r}\approx -\left(\frac{72U_{0}}{R_{0}^{2}}\right)x=12\frac{U_{0}}{R_{0}}\left[\frac{1}{R^{13}_{0}}\left(1-13\frac{x}{R_{0}}\right)-\frac{1}{R^{7}_{0}}\left(1-7\frac{x}{R_{0}}\right)\right]$

If so, could someone show me some working because I cant quite get to the right answer. Any help will be greatly appreciated, thanks

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1 Answer 1

If we apply the binomial series expansion to the two respective terms, we end up with $$\left(\frac{R_0}{R_0 + x}\right)^{13} = \left(1+\frac{x}{R_0}\right)^{-13} = 1 - 13\frac{x}{R_0} + \mathcal{O}\left(\frac{x^2}{R_0^2}\right)$$ and $$\left(\frac{R_0}{R_0 + x}\right)^7 = \left(1+\frac{x}{R_0}\right)^{-7} = 1 - 7\frac{x}{R_0} + \mathcal{O}\left(\frac{x^2}{R_0^2}\right)$$ so their difference simplifies to $$\left(\frac{R_0}{R_0 + x}\right)^{13}-\left(\frac{R_0}{R_0 + x}\right)^7 = -6\frac{x}{R_0} + \mathcal{O}\left(\frac{x^2}{R_0^2}\right)$$ If $\frac{x}{R_0}$ is very small, then we can ignore the higher order terms to get $$F_r \approx \frac{12U_0}{R_0}\left[-6\frac{x}{R_0}\right] = -\frac{72U_0}{R_0^2}x$$ Your approach is very close. I'm not sure where the extra $\frac{1}{R_0^{13}}$ and $\frac{1}{R_0^7}$ terms sudden came from. Otherwise you probably could've simplified to the answer.

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