Precise Free Fall

If $r$ is the distance of the object from the centre of the earth, then $\frac{d}{dr}\big(\tfrac12\dot{r}^2\big) \; = \; \ddot{r} \; = \; -\frac{gR^2}{r^2}$ and hence $\tfrac12\dot{r}^2 \; = \; \frac{gR^2}{r} - \frac{gR^2}{R+h} \; = \; \frac{gR^2(R+h-r)}{(R+h)r}$ so that the time $T$ to hit the ground is (using the substitution $r = (R+h)\sin^2\theta$ ) $\begin{aligned} T & = \; \int_R^{R+h}\sqrt{\frac{(R+h)r}{2gR^2(R+h-r)}}\,dr \; = \; \sqrt{\frac{R+h}{2gR^2}}\int_{\sin^{-1}\sqrt{\frac{R}{R+h}}}^{\frac12\pi} \tan\theta\,2(R+h)\sin\theta\cos\theta\,d\theta \\ & = \; \frac{(R+h)^{\frac32}}{\sqrt{2gR^2}}\int_{\tan^{-1}\sqrt{\frac{R}{h}}}^{\frac12\pi}(1 - \cos2\theta)\,d\theta \; = \; \frac{(R+h)^{\frac32}}{\sqrt{2gR^2}}\left\{ \tfrac12\pi - \tan^{-1}\sqrt{\tfrac{R}{h}} + \frac{\sqrt{Rh}}{R+h}\right\} \; = \; \frac{(R+h)^{\frac32}}{\sqrt{2gR^2}}\left\{ \tan^{-1}\sqrt{\tfrac{h}{R}} + \frac{\sqrt{Rh}}{R+h}\right\} \end{aligned}$ Putting $k = \tfrac{h}{R}$ we see that $\frac{T}{T_0} \; = \; \tfrac12(1+k)^{\frac32}\left\{ \frac{1}{\sqrt{k}}\tan^{-1}\sqrt{k} + \frac{1}{1+k}\right\}$ Writing down the Maclaurin expansion of $\tfrac{T}{T_0}$ , we see that $\begin{aligned} \frac{T}{T_0} & = \; \tfrac12\left(1 + \tfrac32k + \tfrac38k^2 + O(k^3)\right)\big\{ \left(1 - \tfrac13k + \tfrac15k^2 + O(k^3)\right) + \left(1 - k + k^2 + O(k^3)\right)\big\} \\ & = \; \tfrac12\left(1 + \tfrac32k + \tfrac38k^2 + O(k^3)\right)\left(2 - \tfrac43k + \tfrac65k^2 + O(k^3)\right) \\ & = \; 1 + \tfrac56k - \tfrac{1}{40}k^2 + O(k^3) \end{aligned}$ and hence $\frac{5}{6k} - \frac{1}{k^2}\left[\frac{T}{T_0} - 1\right] \; = \; \frac{1}{40} + O(k)$ where all these asymptotic statements as as $k \to 0$ . Thus we deduce that $\lim_{h to 0} \left[\frac{5R}{6h} - \frac{R^2}{h^2}\left(\frac{T}{T_0} - 1\right)\right] \; = \; \frac{1}{40}$ which makes the answer $\boxed{40}$ .