Flat Earth Theory

Diagram of flat Earth situation It is immediately obvious that, due to symmetry, the horizontal component of the gravitational acceleration will be null. To set up the process of solving, we will divide the flat earth into infinitesimally small circular rings of inner radius $x$ , mass $dM$ and width $dx$ (with $x$ ranging from $0$ to $R$ ). Then, we are going to divide each ring into infinitesimal pieces of mass $ddM$ , delimiting a central angle $d\theta$ . Let us use the following notations: $\Gamma$ – for the strength of the total gravitational field, $d\Gamma$ – for the strength gravitational field generated by a full ring, $dd\Gamma$ – for the strength of the vertical component of gravitational field generated by each piece of a ring. Yikes, enough formalities - let's start integrating! $dd\Gamma=\dfrac{Gddm\cos\gamma }{\sqrt{R^2+x^2}}=\dfrac{GRddm}{(R^2+x^2)^{3/2}},\:\:\:\: ddm=\dfrac{dmd\theta}{2\pi},\:\:\:\: dm=\dfrac{2\pi Mxdx}{\pi R^2}$ $d\Gamma=\dfrac{GRdm}{2\pi(R^2+x^2)}\int_0^{2\pi}d\theta \implies \Gamma=\dfrac{2\pi GRM}{\pi R^2}\int_0^R\dfrac{xdx}{(R^2+x^2)^{3/2}}$ $\Gamma=\dfrac{2GM}{R}\left[\int\dfrac{d(R^2+x^2)}{2(R^2+x^2)^{3/2}}\right]_{x=0}^R=\left.\dfrac{2\Gamma M}{R\sqrt{R^2+x^2}}\right\rvert_{x=0}^R$ $\boxed{|\Gamma|=\frac{GM}{R^2}(2-\sqrt{2})\implies |\Gamma|\approx 5.779\:\text{m\:s}^{-2}}$

With $\rho = \tfrac{M}{\pi R^2}$ being the mass density of the disk, the force acting on a particular of mass $m$ is $\begin{aligned} Gm\int_0^R \frac{2\pi r\rho}{R^2+r^2} \,\frac{R}{\sqrt{R^2+r^2}}\,dr & = \; 2\pi Gm\rho R \int_0^R \frac{r\,dr}{(R^2+r^2)^{\frac32}} \; = \; 2\pi Gm \rho R\left[-\frac{1}{\sqrt{R^2+r^2}}\right]_0^R \\ & = \; 2\pi Gm \frac{M}{\pi R^2}R\left(\frac{1}{R} - \frac{1}{R\sqrt{2}}\right) \; = \; \frac{GmM}{R^2}(2 - \sqrt{2}) \end{aligned}$ acting along the central axis of the disk, and hence the acceleration felt by the particle is $\tfrac{GM}{R^2}(2-\sqrt{2}) = \boxed{5.779}$ m s ${}^{-2}$ .