Ring Gravity (11-14-2020)

Take any general point on the ring as $\vec{r} = (\cos{t},\sin{t})$ and define the location of the point mass as $\vec{r}_p = (2,0)$ . By applying Newton's law of gravity, for an elementary arc of length $dt$ in the neighbourhood of $\vec{r}$ leads to the following. Based on symmetry, only the X component of force would be nonzero. After simplifications, the required integrand for the X force component is also shown in the following step.

$dF_x = d\vec{F} \cdot \hat{i} = G\left(\frac{M \ dt}{2\pi} \right)m \left(\frac{(\vec{r}-\vec{r}_p) \cdot \hat{i}}{\lvert \vec{r}-\vec{r}_p \rvert^3}\right) \implies F_x =\frac{1}{2\pi} \int_{0}^{2 \pi}\frac{(\cos{t}-2) \ dt}{(5-4\cos{t})^{3/2}} \approx -0.311405152555898$