Broken Ring Attraction

Arbitrary points on the right half and left half of the ring are:

$\boxed{\vec{r}_1 = (1 + \cos{\theta_1}) \hat{i} + \sin{\theta_1}\hat{j}} \ | \ \boxed{\vec{r}_2 = -(1 + \cos{\theta_2}) \hat{i} - \sin{\theta_2}\hat{j}}$

The gravitational force between arc length elements on each half is (According to the law of gravitation, considering unit radius and unit mass per unit length):

$d\vec{F} = \left(\frac{Gd\theta_1 d\theta_2}{\mid \vec{r}_1-\vec{r}_2 \mid^3}\right)(\vec{r}_1-\vec{r}_2)$

Substituting expressions and recognising by virtue of symmetry that the net y-component is zero, the net magnitude of the force in the x-direction is:

$\boxed{F_x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(2 + \cos{\theta_1} + \cos{\theta_2}\right)d\theta_1 d\theta_2 }{\left(\left(2 + \cos{\theta_1} + \cos{\theta_2}\right)^2 + \left( \sin{\theta_1} + \sin{\theta_2}\right)^2\right)^{3/2}} \approx 0.8374}$

I used Wolfram-Alpha to get the answer to the double integral. This is slightly different from the mentioned answer of 0.8386.