I'm working on a problem and can't seem to get it.
Find the gravitational force exerted by a thin uniform ring of mass M and radius a on a particle of mass m lying on a line perpendicular to the ring through its center. Assume m is at a distance y from the center of the ring.
I'm going to assume the center of the ring is the origin and the particle is at the point (0, y). Due to symmetry we can ignore the gravitational force in the x direction. It's the y-component of the force that is throwing me off.
Any help would be appreciated.
The force exerted by a small arc of angular width $d\theta$ is, by the Inverse Square Law, equal to $km\frac{M}{2\pi}\frac{1}{y^2+a^2}\,d\theta$, where $k$ is a constant. Integrate from $0$ to $2\pi$. We get $\frac{kmM}{y^2+a^2}$. For the component in the $y$-direction, multiply by $\frac{y}{\sqrt{y^2+a^2}}$.