3.4 Gravitatioal field due to a uniform circular ring at a point on its axis

The mass per unit length of ring is, $$\lambda = \frac{M}{{2\pi R}}$$

Mass of infinitesimally small length $dx$ is, $$dm = \lambda dx = \left( {\frac{M}{{2\pi R}}} \right)dx$$

Gravitational field at point $P$ due to mass $dm$ is, $$dE = \frac{{Gdm}}{{{{\left( {\sqrt {{R^2} + {x^2}} } \right)}^2}}}$$ or $$dE = \frac{{Gdm}}{{{R^2} + {x^2}}}$$

$d{E_y} = dE\sin \theta $: Component is in the vertical direction

$$\int {d{E_y}} = \int {dE\sin \theta = 0} $$

Due to the symmetry of the ring.

$d{E_x} = dE\cos \theta $: Component is in the horizontal direction and towards the centre.

$$d{E_x} = \frac{{Gdm}}{{\left( {{R^2} + {x^2}} \right)}}\cos \theta $$

As $\left( {\cos \theta = \frac{r}{{\sqrt {{R^2} + {r^2}} }}} \right)$ and $\left( {dm = \frac{M}{{2\pi R}}} \right)$,

$$d{E_x} = \frac{G}{{\left( {{R^2} + {x^2}} \right)}} \times \frac{{Mdx}}{{2\pi R}} \times \frac{r}{{\sqrt {{R^2} + {r^2}} }}$$$$d{E_x} = \frac{{GMr}}{{2\pi R{{\left( {{R^2} + {x^2}} \right)}^{\frac{3}{2}}}}}dx$$

Integrating with proper limits we get,

$$\int\limits_0^{{E_x}} {d{E_x}} = \int\limits_0^{2\pi R} {\frac{{GMr}}{{2\pi R{{\left( {{R^2} + {x^2}} \right)}^{\frac{3}{2}}}}}dx} $$$${E_x} = \frac{{GMr}}{{2\pi R{{\left( {{R^2} + {x^2}} \right)}^{\frac{3}{2}}}}}\left[ {2\pi R - 0} \right]$$$${E_x} = \frac{{GMr}}{{{{\left( {{R^2} + {x^2}} \right)}^{\frac{3}{2}}}}}$$