Gravitational potential and gravitational intensity at a point on the axis of the uniform ring:
Let us consider a uniform ring of radius \( a \). There is a point \( P \) on the axis of the ring at a distance \( r \) from the centre \( o \) of the uniform ring as shown in the adjoining Fig. 1. We want to calculate the gravitational potential and gravitational intensity on the point \( P \) due to this ring.
Let us consider a small amount of the ring of mass \( \delta{m} \) at the point \( Q \).
The gravitational potential at \( P \) due to this element \( \delta{m} \) is
\( \delta{v}=-frac{G\delta{m}}{PR} \\=-frac{G\delta{m}}{\sqrt{a^2+r^2}} \)
Hence the gravitational potential at \( P \) due to the whole ring is given by,
\( V=-G\frac{\sum\delta{m}}{\sqrt{a^2+r^2}} \\=-\frac{GM}{\sqrt{a^2+r^2}} \)Where, \( \sum\delta{m}=M \) is the total mass of the ring.
Gravitational intensity at \( P \) due to the ring is,
\( E=-\frac{\,dv}{\,dr} \\=GM\frac{d}{dr}\left(\frac{1}{\sqrt{r^2+a^2}}\right) \\=-GM\frac{2r}{2(r^2+a^2)^{3/2}} \\=-\frac{GMr}{(r^2+a^2)^{(3/2)}} \)Hence the intensity at \( P \) along PO is,
\( \displaystyle{E=\frac{GMr}{(r^2+a^2)^{3/2}}} \)