Find the intensity of gravitational field at a point lying at a distance `x` from the centre on the axis of a ring of radius `a` and mass `M`.

Explanation

Consider a ring of radius a, mass M and centre at O.
Mass per unit length `= (M)/(2pi a)`
P is the point at which gravitational intensity is to be measured.
Let dl is the small element of the ring.
Mass of this element `= (M)/(2pi a) dl = dM`
Let a particle of mass m be placed at point P.
Force on this mass due to small element dl = dF
`= (GmdM)/(r^(2))`
`= (GmMdl)/(2pi a r^(2))`
`:. r = PB = sqrt((a^(2) + x^(2))`
Resolve dF into two perpendicular components, one along the axis of ring and the other perpendicular to the plane of ring. We will find that all components perpendicular to axis due to all elements will cancel each other, whereas the component along the axis of ring will add up.
Therefore, the resultant gravitational force on particle at P due to all elements will be
`F = int dF cos theta`
`= int(GMm)/(2pi a) = (dl)/(r^(2)) cos theta`
`= int(GMm)/(2pi a)(dl)/(r^(2)).(x)/(r ) " "(cos theta = (x)/(r ))`
`= (GMm)/(2pi a )(x)/(r^(3))int dl`
`= (GMm)/(2pi a(a^(2) + x^(2))^((3)/(2)).2pi a`
`= (GMm x)/((a^(2) + x^(2))^((3)/(2))`
Gravitational intensity at `P = I = (F)/(m) = (GMmx)/(m(a^(2) + x^(2))^((3)/(2))`
`I = (GMx)/((a^(2) + x^(2))^((3)/(2))`